# Solution to Schrodinger equation with non-Schwartz initial data

For $\xi_0\in\mathbf{R}^n$ compute the solution of the Schrodinger equation with initial data $$i\partial_tu-\Delta u=0 \text{ in } (0,t)\times\mathbf{R}^n\\ u(0,x)=e^{i\xi_0\cdot x} \text{ for all } x\in\mathbf{R}^n.$$ Give an interpretation of the solution.

The only way that I know of finding the solutions to the Schrodinger equation is by taking the Fourier transform with respect to $x$ of the equation, solving it for $\hat{u}$, then using the convolution of the transform of the initial data with the transform of the complex heat kernel (Schrodinger kernel?). So when I try to do that I get this far.

$$\mathcal{F}\{u(t,x)\}(k)=e^{i|k|^2t}\mathcal{F}\{e^{i\xi_0\cdot x}\}$$

My issue now is that the initial data is not Schwartz, and so its Fourier transform does not exist. When I try to take the Fourier transform, I have to evaluate a circular function at $\pm\infty$, which doesn't make sense to me.

Any advice would be much appreciated!

Tom

• A wave function is supposed to be square integrable at every $t$ with a value of 1. Sounds like a strange exercise. Commented May 31, 2015 at 21:11
• @T.A.E.: It is common in PDEs to consider the Schrödinger equation in the tempered distributional setting. Commented Jun 2, 2015 at 8:54
• @GiuseppeNegro : It's not physical unless the solution is a unit vector in $L^{2}$ for all $t \ge 0$. So asking for an interpretation seems strange. "Interpretation" in what sense? Commented Jun 2, 2015 at 9:09
• @T.A.E.: See my answer below for my interpretation of the word "interpretation". :-) Commented Jun 2, 2015 at 9:26
• @GiuseppeNegro : My idea of an interpretation would be a physical interpretation, not a Mathematical description. Commented Jun 2, 2015 at 9:30

HINTS:

The Fourier transform is defined for all tempered distributions. In this generalized sense you have $$\mathcal{F}_{x\to k}\left(e^{i\xi_0\cdot x}\right)=(2\pi)^{d}\delta(k-\xi_0),$$ where $\delta$ is the Dirac delta.

As for the interpretation, try thinking at the result of applying a Galilean transformation to the initial datum $u_0(x)=1$.

Note 1: A Galilean transformation, aka "Galilean boost", is the map that sends the initial datum $u_0(x)$ to $$\mathcal{F}^{-1}_{k\to x}\left( \hat{u}_0(k+\xi_0)\right).$$

Note 2: I am using the following convention for the Fourier transform: $$\mathcal{F}_{x\to k}(f(x))=\int_{\mathbb{R}^d}f(x)e^{-ix\cdot k}\, dx.$$

• Thanks! So can I say $$\mathcal{F}\{u(t,x)\}=e^{i|k|^2t}(2\pi)^n\delta(k-\xi_0)\\ =\mathcal{F}\{\Gamma(t,x)\}\mathcal{F}\{e^{i\xi_0 \cdot x}\}\\ =\mathcal{F}\left\lbrace \int_{\mathbf{R^n}}\Gamma(t,x-y)e^{i\xi_0\cdot y} dy\right\rbrace$$ where $\Gamma(t,x)=\frac{1}{(4\pi t)^{\frac{n}{2}}}e^{\frac{-|x|^2}{4t}}$. Commented May 31, 2015 at 12:01
• It looks right, but you can give a much more explicit formula. Commented May 31, 2015 at 12:03
• I've been using $\mathcal{F}(f(x))=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} f(x)e^{-ix\cdot k} dx$ for Fourier transform, and $\mathcal{F^{-1}}(f(k))=\int_{\mathbb{R}^n} f(k)e^{ik\cdot k} dk$. So taking the FT and solving it I get $$\mathcal{F}(u(t,x))=e^{i|k|^2t}\delta(k-\xi_0)$$ Then taking the inverse FT I get $$u=\mathcal{F}^{-1}(\mathcal{F}(u))=\int_{\mathbb{R}^n} e^{i|k|^2t\delta(k-\xi_0)e^{ik\cdot x}}dx\\ =e^{i|\xi_0|^2t}e^{i\xi_0\cdot x}$$ How does that look? I tried it for $x\in\mathbb{R}$ and it worked out. Commented May 31, 2015 at 16:10