Difference between revisions of "Talk:Aufgaben:Problem 13"

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(→‎Prove for used exchange)
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It is the same. <br />
 
It is the same. <br />
  
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== Prove for used exchange==
 
 
Prove why we can exchange the derivation and the integral:
 
Prove why we can exchange the derivation and the integral:
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Define
 
$$  
 
$$  
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\frac{\partial}{\partial t} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \\
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h(x,t,k)=\hat g(k)e^{ikx-i\frac{k^2}{2}t}
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= \lim\limits_{h \rightarrow 0} \int_{\mathbb{R}} \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-i\frac{k^2}{2}h}-1)dk
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$$
 
$$
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because
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$$
 
$$
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\lvert \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-i\frac{k^2}{2}h}-1) \rvert \\
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\lvert h(x,t,k)\rvert =
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\le \lvert \frac{1}{h}\hat g(k) 2i \sin(\frac{k^2}{4}h) \rvert \\
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\le \hat g(k) \rvert  
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\le \lvert \frac{1}{h}\hat g(k) 2 \frac{k^2}{4}h \rvert \\
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\le \lvert \hat g(k) \rvert  
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\le \lvert \hat g(k) \frac{k^2}{2} \rvert
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$$
 
$$
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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
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and \( \hat g(k) \in S({\mathbb{R}}) \) and therefore \( \hat g(k) \in L^1 \) so it follows that
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theorem. Further
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−
$$  -\frac{1}{2}\frac{\partial}{\partial x}\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk =  -\frac{1}{2}\frac{1}{\sqrt{2\pi}} \lim\limits_{h \rightarrow 0} \int_{\mathbb{R}} \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-ikh}-1)dk $$
+
−
because
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$$
 
$$
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\lvert \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-ikh}-1) \rvert \\
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\lvert \frac{\partial^2}{\partial x^2}h(x,t,k)\rvert =
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\le \lvert \frac{1}{h}\hat g(k) 2i \sin(kh/2) \rvert \\
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\le (tik)^2 h(x,t,k) \rvert \\
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\le \lvert \frac{1}{h}\hat g(k) 2 kh/2 \rvert \\
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\le \lvert \hat g(k) \rvert \lvert k^2 \rvert
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\le \lvert \hat g(k) k \rvert
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$$
 
$$
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and \( \hat g(k) k \in L^1 \)
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−
and since
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and
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$$ \frac{\partial}{\partial x}\hat g(k) e^{-i\frac{k^2}{2}t + ikx} = k \hat g(k) e^{-i\frac{k^2}{2}t} $$
+
 
−
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.(\( k * \hat g(k)= \hat g'(k)\in S(\mathbb{R})\))
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$$
 +
\lvert \frac{\partial}{\partial t}h(x,t,k)\rvert =
 +
\le \lvert \frac{-ik^2}{2} h(x,t,k) \rvert  \\
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\le \lvert \hat g(k) \rvert \lvert  \frac{k^2}{2} \rvert
 +
$$
 +
 
 +
and therefore we are allowed to exchange integral and differential

Revision as of 19:37, 12 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 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

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