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I derived some partial differential equation.

$$V(x)-u_{xx}-u_x^2=i u_t,$$ where $u=u(x,t)$.

I do not know even if this differential equation has some special name. I also do not know how to solve it.

I am looking for numerical solution for the following initial condition $u(x,0)=-(x-x_0)^2$ and potential $V(x)=x^2$ in range $x \in [-8,8]$, $t \in [0,t_{max}]$.

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  • $\begingroup$ Look into NDSolve $\endgroup$ – That Gravity Guy Sep 21 '18 at 22:21
  • $\begingroup$ I know Mathematica functions. Problem is that simple direct solution does not work. Well, it requires specification of boundary conditions which I do not know. $\endgroup$ – QuantumNik Sep 21 '18 at 22:24
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Consider the equation

$-if_t-f_{xx}+V(x)f=0$

We make a substitution $f=e^{u(x,t)}$.

f = Exp[u[x, t]];
FullSimplify[D[f, t]/I - D[f, x, x] + f*V[x]]
E^u[x, t](V[x] - I Derivative[0, 1][u][x, t] - Derivative[1, 0][u][x, t]^2 - Derivative[2, 0][u][x, t])

fig2

Thus, we have obtained an equation that the author considers. But the original equation is the Schrödinger equation. Let us consider a problem with initial data and with boundary conditions. Using the NDSolve we find a numerical solution

L = 4; t0 = 10; x0 = 1; 
F0[x_] := Exp[-(x - x0)^2]
sol = 
  NDSolveValue[
    {D[F[x, t], t]/I - D[F[x, t], x, x] + F[x, t]*x^2 == 0, 
     F[x, 0] == F0[x], F[L, t] == F0[L], F[-L, t] == F0[-L]}, 
    F, {x, -L, L}, {t, 0, t0}]

Plot3D[Abs[sol[x, t]], {x, -L, L}, {t, 0, t0}, 
  Mesh -> None, ColorFunction -> Hue]

fig1

Phase of the wave function in the plane (x,t). In this case we set L = 8

DensityPlot[Arg[sol[x, t]], {x, -L, L}, {t, 0, t0}, Mesh -> None, 
 ColorFunction -> Hue, PlotRange -> All, PlotPoints -> 100, 
 PlotLegends -> Automatic]

fig2

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  • $\begingroup$ Thanks for your comment. The problem is that I deliberately consider the presented equation which was indeed obtained by substitution of $e^u$ into the Schrodinger equation. My initial problem is to obtain the phase of the wavefunction over whole $x$ domain. Try to extract the $u$ function from your calculation. You will see that due to numerical errors it is possible only in the region where the wave packet is large enough. I was trying to avoid this limitation by transforming the SE into the presented one. Thus I am looking for a way to solve the presented equation in the original form. $\endgroup$ – QuantumNik Sep 24 '18 at 11:13
  • $\begingroup$ @QuantumNik I published a picture with the phase of the wave function. Please specify a problem. $\endgroup$ – Alex Trounev Sep 24 '18 at 13:32
  • $\begingroup$ Yes, it works in certain situations. Now increase your $x$ domain, so $L=40$. You will see that the phase of the solution will be just a noise outside of a small region near the maximum of the wave packet. Of course, you can increase precision but this is a dead end definitely. I am looking for a method which will allow me to deal with the phase in a different way. Thus I tried to represent it in a form of the exponential of a function. The original problem is to obtain the phase from the solution of the Schrodinger equation. $\endgroup$ – QuantumNik Sep 24 '18 at 15:15
  • $\begingroup$ @QuantumNik We will not improve this situation with a numerical solution when solving a nonlinear equation. It will only get worse. $\endgroup$ – Alex Trounev Sep 24 '18 at 16:33

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