For $f\in L^1(\mathbb{R}^n)$ a solution to the heat equation $\frac{1}{2}\Delta u=\frac{\partial}{\partial t} u$ is given by $$u(x,t)=(2\pi t)^{-n/2} \int\exp{\left(-\frac{\lVert x-y\lVert^2}{2t}\right)}f(y)~dy.$$

I tried to prove this but got stuck at differentiating under the integral sign. With $$f_x(y,t):=\exp{\left(-\frac{\lVert x-y\lVert^2}{2t}\right)}f(y)$$ I tried to compute $$\frac{\partial u}{\partial t}=-\frac{n}{2} (2\pi)^{-n/2}t^{-n/2-1}\int f_x(y,t)~dy+(2\pi t)^{-n/2}\frac{\partial}{\partial t} \int f_x(y,t)~dy.$$ Then I would like to exchange integration and differentiation in the last term. To do so I would use a corollary to Lebesgue's dominated convergence theorem. Obviously $y\mapsto f_x(y,t)$ is integrable for all $t>0$ and $f_x(y,t)$ is differentiable in $t$ for all $y\in\mathbb{R}^n$. But I would still need some integrable $t$-independent $g_x\colon \mathbb{R}^n\to\mathbb{[0,\infty]}$ dominating $$\left\lvert\frac{\partial f_x(y,t)}{\partial t}\right\lvert=\left\lvert\frac{\lVert x-y\lVert^2}{2t^2}\exp{\left(-\frac{\lVert x-y\lVert^2}{2t}\right)}f(y)\right\lvert\leq g_x(y).$$

It seems that this is possible due to the exponential decay but I can't come up with an explicit $g_x$.

  • $\begingroup$ Do you know about the method of Fourier transformation of the PDE? (That's how this integral representation can be obtained) $\endgroup$ – AlexR Nov 25 '13 at 9:42
  • $\begingroup$ its a convolution..... $\endgroup$ – Alexander Grothendieck Nov 25 '13 at 13:07

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