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Authors: Jacob Cable, Hendrik Suess
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Description: On the classification of Kähler-Ricci solitons on Gorenstein del Pezzo surfaces: Calculations for surfaces
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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{u45,0},  Φ(u)=u23,  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(15u45,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{5} \, u - \frac{4}{5}, 0\right)
u  13u23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}
u  12u12\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 udegΦˉ(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))
25ξe(5ξ)12ξe(4ξ)10e(5ξ)+6e(4ξ)30ξ312ξe(4ξ)+ξe(ξ)6e(4ξ)+2e(ξ)30ξ3\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 DFξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of yP1y \in \mathbb P^1 we have to symbolically solve the integrals euξ((1+Φy(u))2(1+zyΦz(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 DFξ(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(5ξ1)e(5ξ)+11)e(ξ)900ξ3+25(6ξ1)e(5ξ)18(5ξ1)e(4ξ)900ξ30.01128570\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(15ξ8)e(5ξ)+19)e(ξ)900ξ3+25(12ξ5)e(5ξ)9(15ξ8)e(4ξ)900ξ30.01949453\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(25ξ13)e(5ξ)+30ξ+31)e(ξ)900ξ325(12ξ5)e(5ξ)6(25ξ13)e(4ξ)900ξ30.02975460\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(25ξ4)e(5ξ)30ξ1)e(ξ)900ξ3+25(6ξ1)e(5ξ)3(25ξ4)e(4ξ)900ξ30.06053484\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{u23,0},  Φ(u)=u34,  Φ1(u)=u12\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(13u23,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{3} \, u - \frac{2}{3}, 0\right)
u  14u34\displaystyle u \ {\mapsto}\ \frac{1}{4} \, u - \frac{3}{4}
u  12u12\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 udegΦˉ(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))
(3ξ2)e(3ξ)4ξ3(8(ξ1)e(3ξ)+ξ+2)e(ξ)12ξ3\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 DFξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of yP1y \in \mathbb P^1 we have to symbolically solve the integrals euξ((1+Φy(u))2(1+zyΦz(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 DFξ(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:
(4ξ1)e(3ξ)16ξ3(16(3ξ1)e(3ξ)+7)e(ξ)144ξ30.04440443\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
(4ξ3)e(3ξ)16ξ3(8(3ξ4)e(3ξ)+5)e(ξ)144ξ30.03181965\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
(4ξ3)e(3ξ)16ξ3+(4(9ξ10)e(3ξ)+12ξ+13)e(ξ)144ξ30.06993168\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
(4ξ1)e(3ξ)16ξ3(4(9ξ2)e(3ξ)12ξ1)e(ξ)144ξ30.1461557\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{u12,0},  Φ(u)=u23,  Φ1(u)=2u23\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(12u12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u  13u23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}
u  23u23\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 udegΦˉ(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(ξ1)e(2ξ)3ξ3(3(ξ2)e(2ξ)+ξ+2)e(ξ)6ξ3\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 DFξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of yP1y \in \mathbb P^1 we have to symbolically solve the integrals euξ((1+Φy(u))2(1+zyΦz(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 DFξ(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:
(3ξ1)e(2ξ)9ξ3(9(2ξ1)e(2ξ)+5)e(ξ)36ξ30.1187884\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
(ξ1)e(2ξ)3ξ3((2ξ5)e(2ξ)+1)e(ξ)12ξ30.07315991\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
(ξ1)e(2ξ)3ξ3+((4ξ7)e(2ξ)+2ξ+3)e(ξ)12ξ30.07315987\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
(3ξ1)e(2ξ)9ξ3(3(4ξ1)e(2ξ)6ξ1)e(ξ)36ξ30.2651082\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)=u23,  Φ1(u)=u23\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
[1\displaystyle -1, 2\displaystyle 2]
u  min(u,0)\displaystyle u \ {\mapsto}\ \min\left(-u, 0\right)
u  13u23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}
u  13u23\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 udegΦˉ(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)3ξ3+2((ξ1)e(2ξ)+3)3ξ3\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 DFξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of yP1y \in \mathbb P^1 we have to symbolically solve the integrals euξ((1+Φy(u))2(1+zyΦz(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 DFξ(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(3ξ+2)e(ξ)9ξ3(3ξ5)e(2ξ)+9ξ+99ξ30.4790611\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
(ξ1)e(2ξ)+ξ+13ξ30.09239867\displaystyle \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + \xi + 1}{3 \, \xi^{3}} \geq 0.09239867
(ξ1)e(2ξ)+ξ+13ξ30.09239867\displaystyle \frac{{\left(\xi - 1\right)} e^{\left(2 \, \xi\right)} + \xi + 1}{3 \, \xi^{3}} \geq 0.09239867
2(3ξ+2)e(ξ)9ξ3+(3ξ1)e(2ξ)3ξ39ξ30.6638587\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{u12,0},  Φ(u)=u34,  Φ1(u)=u12\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(12u12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u  14u34\displaystyle u \ {\mapsto}\ \frac{1}{4} \, u - \frac{3}{4}
u  12u12\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 udegΦˉ(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))
(3ξ2)e(3ξ)4ξ3(2(ξ2)e(2ξ)+ξ+2)e(ξ)4ξ3\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 DFξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of yP1y \in \mathbb P^1 we have to symbolically solve the integrals euξ((1+Φy(u))2(1+zyΦz(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 DFξ(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:
(4ξ1)e(3ξ)16ξ3(4(2ξ1)e(2ξ)+3)e(ξ)16ξ30.1624897\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
(4ξ3)e(3ξ)16ξ3+(4e(2ξ)1)e(ξ)16ξ30.05832627\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
(4ξ3)e(3ξ)16ξ3+(4(ξ2)e(2ξ)+4ξ+5)e(ξ)16ξ30.1687342\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
(4ξ1)e(3ξ)16ξ3(4ξe(2ξ)4ξ1)e(ξ)16ξ30.3895503\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{u34,0},  Φ(u)=u23,  Φ1(u)=u12\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(14u34,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{4} \, u - \frac{3}{4}, 0\right)
u  13u23\displaystyle u \ {\mapsto}\ \frac{1}{3} \, u - \frac{2}{3}
u  12u12\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 udegΦˉ(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))
(5ξ2)e(5ξ)6ξ3(3(3ξ2)e(4ξ)+ξ+2)e(ξ)12ξ3\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 DFξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of yP1y \in \mathbb P^1 we have to symbolically solve the integrals euξ((1+Φy(u))2(1+zyΦz(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 DFξ(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:
(6ξ1)e(5ξ)36ξ3(9(4ξ1)e(4ξ)+5)e(ξ)144ξ30.03202411\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
(12ξ5)e(5ξ)36ξ3(9(4ξ3)e(4ξ)+7)e(ξ)144ξ30.04486743\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
(12ξ5)e(5ξ)36ξ3+(3(16ξ11)e(4ξ)+12ξ+13)e(ξ)144ξ30.07049793\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
(6ξ1)e(5ξ)36ξ3(3(8ξ1)e(4ξ)12ξ1)e(ξ)144ξ30.1473894\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{u23,0},  Φ(u)=min{u23,0},  Φ1(u)=u12\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(13u23,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{3} \, u - \frac{2}{3}, 0\right)
u  min(13u23,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{3} \, u - \frac{2}{3}, 0\right)
u  12u12\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 udegΦˉ(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))
(3ξ2)e(3ξ)2ξ3(8(ξ1)e(3ξ)+ξ+2)e(ξ)6ξ3\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 DFξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of yP1y \in \mathbb P^1 we have to symbolically solve the integrals euξ((1+Φy(u))2(1+zyΦz(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 DFξ(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:
(2ξ1)e(3ξ)4ξ3(2(3ξ2)e(3ξ)+1)e(ξ)12ξ30.07622408\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
(2ξ1)e(3ξ)4ξ3(2(3ξ2)e(3ξ)+1)e(ξ)12ξ30.07622408\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
(2ξ1)e(3ξ)4ξ3+(8(3ξ2)e(3ξ)+6ξ+7)e(ξ)36ξ30.1272785\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
(2ξ1)e(3ξ)4ξ3(4(3ξ2)e(3ξ)6ξ1)e(ξ)36ξ30.2797267\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{u12,0},  Φ(u)=min{u12,0},  Φ1(u)=2u23\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(12u12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u  min(12u12,0)\displaystyle u \ {\mapsto}\ \min\left(\frac{1}{2} \, u - \frac{1}{2}, 0\right)
u  23u23\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 udegΦˉ(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(ξ1)e(2ξ)3ξ3(3(ξ2)e(2ξ)+ξ+2)e(ξ)3ξ3\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 DFξ(Xy,0,1)\operatorname{DF}_\xi(X_{y,0,1}) and plug in the estimate for ξ\xi

