Difference between revisions of "Talk:Aufgaben:Problem 13"
| Line 25: | Line 25: | ||
\end{align} | \end{align} | ||
For this, we used again '''''(1)''''' and the identity \(\check{\hat\phi}=\phi\). | For this, we used again '''''(1)''''' and the identity \(\check{\hat\phi}=\phi\). | ||
| + | |||
| + | Check if the solution satisfies the differential equation: <br /> | ||
| + | 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{ik^2}{4\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk | ||
| + | \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{k^2}{2\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk | ||
| + | \end{align} | ||
| + | |||
| + | Compare it with the differential equation: | ||
| + | \begin{align} | ||
| + | i\frac{\partial}{\partial t} f(x,t) &= -\frac{i^2k^2}{4\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk &=\frac{k^2}{4\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk | ||
| + | \end{align} | ||
| + | \begin{align} | ||
| + | -\frac{1}{2}\frac{\partial^2}{\partial x^2} f(x,t)&= (-\frac{1}{2})*(-\frac{k^2}{2\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk) &= \frac{k^2}{4\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk) | ||
| + | \end{align} | ||
| + | It is the same. <br /> | ||
| + | Prove why we can exchange the derivation and the integral: | ||
Revision as of 15:39, 30 December 2014
Consider the initial value problem (IVP) $$ \begin{align} \frac{\partial}{\partial t} f(x,t) &=-\frac{1}{2} \frac{\partial^2}{\partial x^2} f(x,t) \\ f(x,0) &= g(x) \in S(\mathbb{R}) \\ \end{align} $$ 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{ik^2}{4\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk
\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{k^2}{2\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \end{align}
Compare it with the differential equation:
\begin{align}
i\frac{\partial}{\partial t} f(x,t) &= -\frac{i^2k^2}{4\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk &=\frac{k^2}{4\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk
\end{align}
\begin{align}
-\frac{1}{2}\frac{\partial^2}{\partial x^2} f(x,t)&= (-\frac{1}{2})*(-\frac{k^2}{2\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk) &= \frac{k^2}{4\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk)
\end{align}
It is the same.
Prove why we can exchange the derivation and the integral:
