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On the classification of Kähler-Ricci solitons on Gorenstein del Pezzo surfaces: Calculations for surfaces

Project: solitons
Views: 377

On the classficiation of Kähler-Ricci solitons on Gorenstein del Pezzo surfaces

Calculation for the preprint arXiv:1705.02920.

List of Gorenstein del Pezzo surfaces with 1-torus action

  1. Degree 1, Sing=2D42D_4

  2. Degree 1, Sing=E8E_8

  3. Degree 1, Sing=E7A1E_7A_1

  4. Degree 1, Sing=E6A2E_6A_2

  5. Degree 2, Sing=2A3A12A_3A_1

  6. Degree 2, Sing=A5A2A_5A_2

  7. Degree 2, Sing=D6A1D_6A_1

  8. Degree 2, Sing=E7E_7

  9. Degree 2, Sing=E6E_6

  10. Degree 2, Sing=2A32A_3

  11. Degree 2, Sing=D5A1D_5A_1

  12. Degree 2, Sing=3A1D43A_1D_4

  13. Degree 3, Sing=A5A1A_5A_1

  14. Degree 3, Sing=E6E_6

  15. Degree 3, Sing=A4A1A_4A_1

  16. Degree 3, Sing=2A2A12A_2A_1

  17. Degree 3, Sing=D5D_5

  18. Degree 3, Sing=2A22A_2

  19. Degree 3, Sing=D4D_4

  20. Degree 3, Sing=A32A1A_32A_1

  21. Degree 4, Sing=D5D_5

  22. Degree 4, Sing=A3A1A_3A_1

  23. Degree 4, Sing=D4D_4

  24. Degree 4, Sing=A4A_4

  25. Degree 4, Sing=3A13A_1

  26. Degree 4, Sing=A3A_3

  27. Degree 4, Sing=2A12A_1

  28. Degree 4, Sing=A2A1A_2A_1

  29. Degree 5, Sing=A4A_4

  30. Degree 5, Sing=A3A_3

  31. Degree 5, Sing=A2A_2

  32. Degree 5, Sing=A1A_1

  33. Degree 6, Sing=A2A_2

  34. Degree 6, Sing=A1A_1

No. 2: Degree 11 / Singularity type E8E_8.

The combinatorial data is is given by □=[−1,5]\Box=[-1,5] and Φ0(u)=min⁡{u−45,0},  Φ∞(u)=u−23,  Φ1(u)=−u+12\Phi_0(u) = \min \{\frac{u-4}{5},0\},\; \Phi_\infty(u)= \frac{u-2}{3},\; \Phi_1(u) = -\frac{u+1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,5]

# 3 PL functions on the interval
Phi1(u)=min_symbolic((u-4)/5,0)
Phi2(u)=1/3*u-2/3
Phi3(u)=-1/2*u-1/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print("The combinatorial data")
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 5\displaystyle 5]
u ↦ min⁡(15 u−45,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{5} \, u - \frac{4}{5}, 0\right)
u ↦ 13 u−23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(x)=integral(degPhi(u)*u*exp(x*u),u,*B)

show(F(xi))
25 ξe(5 ξ)−12 ξe(4 ξ)−10 e(5 ξ)+6 e(4 ξ)30 ξ3−12 ξe(4 ξ)+ξe(−ξ)−6 e(4 ξ)+2 e(−ξ)30 ξ3\displaystyle \frac{25 \, \xi e^{\left(5 \, \xi\right)} - 12 \, \xi e^{\left(4 \, \xi\right)} - 10 \, e^{\left(5 \, \xi\right)} + 6 \, e^{\left(4 \, \xi\right)}}{30 \, \xi^{3}} - \frac{12 \, \xi e^{\left(4 \, \xi\right)} + \xi e^{\left(-\xi\right)} - 6 \, e^{\left(4 \, \xi\right)} + 2 \, e^{\left(-\xi\right)}}{30 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F(xi),-10,10)
print ("Numerical solution:",sol)

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print ("Interval containing solution:", xi0.str(style='brackets'))
Numerical solution: -1.9976195991011914

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
−(18 (5 ξ−1)e(5 ξ)+11)e(−ξ)900 ξ3+25 (6 ξ−1)e(5 ξ)−18 (5 ξ−1)e(4 ξ)900 ξ3≥0.01128570\displaystyle -\frac{{\left(18 \, {\left(5 \, \xi - 1\right)} e^{\left(5 \, \xi\right)} + 11\right)} e^{\left(-\xi\right)}}{900 \, \xi^{3}} + \frac{25 \, {\left(6 \, \xi - 1\right)} e^{\left(5 \, \xi\right)} - 18 \, {\left(5 \, \xi - 1\right)} e^{\left(4 \, \xi\right)}}{900 \, \xi^{3}} \geq 0.01128570
−(9 (15 ξ−8)e(5 ξ)+19)e(−ξ)900 ξ3+25 (12 ξ−5)e(5 ξ)−9 (15 ξ−8)e(4 ξ)900 ξ3≥0.01949453\displaystyle -\frac{{\left(9 \, {\left(15 \, \xi - 8\right)} e^{\left(5 \, \xi\right)} + 19\right)} e^{\left(-\xi\right)}}{900 \, \xi^{3}} + \frac{25 \, {\left(12 \, \xi - 5\right)} e^{\left(5 \, \xi\right)} - 9 \, {\left(15 \, \xi - 8\right)} e^{\left(4 \, \xi\right)}}{900 \, \xi^{3}} \geq 0.01949453
(6 (25 ξ−13)e(5 ξ)+30 ξ+31)e(−ξ)900 ξ3−25 (12 ξ−5)e(5 ξ)−6 (25 ξ−13)e(4 ξ)900 ξ3≥0.02975460\displaystyle \frac{{\left(6 \, {\left(25 \, \xi - 13\right)} e^{\left(5 \, \xi\right)} + 30 \, \xi + 31\right)} e^{\left(-\xi\right)}}{900 \, \xi^{3}} - \frac{25 \, {\left(12 \, \xi - 5\right)} e^{\left(5 \, \xi\right)} - 6 \, {\left(25 \, \xi - 13\right)} e^{\left(4 \, \xi\right)}}{900 \, \xi^{3}} \geq 0.02975460
−(3 (25 ξ−4)e(5 ξ)−30 ξ−1)e(−ξ)900 ξ3+25 (6 ξ−1)e(5 ξ)−3 (25 ξ−4)e(4 ξ)900 ξ3≥0.06053484\displaystyle -\frac{{\left(3 \, {\left(25 \, \xi - 4\right)} e^{\left(5 \, \xi\right)} - 30 \, \xi - 1\right)} e^{\left(-\xi\right)}}{900 \, \xi^{3}} + \frac{25 \, {\left(6 \, \xi - 1\right)} e^{\left(5 \, \xi\right)} - 3 \, {\left(25 \, \xi - 4\right)} e^{\left(4 \, \xi\right)}}{900 \, \xi^{3}} \geq 0.06053484
Surface is stable!

No. 3: Degree 11 / Singularity type E7A1E_7A_1.

The combinatorial data is is given by □=[−1,3]\Box=[-1,3] and Φ0(u)=min⁡{u−23,0},  Φ∞(u)=u−34,  Φ1(u)=−u−12\Phi_0(u) = \min \{\frac{u-2}{3},0\},\; \Phi_\infty(u)= \frac{u-3}{4},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=min_symbolic((u-2)/3,0)
Phi2(u)=1/4*u-3/4
Phi3(u)=-1/2*u-1/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ min⁡(13 u−23,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{3} \, u - \frac{2}{3}, 0\right)
u ↦ 14 u−34\displaystyle u \ {\mapsto}\ \frac{1}{4} \, u - \frac{3}{4}
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
(3 ξ−2)e(3 ξ)4 ξ3−(8 (ξ−1)e(3 ξ)+ξ+2)e(−ξ)12 ξ3\displaystyle \frac{{\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(8 \, {\left(\xi - 1\right)} e^{\left(3 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{12 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.9402428939 Interval containing solution: [-1.94024301 .. -1.94024276]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
(4 ξ−1)e(3 ξ)16 ξ3−(16 (3 ξ−1)e(3 ξ)+7)e(−ξ)144 ξ3≥0.04440443\displaystyle \frac{{\left(4 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} - \frac{{\left(16 \, {\left(3 \, \xi - 1\right)} e^{\left(3 \, \xi\right)} + 7\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.04440443
(4 ξ−3)e(3 ξ)16 ξ3−(8 (3 ξ−4)e(3 ξ)+5)e(−ξ)144 ξ3≥0.03181965\displaystyle \frac{{\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} - \frac{{\left(8 \, {\left(3 \, \xi - 4\right)} e^{\left(3 \, \xi\right)} + 5\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.03181965
−(4 ξ−3)e(3 ξ)16 ξ3+(4 (9 ξ−10)e(3 ξ)+12 ξ+13)e(−ξ)144 ξ3≥0.06993168\displaystyle -\frac{{\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} + \frac{{\left(4 \, {\left(9 \, \xi - 10\right)} e^{\left(3 \, \xi\right)} + 12 \, \xi + 13\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.06993168
(4 ξ−1)e(3 ξ)16 ξ3−(4 (9 ξ−2)e(3 ξ)−12 ξ−1)e(−ξ)144 ξ3≥0.1461557\displaystyle \frac{{\left(4 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} - \frac{{\left(4 \, {\left(9 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} - 12 \, \xi - 1\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.1461557
Surface is stable!

No. 4: Degree 22 / Singularity type E6A2E_6A_2.

