Okay so I am studying for my PDE course and I am convering Fourier transforms. In fact I am using fourier transforms to find a solution to the heat equation on an infinite length rod.
After going through the derivation of the fourier transform and applying it to the heat equation we get the following solution: $u(x,t) = a(x) \ast K(x,t)$ where K(x,t) is the heat kernel so $K(x,t) = \frac{1}{\sqrt{4\pi kt}}e^{\frac{-x^2}{4kt}}$
Okay so given an initial condition $u(x,0) = F(x)$ we can not simply plug it in since when $t=0$ the denominator is undefined. It is at this point where I am getting confused
Our professor told us that to solve the heat equation using the IC (initial condition) we have to study what the convolution is doing. So lets see what happens when we take the limit as $t \rightarrow 0$. \lim_{t\rightarrow 0}\ \ a \ast K(x,t) = \lim_{t\rightarrow 0} \int_{-\infty}^{\infty} a(x')K(x-x', t) dt
okay so what is x'? Where did it come from? and why does our professor do this? I do know for a fact that the integration over the infinite bounds means the kernal has area 1 due to the fact that it is a Guassian function.
After a little bit of notes he ends up with the following \lim_{t\rightarrow 0}\int_{-\infty}^\infty a(x')K(x-x',t) dx' = a(x) and then he says that we can now replace the $a(x)$ with $F(x)$ (our initial condition). Okay well the last line makes sense but the limit of the integral dosn't make sense.
But how does this relate to solving the heat equation and ultimately how does this end up as the Dirac-Delta function ?
So if anyone can explain with a little intuition what it happening. Maybe a graph (i have a graph in my notes but I am confused about it). Thanks