Mathematica Asked by pbdiddy on January 11, 2021
I’m trying to solve:
$partial _{z} U(z,t) = i sqrt{d} P(z,t)$
$partial _{t} P(z,t) = -P(z, t) + i sqrt{d} U(z,t) + i Omega_{c}(t)S(z, t)$
$partial _{t} S(z,t) = i Omega_{c}(t)P(z, t)$
For the initial conditions
$P(z, 0) = 0$
$S(z, 0) = 0$
$U(0, t) = A expBigg(-4 ln(2) Bigg(frac{t-t_{0}}{tau}Bigg)Bigg) ,,$
where A, $tau$ and d are all constants.
As I understand it, NDSolve cannot deal with this problem due to there being a derivative of only one dimension on each of the equations. I’ve attempted to solve using the method of lines, as shown in answers such as this: NDSolve:Coupled PDE's, initial-boundary value problem: unreasonable "insufficient number of boundary conditions" error. However while the solver shows no errors, plotting the result just produces a blank box.
My code is shown below. I have removed the part which computes the constants to make it more legible. Capital "I" below refers to the complex number i.
d = 84.9601
tau = 0.1
OmegaC = 27.7259
pulseTime = 2 Pi/OmegaC
storageTime = 0.2;
twriteon = t0 - pulseTime/2;
twriteoff = t0 + pulseTime/2;
treadon = t0 + pulseTime/2 + storageTime;
treadoff = t0 + pulseTime/2 + storageTime + pulseTime;
Clear[OmegaFunc]
OmegaFunc[t_?NumericQ /; t < twriteon] = 0;
OmegaFunc[t_?NumericQ /; twriteoff >= t >= twriteon] = OmegaC;
OmegaFunc[t_?NumericQ /; treadon > t > twriteoff] = 0;
OmegaFunc[t_?NumericQ /; treadoff >= t >= treadon] = OmegaC;
OmegaFunc[t_?NumericQ /; t > treadoff] = 0;
(*Plot[OmegaFunc[t], {t, 0, tend}, PlotRange -> All]*)
(*Initial conditions*)
p0 [z_] = 0;
s0[z_] = 0;
A = 9.39437
u0[t_] = A* Exp[-4*Log[2]*((t - t0 )/tau)^2];
(*Plot[u0[t], {t, 0, tend}, PlotRange -> All]*)
(*Define arrays in z to discretize problem in z*)
zmax = 1;
n = 4;
h = zmax/n;
P[t_] = Table[p[i][t], {i, 1, n}];
S[t_] = Table[s[i][t], {i, 1, n}];
integrateP = Join[{p[1][t]}, Table[p[i - 2][t] + p[i - 1][t], {i, 3, n}]];
U[t_] = Join[{u0[t]}, integrateP];
(*Construct equations*)
eqP = Thread[
D[P[t], t] == -P[t] + I Sqrt[d] U[t] + I OmegaFunc[t] S[t]];
eqS = Thread[D[S[t], t] == I OmegaFunc[t] P[t]];
initP = Thread[P[0] == Table[p0[(i - 1) h], {i, 1, n}]];
initS = Thread[S[0] == Table[s0[(i - 1) h], {i, 1, n}]];
(*Solve*)
lines = NDSolve[{eqP, eqS, initP, initS}, {P[t], S[t]}, {t, 0, tend}];
(*Plot*)
ztab = Table[(i - 1) h, {i, 1, n}];
ParametricPlot3D[
Evaluate@Thread[{ztab, t, lines[[1, 1]]}], {t, 0, tend},
PlotRange -> All, AxesLabel -> {"z", "t", "P"},
BoxRatios -> {2, 2, 1}, ImageSize -> Large,
LabelStyle -> {Black, Bold, Medium}]
ParametricPlot3D[
Evaluate@Thread[{ztab, t, lines[[1, 2]]}], {t, 0, tend},
PlotRange -> All, AxesLabel -> {"z", "t", "S"},
BoxRatios -> {2, 2, 1}, ImageSize -> Large,
LabelStyle -> {Black, Bold, Medium}]
I’ve shown the result of plotting P(z, t) below. S(z, t) is the same.
Could anyone help with what I’m doing wrong? Thanks in advance.