The combinatorial data is is given by □=[−1,2]\Box=[-1,2] and Φ0(u)=min⁡{u−12,0},  Φ∞(u)=u−23,  Φ1(u)=−2u−23\Phi_0(u) = \min \{\frac{u-1}{2},0\},\; \Phi_\infty(u)= \frac{u-2}{3},\; \Phi_1(u) = \frac{-2u-2}{3}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,2]

# 3 PL functions on the interval
Phi1(u)=min_symbolic((u-1)/2,0)
Phi2(u)=1/3*u-2/3
Phi3(u)=-2/3*u-2/3

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 2\displaystyle 2]
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u ↦ 13 u−23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}
u ↦ −23 u−23\displaystyle u \ {\mapsto}\ -\frac{2}{3} \, u - \frac{2}{3}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
2 (ξ−1)e(2 ξ)3 ξ3−(3 (ξ−2)e(2 ξ)+ξ+2)e(−ξ)6 ξ3\displaystyle \frac{2 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)}}{3 \, \xi^{3}} - \frac{{\left(3 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{6 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.69131741884 Interval containing solution: [-1.69131753 .. -1.69131729]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
(3 ξ−1)e(2 ξ)9 ξ3−(9 (2 ξ−1)e(2 ξ)+5)e(−ξ)36 ξ3≥0.1187884\displaystyle \frac{{\left(3 \, \xi - 1\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} - \frac{{\left(9 \, {\left(2 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} + 5\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.1187884
(ξ−1)e(2 ξ)3 ξ3−((2 ξ−5)e(2 ξ)+1)e(−ξ)12 ξ3≥0.07315991\displaystyle \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)}}{3 \, \xi^{3}} - \frac{{\left({\left(2 \, \xi - 5\right)} e^{\left(2 \, \xi\right)} + 1\right)} e^{\left(-\xi\right)}}{12 \, \xi^{3}} \geq 0.07315991
−(ξ−1)e(2 ξ)3 ξ3+((4 ξ−7)e(2 ξ)+2 ξ+3)e(−ξ)12 ξ3≥0.07315987\displaystyle -\frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)}}{3 \, \xi^{3}} + \frac{{\left({\left(4 \, \xi - 7\right)} e^{\left(2 \, \xi\right)} + 2 \, \xi + 3\right)} e^{\left(-\xi\right)}}{12 \, \xi^{3}} \geq 0.07315987
(3 ξ−1)e(2 ξ)9 ξ3−(3 (4 ξ−1)e(2 ξ)−6 ξ−1)e(−ξ)36 ξ3≥0.2651082\displaystyle \frac{{\left(3 \, \xi - 1\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} - \frac{{\left(3 \, {\left(4 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} - 6 \, \xi - 1\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.2651082
Surface is stable!

No. 6: Degree 22 / Singularity type A5A2A_5A_2.

The combinatorial data is is given by □=[−1,2]\Box=[-1,2] and Φ0(u)=min⁡{−u,0},  Φ∞(u)=u−23,  Φ1(u)=u−23\Phi_0(u) = \min \{-u,0\},\; \Phi_\infty(u)= \frac{u-2}{3},\; \Phi_1(u) = \frac{u-2}{3}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,2]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(-u,0)
Phi2(u)=1/3*u-2/3
Phi3(u)=1/3*u-2/3

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
plot(Phi2+1,*B)+plot(-Phi1-Phi3-1,*B)

-Phi1(0)-Phi3(0)-1
The combinatorial data
[−1\displaystyle -1, 2\displaystyle 2]
u ↦ min⁡(−u,0)\displaystyle u \ {\mapsto}\ \min\left(-u, 0\right)
u ↦ 13 u−23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}
u ↦ 13 u−23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}
-1/3

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
−2 (ξ+2)e(−ξ)3 ξ3+2 ((ξ−1)e(2 ξ)+3)3 ξ3\displaystyle -\frac{2 \, {\left(\xi + 2\right)} e^{\left(-\xi\right)}}{3 \, \xi^{3}} + \frac{2 \, {\left({\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 3\right)}}{3 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -0.97052220918 Interval containing solution: [-0.970522315 .. -0.970522105]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
2 (3 ξ+2)e(−ξ)9 ξ3−(3 ξ−5)e(2 ξ)+9 ξ+99 ξ3≥0.4790611\displaystyle \frac{2 \, {\left(3 \, \xi + 2\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} - \frac{{\left(3 \, \xi - 5\right)} e^{\left(2 \, \xi\right)} + 9 \, \xi + 9}{9 \, \xi^{3}} \geq 0.4790611
(ξ−1)e(2 ξ)+ξ+13 ξ3≥0.09239867\displaystyle \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + \xi + 1}{3 \, \xi^{3}} \geq 0.09239867
(ξ−1)e(2 ξ)+ξ+13 ξ3≥0.09239867\displaystyle \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + \xi + 1}{3 \, \xi^{3}} \geq 0.09239867
2 (3 ξ+2)e(−ξ)9 ξ3+(3 ξ−1)e(2 ξ)−3 ξ−39 ξ3≥0.6638587\displaystyle \frac{2 \, {\left(3 \, \xi + 2\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} + \frac{{\left(3 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} - 3 \, \xi - 3}{9 \, \xi^{3}} \geq 0.6638587
Surface is stable!

No. 7: Degree 22 / Singularity type D6A1D_6A_1.

The combinatorial data is is given by □=[−1,3]\Box=[-1,3] and Φ0(u)=min⁡{u−12,0},  Φ∞(u)=u−34,  Φ1(u)=−u−12\Phi_0(u) = \min \{\frac{u-1}{2},0\},\; \Phi_\infty(u)= \frac{u-3}{4},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=min_symbolic((u-1)/2,0)
Phi2(u)=1/4*u-3/4
Phi3(u)=-1/2*u-1/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u ↦ 14 u−34\displaystyle u \ {\mapsto}\ \frac{1}{4} \, u - \frac{3}{4}
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
(3 ξ−2)e(3 ξ)4 ξ3−(2 (ξ−2)e(2 ξ)+ξ+2)e(−ξ)4 ξ3\displaystyle \frac{{\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.79675598472 Interval containing solution: [-1.79675609 .. -1.79675588]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
(4 ξ−1)e(3 ξ)16 ξ3−(4 (2 ξ−1)e(2 ξ)+3)e(−ξ)16 ξ3≥0.1624897\displaystyle \frac{{\left(4 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} - \frac{{\left(4 \, {\left(2 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} + 3\right)} e^{\left(-\xi\right)}}{16 \, \xi^{3}} \geq 0.1624897
(4 ξ−3)e(3 ξ)16 ξ3+(4 e(2 ξ)−1)e(−ξ)16 ξ3≥0.05832627\displaystyle \frac{{\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} + \frac{{\left(4 \, e^{\left(2 \, \xi\right)} - 1\right)} e^{\left(-\xi\right)}}{16 \, \xi^{3}} \geq 0.05832627
−(4 ξ−3)e(3 ξ)16 ξ3+(4 (ξ−2)e(2 ξ)+4 ξ+5)e(−ξ)16 ξ3≥0.1687342\displaystyle -\frac{{\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} + \frac{{\left(4 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + 4 \, \xi + 5\right)} e^{\left(-\xi\right)}}{16 \, \xi^{3}} \geq 0.1687342
(4 ξ−1)e(3 ξ)16 ξ3−(4 ξe(2 ξ)−4 ξ−1)e(−ξ)16 ξ3≥0.3895503\displaystyle \frac{{\left(4 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} - \frac{{\left(4 \, \xi e^{\left(2 \, \xi\right)} - 4 \, \xi - 1\right)} e^{\left(-\xi\right)}}{16 \, \xi^{3}} \geq 0.3895503
Surface is stable!

No. 8: Degree 22 / Singularity type E7E_7.

The combinatorial data is is given by □=[−1,5]\Box=[-1,5] and Φ0(u)=min⁡{u−34,0},  Φ∞(u)=u−23,  Φ1(u)=−u−12\Phi_0(u) = \min \{\frac{u-3}{4},0\},\; \Phi_\infty(u)= \frac{u-2}{3},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,5]

# 3 PL functions on the interval
Phi1(u)=min_symbolic((u-3)/4,0)
Phi2(u)=1/3*(u-2)
Phi3(u)=1/2*(-u-1)

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 5\displaystyle 5]
u ↦ min⁡(14 u−34,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{4} \, u - \frac{3}{4}, 0\right)
u ↦ 13 u−23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
(5 ξ−2)e(5 ξ)6 ξ3−(3 (3 ξ−2)e(4 ξ)+ξ+2)e(−ξ)12 ξ3\displaystyle \frac{{\left(5 \, \xi - 2\right)} e^{\left(5 \, \xi\right)}}{6 \, \xi^{3}} - \frac{{\left(3 \, {\left(3 \, \xi - 2\right)} e^{\left(4 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{12 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.99186220915 Interval containing solution: [-1.99186233 .. -1.99186208]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
(6 ξ−1)e(5 ξ)36 ξ3−(9 (4 ξ−1)e(4 ξ)+5)e(−ξ)144 ξ3≥0.03202411\displaystyle \frac{{\left(6 \, \xi - 1\right)} e^{\left(5 \, \xi\right)}}{36 \, \xi^{3}} - \frac{{\left(9 \, {\left(4 \, \xi - 1\right)} e^{\left(4 \, \xi\right)} + 5\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.03202411
(12 ξ−5)e(5 ξ)36 ξ3−(9 (4 ξ−3)e(4 ξ)+7)e(−ξ)144 ξ3≥0.04486743\displaystyle \frac{{\left(12 \, \xi - 5\right)} e^{\left(5 \, \xi\right)}}{36 \, \xi^{3}} - \frac{{\left(9 \, {\left(4 \, \xi - 3\right)} e^{\left(4 \, \xi\right)} + 7\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.04486743
−(12 ξ−5)e(5 ξ)36 ξ3+(3 (16 ξ−11)e(4 ξ)+12 ξ+13)e(−ξ)144 ξ3≥0.07049793\displaystyle -\frac{{\left(12 \, \xi - 5\right)} e^{\left(5 \, \xi\right)}}{36 \, \xi^{3}} + \frac{{\left(3 \, {\left(16 \, \xi - 11\right)} e^{\left(4 \, \xi\right)} + 12 \, \xi + 13\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.07049793
(6 ξ−1)e(5 ξ)36 ξ3−(3 (8 ξ−1)e(4 ξ)−12 ξ−1)e(−ξ)144 ξ3≥0.1473894\displaystyle \frac{{\left(6 \, \xi - 1\right)} e^{\left(5 \, \xi\right)}}{36 \, \xi^{3}} - \frac{{\left(3 \, {\left(8 \, \xi - 1\right)} e^{\left(4 \, \xi\right)} - 12 \, \xi - 1\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.1473894
Surface is stable!

No. 9: Degree 22 / Singularity type E6E_6.

The combinatorial data is is given by □=[−1,3]\Box=[-1,3] and Φ0(u)=min⁡{u−23,0},  Φ∞(u)=min⁡{u−23,0},  Φ1(u)=−u−12\Phi_0(u) = \min \{\frac{u-2}{3},0\},\; \Phi_\infty(u)= \min \{\frac{u-2}{3},0\},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=min_symbolic((u-2)/3,0)
Phi2(u)=min_symbolic((u-2)/3,0)
Phi3(u)=1/2*(-u-1)

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ min⁡(13 u−23,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{3} \, u - \frac{2}{3}, 0\right)
u ↦ min⁡(13 u−23,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{3} \, u - \frac{2}{3}, 0\right)
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
(3 ξ−2)e(3 ξ)2 ξ3−(8 (ξ−1)e(3 ξ)+ξ+2)e(−ξ)6 ξ3\displaystyle \frac{{\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)}}{2 \, \xi^{3}} - \frac{{\left(8 \, {\left(\xi - 1\right)} e^{\left(3 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{6 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.9402428939 Interval containing solution: [-1.94024301 .. -1.94024276]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
(2 ξ−1)e(3 ξ)4 ξ3−(2 (3 ξ−2)e(3 ξ)+1)e(−ξ)12 ξ3≥0.07622408\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, {\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} + 1\right)} e^{\left(-\xi\right)}}{12 \, \xi^{3}} \geq 0.07622408
(2 ξ−1)e(3 ξ)4 ξ3−(2 (3 ξ−2)e(3 ξ)+1)e(−ξ)12 ξ3≥0.07622408\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, {\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} + 1\right)} e^{\left(-\xi\right)}}{12 \, \xi^{3}} \geq 0.07622408
−(2 ξ−1)e(3 ξ)4 ξ3+(8 (3 ξ−2)e(3 ξ)+6 ξ+7)e(−ξ)36 ξ3≥0.1272785\displaystyle -\frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} + \frac{{\left(8 \, {\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} + 6 \, \xi + 7\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.1272785
(2 ξ−1)e(3 ξ)4 ξ3−(4 (3 ξ−2)e(3 ξ)−6 ξ−1)e(−ξ)36 ξ3≥0.2797267\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(4 \, {\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} - 6 \, \xi - 1\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.2797267
Surface is stable!

