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We call a function that assigns a starting value of a time-dependent differentialfunction to a solution of a later timevalue as the evolution operator $E(t)$.

Look at the thermal equation

$$ u_t=u_{xx},~~~u_0(x)=u(0,x) $$

where $u\colon [0,T]\times\mathbb{R}\to\mathbb{R}$ with the evolution operator $E(T)u_0=u(T,0)$.

Show that $u_0\star G_t$ with

$$ G_t(x)=\frac{1}{\sqrt{2\pi t}}\exp\left(-\frac{x^2}{t}\right) $$ is a solution of the above equation.

Advice: You therefore have to show $E(T)u_0=u_0\star G_T$.

I am a little bit helpless.

Did I understand the evolution operator right, when thinking that

$E(T)(x)=u(T,0)$ for every $x$?

Then I have to show $(u_0\star G_T)(x)=u(T,0)$?

If yes: How can I do so? Does the Fourier transformation play any role? I thought of that because of the convolution but it was only a quick thought...

I would be very thankful if anyone could help me to solve this question.

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Yes, your suspicions are correct. Try the Fourier transform: if $U$ is the Fourier transform of $u$, then Fourier transforming your PDE yields: $$ \frac{\partial U(k,t)}{\partial t} = -k^2 U(k,t) $$ See if you can work with this new PDE.

Note: I am use the convention that the Fourier transform is: $$ U(k,t) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} e^{-ikx} u(x,t) \, dx$$

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My problem is the following: I do not understand the advice... why does $E(T)u_0=u_0\star G_T$ proof, that $u_0\star G_t$ is a solution of (1)? – math12 Jan 11 '13 at 23:00
Thank you. As I said my problem is the advice. If I do not consider the advice and just see it as solving a differentialequation, i think i can handle it. But the advice confuses me... i do not know how to start with this advice... – math12 Jan 11 '13 at 23:26
When you solve for $U(k,t)$, you will obtain an expression that involves the Fourier transform of $u_0$. Then you can use the convolution identity when solving for $u$ by inverse Fourier transform, and you will get an expression of the desired form. – Christopher A. Wong Jan 12 '13 at 8:30
Hm, if i do so, i get $G_t(x)=\frac{1}{\sqrt{4\pi t}}e^{-\lvert x^2\rvert/4t}$ and not $G_t(x)=\frac{1}{\sqrt{2\pi t}}e^{-x^2/t}$ – math12 Jan 12 '13 at 11:14
I got this: $\hat{u}(t,k)=\hat{u_0}e^{-\lvert k\rvert^2 t}$. Then with inverse formula $u(t,x)=\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb{R}}\hat{u_0}(k)e^{-\lvert k\rvert^2t}e^{ikx}\, dk$. Setting $\omega=\mathcal{F}^{-1}(e^{-\lvert k\rvert^2 t})$, I got $u(t,x)=\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb{R}}\hat{(u_0\star \omega)}(k,t)e^{ikx}\, dk=(u_0\star \omega)(t,x)$. Then $\omega(t,x)=\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb{R}}e^{-\lvert k\rvert^2 t}e^{ikx}\, dx=(4\pi t)^{-1/2}e^{-\lvert x\rvert^2/4t}$ – math12 Jan 12 '13 at 12:31

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