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So this question says to take $u(x,t) = v(x^2/t)$ to solve the 1d heat equation. That is, $$ u_t = u_{xx} $$ and it gives the general solution in the form $$ v(z) = c\int_{0}^z e^{-s/4}s^{-1/2} ds + d $$ (you can verify that this indeed is a solution to the heat equation). Anyway, the question involves choosing $c$ such that this solution is the fundamental solution, so that the initial condition $\Phi(x,0) = \delta(x)$ holds. I'm not sure how to go about doing this, can you perhaps point me in the right direction?

I think that $c$ solves $$ c = \frac{1}{\int_{-\infty}^\infty e^{-s^2/4t}\left(\frac{s^2}{t}\right)^{-1/2} ds } $$ Can someone verify this/explain why this?

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The heat equation (also known as diffusion equation) conserves total mass, which by definition is the integral $M (t) = \int_{-\infty}^\infty u(x,t)\,dx$. (This can be proved by taking time derivative of $M$ and using the PDE.) Since $\delta$ is a unit mass, in order to satisfy $u(x,0)=\delta(x)$ we need $M$ to be $1$.

This is why we normalize the fundamental solution $$u(x,t)=\frac{c}{\sqrt{t}}e^{-x^2/(4t)}$$ so that its mass is $1$. Which leads to $$c^{-1} = \int_{-\infty}^{\infty} e^{-x^2/4}\,dt =\sqrt{4\pi}$$ Above, the mass is computed at time $t=1$ and equated to $1$. It could be computed in any other moment $t_0>0$; the result is the same.

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  • $\begingroup$ Shouldn't the last integral, be an integral over space and not time? $\endgroup$ – Tucker May 29 '17 at 5:04

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