No. 11: Degree 22 / Singularity type D5A1D_5A_1.

The combinatorial data is is given by □=[−1,2]\Box=[-1,2] and Φ0(u)=min⁡{u−12,0},  Φ∞(u)=min⁡{u−12,0},  Φ1(u)=−2u−23\Phi_0(u) = \min \{\frac{u-1}{2},0\},\; \Phi_\infty(u)= \min \{\frac{u-1}{2},0\},\; \Phi_1(u) = \frac{-2u-2}{3}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,2]

# 3 PL functions on the interval
Phi1(u)=min_symbolic((u-1)/2,0)
Phi2(u)=min_symbolic((u-1)/2,0)
Phi3(u)=1/3*(-2*u-2)

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 2\displaystyle 2]
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u ↦ −23 u−23\displaystyle u \ {\mapsto}\ -\frac{2}{3} \, u - \frac{2}{3}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
4 (ξ−1)e(2 ξ)3 ξ3−(3 (ξ−2)e(2 ξ)+ξ+2)e(−ξ)3 ξ3\displaystyle \frac{4 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)}}{3 \, \xi^{3}} - \frac{{\left(3 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{3 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.00001
xi_2=xi_1+0.00002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.69131741884 Interval containing solution: [-1.69132743 .. -1.69130739]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
2 (3 ξ−2)e(2 ξ)9 ξ3−2 (3 (ξ−1)e(2 ξ)+1)e(−ξ)9 ξ3≥0.1919413\displaystyle \frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} - \frac{2 \, {\left(3 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 1\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \geq 0.1919413
2 (3 ξ−2)e(2 ξ)9 ξ3−2 (3 (ξ−1)e(2 ξ)+1)e(−ξ)9 ξ3≥0.1919413\displaystyle \frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} - \frac{2 \, {\left(3 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 1\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \geq 0.1919413
−2 (3 ξ−2)e(2 ξ)9 ξ3+(9 (ξ−1)e(2 ξ)+3 ξ+5)e(−ξ)9 ξ3≥0.1006816\displaystyle -\frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} + \frac{{\left(9 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 3 \, \xi + 5\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \geq 0.1006816
2 (3 ξ−2)e(2 ξ)9 ξ3−(3 (ξ−1)e(2 ξ)−3 ξ−1)e(−ξ)9 ξ3≥0.4845700\displaystyle \frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} - \frac{{\left(3 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} - 3 \, \xi - 1\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \geq 0.4845700
Surface is stable!

No. 12: Degree 22 / Singularity type 3A1D43A_1D_4.

The combinatorial data is is given by □=[−1,1]\Box=[-1,1] and Φ0(u)=u−12,  Φ∞(u)=u−12,  Φ1(u)=−u−12\Phi_0(u) = \frac{u-1}{2},\; \Phi_\infty(u)= \frac{u-1}{2},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=1/2*(u-1)
Phi2(u)=1/2*(u-1)
Phi3(u)=1/2*(-u-1)

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print ("The combinatorial data")
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ 12 u−12\displaystyle u \ {\mapsto}\ \frac{1}{2} \, u - \frac{1}{2}
u ↦ 12 u−12\displaystyle u \ {\mapsto}\ \frac{1}{2} \, u - \frac{1}{2}
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
(12 ξ2−7 ξ+2)e(3 ξ)2 ξ3−(ξ+2)e(−ξ)2 ξ3\displaystyle \frac{{\left(12 \, \xi^{2} - 7 \, \xi + 2\right)} e^{\left(3 \, \xi\right)}}{2 \, \xi^{3}} - \frac{{\left(\xi + 2\right)} e^{\left(-\xi\right)}}{2 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print ("Numerical solution:",sol)

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print ("Interval containing solution:", xi0.str(style='brackets'))
Numerical solution: -1.9769185047809856 Interval containing solution: [-1.97691861 .. -1.97691839]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
(8 ξ2−4 ξ+1)e(3 ξ)4 ξ3−e(−ξ)4 ξ3≥0.2301820\displaystyle \frac{{\left(8 \, \xi^{2} - 4 \, \xi + 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.2301820
(8 ξ2−4 ξ+1)e(3 ξ)4 ξ3−e(−ξ)4 ξ3≥0.2301820\displaystyle \frac{{\left(8 \, \xi^{2} - 4 \, \xi + 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.2301820
−(16 ξ2−10 ξ+3)e(3 ξ)4 ξ3+(2 ξ+3)e(−ξ)4 ξ3≥0.2301820\displaystyle -\frac{{\left(16 \, \xi^{2} - 10 \, \xi + 3\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} + \frac{{\left(2 \, \xi + 3\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.2301820
(2 ξ−1)e(3 ξ)4 ξ3+(2 ξ+1)e(−ξ)4 ξ3≥0.6905461\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} + \frac{{\left(2 \, \xi + 1\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.6905461
Surface is stable!

No. 13: Degree 33 / Singularity type A5A1A_5A_1.

The combinatorial data is is given by □=[−1,3]\Box=[-1,3] and Φ0(u)=min⁡{−u,0},  Φ∞(u)=u−34,  Φ1(u)=u−12\Phi_0(u) = \min \{-u,0\},\; \Phi_\infty(u)= \frac{u-3}{4},\; \Phi_1(u) = \frac{u-1}{2}

# We are working with interval arithmetic with precision of 16 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(11)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(-u,0)
Phi2(u)=1/4*u-3/4
Phi3(u)=1/2*u-1/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ min⁡(−u,0)\displaystyle u \ {\mapsto}\ \min\left(-u, 0\right)
u ↦ 14 u−34\displaystyle u \ {\mapsto}\ \frac{1}{4} \, u - \frac{3}{4}
u ↦ 12 u−12\displaystyle u \ {\mapsto}\ \frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])

show(F(xi))
−3 (ξ+2)e(−ξ)4 ξ3+(3 ξ−2)e(3 ξ)+84 ξ3\displaystyle -\frac{3 \, {\left(\xi + 2\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} + \frac{{\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} + 8}{4 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0001
xi_2=sol+0.0001

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0: 
    print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.24607304295 Interval containing solution: [-1.2471 .. -1.2451]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
3 (4 ξ+3)e(−ξ)16 ξ3−(8 ξ−7)e(3 ξ)+16 ξ+1616 ξ3≥0.521\displaystyle \frac{3 \, {\left(4 \, \xi + 3\right)} e^{\left(-\xi\right)}}{16 \, \xi^{3}} - \frac{{\left(8 \, \xi - 7\right)} e^{\left(3 \, \xi\right)} + 16 \, \xi + 16}{16 \, \xi^{3}} \geq 0.521
(4 ξ−3)e(3 ξ)+8 ξ16 ξ3+3 e(−ξ)16 ξ3≥(−0.0120)\displaystyle \frac{{\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)} + 8 \, \xi}{16 \, \xi^{3}} + \frac{3 \, e^{\left(-\xi\right)}}{16 \, \xi^{3}} \geq \left(-0.0120\right)
(8 ξ−5)e(3 ξ)+4 ξ+816 ξ3−3 e(−ξ)16 ξ3≥0.248\displaystyle \frac{{\left(8 \, \xi - 5\right)} e^{\left(3 \, \xi\right)} + 4 \, \xi + 8}{16 \, \xi^{3}} - \frac{3 \, e^{\left(-\xi\right)}}{16 \, \xi^{3}} \geq 0.248
3 (4 ξ+3)e(−ξ)16 ξ3+(4 ξ−1)e(3 ξ)−4 ξ−816 ξ3≥0.764\displaystyle \frac{3 \, {\left(4 \, \xi + 3\right)} e^{\left(-\xi\right)}}{16 \, \xi^{3}} + \frac{{\left(4 \, \xi - 1\right)} e^{\left(3 \, \xi\right)} - 4 \, \xi - 8}{16 \, \xi^{3}} \geq 0.764
Surface is not stable!
(4 ξ−3)e(3 ξ)+8 ξ16 ξ3+3 e(−ξ)16 ξ3≤(−0.00537)\displaystyle \frac{{\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)} + 8 \, \xi}{16 \, \xi^{3}} + \frac{3 \, e^{\left(-\xi\right)}}{16 \, \xi^{3}} \leq \left(-0.00537\right)

No. 14: Degree 33 / Singularity type E6E_6.

