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Given $u_0 \in C_c(\mathbb{R}^n)$, consider the solution of the Cauchy problem $$u(x,t) = \int_{\mathbb{R}^n} \Gamma (x - y,t)u_0(y) dy \qquad x \in \mathbb{R}^n,t>0\, \, .$$ Given $0<s<t$ , define $ v:\mathbb{R}^n \to \mathbb{R}$ as $$v(x) = \int_{\mathbb{R}^n} \Gamma(x-z,t-s)u(z,s)dz \, \, .$$ How would you prove that $v(x) = u(x,t)$?

In the help sheet it says notice notice that $$v(x) = \Gamma(t-s)\star u(s) = \Gamma(t-s)\star (\Gamma(s) \star u_0)$$ $$ = (\Gamma(t-s) \star \Gamma(s) )\star u_0$$ $$= \Gamma(t) \star u_0$$

but I don't understand how they get from the second last like to the last line? Could someone please explain?

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  • $\begingroup$ It's very difficult to understand what's going on: you're notation is wildly inconsistent. Why is $v$ not a function of $t$? And if it is not, then how can it equal $u$, which is? And in the final set of equations, is $\Gamma$ a function of one, or two variables? And further, how is a function of $x$ equal to a function of $t$? $\endgroup$
    – Ron Gordon
    May 24 '13 at 17:53
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I assume $u$ is a solution of the heat equation and $\Gamma$ is its fundamental solution. The property $\Gamma(x,t)=\Gamma(\cdot,t-s) \star \Gamma(\cdot,s)\,$, $0<s<t\,$, can be verified directly by computing the integral for convolution. Or it can be done using the uniqueness theorem for bounded solutions of the Cauchy problem. Namely, function $\Gamma(x,t)$ is evidently a solution of $$ u_t-\Delta u=0,\ \ x\in \mathbb R,\ t>s,\quad u|_{t=s}=\Gamma(x,s). $$ And $$ u(x,t) = \int_{\mathbb{R}^n} \Gamma (x - y,t-s)\Gamma(y,s)\,dy $$ is also a solution.

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