CoCalc Public Filessurfaces.sagews
Authors: Jacob Cable, Hendrik Suess
Views : 183
Description: On the classification of Kähler-Ricci solitons on Gorenstein del Pezzo surfaces: Calculations for surfaces
Compute Environment: Ubuntu 18.04 (Deprecated)

# 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=$2D_4$
2. Degree 1, Sing=$E_8$
3. Degree 1, Sing=$E_7A_1$
4. Degree 1, Sing=$E_6A_2$
5. Degree 2, Sing=$2A_3A_1$
6. Degree 2, Sing=$A_5A_2$
7. Degree 2, Sing=$D_6A_1$
8. Degree 2, Sing=$E_7$
9. Degree 2, Sing=$E_6$
10. Degree 2, Sing=$2A_3$
11. Degree 2, Sing=$D_5A_1$
12. Degree 2, Sing=$3A_1D_4$
13. Degree 3, Sing=$A_5A_1$
14. Degree 3, Sing=$E_6$
15. Degree 3, Sing=$A_4A_1$
16. Degree 3, Sing=$2A_2A_1$
17. Degree 3, Sing=$D_5$
18. Degree 3, Sing=$2A_2$
19. Degree 3, Sing=$D_4$
20. Degree 3, Sing=$A_32A_1$
21. Degree 4, Sing=$D_5$
22. Degree 4, Sing=$A_3A_1$
23. Degree 4, Sing=$D_4$
24. Degree 4, Sing=$A_4$
25. Degree 4, Sing=$3A_1$
26. Degree 4, Sing=$A_3$
27. Degree 4, Sing=$2A_1$
28. Degree 4, Sing=$A_2A_1$
29. Degree 5, Sing=$A_4$
30. Degree 5, Sing=$A_3$
31. Degree 5, Sing=$A_2$
32. Degree 5, Sing=$A_1$
33. Degree 6, Sing=$A_2$
34. Degree 6, Sing=$A_1$

#### No. 2: Degree $1$ / Singularity type $E_8$.

The combinatorial data is is given by $\Box=[-1,5]$ and $\Phi_0(u) = \min \{\frac{u-4}{5},0\},\; \Phi_\infty(u)= \frac{u-2}{3},\; \Phi_1(u) = -\frac{u+1}{2}$

# 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
[$\displaystyle -1$, $\displaystyle 5$]
$\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{5} \, u - \frac{4}{5}, 0\right)$
$\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}$
$\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}$

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

For this we have to analytically solve the integral $\int_\Box u \deg \bar \Phi (u) e^{\xi u} du$.

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

show(F(xi))

$\displaystyle \frac{25 \, \xi e^{\left(5 \, \xi\right)} - 12 \, \xi e^{\left(4 \, \xi\right)} - 10 \, e^{\left(5 \, \xi\right)} + 6 \, e^{\left(4 \, \xi\right)}}{30 \, \xi^{3}} - \frac{12 \, \xi e^{\left(4 \, \xi\right)} + \xi e^{\left(-\xi\right)} - 6 \, e^{\left(4 \, \xi\right)} + 2 \, e^{\left(-\xi\right)}}{30 \, \xi^{3}}$

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

# 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 $\operatorname{DF}_\xi(X_{y,0,1})$ and plug in the estimate for $\xi$

