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)=(\hat g(k)e^{-i\frac{k^2}{2}t})\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: \begin{align} 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)\\ \end{align} 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} (\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk) &= \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} (\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk) &= \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}}i^2k^2\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^2dk \end{align}

Compare it with the differential equation: \begin{align} 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^2dk &=\frac{1}{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{1}{2}\frac{\partial^2}{\partial x^2} f(x,t)&= (-\frac{1}{2})*(-\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk) &= \frac{1}{2\sqrt{2\pi}}\int_{\mathbb{R}} \hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk) \end{align} It is the same.

Prove why we can exchange the derivation and the integral: Define $$ h(x,t,k)=\hat g(k)e^{ikx-i\frac{k^2}{2}t} $$

$$ \lvert h(x,t,k)\rvert = \le \hat g(k) \rvert \le \lvert \hat g(k) \rvert $$ and \( \hat g(k) \in S({\mathbb{R}}) \) 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