Difference between revisions of "Talk:Aufgaben:Problem 13"
(Created page with "Consider the initial value problem (IVP) $$ \begin{align} \frac{\partial}{\partial t} f(x,t) &=-\frac{1}{2} \frac{\partial^2}{\partial x^2} f(x,t) \\ f(x,0) &= g(x) \in S(\ma...") |
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\end{align} | \end{align} | ||
$$ | $$ | ||
− | with \(\check{\hat\phi}=\phi\) and \(\check\phi (x)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\phi(k)e^{ikx}dk\), is a solution of (IVP). | + | 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} |
Revision as of 13:44, 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}