I have a PDE (1D Schrödinger equation) as follows:
$$iu_t-\frac12 u_{xx}-\left(|u|^2-A\right)u=0$$
Applying the following transformations to the PDE,
$$\operatorname{u}(x,y,t)=\operatorname{u}(\xi) e^{i\theta},\quad \xi=x+y+\alpha t,\;\theta=\alpha x+\beta y+\gamma t$$
we get the following ODE:
$$\frac12 u^{\prime\prime}+\left(\gamma-\frac12\alpha^2\right)u+u\left(u^2-A\right)=0$$
How to achieve the transformation above using DSolveChangeVariables
introduced in Mathematica 13.1?
My Try:
eq = I*D[u[x, y, t], t] -
1/2 D[u[x, y, t], {x, 2}] - (Abs[u[x, y, t]]^2 - A)*u[x, y, t] ==
0;
deq = Inactive[DSolve][eq, u, {x, y, t}];
DSolveChangeVariables[deq,
U, ξ, {u[x, y, t] == U[ξ]*E^(I*θ), ξ ==
x + y + α*t, θ == α*x + β*y + γ*t}] // Simplify
DChange
code. $\endgroup$pde
there while you already haveeq
? Also, as pointed out in the original post, the given ODE is not correct unless $u(\xi)\geq 0$, but this is not clarified at all. $\endgroup$