For every choice of yP1y \in \mathbb P^1 we have to symbolically solve the integrals euξ((1+Φy(u))2(1+zyΦz(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 DFξ(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(3ξ2)e(2ξ)9ξ32(3(ξ1)e(2ξ)+1)e(ξ)9ξ30.1919413\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(3ξ2)e(2ξ)9ξ32(3(ξ1)e(2ξ)+1)e(ξ)9ξ30.1919413\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(3ξ2)e(2ξ)9ξ3+(9(ξ1)e(2ξ)+3ξ+5)e(ξ)9ξ30.1006816\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(3ξ2)e(2ξ)9ξ3(3(ξ1)e(2ξ)3ξ1)e(ξ)9ξ30.4845700\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)=u12,  Φ(u)=u12,  Φ1(u)=u12\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  12u12\displaystyle u \ {\mapsto}\ \frac{1}{2} \, u - \frac{1}{2}
u  12u12\displaystyle u \ {\mapsto}\ \frac{1}{2} \, u - \frac{1}{2}
u  12u12\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 udegΦˉ(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))
(12ξ27ξ+2)e(3ξ)2ξ3(ξ+2)e(ξ)2ξ3\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: -1.97691850478 Interval containing solution: [-1.97691861 .. -1.97691839]

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

For every choice of yP1y \in \mathbb P^1 we have to symbolically solve the integrals euξ((1+Φy(u))2(1+zyΦz(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 DFξ(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:
(8ξ24ξ+1)e(3ξ)4ξ3e(ξ)4ξ30.2301820\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
(8ξ24ξ+1)e(3ξ)4ξ3e(ξ)4ξ30.2301820\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
(16ξ210ξ+3)e(3ξ)4ξ3+(2ξ+3)e(ξ)4ξ30.2301820\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
(2ξ1)e(3ξ)4ξ3+(2ξ+1)e(ξ)4ξ30.6905461\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)=u34,  Φ1(u)=u12\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  14u34\displaystyle u \ {\mapsto}\ \frac{1}{4} \, u - \frac{3}{4}
u  12u12\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 udegΦˉ(u)eξudu\int_\Box u \deg \bar \Phi (u) e^{\xi u} du