For every choice of $y \in \mathbb P^1$ we have to symbolically solve the integrals $\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du$, which up to scaling with a positive constant coincide with $\operatorname{DF}_\xi(\mathcal{X}_{y,0,1})$. Then we plug in the estimate for $\xi$ 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:
$\displaystyle -\frac{{\left(18 \, {\left(5 \, \xi - 1\right)} e^{\left(5 \, \xi\right)} + 11\right)} e^{\left(-\xi\right)}}{900 \, \xi^{3}} + \frac{25 \, {\left(6 \, \xi - 1\right)} e^{\left(5 \, \xi\right)} - 18 \, {\left(5 \, \xi - 1\right)} e^{\left(4 \, \xi\right)}}{900 \, \xi^{3}} \geq 0.01128570$
$\displaystyle -\frac{{\left(9 \, {\left(15 \, \xi - 8\right)} e^{\left(5 \, \xi\right)} + 19\right)} e^{\left(-\xi\right)}}{900 \, \xi^{3}} + \frac{25 \, {\left(12 \, \xi - 5\right)} e^{\left(5 \, \xi\right)} - 9 \, {\left(15 \, \xi - 8\right)} e^{\left(4 \, \xi\right)}}{900 \, \xi^{3}} \geq 0.01949453$
$\displaystyle \frac{{\left(6 \, {\left(25 \, \xi - 13\right)} e^{\left(5 \, \xi\right)} + 30 \, \xi + 31\right)} e^{\left(-\xi\right)}}{900 \, \xi^{3}} - \frac{25 \, {\left(12 \, \xi - 5\right)} e^{\left(5 \, \xi\right)} - 6 \, {\left(25 \, \xi - 13\right)} e^{\left(4 \, \xi\right)}}{900 \, \xi^{3}} \geq 0.02975460$
$\displaystyle -\frac{{\left(3 \, {\left(25 \, \xi - 4\right)} e^{\left(5 \, \xi\right)} - 30 \, \xi - 1\right)} e^{\left(-\xi\right)}}{900 \, \xi^{3}} + \frac{25 \, {\left(6 \, \xi - 1\right)} e^{\left(5 \, \xi\right)} - 3 \, {\left(25 \, \xi - 4\right)} e^{\left(4 \, \xi\right)}}{900 \, \xi^{3}} \geq 0.06053484$
Surface is stable!

#### No. 3: Degree $1$ / Singularity type $E_7A_1$.

The combinatorial data is is given by $\Box=[-1,3]$ and $\Phi_0(u) = \min \{\frac{u-2}{3},0\},\; \Phi_\infty(u)= \frac{u-3}{4},\; \Phi_1(u) = \frac{-u-1}{2}$

# 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
[$\displaystyle -1$, $\displaystyle 3$]
$\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{3} \, u - \frac{2}{3}, 0\right)$
$\displaystyle u \ {\mapsto}\ \frac{1}{4} \, u - \frac{3}{4}$
$\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}$

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

For this we have to analytically solve the integral $\int_\Box u \deg \bar \Phi (u) e^{\xi u} du$.

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

$\displaystyle \frac{{\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(8 \, {\left(\xi - 1\right)} e^{\left(3 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{12 \, \xi^{3}}$

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

# 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 $\operatorname{DF}_\xi(X_{y,0,1})$ and plug in the estimate for $\xi$

For every choice of $y \in \mathbb P^1$ we have to symbolically solve the integrals $\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du$, which up to scaling with a positive constant coincide with $\operatorname{DF}_\xi(\mathcal{X}_{y,0,1})$. Then we plug in the estimate for $\xi$ 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:
$\displaystyle \frac{{\left(4 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} - \frac{{\left(16 \, {\left(3 \, \xi - 1\right)} e^{\left(3 \, \xi\right)} + 7\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.04440443$
$\displaystyle \frac{{\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} - \frac{{\left(8 \, {\left(3 \, \xi - 4\right)} e^{\left(3 \, \xi\right)} + 5\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.03181965$
$\displaystyle -\frac{{\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} + \frac{{\left(4 \, {\left(9 \, \xi - 10\right)} e^{\left(3 \, \xi\right)} + 12 \, \xi + 13\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.06993168$
$\displaystyle \frac{{\left(4 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} - \frac{{\left(4 \, {\left(9 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} - 12 \, \xi - 1\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.1461557$
Surface is stable!

#### No. 4: Degree $2$ / Singularity type $E_6A_2$.

The combinatorial data is is given by $\Box=[-1,2]$ and $\Phi_0(u) = \min \{\frac{u-1}{2},0\},\; \Phi_\infty(u)= \frac{u-2}{3},\; \Phi_1(u) = \frac{-2u-2}{3}$

# 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
[$\displaystyle -1$, $\displaystyle 2$]
$\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)$
$\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}$
$\displaystyle u \ {\mapsto}\ -\frac{2}{3} \, u - \frac{2}{3}$

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

For this we have to analytically solve the integral $\int_\Box u \deg \bar \Phi (u) e^{\xi u} du$.

