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

Revision as of 22:04, 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): $$ \begin{align} f(x,t)=\left( \hat g(k) e^{-i\frac{k^2}{2}t} \right) \check (x) &=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat{g(k)} e^{-i\frac{k^2}{2}t}e^{ikx} \, dk \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \end{align} $$

Check if the solution satisfies the initial condition: $$ f(x,0) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-0}dk = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx} \, dk =\check{\hat g}(x) = g(x) $$ For this, we used again (1) and the identity \(\check{\hat\phi}=\phi\).

Check if the solution satisfies the differential equation:
Calculate de derivations: $$ \begin{align} \frac{\partial}{\partial t} f(x,t) &= \frac{\partial}{\partial t} \left( \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \right) \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial}{\partial t}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}-i\frac{k^2}{2}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \\ &= -\frac{i}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk \end{align} $$

$$ \begin{align} \frac{\partial^2}{\partial x^2} f(x,t) &= \frac{\partial^2}{\partial x^2} \left( \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \right) \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial^2}{\partial x^2}\hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\left(i^2k^2 \right) \hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \\ &= -\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2 \, dk \end{align} $$

Put the whole thing in the differential equation and we see: $$ i\frac{\partial}{\partial t} f(x,t) = -\frac{i^2}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2 \, dk =\frac{1}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk = \left( -\frac{1}{2}\right) \cdot \left( -\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk\right) = -\frac{1}{2}\frac{\partial^2}{\partial x^2} f(x,t) $$ it is satisfied.

Proof of exchanging the derivation and the integral:

We want to apply the Lebesgue dominated convergence theorem. This is a standard argument being used in some other problems.

First, define \( h(x,t,k)=\hat g(k)e^{ikx-i\frac{k^2}{2}t} \) and since \( g(x) \in \mathcal{S} ({\mathbb{R}}) \rightarrow \hat g(k) \in \mathcal{S} ({\mathbb{R}}) \), we get considering the following argument:

$$ \left| f(x) \right| \leq \frac{C}{(1 + \vert x \vert^2)} $$

for some \( C \in \mathbb{R} \) and thus:

$$ \int_{\mathbb{R}} \left| f(x) \right| dx \leq \int_{\mathbb{R}} \left| \frac{C}{(1 + \vert x \vert^2)} \right| dx < \infty $$

that \( \hat g(k) \in L^1 \).

We then want to consider

$$ \lim_{n \rightarrow \infty} \int_{\mathbb{R}} \frac{h(x + s_n, t, k) - h(x,t,k)}{s_n} \, dk $$

with \( s_n \) an arbitrary zero-sequence and see that

$$ \left| \frac{1}{s_n} \hat g (k) e^{ikx} e^{-i\frac{k^2}{2}t} \left( e^{iks_n} -1 \right) \right| = \left| \frac{1}{s_n} \hat g (k) \sin \left( \frac{ks_n}{2} \right)\right| \leq \left| \frac{2}{s_n} \hat g (k) \frac{ks_n}{2} \right| = \left| k \cdot \hat g(k) \right| \in \mathcal{S} ({\mathbb{R}}) \subset L^1(\mathbb{R} ) $$

So \( \frac{d}{dx} \int_{\mathbb{R}} h(x,t,k) \, dk = \int_{\mathbb{R}} \frac{d}{dx} h(x,t,k) \, dk \), and since \( \frac{d}{dx} h(x,t,k) \) results in a Schwartz-function depending on \( k \) times \( h \) we can apply the Lebesgue-dominated-convergence theorem twice.

It remains to be shown for \( \frac{d}{dt} \) which is basically just repeating the argument:

$$ \lim_{n \rightarrow \infty} \int_{\mathbb{R}} \frac{h(x, t + s_n, k) - h(x,t,k)}{s_n} \, dk $$

again with \( s_n \) an arbitrary zero-sequence, and then we go on:

$$ \left| \frac{1}{s_n} \hat g (k) e^{ikx} e^{-i\frac{k^2}{2}t} \left( e^{i\frac{k^2}{2}s_n} -1 \right) \right| = \left| \frac{1}{s_n} \hat g (k) \sin \left( \frac{k^2s_n}{4} \right)\right| \leq \left| \frac{2}{s_n} \hat g (k) \frac{k^2s_n}{4} \right| = \left| \frac{k^2}{2} \cdot \hat g(k) \right| \in \mathcal{S} ({\mathbb{R}}) \subset L^1(\mathbb{R} ) $$.

Ende der Vorstellung.