# Return to Answer

 4 replaced http://mathematica.stackexchange.com/ with https://mathematica.stackexchange.com/ edited Apr 13 '17 at 12:55 I'm so excited now! I might found a solution for a certain type of PDE related problem! The key point is choosing a odd "DifferenceOrder"! Let's define a auxiliary function first: mol[n_, o_] := {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n, "MinPoints" -> n, "DifferenceOrder" -> o}}  Then try this: Tmax = 1; L = 60; gamma = 5000; A = 1; xmin = 0; xmax = π L; Subscript[ρ, 0] = 1/(1 + A); mi[x_] := 8 π Subscript[ρ, 0] A L Sin[x/L]; eq1nonstandard = D[m[x, t], t] + v[x, t] D[m[x, t], x] + gamma D[v[x, t], x, x]; eq2nonstandard = D[v[x, t], t] + v[x, t] D[v[x, t], x] + 1/2 m[x, t]; Vnonstandard = First[v /. NDSolve[{eq1nonstandard == 0, eq2nonstandard == 0, v[x, 0] == 0, m[x, 0] == mi[x], v[xmin, t] == 0, v[xmax, t] == 0, m[xmin, t] == 0, m[xmax, t] == mi[xmax]}, {v}, {x, xmin, xmax}, {t, 0, Tmax}, Method -> mol[25, 9]]] (* The following line allows you to plot the result easily when NDSolve stops at the half-way. *) {{xl, xr}, {tl, tr}} = Vnonstandard["Domain"]; Plot3D[Vnonstandard[x, t], {x, xl, xr}, {t, tl, tr}]  Some observation: The number of grid points can't be too large, I guess it's because something similar to thisthis happens. The bigger the "DifferenceOrder" is, the better. This is the result under mol[12, 3]: and this is under mol[16, 5]: Under some "DifferenceOrder", Mathematica can choose suitable number of grid points automatically. For example mol[Automatic, 9]. Nevertheless, I'm not sure why this Method works, this is just a rare victory among my numerous failures when trying to solve the PDE related problem in this site by trial and error. I'm so excited now! I might found a solution for a certain type of PDE related problem! The key point is choosing a odd "DifferenceOrder"! Let's define a auxiliary function first: mol[n_, o_] := {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n, "MinPoints" -> n, "DifferenceOrder" -> o}}  Then try this: Tmax = 1; L = 60; gamma = 5000; A = 1; xmin = 0; xmax = π L; Subscript[ρ, 0] = 1/(1 + A); mi[x_] := 8 π Subscript[ρ, 0] A L Sin[x/L]; eq1nonstandard = D[m[x, t], t] + v[x, t] D[m[x, t], x] + gamma D[v[x, t], x, x]; eq2nonstandard = D[v[x, t], t] + v[x, t] D[v[x, t], x] + 1/2 m[x, t]; Vnonstandard = First[v /. NDSolve[{eq1nonstandard == 0, eq2nonstandard == 0, v[x, 0] == 0, m[x, 0] == mi[x], v[xmin, t] == 0, v[xmax, t] == 0, m[xmin, t] == 0, m[xmax, t] == mi[xmax]}, {v}, {x, xmin, xmax}, {t, 0, Tmax}, Method -> mol[25, 9]]] (* The following line allows you to plot the result easily when NDSolve stops at the half-way. *) {{xl, xr}, {tl, tr}} = Vnonstandard["Domain"]; Plot3D[Vnonstandard[x, t], {x, xl, xr}, {t, tl, tr}]  Some observation: The number of grid points can't be too large, I guess it's because something similar to this happens. The bigger the "DifferenceOrder" is, the better. This is the result under mol[12, 3]: and this is under mol[16, 5]: Under some "DifferenceOrder", Mathematica can choose suitable number of grid points automatically. For example mol[Automatic, 9]. Nevertheless, I'm not sure why this Method works, this is just a rare victory among my numerous failures when trying to solve the PDE related problem in this site by trial and error. I'm so excited now! I might found a solution for a certain type of PDE related problem! The key point is choosing a odd "DifferenceOrder"! Let's define a auxiliary function first: mol[n_, o_] := {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n, "MinPoints" -> n, "DifferenceOrder" -> o}}  Then try this: Tmax = 1; L = 60; gamma = 5000; A = 1; xmin = 0; xmax = π L; Subscript[ρ, 0] = 1/(1 + A); mi[x_] := 8 π Subscript[ρ, 0] A L Sin[x/L]; eq1nonstandard = D[m[x, t], t] + v[x, t] D[m[x, t], x] + gamma D[v[x, t], x, x]; eq2nonstandard = D[v[x, t], t] + v[x, t] D[v[x, t], x] + 1/2 m[x, t]; Vnonstandard = First[v /. NDSolve[{eq1nonstandard == 0, eq2nonstandard == 0, v[x, 0] == 0, m[x, 0] == mi[x], v[xmin, t] == 0, v[xmax, t] == 0, m[xmin, t] == 0, m[xmax, t] == mi[xmax]}, {v}, {x, xmin, xmax}, {t, 0, Tmax}, Method -> mol[25, 9]]] (* The following line allows you to plot the result easily when NDSolve stops at the half-way. *) {{xl, xr}, {tl, tr}} = Vnonstandard["Domain"]; Plot3D[Vnonstandard[x, t], {x, xl, xr}, {t, tl, tr}]  Some observation: The number of grid points can't be too large, I guess it's because something similar to this happens. The bigger the "DifferenceOrder" is, the better. This is the result under mol[12, 3]: and this is under mol[16, 5]: Under some "DifferenceOrder", Mathematica can choose suitable number of grid points automatically. For example mol[Automatic, 9]. Nevertheless, I'm not sure why this Method works, this is just a rare victory among my numerous failures when trying to solve the PDE related problem in this site by trial and error. 3 a new observation added. edited Mar 21 '15 at 7:33 xzczd 29.8k66 gold badges8484 silver badges276276 bronze badges I'm so excited now! I might found a solution for a certain type of PDE related problem! The key point is choosing a odd "DifferenceOrder"! Let's define a auxiliary function first: mol[n_, o_] := {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n, "MinPoints" -> n, "DifferenceOrder" -> o}}  Then try this: Tmax = 1; L = 60; gamma = 5000; A = 1; xmin = 0; xmax = π L; Subscript[ρ, 0] = 1/(1 + A); mi[x_] := 8 π Subscript[ρ, 0] A L Sin[x/L]; eq1nonstandard = D[m[x, t], t] + v[x, t] D[m[x, t], x] + gamma D[v[x, t], x, x]; eq2nonstandard = D[v[x, t], t] + v[x, t] D[v[x, t], x] + 1/2 m[x, t]; Vnonstandard = First[v /. NDSolve[{eq1nonstandard == 0, eq2nonstandard == 0, v[x, 0] == 0, m[x, 0] == mi[x], v[xmin, t] == 0, v[xmax, t] == 0, m[xmin, t] == 0, m[xmax, t] == mi[xmax]}, {v}, {x, xmin, xmax}, {t, 0, Tmax}, Method -> mol[25, 9]]] (* The following line allows you to plot the result easily when NDSolve stops at the half-way. *) {{xl, xr}, {tl, tr}} = Vnonstandard["Domain"]; Plot3D[Vnonstandard[x, t], {x, xl, xr}, {t, tl, tr}]  Some observation: The number of grid points can't be too large, I guess it's because something similar to this happens. The bigger the "DifferenceOrder" is, the better. This is the result under mol[12, 3]: and this is under mol[16, 5]: Under some "DifferenceOrder", Mathematica can choose suitable number of grid points automatically. For example mol[Automatic, 9]. Nevertheless, I'm not sure why this Method works, this is just a rare victory among my numerous failures when trying to solve the PDE related problem in this site by trial and error. I'm so excited now! I might found a solution for a certain type of PDE related problem! The key point is choosing a odd "DifferenceOrder"! Let's define a auxiliary function first: mol[n_, o_] := {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n, "MinPoints" -> n, "DifferenceOrder" -> o}}  Then try this: Tmax = 1; L = 60; gamma = 5000; A = 1; xmin = 0; xmax = π L; Subscript[ρ, 0] = 1/(1 + A); mi[x_] := 8 π Subscript[ρ, 0] A L Sin[x/L]; eq1nonstandard = D[m[x, t], t] + v[x, t] D[m[x, t], x] + gamma D[v[x, t], x, x]; eq2nonstandard = D[v[x, t], t] + v[x, t] D[v[x, t], x] + 1/2 m[x, t]; Vnonstandard = First[v /. NDSolve[{eq1nonstandard == 0, eq2nonstandard == 0, v[x, 0] == 0, m[x, 0] == mi[x], v[xmin, t] == 0, v[xmax, t] == 0, m[xmin, t] == 0, m[xmax, t] == mi[xmax]}, {v}, {x, xmin, xmax}, {t, 0, Tmax}, Method -> mol[25, 9]]] (* The following line allows you to plot the result easily when NDSolve stops at the half-way. *) {{xl, xr}, {tl, tr}} = Vnonstandard["Domain"]; Plot3D[Vnonstandard[x, t], {x, xl, xr}, {t, tl, tr}]  Some observation: The number of grid points can't be too large, I guess it's because something similar to this happens. The bigger the "DifferenceOrder" is, the better. This is the result under mol[12, 3]: and this is under mol[16, 5]: Nevertheless, I'm not sure why this Method works, this is just a rare victory among my numerous failures when trying to solve the PDE related problem in this site by trial and error. I'm so excited now! I might found a solution for a certain type of PDE related problem! The key point is choosing a odd "DifferenceOrder"! Let's define a auxiliary function first: mol[n_, o_] := {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n, "MinPoints" -> n, "DifferenceOrder" -> o}}  Then try this: Tmax = 1; L = 60; gamma = 5000; A = 1; xmin = 0; xmax = π L; Subscript[ρ, 0] = 1/(1 + A); mi[x_] := 8 π Subscript[ρ, 0] A L Sin[x/L]; eq1nonstandard = D[m[x, t], t] + v[x, t] D[m[x, t], x] + gamma D[v[x, t], x, x]; eq2nonstandard = D[v[x, t], t] + v[x, t] D[v[x, t], x] + 1/2 m[x, t]; Vnonstandard = First[v /. NDSolve[{eq1nonstandard == 0, eq2nonstandard == 0, v[x, 0] == 0, m[x, 0] == mi[x], v[xmin, t] == 0, v[xmax, t] == 0, m[xmin, t] == 0, m[xmax, t] == mi[xmax]}, {v}, {x, xmin, xmax}, {t, 0, Tmax}, Method -> mol[25, 9]]] (* The following line allows you to plot the result easily when NDSolve stops at the half-way. *) {{xl, xr}, {tl, tr}} = Vnonstandard["Domain"]; Plot3D[Vnonstandard[x, t], {x, xl, xr}, {t, tl, tr}]  Some observation: The number of grid points can't be too large, I guess it's because something similar to this happens. The bigger the "DifferenceOrder" is, the better. This is the result under mol[12, 3]: and this is under mol[16, 5]: Under some "DifferenceOrder", Mathematica can choose suitable number of grid points automatically. For example mol[Automatic, 9]. Nevertheless, I'm not sure why this Method works, this is just a rare victory among my numerous failures when trying to solve the PDE related problem in this site by trial and error. 2 add a note. edited Mar 21 '15 at 7:17 xzczd 29.8k66 gold badges8484 silver badges276276 bronze badges I'm so excited now! I might found a solution for a certain type of PDE related problem! The key point is choosing a odd "DifferenceOrder"! Let's define a auxiliary function first: mol[n_, o_] := {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n, "MinPoints" -> n, "DifferenceOrder" -> o}}  Then try this: Tmax = 1; L = 60; gamma = 5000; A = 1; xmin = 0; xmax = π L; Subscript[ρ, 0] = 1/(1 + A); mi[x_] := 8 π Subscript[ρ, 0] A L Sin[x/L]; eq1nonstandard = D[m[x, t], t] + v[x, t] D[m[x, t], x] + gamma D[v[x, t], x, x]; eq2nonstandard = D[v[x, t], t] + v[x, t] D[v[x, t], x] + 1/2 m[x, t]; Vnonstandard = First[v /. NDSolve[{eq1nonstandard == 0, eq2nonstandard == 0, v[x, 0] == 0, m[x, 0] == mi[x], v[xmin, t] == 0, v[xmax, t] == 0, m[xmin, t] == 0, m[xmax, t] == mi[xmax]}, {v}, {x, xmin, xmax}, {t, 0, Tmax}, Method -> mol[25, 