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Suppose $\Omega = [0, 1]^3$ is a cube in $R^3$. Consider the problem $$u_t = \Delta u$$ $$u(0,x) = f(x)$$ $$u \mid_{\partial \Omega} = 0$$

Then by taking the convolution of $f$ with the fundamental solution we see that $$u(t,x) = \int_\Omega \Phi(x-y)f(y) \, dy$$

where $\Phi(x) = \frac{1}{(4\pi t)^{3/2}}e^{-\|x\|^2/4t}. $

Is there a similar formula for the situation with the non-homogenous boundary condition $u \mid_{\partial \Omega} = g(t,x)$?


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You definition of $u(x,t)$ is not entirely correct. You need to extend $f$ to a function on ${\mathbb R}^3$ so that the convolution makes sense. For homogeneous boundary conditions, an odd reflection about the boundary will ensure that $u=0$ on $\partial \Omega \times (0,\infty)$. It may be possible to get other boundary conditions by taking other extensions of $f$, but I am not too sure about this. Maybe someone else will know. –  Jeff Nov 17 '11 at 21:33

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