The combinatorial data is is given by □=[−1,5]\Box=[-1,5] and Φ0(u)=min⁡{u−13,0},  Φ∞(u)=u−23,  Φ1(u)=−u−12\Phi_0(u) = \min \{\frac{u-1}{3},0\},\; \Phi_\infty(u)= \frac{u-2}{3},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=min_symbolic((u-2)/3,0)
Phi2(u)=(u-2)/3
Phi3(u)=(-u-1)/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ min⁡(13 u−23,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{3} \, u - \frac{2}{3}, 0\right)
u ↦ 13 u−23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
(6 ξ2+ξ−2)e(3 ξ)6 ξ3−(4 (ξ−1)e(3 ξ)+ξ+2)e(−ξ)6 ξ3\displaystyle \frac{{\left(6 \, \xi^{2} + \xi - 2\right)} e^{\left(3 \, \xi\right)}}{6 \, \xi^{3}} - \frac{{\left(4 \, {\left(\xi - 1\right)} e^{\left(3 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{6 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.95932042376 Interval containing solution: [-1.95932055 .. -1.95932030]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
(10 ξ2+4 ξ−1)e(3 ξ)36 ξ3−(4 (3 ξ−1)e(3 ξ)+3)e(−ξ)36 ξ3≥0.07627603\displaystyle \frac{{\left(10 \, \xi^{2} + 4 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{36 \, \xi^{3}} - \frac{{\left(4 \, {\left(3 \, \xi - 1\right)} e^{\left(3 \, \xi\right)} + 3\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.07627603
(14 ξ2+2 ξ−5)e(3 ξ)36 ξ3−(2 (3 ξ−4)e(3 ξ)+3)e(−ξ)36 ξ3≥0.07668713\displaystyle \frac{{\left(14 \, \xi^{2} + 2 \, \xi - 5\right)} e^{\left(3 \, \xi\right)}}{36 \, \xi^{3}} - \frac{{\left(2 \, {\left(3 \, \xi - 4\right)} e^{\left(3 \, \xi\right)} + 3\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.07668713
−(14 ξ2+2 ξ−5)e(3 ξ)36 ξ3+(12 (ξ−1)e(3 ξ)+6 ξ+7)e(−ξ)36 ξ3≥0.1276748\displaystyle -\frac{{\left(14 \, \xi^{2} + 2 \, \xi - 5\right)} e^{\left(3 \, \xi\right)}}{36 \, \xi^{3}} + \frac{{\left(12 \, {\left(\xi - 1\right)} e^{\left(3 \, \xi\right)} + 6 \, \xi + 7\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.1276748
(10 ξ2+4 ξ−1)e(3 ξ)36 ξ3−(6 ξe(3 ξ)−6 ξ−1)e(−ξ)36 ξ3≥0.2806380\displaystyle \frac{{\left(10 \, \xi^{2} + 4 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{36 \, \xi^{3}} - \frac{{\left(6 \, \xi e^{\left(3 \, \xi\right)} - 6 \, \xi - 1\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.2806380
Surface is stable!

No. 15: Degree 33 / Singularity type A4A1A_4A_1.

The combinatorial data is is given by □=[−1,2]\Box=[-1,2] and Φ0(u)=min⁡{u−12,0},  Φ∞(u)=min⁡{−u,0},  Φ1(u)=u−23\Phi_0(u) = \min \{\frac{u-1}{2},0\},\; \Phi_\infty(u)= \min \{-u,0\},\; \Phi_1(u) = \frac{u-2}{3}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,2]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(1/2*u-1/2,0)
Phi2(u)=min_symbolic(0,-u)
Phi3(u)=1/3*u-2/3

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 2\displaystyle 2]
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u ↦ min⁡(0,−u)\displaystyle u \ {\mapsto}\ \min\left(0, -u\right)
u ↦ 13 u−23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
−(3 (ξ−2)e(2 ξ)+5 ξ+10)e(−ξ)6 ξ3+2 (2 (ξ−1)e(2 ξ)+3)3 ξ3\displaystyle -\frac{{\left(3 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + 5 \, \xi + 10\right)} e^{\left(-\xi\right)}}{6 \, \xi^{3}} + \frac{2 \, {\left(2 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 3\right)}}{3 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.19618719651 Interval containing solution: [-1.19618732 .. -1.19618707]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
−(9 (2 ξ−1)e(2 ξ)+5)e(−ξ)36 ξ3+2 (3 ξ−2)e(2 ξ)+3 ξ+39 ξ3≥0.2231133\displaystyle -\frac{{\left(9 \, {\left(2 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} + 5\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} + \frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)} + 3 \, \xi + 3}{9 \, \xi^{3}} \geq 0.2231133
(3 (4 ξ−7)e(2 ξ)+30 ξ+25)e(−ξ)36 ξ3−2 (3 ξ−4)e(2 ξ)+9 ξ+99 ξ3≥0.5531086\displaystyle \frac{{\left(3 \, {\left(4 \, \xi - 7\right)} e^{\left(2 \, \xi\right)} + 30 \, \xi + 25\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} - \frac{2 \, {\left(3 \, \xi - 4\right)} e^{\left(2 \, \xi\right)} + 9 \, \xi + 9}{9 \, \xi^{3}} \geq 0.5531086
(27 ξe(2 ξ)−18 (ξ+2)eξ+10)e(−ξ)72 ξ3+16 (3 ξ−4)e(2 ξ)−27 (ξ−2)eξ+54 ξ+3672 ξ3≥0.03862803\displaystyle \frac{{\left(27 \, \xi e^{\left(2 \, \xi\right)} - 18 \, {\left(\xi + 2\right)} e^{\xi} + 10\right)} e^{\left(-\xi\right)}}{72 \, \xi^{3}} + \frac{16 \, {\left(3 \, \xi - 4\right)} e^{\left(2 \, \xi\right)} - 27 \, {\left(\xi - 2\right)} e^{\xi} + 54 \, \xi + 36}{72 \, \xi^{3}} \geq 0.03862803
(3 (5 ξ−8)e(2 ξ)−18 (ξ+2)eξ+60 ξ+50)e(−ξ)72 ξ3+16 (3 ξ−2)e(2 ξ)−27 (ξ−2)eξ+6 ξ−1272 ξ3≥0.8148505\displaystyle \frac{{\left(3 \, {\left(5 \, \xi - 8\right)} e^{\left(2 \, \xi\right)} - 18 \, {\left(\xi + 2\right)} e^{\xi} + 60 \, \xi + 50\right)} e^{\left(-\xi\right)}}{72 \, \xi^{3}} + \frac{16 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)} - 27 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi - 12}{72 \, \xi^{3}} \geq 0.8148505
Surface is stable!

No. 17: Degree 33 / Singularity type D5D_5.

The combinatorial data is is given by □=[−1,3]\Box=[-1,3] and Φ0(u)=min⁡{u−23,0},  Φ∞(u)=min⁡{u−12,0},  Φ1(u)=−u−12\Phi_0(u) = \min \{\frac{u-2}{3},0\},\; \Phi_\infty(u)= \min \{\frac{u-1}{2},0\},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=min_symbolic((u-2)/3,0)
Phi2(u)=min_symbolic((u-1)/2,0)
Phi3(u)=(-u-1)/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ min⁡(13 u−23,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{3} \, u - \frac{2}{3}, 0\right)
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
(3 ξ−2)e(3 ξ)2 ξ3−(4 (ξ−1)e(3 ξ)+3 (ξ−2)e(2 ξ)+2 ξ+4)e(−ξ)6 ξ3\displaystyle \frac{{\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)}}{2 \, \xi^{3}} - \frac{{\left(4 \, {\left(\xi - 1\right)} e^{\left(3 \, \xi\right)} + 3 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + 2 \, \xi + 4\right)} e^{\left(-\xi\right)}}{6 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.83879135601 Interval containing solution: [-1.83879146 .. -1.83879125]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
(2 ξ−1)e(3 ξ)4 ξ3−(4 (3 ξ−1)e(3 ξ)−9 e(2 ξ)+4)e(−ξ)36 ξ3≥0.1038101\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(4 \, {\left(3 \, \xi - 1\right)} e^{\left(3 \, \xi\right)} - 9 \, e^{\left(2 \, \xi\right)} + 4\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.1038101
(2 ξ−1)e(3 ξ)4 ξ3−(2 (3 ξ−4)e(3 ξ)+9 (2 ξ−1)e(2 ξ)+8)e(−ξ)36 ξ3≥0.1934829\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, {\left(3 \, \xi - 4\right)} e^{\left(3 \, \xi\right)} + 9 \, {\left(2 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} + 8\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.1934829
((27 ξ−14)e(3 ξ)+30 (ξ−1)e(2 ξ)+24 ξ+32)e(−ξ)72 ξ3−6 (2 ξ−1)e(3 ξ)+(ξ−2)e(2 ξ)+2 (ξ+2)eξ24 ξ3≥0.2032122\displaystyle \frac{{\left({\left(27 \, \xi - 14\right)} e^{\left(3 \, \xi\right)} + 30 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 24 \, \xi + 32\right)} e^{\left(-\xi\right)}}{72 \, \xi^{3}} - \frac{6 \, {\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)} + {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + 2 \, {\left(\xi + 2\right)} e^{\xi}}{24 \, \xi^{3}} \geq 0.2032122
−((9 ξ−10)e(3 ξ)+6 (ξ−1)e(2 ξ)−24 ξ−8)e(−ξ)72 ξ3+6 (2 ξ−1)e(3 ξ)−(ξ−2)e(2 ξ)−2 (ξ+2)eξ24 ξ3≥0.5005054\displaystyle -\frac{{\left({\left(9 \, \xi - 10\right)} e^{\left(3 \, \xi\right)} + 6 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} - 24 \, \xi - 8\right)} e^{\left(-\xi\right)}}{72 \, \xi^{3}} + \frac{6 \, {\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)} - {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} - 2 \, {\left(\xi + 2\right)} e^{\xi}}{24 \, \xi^{3}} \geq 0.5005054
Surface is stable!

No. 19: Degree 33 / Singularity type D4D_4.

The combinatorial data is is given by □=[−1,2]\Box=[-1,2] and Φ0(u)=min⁡{−u,−u−12},  Φ∞(u)=min⁡{0,u−12},  Φ1(u)=min⁡{0,u−12}\Phi_0(u) = \min \{-u,\frac{-u-1}{2}\},\; \Phi_\infty(u)= \min \{0,\frac{u-1}{2}\},\; \Phi_1(u) = \min \{0,\frac{u-1}{2}\}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,2]

# 3 PL functions on the interval
Phi1(u)=min_symbolic((-u-1)/2,-u)
Phi2(u)=min_symbolic((u-1)/2,0)
Phi3(u)=min_symbolic((u-1)/2,0)

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print ("The combinatorial data")
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 2\displaystyle 2]
u ↦ min⁡(−12 u−12,−u)\displaystyle u \ {\mapsto}\ \min\left(-\frac{1}{2} \, u - \frac{1}{2}, -u\right)
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
−(3 (ξ−2)e(2 ξ)+2 ξ+4)e(−ξ)4 ξ3+8 (ξ−1)e(2 ξ)−3 (ξ−2)eξ4 ξ3\displaystyle -\frac{{\left(3 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + 2 \, \xi + 4\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} + \frac{8 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} - 3 \, {\left(\xi - 2\right)} e^{\xi}}{4 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print ("Numerical solution:",sol)

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print ("Interval containing solution:", xi0.str(style='brackets'))
('Numerical solution:', -1.6913174188369944) ('Interval containing solution:', '[-1.69131753 .. -1.69131729]')