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

$\displaystyle \frac{2 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)}}{3 \, \xi^{3}} - \frac{{\left(3 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{6 \, \xi^{3}}$

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

# 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 $\operatorname{DF}_\xi(X_{y,0,1})$ and plug in the estimate for $\xi$

For every choice of $y \in \mathbb P^1$ we have to symbolically solve the integrals $\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du$, which up to scaling with a positive constant coincide with $\operatorname{DF}_\xi(\mathcal{X}_{y,0,1})$. Then we plug in the estimate for $\xi$ 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:
$\displaystyle \frac{{\left(3 \, \xi - 1\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} - \frac{{\left(9 \, {\left(2 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} + 5\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.1187884$
$\displaystyle \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)}}{3 \, \xi^{3}} - \frac{{\left({\left(2 \, \xi - 5\right)} e^{\left(2 \, \xi\right)} + 1\right)} e^{\left(-\xi\right)}}{12 \, \xi^{3}} \geq 0.07315991$
$\displaystyle -\frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)}}{3 \, \xi^{3}} + \frac{{\left({\left(4 \, \xi - 7\right)} e^{\left(2 \, \xi\right)} + 2 \, \xi + 3\right)} e^{\left(-\xi\right)}}{12 \, \xi^{3}} \geq 0.07315987$
$\displaystyle \frac{{\left(3 \, \xi - 1\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} - \frac{{\left(3 \, {\left(4 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} - 6 \, \xi - 1\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.2651082$
Surface is stable!

#### No. 6: Degree $2$ / Singularity type $A_5A_2$.

The combinatorial data is is given by $\Box=[-1,2]$ and $\Phi_0(u) = \min \{-u,0\},\; \Phi_\infty(u)= \frac{u-2}{3},\; \Phi_1(u) = \frac{u-2}{3}$

# 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)


The combinatorial data
[$\displaystyle -1$, $\displaystyle 2$]
$\displaystyle u \ {\mapsto}\ \min\left(-u, 0\right)$
$\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}$
$\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}$

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

For this we have to analytically solve the integral $\int_\Box u \deg \bar \Phi (u) e^{\xi u} du$.

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

$\displaystyle -\frac{2 \, {\left(\xi + 2\right)} e^{\left(-\xi\right)}}{3 \, \xi^{3}} + \frac{2 \, {\left({\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 3\right)}}{3 \, \xi^{3}}$

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

# 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 $\operatorname{DF}_\xi(X_{y,0,1})$ and plug in the estimate for $\xi$

For every choice of $y \in \mathbb P^1$ we have to symbolically solve the integrals $\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du$, which up to scaling with a positive constant coincide with $\operatorname{DF}_\xi(\mathcal{X}_{y,0,1})$. Then we plug in the estimate for $\xi$ 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:
$\displaystyle \frac{2 \, {\left(3 \, \xi + 2\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} - \frac{{\left(3 \, \xi - 5\right)} e^{\left(2 \, \xi\right)} + 9 \, \xi + 9}{9 \, \xi^{3}} \geq 0.4790611$
$\displaystyle \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + \xi + 1}{3 \, \xi^{3}} \geq 0.09239867$
$\displaystyle \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + \xi + 1}{3 \, \xi^{3}} \geq 0.09239867$
$\displaystyle \frac{2 \, {\left(3 \, \xi + 2\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} + \frac{{\left(3 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} - 3 \, \xi - 3}{9 \, \xi^{3}} \geq 0.6638587$
Surface is stable!

