# solving sets of partial Differential equations

Considering two functions $$\psi_{1}(u,v)$$ and $$\psi_{4}(u,v)$$. we have these two parial differential equation for them

$$(-2 i Sech[\frac{u}{\alpha}] \frac{\partial\psi_{4}}{\partial u}+2 i Sech[\frac{v}{\alpha}] \frac{\partial\psi_{4}}{\partial v})+(2 i Sech[\frac{u}{\alpha}] \frac{\partial\psi_{1}}{\partial u}+2 i Sech[\frac{v}{\alpha}] \frac{\partial\psi_{1}}{\partial v}-m\psi_{1})=0$$

$$(2 i Sech[\frac{u}{\alpha}] \frac{\partial\psi_{1}}{\partial u}-2 i Sech[\frac{v}{\alpha}] \frac{\partial\psi_{1}}{\partial v})+(-2 i Sech[\frac{u}{\alpha}] \frac{\partial\psi_{4}}{\partial u}-2 i Sech[\frac{v}{\alpha}] \frac{\partial\psi_{4}}{\partial v}-m\psi_{4})=0$$

I need to find $$\psi_{1}(u,v)$$ and $$\psi_{4}(u,v)$$ I wrote these code for them but I don't know why is the problem and why it doesn't give the answer.

DSolve[{((-2 I Sech[u/α] D[ψ4[u, v], u]+2 ISech[v/α] D[ψ4[u, v], v]) + (-m ψ1 [u, v] + 2 I  Sech[u/α] D[ψ1[u, v], u] +2 I  Sech[v/α] D[ψ1[u, v], v])) ==0, ((2I Sech[u/α] D[ψ1[u, v], u]-2 I  Sech[v/α] D[ψ1[u, v], v]) + (-m ψ4[u, v] - 2 I  Sech[u/α] D[ψ4[u, v], u]-2 I  Sech[v/α] D[ψ4[u, v], v])) == 0}, {ψ1[u,v],ψ4[u, v]}, {u, v}]


I also tried NDSolve but still didn't get the answer

• For a numerical solution, it is necessary to determine the initial and boundary conditions. Oct 13 '18 at 9:52
• The code contains several errors. I corrected them in my answer. Oct 13 '18 at 13:14

This system of equations also can be solved symbolically, although not with DSolve. Beginning with

eq = {((-2 I Sech[u/α] D[ψ4[u, v], u] + 2 I Sech[v/α] D[ψ4[u, v], v]) +
(-m ψ1[u, v] + 2 I Sech[u/α] D[ψ1[u, v], u] + 2 I Sech[v/α] D[ψ1[u, v], v])),
(( 2 I Sech[u/α] D[ψ1[u, v], u] - 2 I Sech[v/α] D[ψ1[u, v], v]) +
(-m ψ4[u, v] - 2 I Sech[u/α] D[ψ4[u, v], u] - 2 I Sech[v/α] D[ψ4[u, v], v]))};


make the transformation,

rulet = {Sinh[u/α] -> ut, Sinh[v/α] -> vt};


with corresponding transformation of derivatives,

ruled = {Derivative[1, 0][ψ_][u, v] -> Derivative[1, 0][ψ][ut, vt] Cosh[u/α]/α,
Derivative[0, 1][ψ_][u, v] -> Derivative[0, 1][ψ][ut, vt] Cosh[v/α]/α,
ψ_[u, v] -> ψ[ut, vt]}


Applying this transformation then eliminates the independent variables from the coefficients of the equations.

Simplify[α eq /. ruled]
(* {((-2 I D[ψ4[ut, vt], ut] + 2 I D[ψ4[ut, vt], vt]) +
(-m α ψ1[ut, vt] + 2 I D[ψ1[ut, vt], ut] + 2 I D[ψ1[ut, vt], vt])),
(( 2 I D[ψ1[ut, vt], ut] - 2 I D[ψ1[ut, vt], vt]) +
(-m ψ4[ut, vt] - 2 I D[ψ4[ut, vt], ut] - 2 I D[ψ4[ut, vt], vt]))} *)


It follows, therefore, that the solution can be decomposed into Fourier modes,

{ψ1 -> Function[{ut, vt}, ψ10 Exp[I ku ut + I kv vt]],
ψ4 -> Function[{ut, vt}, ψ40 Exp[I ku ut + I kv vt]]}


or, in terms of the original variables,

{ψ1 -> Function[{u, v}, ψ10 Exp[I ku Sinh[u/α] + I kv Sinh[v/α]]],
ψ4 -> Function[{u, v}, ψ40 Exp[I ku Sinh[u/α] + I kv Sinh[v/α]]]}


The wavenumbers, {ku, kv} can be determined in the usual manner,

Simplify[eq Exp[-I ku Sinh[u/α] - I kv Sinh[v/α]]/.%];
CoefficientArrays[% I α/4, {ψ10, ψ40}] // Last // Normal // Det
(* ku kv - (m^2 α^2)/16 *)

– aber
Oct 16 '18 at 3:15
• I have a more general solution and shall post it later this week. Oct 16 '18 at 3:46

An example of a numerical solution

eq = {((-2 I Sech[u/α] D[ψ4[u, v], u] + 2 I*Sech[v/α] D[ψ4[u, v], v])
+ (-m ψ1[u, v] + 2 I Sech[u/α] D[ψ1[u, v], u]
+ 2 I Sech[v/α] D[ψ1[u, v], v])) == 0, ((2 I*Sech[u/α] D[ψ1[u, v], u] -
2 I Sech[v/α] D[ψ1[u, v], v]) + (-m ψ4[u, v] - 2 I Sech[u/α] D[ψ4[u, v], u] -
2 I Sech[v/α] D[ψ4[u, v], v])) == 0};

bc = {ψ1[0, v] == Exp[-v^2], ψ4[0, v] == 0, ψ1[L, v] == Exp[-L^2], ψ4[L, v] == 0};
L = 5; α = 1; m = 1;
s = NDSolve[{eq, bc}, {ψ1[u, v], ψ4[u, v]}, {u, 0, L}, {v, 0, L}];

Plot3D[Evaluate[Abs[ψ1[u, v]] /. s], {u, 0, L}, {v, 0, L},
PlotRange -> All, PlotPoints -> 50, Mesh -> None, ColorFunction -> Hue] • Dear Alex Thank you very much. You helped a lot
– aber
Oct 13 '18 at 15:18
• @aber You're welcome! Oct 13 '18 at 17:00