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
((4 ξ−7)e(2 ξ)+4 ξ+6)e(−ξ)8 ξ3−8 (ξ−1)e(2 ξ)−(4 ξ−7)eξ8 ξ3≥0.2194796\displaystyle \frac{{\left({\left(4 \, \xi - 7\right)} e^{\left(2 \, \xi\right)} + 4 \, \xi + 6\right)} e^{\left(-\xi\right)}}{8 \, \xi^{3}} - \frac{8 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} - {\left(4 \, \xi - 7\right)} e^{\xi}}{8 \, \xi^{3}} \geq 0.2194796
−((2 ξ−5)e(2 ξ)+2)e(−ξ)8 ξ3+8 (ξ−1)e(2 ξ)−(2 ξ−5)eξ8 ξ3≥0.2194797\displaystyle -\frac{{\left({\left(2 \, \xi - 5\right)} e^{\left(2 \, \xi\right)} + 2\right)} e^{\left(-\xi\right)}}{8 \, \xi^{3}} + \frac{8 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} - {\left(2 \, \xi - 5\right)} e^{\xi}}{8 \, \xi^{3}} \geq 0.2194797
−((2 ξ−5)e(2 ξ)+2)e(−ξ)8 ξ3+8 (ξ−1)e(2 ξ)−(2 ξ−5)eξ8 ξ3≥0.2194797\displaystyle -\frac{{\left({\left(2 \, \xi - 5\right)} e^{\left(2 \, \xi\right)} + 2\right)} e^{\left(-\xi\right)}}{8 \, \xi^{3}} + \frac{8 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} - {\left(2 \, \xi - 5\right)} e^{\xi}}{8 \, \xi^{3}} \geq 0.2194797
(4 ξ+3 e(2 ξ)+2)e(−ξ)8 ξ3+8 (ξ−1)e(2 ξ)+3 eξ8 ξ3≥0.6584392\displaystyle \frac{{\left(4 \, \xi + 3 \, e^{\left(2 \, \xi\right)} + 2\right)} e^{\left(-\xi\right)}}{8 \, \xi^{3}} + \frac{8 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 3 \, e^{\xi}}{8 \, \xi^{3}} \geq 0.6584392
Surface is stable!

No. 20: Degree 33 / Singularity type 2A1A32A_1A_3.

The combinatorial data is is given by □=[−1,1]\Box=[-1,1] and Φ0(u)=min⁡{0,u},  Φ∞(u)=u−12,  Φ1(u)=−u−12\Phi_0(u) = \min \{0,u\},\; \Phi_\infty(u)= \frac{u-1}{2},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,1]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(0,u)
Phi2(u)=(u-1)/2
Phi3(u)=(-u-1)/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 1\displaystyle 1]
u ↦ min⁡(0,u)\displaystyle u \ {\mapsto}\ \min\left(0, u\right)
u ↦ 12 u−12\displaystyle u \ {\mapsto}\ \frac{1}{2} \, u - \frac{1}{2}
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
(ξ−1)eξξ2−(ξ−2 eξ+2)e(−ξ)ξ3\displaystyle \frac{{\left(\xi - 1\right)} e^{\xi}}{\xi^{2}} - \frac{{\left(\xi - 2 \, e^{\xi} + 2\right)} e^{\left(-\xi\right)}}{\xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -0.944683075826 Interval containing solution: [-0.944683180 .. -0.944682970]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
eξ2 ξ−((ξ−1)eξ+1)e(−ξ)ξ3≥0.5383074\displaystyle \frac{e^{\xi}}{2 \, \xi} - \frac{{\left({\left(\xi - 1\right)} e^{\xi} + 1\right)} e^{\left(-\xi\right)}}{\xi^{3}} \geq 0.5383074
(ξ−1)eξ2 ξ2+12 ξ2≥0.1366506\displaystyle \frac{{\left(\xi - 1\right)} e^{\xi}}{2 \, \xi^{2}} + \frac{1}{2 \, \xi^{2}} \geq 0.1366506
−(ξ−1)eξ2 ξ2+((ξ−4)eξ+2 ξ+4)e(−ξ)2 ξ3≥0.1366498\displaystyle -\frac{{\left(\xi - 1\right)} e^{\xi}}{2 \, \xi^{2}} + \frac{{\left({\left(\xi - 4\right)} e^{\xi} + 2 \, \xi + 4\right)} e^{\left(-\xi\right)}}{2 \, \xi^{3}} \geq 0.1366498
eξ2 ξ+(ξ−eξ+1)e(−ξ)ξ3≥0.8116092\displaystyle \frac{e^{\xi}}{2 \, \xi} + \frac{{\left(\xi - e^{\xi} + 1\right)} e^{\left(-\xi\right)}}{\xi^{3}} \geq 0.8116092
Surface is stable!

No. 21: Degree 44 / Singularity type D5D_5.

The combinatorial data is is given by □=[−1,5]\Box=[-1,5] and Φ0(u)=min⁡{u−12,0},  Φ∞(u)=u−23,  Φ1(u)=−u−12\Phi_0(u) = \min \{\frac{u-1}{2},0\},\; \Phi_\infty(u)= \frac{u-2}{3},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,5]

# 3 PL functions on the interval
Phi1(u)=min_symbolic((u-1)/2,0)
Phi2(u)=(u-2)/3
Phi3(u)=(-u-1)/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 5\displaystyle 5]
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u ↦ 13 u−23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
(5 ξ−2)e(5 ξ)6 ξ3−(3 (ξ−2)e(2 ξ)+2 ξ+4)e(−ξ)6 ξ3\displaystyle \frac{{\left(5 \, \xi - 2\right)} e^{\left(5 \, \xi\right)}}{6 \, \xi^{3}} - \frac{{\left(3 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + 2 \, \xi + 4\right)} e^{\left(-\xi\right)}}{6 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.85969017295 Interval containing solution: [-1.85969028 .. -1.85969007]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations: 
# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,2]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(0,u)
Phi2(u)=min_symbolic(0,u)
Phi3(u)=-2/3*u-2/3

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 2\displaystyle 2]
u ↦ min⁡(0,u)\displaystyle u \ {\mapsto}\ \min\left(0, u\right)
u ↦ min⁡(0,u)\displaystyle u \ {\mapsto}\ \min\left(0, u\right)
u ↦ −23 u−23\displaystyle u \ {\mapsto}\ -\frac{2}{3} \, u - \frac{2}{3}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
4 (ξ−1)e(2 ξ)3 ξ3−4 (ξ−3 eξ+2)e(−ξ)3 ξ3\displaystyle \frac{4 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)}}{3 \, \xi^{3}} - \frac{4 \, {\left(\xi - 3 \, e^{\xi} + 2\right)} e^{\left(-\xi\right)}}{3 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -0.97052220918 Interval containing solution: [-0.970522315 .. -0.970522105]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
2 (3 ξ−2)e(2 ξ)9 ξ3−2 (3 (ξ−2)eξ+4)e(−ξ)9 ξ3≥0.5714598\displaystyle \frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} - \frac{2 \, {\left(3 \, {\left(\xi - 2\right)} e^{\xi} + 4\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \geq 0.5714598
2 (3 ξ−2)e(2 ξ)9 ξ3−2 (3 (ξ−2)eξ+4)e(−ξ)9 ξ3≥0.5714598\displaystyle \frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} - \frac{2 \, {\left(3 \, {\left(\xi - 2\right)} e^{\xi} + 4\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \geq 0.5714598
−2 (3 ξ−2)e(2 ξ)9 ξ3+2 (9 (ξ−2)eξ+6 ξ+16)e(−ξ)9 ξ3≥(−0.2018674)\displaystyle -\frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} + \frac{2 \, {\left(9 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi + 16\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \geq \left(-0.2018674\right)
2 (3 ξ−2)e(2 ξ)9 ξ3+2 (3 (ξ−2)eξ+6 ξ+8)e(−ξ)9 ξ3≥0.9410544\displaystyle \frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} + \frac{2 \, {\left(3 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi + 8\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \geq 0.9410544
Surface is not stable!
−2 (3 ξ−2)e(2 ξ)9 ξ3+2 (9 (ξ−2)eξ+6 ξ+16)e(−ξ)9 ξ3≤(−0.2018636)\displaystyle -\frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} + \frac{2 \, {\left(9 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi + 16\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \leq \left(-0.2018636\right)

No. 23: Degree 44 / Singularity type D4D_4.

The combinatorial data is is given by □=[−1,3]\Box=[-1,3] and Φ0(u)=min⁡{u−12,0},  Φ∞(u)=min⁡{u−12,0},  Φ1(u)=−u−12\Phi_0(u) = \min \{\frac{u-1}{2},0\},\; \Phi_\infty(u)= \min \{\frac{u-1}{2},0\},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=min_symbolic((u-1)/2,0)
Phi2(u)=min_symbolic((u-1)/2,0)
Phi3(u)=(-u-1)/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
(3 ξ−2)e(3 ξ)2 ξ3−(2 (ξ−2)e(2 ξ)+ξ+2)e(−ξ)2 ξ3\displaystyle \frac{{\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)}}{2 \, \xi^{3}} - \frac{{\left(2 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{2 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.79675598472 Interval containing solution: [-1.79675609 .. -1.79675588]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
(2 ξ−1)e(3 ξ)4 ξ3−(2 (ξ−1)e(2 ξ)+1)e(−ξ)4 ξ3≥0.2208160\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 1\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.2208160
(2 ξ−1)e(3 ξ)4 ξ3−(2 (ξ−1)e(2 ξ)+1)e(−ξ)4 ξ3≥0.2208160\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 1\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.2208160
−(2 ξ−1)e(3 ξ)4 ξ3+(4 (ξ−1)e(2 ξ)+2 ξ+3)e(−ξ)4 ξ3≥0.2333050\displaystyle -\frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} + \frac{{\left(4 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 2 \, \xi + 3\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.2333050
(2 ξ−1)e(3 ξ)4 ξ3+(2 ξ+1)e(−ξ)4 ξ3≥0.6749371\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} + \frac{{\left(2 \, \xi + 1\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.6749371
Surface is stable!

No. 24: Degree 44 / Singularity type A4A_4.

The combinatorial data is is given by □=[−1,3]\Box=[-1,3] and Φ0(u)=min⁡{u−23,0},  Φ∞(u)=min⁡{−u,0},  Φ1(u)=−u−12\Phi_0(u) = \min \{\frac{u-2}{3},0\},\; \Phi_\infty(u)= \min \{-u,0\},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(0,1/3*u-2/3)
Phi2(u)=min_symbolic(-u,0)
Phi3(u)=1/2*u-1/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ min⁡(0,13 u−23)\displaystyle u \ {\mapsto}\ \min\left(0, \frac{1}{3} \, u - \frac{2}{3}\right)
u ↦ min⁡(−u,0)\displaystyle u \ {\mapsto}\ \min\left(-u, 0\right)
u ↦ 12 u−12\displaystyle u \ {\mapsto}\ \frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
−(4 (ξ−1)e(3 ξ)+5 ξ+10)e(−ξ)6 ξ3+(3 ξ−2)e(3 ξ)+42 ξ3\displaystyle -\frac{{\left(4 \, {\left(\xi - 1\right)} e^{\left(3 \, \xi\right)} + 5 \, \xi + 10\right)} e^{\left(-\xi\right)}}{6 \, \xi^{3}} + \frac{{\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} + 4}{2 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.38176017478 Interval containing solution: [-1.38176030 .. -1.38176006]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
−(4 (3 ξ−1)e(3 ξ)−5)e(−ξ)36 ξ3+(2 ξ−1)e(3 ξ)+2 ξ4 ξ3≥0.04422930\displaystyle -\frac{{\left(4 \, {\left(3 \, \xi - 1\right)} e^{\left(3 \, \xi\right)} - 5\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} + \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)} + 2 \, \xi}{4 \, \xi^{3}} \geq 0.04422930
(2 (9 ξ−8)e(3 ξ)+30 ξ+25)e(−ξ)36 ξ3−(4 ξ−3)e(3 ξ)+4 ξ+44 ξ3≥0.5594457\displaystyle \frac{{\left(2 \, {\left(9 \, \xi - 8\right)} e^{\left(3 \, \xi\right)} + 30 \, \xi + 25\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} - \frac{{\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)} + 4 \, \xi + 4}{4 \, \xi^{3}} \geq 0.5594457
(2 (3 ξ+1)e(3 ξ)−12 (ξ+1)eξ−5)e(−ξ)36 ξ3+3 (4 ξ−3)e(3 ξ)−6 (ξ−1)e(2 ξ)+8 ξ+812 ξ3≥0.2463752\displaystyle \frac{{\left(2 \, {\left(3 \, \xi + 1\right)} e^{\left(3 \, \xi\right)} - 12 \, {\left(\xi + 1\right)} e^{\xi} - 5\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} + \frac{3 \, {\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)} - 6 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 8 \, \xi + 8}{12 \, \xi^{3}} \geq 0.2463752
(2 (6 ξ−5)e(3 ξ)−12 (ξ+1)eξ+30 ξ+25)e(−ξ)36 ξ3+3 (2 ξ−1)e(3 ξ)−6 (ξ−1)e(2 ξ)+2 ξ−412 ξ3≥0.8500507\displaystyle \frac{{\left(2 \, {\left(6 \, \xi - 5\right)} e^{\left(3 \, \xi\right)} - 12 \, {\left(\xi + 1\right)} e^{\xi} + 30 \, \xi + 25\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} + \frac{3 \, {\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)} - 6 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 2 \, \xi - 4}{12 \, \xi^{3}} \geq 0.8500507
Surface is stable!

No. 26: Degree 44 / Singularity type A3A_3.

The combinatorial data is is given by □=[−1,2]\Box=[-1,2] and Φ0(u)=min⁡{−u,0},  Φ∞(u)=min⁡{0,u−12},  Φ1(u)=min⁡{0,u−12}\Phi_0(u) = \min \{-u,0\},\; \Phi_\infty(u)= \min \{0,\frac{u-1}{2}\},\; \Phi_1(u) = \min \{0,\frac{u-1}{2}\}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,2]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(-u,0)
Phi2(u)=min_symbolic((u-1)/2,0)
Phi3(u)=min_symbolic((u-1)/2,0)

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 2\displaystyle 2]
u ↦ min⁡(−u,0)\displaystyle u \ {\mapsto}\ \min\left(-u, 0\right)
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u ↦ min⁡(12 u−12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
−((ξ−2)e(2 ξ)+ξ+2)e(−ξ)ξ3+2 ((ξ−1)e(2 ξ)+1)ξ3\displaystyle -\frac{{\left({\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{\xi^{3}} + \frac{2 \, {\left({\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 1\right)}}{\xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.31047894021 Interval containing solution: [-1.31047905 .. -1.31047883]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
((ξ−1)e(2 ξ)+ξ+1)e(−ξ)ξ3−(ξ−1)e(2 ξ)+ξ+1ξ3≥0.5757800\displaystyle \frac{{\left({\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + \xi + 1\right)} e^{\left(-\xi\right)}}{\xi^{3}} - \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + \xi + 1}{\xi^{3}} \geq 0.5757800
8 (ξ−1)e(2 ξ)−3 (ξ−2)eξ+6 ξ+48 ξ3−(ξ−2)eξ+2 ξ+48 ξ3≥0.1674627\displaystyle \frac{8 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} - 3 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi + 4}{8 \, \xi^{3}} - \frac{{\left(\xi - 2\right)} e^{\xi} + 2 \, \xi + 4}{8 \, \xi^{3}} \geq 0.1674627
8 (ξ−1)e(2 ξ)−3 (ξ−2)eξ+6 ξ+48 ξ3−(ξ−2)eξ+2 ξ+48 ξ3≥0.1674627\displaystyle \frac{8 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} - 3 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi + 4}{8 \, \xi^{3}} - \frac{{\left(\xi - 2\right)} e^{\xi} + 2 \, \xi + 4}{8 \, \xi^{3}} \geq 0.1674627
((3 ξ−2)e(2 ξ)−2 (ξ+2)eξ+4 ξ+4)e(−ξ)4 ξ3+4 (ξ−1)e(2 ξ)−3 (ξ−2)eξ+2 ξ4 ξ3≥0.9107058\displaystyle \frac{{\left({\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)} - 2 \, {\left(\xi + 2\right)} e^{\xi} + 4 \, \xi + 4\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} + \frac{4 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} - 3 \, {\left(\xi - 2\right)} e^{\xi} + 2 \, \xi}{4 \, \xi^{3}} \geq 0.9107058
Surface is stable!

No. 28: Degree 44 / Singularity type A1A2A_1A_2.

The combinatorial data is is given by □=[−1,1]\Box=[-1,1] and Φ0(u)=min⁡{0,u},  Φ∞(u)=min⁡{0,u},  Φ1(u)=−u−12\Phi_0(u) = \min \{0,u\},\; \Phi_\infty(u)= \min \{0,u\},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,1]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(0,u)
Phi2(u)=min_symbolic(0,u)
Phi3(u)=-1/2*u-1/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 1\displaystyle 1]
u ↦ min⁡(0,u)\displaystyle u \ {\mapsto}\ \min\left(0, u\right)
u ↦ min⁡(0,u)\displaystyle u \ {\mapsto}\ \min\left(0, u\right)
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
−(3 ξ−8 eξ+6)e(−ξ)2 ξ3+(2 ξ2−ξ−2)eξ2 ξ3\displaystyle -\frac{{\left(3 \, \xi - 8 \, e^{\xi} + 6\right)} e^{\left(-\xi\right)}}{2 \, \xi^{3}} + \frac{{\left(2 \, \xi^{2} - \xi - 2\right)} e^{\xi}}{2 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -0.743738705959 Interval containing solution: [-0.743738816 .. -0.743738591]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
−(2 (ξ−2)eξ+3)e(−ξ)4 ξ3+(2 ξ2−1)eξ4 ξ3≥0.4699562\displaystyle -\frac{{\left(2 \, {\left(\xi - 2\right)} e^{\xi} + 3\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} + \frac{{\left(2 \, \xi^{2} - 1\right)} e^{\xi}}{4 \, \xi^{3}} \geq 0.4699562
−(2 (ξ−2)eξ+3)e(−ξ)4 ξ3+(2 ξ2−1)eξ4 ξ3≥0.4699562\displaystyle -\frac{{\left(2 \, {\left(\xi - 2\right)} e^{\xi} + 3\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} + \frac{{\left(2 \, \xi^{2} - 1\right)} e^{\xi}}{4 \, \xi^{3}} \geq 0.4699562
(8 (ξ−2)eξ+6 ξ+15)e(−ξ)4 ξ3−(2 ξ2−1)eξ4 ξ3≥(−0.1023107)\displaystyle \frac{{\left(8 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi + 15\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, \xi^{2} - 1\right)} e^{\xi}}{4 \, \xi^{3}} \geq \left(-0.1023107\right)
(4 (ξ−2)eξ+6 ξ+9)e(−ξ)4 ξ3+(2 ξ2−1)eξ4 ξ3≥0.8376046\displaystyle \frac{{\left(4 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi + 9\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} + \frac{{\left(2 \, \xi^{2} - 1\right)} e^{\xi}}{4 \, \xi^{3}} \geq 0.8376046
Surface is not stable!
(8 (ξ−2)eξ+6 ξ+15)e(−ξ)4 ξ3−(2 ξ2−1)eξ4 ξ3≤(−0.1023035)\displaystyle \frac{{\left(8 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi + 15\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, \xi^{2} - 1\right)} e^{\xi}}{4 \, \xi^{3}} \leq \left(-0.1023035\right)

No. 29: Degree 55 / Singularity type A4A_4.

The combinatorial data is is given by □=[−1,5]\Box=[-1,5] and Φ0(u)=min⁡{−u,0},  Φ∞(u)=u−23,  Φ1(u)=u−12\Phi_0(u) = \min \{-u,0\},\; \Phi_\infty(u)= \frac{u-2}{3},\; \Phi_1(u) = \frac{u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,1]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(0,-u)
Phi2(u)=1/3*u-2/3
Phi3(u)=1/2*u-1/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 1\displaystyle 1]
u ↦ min⁡(0,−u)\displaystyle u \ {\mapsto}\ \min\left(0, -u\right)
u ↦ 13 u−23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}
u ↦ 12 u−12\displaystyle u \ {\mapsto}\ \frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
−5 (ξ+2)e(−ξ)6 ξ3+(4 ξ2−3 ξ−2)eξ+126 ξ3\displaystyle -\frac{5 \, {\left(\xi + 2\right)} e^{\left(-\xi\right)}}{6 \, \xi^{3}} + \frac{{\left(4 \, \xi^{2} - 3 \, \xi - 2\right)} e^{\xi} + 12}{6 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -0.830799974476 Interval containing solution: [-0.830800087 .. -0.830799862]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
5 (6 ξ+5)e(−ξ)36 ξ3−(8 ξ2−20 ξ−11)eξ+36 ξ+3636 ξ3≥0.5216795\displaystyle \frac{5 \, {\left(6 \, \xi + 5\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} - \frac{{\left(8 \, \xi^{2} - 20 \, \xi - 11\right)} e^{\xi} + 36 \, \xi + 36}{36 \, \xi^{3}} \geq 0.5216795
(8 ξ2−8 ξ−5)eξ+18 ξ36 ξ3+5 e(−ξ)36 ξ3≥0.01721630\displaystyle \frac{{\left(8 \, \xi^{2} - 8 \, \xi - 5\right)} e^{\xi} + 18 \, \xi}{36 \, \xi^{3}} + \frac{5 \, e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.01721630
(16 ξ2−10 ξ−7)eξ+12 ξ+1236 ξ3−5 e(−ξ)36 ξ3≥0.1968489\displaystyle \frac{{\left(16 \, \xi^{2} - 10 \, \xi - 7\right)} e^{\xi} + 12 \, \xi + 12}{36 \, \xi^{3}} - \frac{5 \, e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.1968489
5 (6 ξ+5)e(−ξ)36 ξ3+(16 ξ2+2 ξ−1)eξ−6 ξ−2436 ξ3≥0.7357463\displaystyle \frac{5 \, {\left(6 \, \xi + 5\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} + \frac{{\left(16 \, \xi^{2} + 2 \, \xi - 1\right)} e^{\xi} - 6 \, \xi - 24}{36 \, \xi^{3}} \geq 0.7357463
Surface is stable!