We can use first order approximation on z
as follows
d = 84.9601;
tau = 0.1; t0 = .1;
OmegaC = 27.7259;
pulseTime = 2 Pi/OmegaC;
storageTime = 0.2;
twriteon = t0 - pulseTime/2;
twriteoff = t0 + pulseTime/2;
treadon = t0 + pulseTime/2 + storageTime;
treadoff = t0 + pulseTime/2 + storageTime + pulseTime;
Clear[OmegaFunc]
OmegaFunc[t_] :=
Piecewise[{{0, t < twriteon}, {OmegaC,
twriteoff >= t >= twriteon}, {0,
treadon > t > twriteoff}, {OmegaC, treadoff >= t >= treadon}, {0,
t > treadoff}}]
(*Plot[OmegaFunc[t],{t,0,tend},PlotRange[Rule]All]*)
(*Initial conditions*)
p0[z_] = 0;
s0[z_] = 0;
A = 9.39437;
u0[t_] = A*Exp[-4*Log[2]*((t - t0)/tau)^2];
(*Plot[u0[t],{t,0,tend},PlotRange[Rule]All]*)
(*Define arrays in z to discretize problem in z*)
zmax = 1;
n = 41;
h = zmax/n;
P[t_] = Table[p[i][t], {i, 1, n}];
S[t_] = Table[s[i][t], {i, 1, n}];
(*Construct equations*)
eqP = Table[
D[p[i][t], t] == -p[i][t] +
I Sqrt[d] (u0[t] + h Sum[p[j][t], {j, 1, i}]) +
I OmegaFunc[t] s[i][t], {i, n}];
eqS = Table[D[s[i][t], t] == I OmegaFunc[t] p[i][t], {i, n}];
initP = Table[p[i][0] == 0, {i, 1, n}];
initS = Table[s[i][0] == 0, {i, 1, n}];
(*Solve*)
tend = 1.4; var =
Join[Table[p[i], {i, n}], Table[s[i], {i, n}]]; sols =
NDSolve[{eqP, eqS, initP, initS}, var, {t, 0, tend}];
(*Plot*)
Since functions P, S, U
are complex we can visualize Re
,Im
, Abs
as
ztab = Table[(i - 1) h, {i, 1, n}]; U =
Join[{u0[t]}, h Table[Sum[p[j][t], {j, i}], {i, n}]];
lst1 = Flatten[
Table[{ztab[[i]], t, Re[p[i][t]] /. First[sols]}, {i, n}, {t, 0,
tend, .01 tend}], 1];
lst2 = Flatten[
Table[{ztab[[i]], t, Re[s[i][t]] /. First[sols]}, {i, n}, {t, 0,
tend}], 1]; lst3 =
Flatten[Table[{ztab[[i]], t, Re[U[[i]]] /. First[sols]}, {i, n}, {t,
0, tend}], 1];
{ListPlot3D[lst1, ColorFunction -> "Rainbow",
AxesLabel -> {"z", "t", "P"}, PlotRange -> All, Mesh -> None],
ListPlot3D[lst2, ColorFunction -> "Rainbow",
AxesLabel -> {"z", "t", "S"}, PlotRange -> All, Mesh -> None],
ListPlot3D[lst3, ColorFunction -> "Rainbow",
AxesLabel -> {"z", "t", "U"}, PlotRange -> All, Mesh -> None]}
lst11 = Flatten[
Table[{ztab[[i]], t, Im[p[i][t]] /. First[sols]}, {i, n}, {t, 0,
tend, .01 tend}], 1];
lst21 = Flatten[
Table[{ztab[[i]], t, Im[s[i][t]] /. First[sols]}, {i, n}, {t, 0,
tend}], 1]; lst31 =
Flatten[Table[{ztab[[i]], t, Im[U[[i]]] /. First[sols]}, {i, n}, {t,
0, tend}], 1]; {ListPlot3D[lst11, ColorFunction -> "Rainbow",
AxesLabel -> {"z", "t", "P"}, PlotRange -> All, Mesh -> None],
ListPlot3D[lst21, ColorFunction -> "Rainbow",
AxesLabel -> {"z", "t", "S"}, PlotRange -> All, Mesh -> None],
ListPlot3D[lst31, ColorFunction -> "Rainbow",
AxesLabel -> {"z", "t", "U"}, PlotRange -> All, Mesh -> None]}
lst12 = Flatten[
Table[{ztab[[i]], t, Abs[p[i][t]] /. First[sols]}, {i, n}, {t, 0,
tend, .01 tend}], 1];
lst22 = Flatten[
Table[{ztab[[i]], t, Abs[s[i][t]] /. First[sols]}, {i, n}, {t, 0,
tend}], 1]; lst32 =
Flatten[Table[{ztab[[i]], t, Im[U[[i]]] /. First[sols]}, {i, n}, {t,
0, tend}], 1]; {ListPlot3D[lst12, ColorFunction -> "Rainbow",
AxesLabel -> {"z", "t", "|P|"}, PlotRange -> All, Mesh -> None],
ListPlot3D[lst22, ColorFunction -> "Rainbow",
AxesLabel -> {"z", "t", "|S|"}, PlotRange -> All, Mesh -> None],
ListPlot3D[lst32, ColorFunction -> "Rainbow",
AxesLabel -> {"z", "t", "|U|"}, PlotRange -> All, Mesh -> None]}
Answered by Alex Trounev on January 11, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP