Difference between revisions of "Talk:Aufgaben:Problem 13"
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\le \lvert \hat g(k) \frac{k^2}{2} \rvert | \le \lvert \hat g(k) \frac{k^2}{2} \rvert | ||
$$ | $$ | ||
| − | and \( \hat g(k) \frac{k^2}{2} \in L^1 \) | + | and \( \hat g(k) \frac{k^2}{2} \in L^1 \) so we are able to exchange integral and differential due to the Lebesque-dominated convergence |
| − | and | + | theorem. Further |
$$ -\frac{1}{2}\frac{\partial}{\partial x}\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk = -\frac{1}{2}\frac{1}{\sqrt{2\pi}} \lim\limits_{h \rightarrow 0} \int_{\mathbb{R}} \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-ikh}-1)dk $$ | $$ -\frac{1}{2}\frac{\partial}{\partial x}\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk = -\frac{1}{2}\frac{1}{\sqrt{2\pi}} \lim\limits_{h \rightarrow 0} \int_{\mathbb{R}} \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-ikh}-1)dk $$ | ||
because | because | ||
$$ | $$ | ||
\lvert \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-ikh}-1) \rvert \\ | \lvert \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-ikh}-1) \rvert \\ | ||
| − | \le \lvert \frac{1}{h}\hat g(k) 2i \sin(kh) \rvert \\ | + | \le \lvert \frac{1}{h}\hat g(k) 2i \sin(kh/2) \rvert \\ |
| − | \le \lvert \frac{1}{h}\hat g(k) 2 kh \rvert \\ | + | \le \lvert \frac{1}{h}\hat g(k) 2 kh/2 \rvert \\ |
\le \lvert \hat g(k) k \rvert | \le \lvert \hat g(k) k \rvert | ||
$$ | $$ | ||
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and since | and since | ||
$$ \frac{\partial}{\partial x}\hat g(k) e^{-i\frac{k^2}{2}t + ikx} = k \hat g(k) e^{-i\frac{k^2}{2}t} $$ | $$ \frac{\partial}{\partial x}\hat g(k) e^{-i\frac{k^2}{2}t + ikx} = k \hat g(k) e^{-i\frac{k^2}{2}t} $$ | ||
| − | and for \( p \in \mathbb{R}[x] \text{ and } | + | and for \( p \in \mathbb{R}[x] \text{ and } g \in S(\mathbb{R}) \text{ and } p\circ g \in S(\mathbb{R}) \) we are also able to exchange the second derivative, because we can make the exact same estimate.(\( k * \hat g(k)= \hat g'(k)\in S(\mathbb{R})\)) |
Revision as of 13:19, 9 January 2015
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 for used exchange
Prove why we can exchange the derivation and the integral:
$$ \frac{\partial}{\partial t} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \\ = \lim\limits_{h \rightarrow 0} \int_{\mathbb{R}} \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-i\frac{k^2}{2}h}-1)dk $$ because $$ \lvert \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-i\frac{k^2}{2}h}-1) \rvert \\ \le \lvert \frac{1}{h}\hat g(k) 2i \sin(\frac{k^2}{4}h) \rvert \\ \le \lvert \frac{1}{h}\hat g(k) 2 \frac{k^2}{4}h \rvert \\ \le \lvert \hat g(k) \frac{k^2}{2} \rvert $$ and \( \hat g(k) \frac{k^2}{2} \in L^1 \) so we are able to exchange integral and differential due to the Lebesque-dominated convergence theorem. Further $$ -\frac{1}{2}\frac{\partial}{\partial x}\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk = -\frac{1}{2}\frac{1}{\sqrt{2\pi}} \lim\limits_{h \rightarrow 0} \int_{\mathbb{R}} \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-ikh}-1)dk $$ because $$ \lvert \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-ikh}-1) \rvert \\ \le \lvert \frac{1}{h}\hat g(k) 2i \sin(kh/2) \rvert \\ \le \lvert \frac{1}{h}\hat g(k) 2 kh/2 \rvert \\ \le \lvert \hat g(k) k \rvert $$ and \( \hat g(k) k \in L^1 \) and since $$ \frac{\partial}{\partial x}\hat g(k) e^{-i\frac{k^2}{2}t + ikx} = k \hat g(k) e^{-i\frac{k^2}{2}t} $$ and for \( p \in \mathbb{R}[x] \text{ and } g \in S(\mathbb{R}) \text{ and } p\circ g \in S(\mathbb{R}) \) we are also able to exchange the second derivative, because we can make the exact same estimate.(\( k * \hat g(k)= \hat g'(k)\in S(\mathbb{R})\))