No. 30: Degree 55 / Singularity type A3A_3.

The combinatorial data is is given by □=[−1,3]\Box=[-1,3] and Φ0(u)=min⁡{u−12,0},  Φ∞(u)=min⁡{0,u},  Φ1(u)=−u−12\Phi_0(u) = \min \{\frac{u-1}{2},0\},\; \Phi_\infty(u)= \min \{0,u\},\; \Phi_1(u) = \frac{-u-1}{2}

# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(0,1/2*u-1/2)
Phi2(u)=min_symbolic(-u,0)
Phi3(u)=1/2*u-1/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ min⁡(0,12 u−12)\displaystyle u \ {\mapsto}\ \min\left(0, \frac{1}{2} \, u - \frac{1}{2}\right)
u ↦ min⁡(−u,0)\displaystyle u \ {\mapsto}\ \min\left(-u, 0\right)
u ↦ 12 u−12\displaystyle u \ {\mapsto}\ \frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
−((ξ−2)e(2 ξ)+2 ξ+4)e(−ξ)2 ξ3+(3 ξ−2)e(3 ξ)+42 ξ3\displaystyle -\frac{{\left({\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + 2 \, \xi + 4\right)} e^{\left(-\xi\right)}}{2 \, \xi^{3}} + \frac{{\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} + 4}{2 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.4388637356 Interval containing solution: [-1.43886385 .. -1.43886363]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
−(2 ξ−1)eξ4 ξ3+(2 ξ−1)e(3 ξ)+2 ξ4 ξ3≥0.1686591\displaystyle -\frac{{\left(2 \, \xi - 1\right)} e^{\xi}}{4 \, \xi^{3}} + \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)} + 2 \, \xi}{4 \, \xi^{3}} \geq 0.1686591
((2 ξ−3)e(2 ξ)+4 ξ+4)e(−ξ)4 ξ3−(4 ξ−3)e(3 ξ)+4 ξ+44 ξ3≥0.5809738\displaystyle \frac{{\left({\left(2 \, \xi - 3\right)} e^{\left(2 \, \xi\right)} + 4 \, \xi + 4\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} - \frac{{\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)} + 4 \, \xi + 4}{4 \, \xi^{3}} \geq 0.5809738
2 (4 ξ−3)e(3 ξ)−3 (ξ−2)eξ+6 ξ+48 ξ3+3 ξeξ−2 ξ−48 ξ3≥0.1915944\displaystyle \frac{2 \, {\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)} - 3 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi + 4}{8 \, \xi^{3}} + \frac{3 \, \xi e^{\xi} - 2 \, \xi - 4}{8 \, \xi^{3}} \geq 0.1915944
((3 ξ−4)e(2 ξ)−2 (ξ+2)eξ+8 ξ+8)e(−ξ)8 ξ3+2 (2 ξ−1)e(3 ξ)−3 (ξ−2)eξ+2 ξ−48 ξ3≥0.9412277\displaystyle \frac{{\left({\left(3 \, \xi - 4\right)} e^{\left(2 \, \xi\right)} - 2 \, {\left(\xi + 2\right)} e^{\xi} + 8 \, \xi + 8\right)} e^{\left(-\xi\right)}}{8 \, \xi^{3}} + \frac{2 \, {\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)} - 3 \, {\left(\xi - 2\right)} e^{\xi} + 2 \, \xi - 4}{8 \, \xi^{3}} \geq 0.9412277
Surface is stable!

No. 31: Degree 55 / Singularity type A2A_2.

The combinatorial data is is given by □=[−1,2]\Box=[-1,2] and Φ0(u)=min⁡{0,u},  Φ∞(u)=min⁡{0,u},  Φ1(u)=min⁡{0,−u−12}\Phi_0(u) = \min \{0,u\},\; \Phi_\infty(u)= \min \{0,u\},\; \Phi_1(u) = \min \{0,\frac{-u-1}{2}\}


# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(0,u)
Phi2(u)=min_symbolic(0,-u)
Phi3(u)=min_symbolic(0,1/2*u-1/2)

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ min⁡(0,u)\displaystyle u \ {\mapsto}\ \min\left(0, u\right)
u ↦ min⁡(0,−u)\displaystyle u \ {\mapsto}\ \min\left(0, -u\right)
u ↦ min⁡(0,12 u−12)\displaystyle u \ {\mapsto}\ \min\left(0, \frac{1}{2} \, u - \frac{1}{2}\right)

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
−((ξ−2)e(2 ξ)+3 ξ−4 eξ+6)e(−ξ)2 ξ3−(3 ξ2−4 ξ+2)e(3 ξ)−2ξ3\displaystyle -\frac{{\left({\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + 3 \, \xi - 4 \, e^{\xi} + 6\right)} e^{\left(-\xi\right)}}{2 \, \xi^{3}} - \frac{{\left(3 \, \xi^{2} - 4 \, \xi + 2\right)} e^{\left(3 \, \xi\right)} - 2}{\xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.000001
xi_2=xi_1+0.000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) > 0 and RIF(F(xi_2)) < 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -0.242677720874 Interval containing solution: [-0.242678725 .. -0.242676720]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
(3 ξe(2 ξ)−2 (5 ξ−2)eξ−6)e(−ξ)8 ξ3−4 (3 ξ2−4 ξ+2)e(3 ξ)+3 (ξ−2)eξ−6 ξ−48 ξ3≥0.4280776\displaystyle \frac{{\left(3 \, \xi e^{\left(2 \, \xi\right)} - 2 \, {\left(5 \, \xi - 2\right)} e^{\xi} - 6\right)} e^{\left(-\xi\right)}}{8 \, \xi^{3}} - \frac{4 \, {\left(3 \, \xi^{2} - 4 \, \xi + 2\right)} e^{\left(3 \, \xi\right)} + 3 \, {\left(\xi - 2\right)} e^{\xi} - 6 \, \xi - 4}{8 \, \xi^{3}} \geq 0.4280776
((5 ξ−4)e(2 ξ)+6 (ξ−2)eξ+12 ξ+18)e(−ξ)8 ξ3+4 (3 ξ2−4 ξ+2)e(3 ξ)−(ξ−2)eξ−10 ξ−128 ξ3≥0.4278600\displaystyle \frac{{\left({\left(5 \, \xi - 4\right)} e^{\left(2 \, \xi\right)} + 6 \, {\left(\xi - 2\right)} e^{\xi} + 12 \, \xi + 18\right)} e^{\left(-\xi\right)}}{8 \, \xi^{3}} + \frac{4 \, {\left(3 \, \xi^{2} - 4 \, \xi + 2\right)} e^{\left(3 \, \xi\right)} - {\left(\xi - 2\right)} e^{\xi} - 10 \, \xi - 12}{8 \, \xi^{3}} \geq 0.4278600
−((2 ξ−1)e(2 ξ)−4 (ξ−1)eξ−3)e(−ξ)4 ξ3−(3 ξ2−4 ξ+2)e(3 ξ)−2 ξ−22 ξ3≥(−0.1469232)\displaystyle -\frac{{\left({\left(2 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} - 4 \, {\left(\xi - 1\right)} e^{\xi} - 3\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} - \frac{{\left(3 \, \xi^{2} - 4 \, \xi + 2\right)} e^{\left(3 \, \xi\right)} - 2 \, \xi - 2}{2 \, \xi^{3}} \geq \left(-0.1469232\right)
((2 ξ−1)e(2 ξ)+2 (ξ−4)eξ+6 ξ+9)e(−ξ)4 ξ3−(3 ξ2−4 ξ+2)e(3 ξ)+(ξ−2)eξ−ξ2 ξ3≥0.7109128\displaystyle \frac{{\left({\left(2 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} + 2 \, {\left(\xi - 4\right)} e^{\xi} + 6 \, \xi + 9\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} - \frac{{\left(3 \, \xi^{2} - 4 \, \xi + 2\right)} e^{\left(3 \, \xi\right)} + {\left(\xi - 2\right)} e^{\xi} - \xi}{2 \, \xi^{3}} \geq 0.7109128
Surface is not stable!
−((2 ξ−1)e(2 ξ)−4 (ξ−1)eξ−3)e(−ξ)4 ξ3−(3 ξ2−4 ξ+2)e(3 ξ)−2 ξ−22 ξ3≤(−0.1458046)\displaystyle -\frac{{\left({\left(2 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} - 4 \, {\left(\xi - 1\right)} e^{\xi} - 3\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} - \frac{{\left(3 \, \xi^{2} - 4 \, \xi + 2\right)} e^{\left(3 \, \xi\right)} - 2 \, \xi - 2}{2 \, \xi^{3}} \leq \left(-0.1458046\right)

No. 32: Degree 55 / Singularity type A1A_1.

