Similar to the question I asked on the Math Stackexchange (in this question), I am interested in solving a nonlinearly coupled system of PDEs.
The system of PDEs look like this:
$$\begin{align*}\partial_t p_{11} &= E_p p_{13} - E_p p_{31} \\
\partial_t p_{12} &= E_p p_{13} - E_p p_{32} - p_{12} (\Delta_{31}-\Delta_{32}) \\
\partial_t p_{13} &= E_p p_{11} + E_c p_{12} - E_p p_{33} + p_{13} \Delta_{31} \\
\partial_t p_{21} &= E_p p_{23} - E_c p_{31} + p_{21}(-\Delta_{31} + \Delta_{32}) \\
\partial_t p_{22} &= E_c p_{23} - E_c p_{32} \\
\partial_t p_{23} &= E_p p_{21} + E_c p_{22} - E_c p_{33} + p_{23} \Delta_{31} + p_{23} (-\Delta_{31} + \Delta_{32}) \\
\partial_t p_{31} &= - E_p p_{11} - E_c p_{21} + E_p p_{33} - p_{31} \Delta_{31} \\
\partial_t p_{32} &= - E_p p_{21} - E_c p_{22} - E_c p_{33} + p_{23} \Delta_{31} + p_{23} (-\Delta_{31} + \Delta_{32}) \\
\partial_t p_{33} &= E_p p_{13} - E_c p_{23} + E_p p_{31} + E_c p_{32}
\end{align*}$$
$$
\frac{\partial E_p}{\partial z } + \frac{1}{c} \frac{\partial E_p}{\partial t } = i k p_{13}(t, z)
$$
Each of these p's ($p_{11}, p_{12}, ..., p_{33}$) are a function of t and z. Additionally $E_p$ is a function of t and z. And $\Delta_{31, 32, 33}$ are just numerical constants.
I'm trying with the boundary conditions:
$$\text{p11}(0,z)=1,\\
\text{p12}(0,z)=\text{p13}(0,z)=\text{p21}(0,z)= \text{p22}(0,z)=\text{p23}(0,z)=\text{p31}(0,z)= \text{p32}(0,z)= \text{p33}(0,z)=0,\\
\text{Ep}(t,0)=e^{-t^2}$$
Is there a way that I can solve this numerically in Mathematica?
When I try the following code:
NDSolve[{
\!\(\*SuperscriptBox[\(p11\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[t,
z] == -(1/2) I Ep[t, z] (p13[t, z] - p31[t, z]),
\!\(\*SuperscriptBox[\(p12\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, z] ==
1/2 I (-10 p13[t, z] + Ep[t, z] p32[t, z]),
\!\(\*SuperscriptBox[\(p13\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, z] == -(1/2) p13[t, z] -
1/2 I (10 p12[t, z] + Ep[t, z] (p11[t, z] - p33[t, z])),
\!\(\*SuperscriptBox[\(p21\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, z] == -(1/2) p21[t, z] -
1/2 I (Ep[t, z] p23[t, z] - 10 p31[t, z]),
\!\(\*SuperscriptBox[\(p22\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, z] == -5 I (p23[t, z] - p32[t, z]),
\!\(\*SuperscriptBox[\(p23\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[t,
z] == -(1/2)
I (Ep[t, z] p21[t, z] + 10 p22[t, z] - I p23[t, z] -
10 p33[t, z]),
\!\(\*SuperscriptBox[\(p31\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, z] ==
1/2 I (10 p21[t, z] + I p31[t, z] +
Ep[t, z] (p11[t, z] - p33[t, z])),
\!\(\*SuperscriptBox[\(p32\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, z] ==
1/2 I (Ep[t, z] p12[t, z] + 10 p22[t, z] + I p32[t, z] -
10 p33[t, z]),
\!\(\*SuperscriptBox[\(p33\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, z] ==
1/2 I (10 p23[t, z] + Ep[t, z] (p13[t, z] - p31[t, z]) -
10 p32[t, z] + 2 I p33[t, z]),
\!\(\*SuperscriptBox[\(Ep\),
TagBox[
RowBox[{"(",
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, z] +
\!\(\*SuperscriptBox[\(Ep\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, z] == 1/2 I p12[t, z], p11[0, z] == 1,
p12[0, z] == 0, p13[0, z] == 0, p21[0, z] == 0, p22[0, z] == 0,
p23[0, z] == 0, p31[0, z] == 0, p32[0, z] == 0, p33[0, z] == 0,
Ep[t, 0] == E^-t^2}, {\[Rho]11[t, z], \[Rho]12[t, z], \[Rho]13[t,
z], \[Rho]21[t, z], \[Rho]22[t, z], \[Rho]23[t, z], \[Rho]31[t,
z], \[Rho]32[t, z], \[Rho]33[t, z], Ep[t, z]}, {z, 0, 1}, {t, 0,
1}]
Which looks like this:
Mathematica returns the error message: NDSolve::femnonlinear: Nonlinear coefficients are not supported in this version of NDSolve.
Any ideas?