#### No. 7: Degree $2$ / Singularity type $D_6A_1$.

The combinatorial data is is given by $\Box=[-1,3]$ and $\Phi_0(u) = \min \{\frac{u-1}{2},0\},\; \Phi_\infty(u)= \frac{u-3}{4},\; \Phi_1(u) = \frac{-u-1}{2}$

# 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
[$\displaystyle -1$, $\displaystyle 3$]
$\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)$
$\displaystyle u \ {\mapsto}\ \frac{1}{4} \, u - \frac{3}{4}$
$\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}$

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

For this we have to analytically solve the integral $\int_\Box u \deg \bar \Phi (u) e^{\xi u} du$.

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

$\displaystyle \frac{{\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}}$

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

# 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 $\operatorname{DF}_\xi(X_{y,0,1})$ and plug in the estimate for $\xi$

For every choice of $y \in \mathbb P^1$ we have to symbolically solve the integrals $\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du$, which up to scaling with a positive constant coincide with $\operatorname{DF}_\xi(\mathcal{X}_{y,0,1})$. Then we plug in the estimate for $\xi$ 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:
$\displaystyle \frac{{\left(4 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} - \frac{{\left(4 \, {\left(2 \, \xi - 1\right)} e^{\left(2 \, \xi\right)} + 3\right)} e^{\left(-\xi\right)}}{16 \, \xi^{3}} \geq 0.1624897$
$\displaystyle \frac{{\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} + \frac{{\left(4 \, e^{\left(2 \, \xi\right)} - 1\right)} e^{\left(-\xi\right)}}{16 \, \xi^{3}} \geq 0.05832627$
$\displaystyle -\frac{{\left(4 \, \xi - 3\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} + \frac{{\left(4 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + 4 \, \xi + 5\right)} e^{\left(-\xi\right)}}{16 \, \xi^{3}} \geq 0.1687342$
$\displaystyle \frac{{\left(4 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{16 \, \xi^{3}} - \frac{{\left(4 \, \xi e^{\left(2 \, \xi\right)} - 4 \, \xi - 1\right)} e^{\left(-\xi\right)}}{16 \, \xi^{3}} \geq 0.3895503$
Surface is stable!

#### No. 8: Degree $2$ / Singularity type $E_7$.

The combinatorial data is is given by $\Box=[-1,5]$ and $\Phi_0(u) = \min \{\frac{u-3}{4},0\},\; \Phi_\infty(u)= \frac{u-2}{3},\; \Phi_1(u) = \frac{-u-1}{2}$

# 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
[$\displaystyle -1$, $\displaystyle 5$]
$\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{4} \, u - \frac{3}{4}, 0\right)$
$\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}$
$\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}$

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

For this we have to analytically solve the integral $\int_\Box u \deg \bar \Phi (u) e^{\xi u} du$.

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

$\displaystyle \frac{{\left(5 \, \xi - 2\right)} e^{\left(5 \, \xi\right)}}{6 \, \xi^{3}} - \frac{{\left(3 \, {\left(3 \, \xi - 2\right)} e^{\left(4 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{12 \, \xi^{3}}$

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

# 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 $\operatorname{DF}_\xi(X_{y,0,1})$ and plug in the estimate for $\xi$

For every choice of $y \in \mathbb P^1$ we have to symbolically solve the integrals $\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du$, which up to scaling with a positive constant coincide with $\operatorname{DF}_\xi(\mathcal{X}_{y,0,1})$. Then we plug in the estimate for $\xi$ 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:
$\displaystyle \frac{{\left(6 \, \xi - 1\right)} e^{\left(5 \, \xi\right)}}{36 \, \xi^{3}} - \frac{{\left(9 \, {\left(4 \, \xi - 1\right)} e^{\left(4 \, \xi\right)} + 5\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.03202411$
$\displaystyle \frac{{\left(12 \, \xi - 5\right)} e^{\left(5 \, \xi\right)}}{36 \, \xi^{3}} - \frac{{\left(9 \, {\left(4 \, \xi - 3\right)} e^{\left(4 \, \xi\right)} + 7\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.04486743$
$\displaystyle -\frac{{\left(12 \, \xi - 5\right)} e^{\left(5 \, \xi\right)}}{36 \, \xi^{3}} + \frac{{\left(3 \, {\left(16 \, \xi - 11\right)} e^{\left(4 \, \xi\right)} + 12 \, \xi + 13\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.07049793$
$\displaystyle \frac{{\left(6 \, \xi - 1\right)} e^{\left(5 \, \xi\right)}}{36 \, \xi^{3}} - \frac{{\left(3 \, {\left(8 \, \xi - 1\right)} e^{\left(4 \, \xi\right)} - 12 \, \xi - 1\right)} e^{\left(-\xi\right)}}{144 \, \xi^{3}} \geq 0.1473894$
Surface is stable!

