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!
 A: 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 
A: 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.
