Skip to main content

differential equations - Numerical solution to PDE seems to contradict initial values


I am trying to solve this PDE system:


EPS = NDSolveValue[{
D[e[z, t], z] == 0,
D[p[z, t], t] == -p[z, t] + I s[z, t],
D[s[z, t], t] == I p[z, t],
e[0, t] == 1, p[z, 0] == 0, s[z, 0] == 0

}, {e, p, s}, {z, 0, 1}, {t, 0, 1}];

Then I seek values for e and p at different z and t=0:


{EPS[[1]][0, 0], EPS[[2]][0, 0]}

{EPS[[1]][1, 0], EPS[[2]][1, 0]}

The result contradicts the condition p[z, 0] == 0:


{1, I}
{1, I}


Furthermore, if I ONLY change the order of the equations


EPS = NDSolveValue[{
D[p[z, t], t] == -p[z, t] + I s[z, t],
D[s[z, t], t] == I p[z, t],
D[e[z, t], z] == 0,
e[0, t] == 1, p[z, 0] == 0, s[z, 0] == 0
}, {e, p, s}, {z, 0, 1}, {t, 0, 1}];

{EPS[[1]][0, 0], EPS[[2]][0, 0]}


{EPS[[1]][1, 0], EPS[[2]][1, 0]}

The result changes, and is wrong again


{1, 0.999909 I}
{1.76683*10^10, - 1.09719*10^6 I}

Does anyone have an idea what is happening?


UPDATE: The original problem arised from this fully coupled equation:


NDSolveValue[{

D[e[z, t], z] == I p[z, t],
D[p[z, t], t] == -p[z, t] + I e[z, t] + I g[t] s[z, t],
D[s[z, t], t] == I Conjugate[g[t]] p[z, t],
s[z, 0] == p[z, 0] == 0, e[0, t] == f[t]
}, {e, p, s}, {z, 0, 1}, {t, 0, 1}]

Here g[t] and f[t] are simple functions, such as const, Gaussian, etc.



Answer



The first thing I tried was to check if this is actually time integrated and not treated as a pure spatial PDE. Just calling a variable t does not tell NDSolve that something is to be considered time dependent.


EPS = NDSolve[{

D[e[z, t], z] == 0,
D[p[z, t], t] == -p[z, t] + I s[z, t],
D[s[z, t], t] == I p[z, t],
e[0, t] == 1, p[z, 0] == 0, s[z, 0] == 0}, {e, p, s}, {z, 0,
1}, {t, 0, 1}, Method -> "MethodOfLines"][[1]]

NDSolve::ivone: Boundary values may only be specified for one independent variable. Initial values may only be specified at one value of the other independent variable.

OK, so the PDE is treated as a pure spatial problem and it used the FEM for that:


EPS = NDSolve[{

D[e[z, t], z] == 0,
D[p[z, t], t] == -p[z, t] + I s[z, t],
D[s[z, t], t] == I p[z, t],
e[0, t] == 1, p[z, 0] == 0, s[z, 0] == 0}, {e, p, s}, {z, 0,
1}, {t, 0, 1}][[1]];
(e /. EPS)["ElementMesh"]

NDSolve`FEM`ElementMesh[{{0., 1.}, {0.,
1.}}, {NDSolve`FEM`QuadElement["<" 400 ">"]}]


Now, we look at the equation once more and when we eliminate the constant derivative condition we get:


EPS = NDSolve[{
(*D[e[z,t],z]\[Equal]0,*)
D[p[z, t], t] == -p[z, t] + I s[z, t],
D[s[z, t], t] == I p[z, t],
e[0, t] == 1, p[z, 0] == 0, s[z, 0] == 0}, {e, p, s}, {z, 0,
1}, {t, 0, 1}][[1]]
{e -> Function[{z, t}, 1],
p -> InterpolatingFunction[{{0., 1.}, {0., 1.}}, <>],
s -> InterpolatingFunction[{{0., 1.}, {0., 1.}}, <>]}


{(e /. EPS)[0, 0], (p /. EPS)[0, 0]}
{(e /. EPS)[1, 0], (p /. EPS)[1, 0]}
{1, 0.}
{1, 0.}

This time a time integration did happen and for the spatial discretization you can use either FEM or TGP by givin the option (if you want): Method -> {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid"}} or "FiniteElement".


In the pure spatial problem a slightly different PDE is solved. Now, the reordering is an unfortunate issue that is documented and there is currently no way to reliably autodetect that and warn about it. For an explanation and workarounds see this tutorial in the section about Ordering of Dependent Variable Names.


Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....