#### No. 9: Degree $2$ / Singularity type $E_6$.

The combinatorial data is is given by $\Box=[-1,3]$ and $\Phi_0(u) = \min \{\frac{u-2}{3},0\},\; \Phi_\infty(u)= \min \{\frac{u-2}{3},0\},\; \Phi_1(u) = \frac{-u-1}{2}$

# 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
[$\displaystyle -1$, $\displaystyle 3$]
$\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{3} \, u - \frac{2}{3}, 0\right)$
$\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{3} \, u - \frac{2}{3}, 0\right)$
$\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}$

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

For this we have to analytically solve the integral $\int_\Box u \deg \bar \Phi (u) e^{\xi u} du$.

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

$\displaystyle \frac{{\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)}}{2 \, \xi^{3}} - \frac{{\left(8 \, {\left(\xi - 1\right)} e^{\left(3 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{6 \, \xi^{3}}$

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

# 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 $\operatorname{DF}_\xi(X_{y,0,1})$ and plug in the estimate for $\xi$

For every choice of $y \in \mathbb P^1$ we have to symbolically solve the integrals $\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du$, which up to scaling with a positive constant coincide with $\operatorname{DF}_\xi(\mathcal{X}_{y,0,1})$. Then we plug in the estimate for $\xi$ 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:
$\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, {\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} + 1\right)} e^{\left(-\xi\right)}}{12 \, \xi^{3}} \geq 0.07622408$
$\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(2 \, {\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} + 1\right)} e^{\left(-\xi\right)}}{12 \, \xi^{3}} \geq 0.07622408$
$\displaystyle -\frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} + \frac{{\left(8 \, {\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} + 6 \, \xi + 7\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.1272785$
$\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{{\left(4 \, {\left(3 \, \xi - 2\right)} e^{\left(3 \, \xi\right)} - 6 \, \xi - 1\right)} e^{\left(-\xi\right)}}{36 \, \xi^{3}} \geq 0.2797267$
Surface is stable!

#### No. 11: Degree $2$ / Singularity type $D_5A_1$.

The combinatorial data is is given by $\Box=[-1,2]$ and $\Phi_0(u) = \min \{\frac{u-1}{2},0\},\; \Phi_\infty(u)= \min \{\frac{u-1}{2},0\},\; \Phi_1(u) = \frac{-2u-2}{3}$

# 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
[$\displaystyle -1$, $\displaystyle 2$]
$\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)$
$\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)$
$\displaystyle u \ {\mapsto}\ -\frac{2}{3} \, u - \frac{2}{3}$

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

For this we have to analytically solve the integral $\int_\Box u \deg \bar \Phi (u) e^{\xi u} du$.

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

$\displaystyle \frac{4 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)}}{3 \, \xi^{3}} - \frac{{\left(3 \, {\left(\xi - 2\right)} e^{\left(2 \, \xi\right)} + \xi + 2\right)} e^{\left(-\xi\right)}}{3 \, \xi^{3}}$

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

# 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 $\operatorname{DF}_\xi(X_{y,0,1})$ and plug in the estimate for $\xi$