9]]] (* The following line allows you to plot the result easily when NDSolve stops at the half-way. *) {{xl, xr}, {tl, tr}} = Vnonstandard["Domain"]; Plot3D[Vnonstandard[x, t], {x, xl, xr}, {t, tl, tr}]  Some observation: The number of grid points can't be too large, I guess it's because something similar to this happens. The bigger the "DifferenceOrder" is, the better. This is the result under mol[12, 3]: and this is under mol[16, 5]: Nevertheless, I'm not sure why this Method works, this is just a rare victory among my numerous failures when trying to solve the PDE related problem in this site by trial and error. I'm so excited now! I might found a solution for a certain type of PDE related problem! The key point is choosing a odd "DifferenceOrder"! Let's define a auxiliary function first: mol[n_, o_] := {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n, "MinPoints" -> n, "DifferenceOrder" -> o}}  Then try this: Tmax = 1; L = 60; gamma = 5000; A = 1; xmin = 0; xmax = π L; Subscript[ρ, 0] = 1/(1 + A); mi[x_] := 8 π Subscript[ρ, 0] A L Sin[x/L]; eq1nonstandard = D[m[x, t], t] + v[x, t] D[m[x, t], x] + gamma D[v[x, t], x, x]; eq2nonstandard = D[v[x, t], t] + v[x, t] D[v[x, t], x] + 1/2 m[x, t]; Vnonstandard = First[v /. NDSolve[{eq1nonstandard == 0, eq2nonstandard == 0, v[x, 0] == 0, m[x, 0] == mi[x], v[xmin, t] == 0, v[xmax, t] == 0, m[xmin, t] == 0, m[xmax, t] == mi[xmax]}, {v}, {x, xmin, xmax}, {t, 0, Tmax}, Method -> mol[25, 9]]] {{xl, xr}, {tl, tr}} = Vnonstandard["Domain"]; Plot3D[Vnonstandard[x, t], {x, xl, xr}, {t, tl, tr}]  Some observation: The number of grid points can't be too large, I guess it's because something similar to this happens. The bigger the "DifferenceOrder" is, the better. This is the result under mol[12, 3]: and this is under mol[16, 5]: Nevertheless, I'm not sure why this Method works, this is just a rare victory among my numerous failures when trying to solve the PDE related problem in this site by trial and error. I'm so excited now! I might found a solution for a certain type of PDE related problem! The key point is choosing a odd "DifferenceOrder"! Let's define a auxiliary function first: mol[n_, o_] := {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n, "MinPoints" -> n, "DifferenceOrder" -> o}}  Then try this: Tmax = 1; L = 60; gamma = 5000; A = 1; xmin = 0; xmax = π L; Subscript[ρ, 0] = 1/(1 + A); mi[x_] := 8 π Subscript[ρ, 0] A L Sin[x/L]; eq1nonstandard = D[m[x, t], t] + v[x, t] D[m[x, t], x] + gamma D[v[x, t], x, x]; eq2nonstandard = D[v[x, t], t] + v[x, t] D[v[x, t], x] + 1/2 m[x, t]; Vnonstandard = First[v /. NDSolve[{eq1nonstandard == 0, eq2nonstandard == 0, v[x, 0] == 0, m[x, 0] == mi[x], v[xmin, t] == 0, v[xmax, t] == 0, m[xmin, t] == 0, m[xmax, t] == mi[xmax]}, {v}, {x, xmin, xmax}, {t, 0, Tmax}, Method -> mol[25, 9]]] (* The following line allows you to plot the result easily when NDSolve stops at the half-way. *) {{xl, xr}, {tl, tr}} = Vnonstandard["Domain"]; Plot3D[Vnonstandard[x, t], {x, xl, xr}, {t, tl, tr}]  Some observation: The number of grid points can't be too large, I guess it's because something similar to this happens. The bigger the "DifferenceOrder" is, the better. This is the result under mol[12, 3]: and this is under mol[16, 5]: Nevertheless, I'm not sure why this Method works, this is just a rare victory among my numerous failures when trying to solve the PDE related problem in this site by trial and error. 1 answered Mar 21 '15 at 7:08 xzczd 29.8k66 gold badges8484 silver badges276276 bronze badges