The combinatorial data is is given by □=[−1,1]\Box=[-1,1] and Φ0(u)=min⁡{0,u},  Φ∞(u)=min⁡{0,u},  Φ1(u)=min⁡{−u,0}\Phi_0(u) = \min \{0,u\},\; \Phi_\infty(u)= \min \{0,u\},\; \Phi_1(u) = \min \{-u,0\}


# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(0,u)
Phi2(u)=min_symbolic(0,u)
Phi3(u)=min_symbolic(-u,0)

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ min⁡(0,u)\displaystyle u \ {\mapsto}\ \min\left(0, u\right)
u ↦ min⁡(0,u)\displaystyle u \ {\mapsto}\ \min\left(0, u\right)
u ↦ min⁡(−u,0)\displaystyle u \ {\mapsto}\ \min\left(-u, 0\right)

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
−2 (ξ−2 eξ+2)e(−ξ)ξ3−(3 ξ2−4 ξ+2)e(3 ξ)−2ξ3\displaystyle -\frac{2 \, {\left(\xi - 2 \, e^{\xi} + 2\right)} e^{\left(-\xi\right)}}{\xi^{3}} - \frac{{\left(3 \, \xi^{2} - 4 \, \xi + 2\right)} e^{\left(3 \, \xi\right)} - 2}{\xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_2=sol-0.0000001
xi_1=xi_2+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -0.276432541617 Interval containing solution: [-0.276432649 .. -0.276432439]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
−(3 ξ2−4 ξ+2)e(3 ξ)−2 ξ−22 ξ3≥0.1917138\displaystyle -\frac{{\left(3 \, \xi^{2} - 4 \, \xi + 2\right)} e^{\left(3 \, \xi\right)} - 2 \, \xi - 2}{2 \, \xi^{3}} \geq 0.1917138
−(3 ξ2−4 ξ+2)e(3 ξ)−2 ξ−22 ξ3≥0.1917138\displaystyle -\frac{{\left(3 \, \xi^{2} - 4 \, \xi + 2\right)} e^{\left(3 \, \xi\right)} - 2 \, \xi - 2}{2 \, \xi^{3}} \geq 0.1917138
2 ((ξ−2)eξ+ξ+2)e(−ξ)ξ3+(3 ξ2−4 ξ+2)e(3 ξ)−2 ξ−22 ξ3≥0.1916581\displaystyle \frac{2 \, {\left({\left(\xi - 2\right)} e^{\xi} + \xi + 2\right)} e^{\left(-\xi\right)}}{\xi^{3}} + \frac{{\left(3 \, \xi^{2} - 4 \, \xi + 2\right)} e^{\left(3 \, \xi\right)} - 2 \, \xi - 2}{2 \, \xi^{3}} \geq 0.1916581
2 ((ξ−2)eξ+ξ+2)e(−ξ)ξ3−(3 ξ2−4 ξ+2)e(3 ξ)−2 ξ−22 ξ3≥0.5751315\displaystyle \frac{2 \, {\left({\left(\xi - 2\right)} e^{\xi} + \xi + 2\right)} e^{\left(-\xi\right)}}{\xi^{3}} - \frac{{\left(3 \, \xi^{2} - 4 \, \xi + 2\right)} e^{\left(3 \, \xi\right)} - 2 \, \xi - 2}{2 \, \xi^{3}} \geq 0.5751315
Surface is stable!

No. 33: Degree 66 / Singularity type A2A_2.

The combinatorial data is is given by □=[−1,3]\Box=[-1,3] and Φ0(u)=min⁡{0,u},  Φ∞(u)=min⁡{0,u},  Φ1(u)=−u−12\Phi_0(u) = \min \{0,u\},\; \Phi_\infty(u)= \min \{0,u\},\; \Phi_1(u) = \frac{-u-1}{2}


# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,3]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(0,u)
Phi2(u)=min_symbolic(0,u)
Phi3(u)=-1/2*u-1/2

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 3\displaystyle 3]
u ↦ min⁡(0,u)\displaystyle u \ {\mapsto}\ \min\left(0, u\right)
u ↦ min⁡(0,u)\displaystyle u \ {\mapsto}\ \min\left(0, u\right)
u ↦ −12 u−12\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
(3 ξ−2)e(3 ξ)2 ξ3−(3 ξ−8 eξ+6)e(−ξ)2 ξ3\displaystyle \frac{{\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)}}{2 \, \xi^{3}} - \frac{{\left(3 \, \xi - 8 \, e^{\xi} + 6\right)} e^{\left(-\xi\right)}}{2 \, \xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -1.24607304295 Interval containing solution: [-1.24607316 .. -1.24607291]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
(2 ξ−1)e(3 ξ)4 ξ3−(2 (ξ−2)eξ+3)e(−ξ)4 ξ3≥0.5195618\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, {\left(\xi - 2\right)} e^{\xi} + 3\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.5195618
(2 ξ−1)e(3 ξ)4 ξ3−(2 (ξ−2)eξ+3)e(−ξ)4 ξ3≥0.5195618\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, {\left(\xi - 2\right)} e^{\xi} + 3\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.5195618
−(2 ξ−1)e(3 ξ)4 ξ3+(8 (ξ−2)eξ+6 ξ+15)e(−ξ)4 ξ3≥(−0.03507074)\displaystyle -\frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} + \frac{{\left(8 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi + 15\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq \left(-0.03507074\right)
(2 ξ−1)e(3 ξ)4 ξ3+(4 (ξ−2)eξ+6 ξ+9)e(−ξ)4 ξ3≥1.004053\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} + \frac{{\left(4 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi + 9\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 1.004053
Surface is not stable!
−(2 ξ−1)e(3 ξ)4 ξ3+(8 (ξ−2)eξ+6 ξ+15)e(−ξ)4 ξ3≤(−0.03506876)\displaystyle -\frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} + \frac{{\left(8 \, {\left(\xi - 2\right)} e^{\xi} + 6 \, \xi + 15\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \leq \left(-0.03506876\right)

No. 34: Degree 66 / Singularity type A1A_1.

The combinatorial data is is given by □=[−1,2]\Box=[-1,2] and Φ0(u)=min⁡{−u,0},  Φ∞(u)=min⁡{0,u},  Φ1(u)=min⁡{0,u}\Phi_0(u) = \min \{-u,0\},\; \Phi_\infty(u)= \min \{0,u\},\; \Phi_1(u) = \min \{0,u\}


# We are working with interval arithmetic with precision of 26 bit
# this fixes the precision e.g. for evaluations of exponential functions
RIF=RealIntervalField(26)

# The base polytope/interval 
B=[-1,2]

# 3 PL functions on the interval
Phi1(u)=min_symbolic(-u,0)
Phi2(u)=min_symbolic(0,u)
Phi3(u)=min_symbolic(0,u)

# the degree of \bar Phi
degPhi=Phi1+Phi2+Phi3+2

print "The combinatorial data"
show(B)
show(Phi1)
show(Phi2)
show(Phi3)
The combinatorial data
[−1\displaystyle -1, 2\displaystyle 2]
u ↦ min⁡(−u,0)\displaystyle u \ {\mapsto}\ \min\left(-u, 0\right)
u ↦ min⁡(0,u)\displaystyle u \ {\mapsto}\ \min\left(0, u\right)
u ↦ min⁡(0,u)\displaystyle u \ {\mapsto}\ \min\left(0, u\right)

Step (i) -- obtain a closed form for FX,ξ(1)F_{X,\xi}(1)

For this we have to analytically solve the integral ∫□udeg⁡Φˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du.

# The integral can be solved symbolically:
F(xi)=integral(degPhi(u)*u*exp(xi*u),u,-1,B[1])
show(F(xi))
−2 (ξ−2 eξ+2)e(−ξ)ξ3+2 ((ξ−1)e(2 ξ)+1)ξ3\displaystyle -\frac{2 \, {\left(\xi - 2 \, e^{\xi} + 2\right)} e^{\left(-\xi\right)}}{\xi^{3}} + \frac{2 \, {\left({\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 1\right)}}{\xi^{3}}

Step (ii) -- find an estimate for the soliton candidate vector field ξ\xi

# Solve numerically
sol=find_root(F,-10,10)
print "Numerical solution:",sol

# Choose upper and lower bound and
# hope that the exact solution lies in between
xi_1=sol-0.0000001
xi_2=xi_1+0.0000002

#define a real value xi0 between xi_1 and xi_2
# representing the exact solution
xi0=RIF(xi_1,xi_2)

# Check whether Intermediate value theorem guarantees a zero beween
# xi_1 and xi_2, i.e evaluate F at xi_1 and xi_2 using interval arithmetic
if RIF(F(xi_1)) < 0 and RIF(F(xi_2)) > 0:
       print "Interval containing solution:", xi0.str(style='brackets')
Numerical solution: -0.97052220918 Interval containing solution: [-0.970522315 .. -0.970522105]

Step (iii) & (iv) -- obtain closed forms for DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of y∈P1y \in \mathbb P^1 we have to symbolically solve the integrals ∫□euξ⋅((1+Φy(u))2−(1+∑z≠yΦz(u))2)  du\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du, which up to scaling with a positive constant coincide with DF⁡ξ(Xy,0,1)\operatorname{DF}_\xi(\mathcal{X}_{y,0,1}). Then we plug in the estimate for ξ\xi into the resulting expression.

#check for positivity of the DF invariant for our vector field xi
print("Stability test for test configurations:")

#storage for DF values (as intervals)
IDF={}

DF1(xi)=integral(exp(xi*u)*((Phi1+1)(u)^2-(Phi3+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[1]=RIF(DF1(xi0)) # evaluate DF1 at xi0 using interval arithmetic^
show(DF1(xi) >= IDF[1].lower())

DF2(xi)=integral(exp(xi*u)*((Phi2+1)(u)^2-(Phi1+Phi3+1)(u)^2)/2,u,-1,B[1])
IDF[2]=RIF(DF2(xi0))
show(DF2(xi) >= IDF[2].lower())

DF3(xi)=integral(exp(xi*u)*((Phi3+1)(u)^2-(Phi1+Phi2+1)(u)^2)/2,u,-1,B[1])
IDF[3]=RIF(DF3(xi0))
show(DF3(xi) >= IDF[3].lower())

DF4(xi)=integral(exp(xi*u)*((1)^2-(Phi3+Phi2+Phi1+1)(u)^2)/2,u,-1,B[1])
IDF[4]=RIF(DF4(xi0))
show(DF4(xi) >= IDF[4].lower())

if all([DF > 0 for DF in IDF.values()]): 
    print("Surface is stable!")
else:
    if any([DF <= 0 for DF in IDF.values()]):
        print("Surface is not stable!")
        
        # print upper bound for destablising DF value
        if IDF[1] <= 0: show(DF1(xi) <= IDF[1].upper())
        if IDF[2] <= 0: show(DF2(xi) <= IDF[2].upper())
        if IDF[3] <= 0: show(DF3(xi) <= IDF[3].upper())
        if IDF[4] <= 0: show(DF4(xi) <= IDF[4].upper())
    else:
        print("Cannot determine stability")
        # Note, that "not a > 0" does not imply "a <= 0" for intervals
Stability test for test configurations:
2 ((ξ−2)eξ+ξ+2)e(−ξ)ξ3−(ξ−1)e(2 ξ)+ξ+1ξ3≥0.2771940\displaystyle \frac{2 \, {\left({\left(\xi - 2\right)} e^{\xi} + \xi + 2\right)} e^{\left(-\xi\right)}}{\xi^{3}} - \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + \xi + 1}{\xi^{3}} \geq 0.2771940
(ξ−1)e(2 ξ)+ξ+1ξ3≥0.2771960\displaystyle \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + \xi + 1}{\xi^{3}} \geq 0.2771960
(ξ−1)e(2 ξ)+ξ+1ξ3≥0.2771960\displaystyle \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + \xi + 1}{\xi^{3}} \geq 0.2771960
2 ((ξ−2)eξ+ξ+2)e(−ξ)ξ3+(ξ−1)e(2 ξ)+ξ+1ξ3≥0.8315867\displaystyle \frac{2 \, {\left({\left(\xi - 2\right)} e^{\xi} + \xi + 2\right)} e^{\left(-\xi\right)}}{\xi^{3}} + \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + \xi + 1}{\xi^{3}} \geq 0.8315867
Surface is stable!