Talk:Aufgaben:Problem 13

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{1}{4\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}{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}{4\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk &=\frac{1}{4\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}{2\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk) &= \frac{1}{4\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: $$ i\frac{\partial}{\partial t} \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk=\frac{i}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial}{\partial t}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk $$ because $$ \lvert\frac{\partial}{\partial t}\hat g(k)e^{ikx-i\frac{k^2}{2}t}\rvert = \lvert\lim\limits_{n \rightarrow 0}{\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-i\frac{k^2}{2}n}-1)}\rvert = \frac{2}{n}\lvert\hat g(k)\rvert\lvert e^{-i\frac{k^2 x^2}{2}}\rvert\lvert \sin(\frac{k^2}{4})\rvert \underset{\overbrace{\lvert\sin(k^2)\rvert < \lvert k \rvert}}{\leqslant} \lvert k \rvert \lvert \hat g(k) \rvert \in \ L^1 (\text{since } g(k) \in S(\mathbb{R})\subset L^1)$$ and $$ -\frac{1}{2}\frac{\partial^2}{\partial x^2}\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk = -\frac{1}{2}\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\frac{\partial^2}{\partial x^2}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk $$ because $$ \lvert\frac{\partial}{\partial x}\hat g(k)e^{ikx-i\frac{k^2}{2}t}\rvert = \lvert\lim\limits_{n \rightarrow 0}{\frac{1}{n} \hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{ikn}-1)}\rvert = \frac{2}{n}\lvert\hat g(k)\rvert\lvert e^{i\frac{k^2 x^2}{2}}\rvert\lvert \sin(\frac{xkn}{2})\rvert \underset{\overbrace{\lvert\sin(x)\rvert \leqslant \lvert x \rvert}}{\leqslant} \lvert k \rvert \lvert \hat g(k) \rvert \underset{=1}{\underbrace{\lvert e^{ik^2 x^2} \rvert}} \in L^1 (\text{since } \hat g(k) \in S(\mathbb{R})\subset L^1) $$ 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 \( p \in \mathbb{R}[x] \text{ and } f \in S(\mathbb{R}) \text{ and } p\circ f \in S(\mathbb{R}) \) we are also able to exchange the second derivative.

Therefore we get $$ i \frac{\partial}{\partial t} \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk = \frac{i}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial}{\partial t}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk = -\frac{(i)^2}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)k^2 e^{ikx-i\frac{k^2}{2}t}dx = \boldsymbol{(1)} $$

$$ -\frac{1}{2}\frac{1}{\sqrt{2\pi}}\frac{\partial^2}{\partial x^2} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dx = -\frac{1}{2} \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\frac{\partial^2}{\partial x^2} \hat g(k)e^{ikx-i\frac{k^2}{2}t}dk = -\frac{(i)^2}{2\sqrt{2\pi}}\int_{\mathbb{R}} g(k)k^2 e^{ikx-i\frac{k^2}{2}t}dk = \boldsymbol{(2)} $$

$$ \boldsymbol{(1)} = \boldsymbol{(2)} $$