# 2D linear schrodinger equation

I am trying to solve a 2D Schrodinger equation of the following form. This is in the context of Partial Differential Equations.

\begin{align} iu_t + \frac{1}{2} (u_{xx}+u_{yy}) & = 0 \\ \end{align} And we are given the following conditions, \begin{align} u(x,y,0) = \phi(x,y) \\ -\infty < x < \infty \\ -\infty < y < \infty \\ 0 < t < \infty \end{align} This is the question.

Where I am right now. Take the Fourier Transform, \begin{align} i\frac{\partial}{\partial t}\hat{u}(k_1, k_2, t) + \frac{1}{2} \big(-k_1^2 \hat{u}(k_1,k_2,t) - k_2^2 \hat{u}(k_1,k_2,t) \big) = 0 \end{align} After shuffling things around for a bit we get, \begin{align} \frac{\partial}{\partial t} \hat{u} = -\frac{i}{2} \big( k_1^2+k_2^2 \big) \hat{u} \end{align} This is an ordinary differential equation. Solve this, and we have, \begin{align} \hat{u} = \hat{\phi} e^{-\frac{i}{2}(k_1^2 + k_2^2)t} \end{align} But from here, I am not sure how to take the inverse Fourier Transform to get the original $u(x,y,t)$. I feel like this question is almost solved, but just need a little more help.

Thank you!

Remember than $\widehat{fg} = \widehat{f} \ast \widehat{g}$, where $\ast$ is the convolution. So you have

$$\mathcal{F}[\mathcal{F}[u]] = \mathcal{F}\left[\mathcal{F}[\phi](t) e^{-\frac{i}{2}(k_1^2+k_2^2)t}\right] = \mathcal{F}[\mathcal{F}[\phi]] \ast \mathcal{F}\left[e^{-\frac{i}{2}(k_1^2+k_2^2)t}\right]$$

But now you use the fact that $\mathcal{F}[\mathcal{F}[u]](x)=u(-x)$ (with a coefficient, that depend of the definition of the Fourier transform you choose) to get that

$$u(x) = \phi(x) \ast \mathcal{F}\left[e^{-\frac{i}{2}(k_1^2+k_2^2)t}\right] (-x)$$

I let you calculate this Fourier transform

• Thank you so much! I am looking at this right now. I am a little confused about taking the inverse fourier tranform in this case, however. $u(x,y,t) = \phi(x-\frac{1}{2}t, y-\frac{1}{2}t)$. Is this the correct way to take the inverse transform? – mathsquestions1 Dec 11 '15 at 4:53
• @mathsquestions1 : No, I don't think it's the correct result. – Tryss Dec 11 '15 at 5:05
• Thanks for the feedback! Is it then possible that the only way is to evaluate this integral? But I am not sure if this can be done. Here it is! $u(x,y,t) = \frac{-x}{(2\pi)^2} \phi(x,y) \iint e^{i(k_1x+k_2y)} e^{-\frac{i}{2} (k_1^2+k_2^2)t} dk_1 dk_2$ ? – mathsquestions1 Dec 11 '15 at 7:19
• my mistake. missed the symbol for convolution! – mathsquestions1 Dec 11 '15 at 17:02

One may first seek solutions to the given SE in the form $u_{\boldsymbol{k}}(x,y,t) = v_{\boldsymbol{k}}(t)\exp(-i\boldsymbol{k.r})$. On substitution one has $i\dot{v_{\boldsymbol{k}}}= -|\boldsymbol{k}|^2v_{\boldsymbol{k}}$. The equation in $v_{\boldsymbol{k}}(t)$ is trivial. With $|\boldsymbol{k}|^2=\frac{1}{2}$, one has $v_{\boldsymbol{k}}(t)= a_{\boldsymbol{k}}\exp (\frac{it}{2})$. Now construct the full solution as $u(\boldsymbol{r}, t)= \Sigma _{\boldsymbol{k}}v_{\boldsymbol{k}}(t)\exp(-i\boldsymbol{k.r})=\big(\Sigma _{\boldsymbol{k}}a_{\boldsymbol{k}}\exp(-i\boldsymbol{k.r})\big)\exp (\frac{it}{2})$. At $t=0$, then one needs to resolve $\phi(\boldsymbol{r})$ in plain waves, so from this condition $a_{\boldsymbol{k}}$ are known.