For every choice of $y \in \mathbb P^1$ we have to symbolically solve the integrals $\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du$, which up to scaling with a positive constant coincide with $\operatorname{DF}_\xi(\mathcal{X}_{y,0,1})$. Then we plug in the estimate for $\xi$ 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:
$\displaystyle \frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} - \frac{2 \, {\left(3 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 1\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \geq 0.1919413$
$\displaystyle \frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} - \frac{2 \, {\left(3 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 1\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \geq 0.1919413$
$\displaystyle -\frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} + \frac{{\left(9 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + 3 \, \xi + 5\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \geq 0.1006816$
$\displaystyle \frac{2 \, {\left(3 \, \xi - 2\right)} e^{\left(2 \, \xi\right)}}{9 \, \xi^{3}} - \frac{{\left(3 \, {\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} - 3 \, \xi - 1\right)} e^{\left(-\xi\right)}}{9 \, \xi^{3}} \geq 0.4845700$
Surface is stable!

#### No. 12: Degree $2$ / Singularity type $3A_1D_4$.

The combinatorial data is is given by $\Box=[-1,1]$ and $\Phi_0(u) = \frac{u-1}{2},\; \Phi_\infty(u)= \frac{u-1}{2},\; \Phi_1(u) = \frac{-u-1}{2}$

# 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
[$\displaystyle -1$, $\displaystyle 3$]
$\displaystyle u \ {\mapsto}\ \frac{1}{2} \, u - \frac{1}{2}$
$\displaystyle u \ {\mapsto}\ \frac{1}{2} \, u - \frac{1}{2}$
$\displaystyle u \ {\mapsto}\ -\frac{1}{2} \, u - \frac{1}{2}$

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

For this we have to analytically solve the integral $\int_\Box u \deg \bar \Phi (u) e^{\xi u} du$.

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


$\displaystyle \frac{{\left(12 \, \xi^{2} - 7 \, \xi + 2\right)} e^{\left(3 \, \xi\right)}}{2 \, \xi^{3}} - \frac{{\left(\xi + 2\right)} e^{\left(-\xi\right)}}{2 \, \xi^{3}}$

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

# 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 $\operatorname{DF}_\xi(X_{y,0,1})$ and plug in the estimate for $\xi$

For every choice of $y \in \mathbb P^1$ we have to symbolically solve the integrals $\int_\Box e^{u\xi} \cdot \left((1+\Phi_y(u))^2 - (1+\sum_{z\neq y} \Phi_z(u))^2\right) \; du$, which up to scaling with a positive constant coincide with $\operatorname{DF}_\xi(\mathcal{X}_{y,0,1})$. Then we plug in the estimate for $\xi$ 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:
$\displaystyle \frac{{\left(8 \, \xi^{2} - 4 \, \xi + 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.2301820$
$\displaystyle \frac{{\left(8 \, \xi^{2} - 4 \, \xi + 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} - \frac{e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.2301820$
$\displaystyle -\frac{{\left(16 \, \xi^{2} - 10 \, \xi + 3\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} + \frac{{\left(2 \, \xi + 3\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.2301820$
$\displaystyle \frac{{\left(2 \, \xi - 1\right)} e^{\left(3 \, \xi\right)}}{4 \, \xi^{3}} + \frac{{\left(2 \, \xi + 1\right)} e^{\left(-\xi\right)}}{4 \, \xi^{3}} \geq 0.6905461$
Surface is stable!

#### No. 13: Degree $3$ / Singularity type $A_5A_1$.

The combinatorial data is is given by $\Box=[-1,3]$ and $\Phi_0(u) = \min \{-u,0\},\; \Phi_\infty(u)= \frac{u-3}{4},\; \Phi_1(u) = \frac{u-1}{2}$

# 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
[$\displaystyle -1$, $\displaystyle 3$]
$\displaystyle u \ {\mapsto}\ \min\left(-u, 0\right)$
$\displaystyle u \ {\mapsto}\ \frac{1}{4} \, u - \frac{3}{4}$
$\displaystyle u \ {\mapsto}\ \frac{1}{2} \, u - \frac{1}{2}$

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

For this we have to analytically solve the integral $\int_\Box u \deg \bar \Phi (u) e^{\xi u} du$