Talk:Aufgaben:Problem 13

Consider the initial value problem (IVP) $$ \begin{align} i\frac{\partial}{\partial t} f(x,t) &=-\frac{1}{2} \frac{\partial^2}{\partial x^2} f(x,t) \\ f(x,0) &= g(x) \in S(\mathbb{R}) \\ \end{align} $$ Show that $$ \begin{align} f(x,t)=(\hat g(k)e^{-i\frac{k^2}{2}t})\check (x) \\ \end{align} $$ with \(\check{\hat\phi}=\phi\) and \(\check\phi (x)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\phi(k)e^{ikx}dk\) (1), is a solution of (IVP).

Solution

Rewrite the possible solution with help of (1): $$ \begin{align} f(x,t)=\left( \hat g(k) e^{-i\frac{k^2}{2}t} \right) \check (x) &=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat{g(k)} e^{-i\frac{k^2}{2}t}e^{ikx} \, dk \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \end{align} $$

Check if the solution satisfies the initial condition: $$ f(x,0) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-0}dk = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx} \, dk =\check{\hat g}(x) = g(x) $$ For this, we used again (1) and the identity \(\check{\hat\phi}=\phi\).

Check if the solution satisfies the differential equation:
Calculate de derivations: $$ \begin{align} \frac{\partial}{\partial t} f(x,t) &= \frac{\partial}{\partial t} \left( \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \right) \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial}{\partial t}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}-i\frac{k^2}{2}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \\ &= -\frac{i}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk \end{align} $$

$$ \begin{align} \frac{\partial^2}{\partial x^2} f(x,t) &= \frac{\partial^2}{\partial x^2} \left( \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \right) \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial^2}{\partial x^2}\hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\left(i^2k^2 \right) \hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \\ &= -\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2 \, dk \end{align} $$

Put the whole thing in the differential equation and we see: $$ i\frac{\partial}{\partial t} f(x,t) = -\frac{i^2}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2 \, dk =\frac{1}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk = \left( -\frac{1}{2}\right) \cdot \left( -\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk\right) = -\frac{1}{2}\frac{\partial^2}{\partial x^2} f(x,t) $$ it is satisfied.

Proof of exchanging the derivation and the integral:

Define \( h(x,t,k)=\hat g(k)e^{ikx-i\frac{k^2}{2}t} \), \( \left| h(x,t,k)\right| \leq \left| \hat g(k) \right| \leq \left| \hat g(k) \right| \) and \( \hat g(k) \in \mathcal{S} ({\mathbb{R}}) \) and therefore \( \hat g(k) \in L^1 \) so it follows that $$ \lvert \frac{\partial^2}{\partial x^2}h(x,t,k)\rvert = \le (tik)^2 h(x,t,k) \rvert \\ \le \lvert \hat g(k) \rvert \lvert k^2 \rvert $$

and

$$ \lvert \frac{\partial}{\partial t}h(x,t,k)\rvert = \le \lvert \frac{-ik^2}{2} h(x,t,k) \rvert \\ \le \lvert \hat g(k) \rvert \lvert \frac{k^2}{2} \rvert $$

and therefore we are allowed to exchange integral and differential