so I've been confused by this before and i still have no idea.
Use the method of fundamental solutions to solve the heat equation $u_t=Du_{xx}$ with $x \in \mathbb{R}$, $t>0$ and the following initial data
For question a) we need to use
$u(x,0)=e^{-x^2}$
from my understanding of the problem i need to use the fundamental solution
$$\Phi(x,t) = \frac{e^{\frac{-x^2}{4Dt}}}{\sqrt{4 \pi Dt}}$$
then the solution $u(x,t)$ is
$$u(x,t) = \int_{\mathbb{R}} \Phi(x-y,t)\phi(y)dy$$
the issue is the notes i have dont explain how to expand or "solve" this in particualar ie in the above we have $\phi(x) = e^{-x^2}$
so we have $$u(x,t) = \int_{\mathbb{R}} \frac{e^{\frac{-(x-y)^2}{4Dt}}}{\sqrt{4 \pi Dt}}e^{-y^2}dy$$
from here i still have no idea where to go with this ie no idea how to evaluate the above. any hints/tips of solutions would be great.
For the record i've yet to start fourier transforms or greens.
any help would be great. thanks!
Edit 2: the above example question came from a work sheet. The solution to the above is apparently.
$$u(x,t) = \frac{e^{\frac{-x^2}{4Dt+1}}}{\sqrt{4 Dt+1}}$$
I dont understand how they came to this value.. what calculations were done or if im just missing a property which allows me to link the two. This is what i mean by evaluate the integral.
Again thank you for taking the time to read this. Any and all help is appreciated.
You can obtain the result by a change of variables inside the integral. $$u(x,t) = \int_{\mathbb{R}} \frac{e^{\frac{-(x-y)^2}{4Dt}}}{\sqrt{4 \pi Dt}}e^{-y^2}dy= \int_{\mathbb{R}} \frac{e^{\frac{-x^2-y^2+2xy-4Dty^2}{4Dt}}}{\sqrt{4 \pi Dt}}dy=\int_{\mathbb{R}} \frac{e^{\frac{-x^2+\frac{x^2}{1+4Dt}-\frac{x^2}{1+4Dt}-y^2(1+4Dt)+2xy}{4Dt}}}{\sqrt{4 \pi Dt}}dy= \int_{\mathbb{R}} \frac{e^{-\frac{x^2}{1+4Dt}-\frac{\left(\frac{x}{\sqrt{1+4Dt}}-y\sqrt{1+4Dt}\right)^2}{4Dt}}}{\sqrt{4 \pi Dt}}dy=e^{-\frac{x^2}{1+4Dt}}\int_{\mathbb{R}}\frac{e^{-z^2}}{\sqrt{1+4Dt}\sqrt{\pi}}dz=\frac{e^{-\frac{x^2}{1+4Dt}}}{\sqrt{1+4Dt}},$$ where $z=\frac{\frac{x}{\sqrt{1+4Dt}}-y\sqrt{1+4Dt}}{\sqrt{4 \pi Dt}}$.