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I read from Evan's PDE book, that we have the explicit solution in $\mathbb{R}^n\times [0,T]$ for the heat equation: $$ \partial_t u-\Delta u=f, $$ $$ u(t=0,x)=u_0, $$ when $f$ is $C^2$ with compact support, but what can we say about the solution in the whole space $\mathbb{R}^n$ if $f$ is only in some $L^p(\mathbb{R}^n)$? Is it possible to have the solution $u\in C([0,T];L^1(\mathbb{R}^n))$?

There're some results for general parabolic equations, but in the book it's discussed in open bounded domain. I don't know if we can extend the results to the whole $\mathbb{R}^n$, since when proving the existence of solutions, it uses Galerkin approximations. Do we still have countable orthogonal base for $H^1_0(\mathbb{R}^n)$ and $L^2(\mathbb{R}^n)$, which are needed in Galerkin approximations? Any helps or references about this would be great. Thanks a lot!

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A solution can be written as $$ \partial_t u - \Delta u=f \\ \partial_t\left(e^{-t\Delta}u\right)=e^{-t\Delta}f \\ e^{-t\Delta}u-u_0=\int_{0}^{t}e^{-t'\Delta}fdt' \\ u = e^{t\Delta}u_0+\int_0^t e^{(t-t')\Delta}fdt'\;\;\;\ (*) $$ Here, $$ e^{s\Delta}g = \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}e^{-s|\xi|^2}\hat{g}(\xi)e^{i\xi\cdot x}d\xi $$ The operator $e^{s\Delta}$ can be written as a convolution integral operator with respect to an $L^1(\mathbb{R})$ function for all $s > 0$. Such an operator is a continuous for fixed $s$ as an operator on $L^p$, for all $1 \le p \le \infty$. The operator $e^{s\Delta}g$ is continuous in $s$ on $[0,\infty)$ into $L^p$ for $1\le p < \infty$, but not for $L^\infty$. The heat semigroup $e^{s\Delta}$ is can be written as a convolution with respect to a Gaussian kernel whose standard deviation is a function of $s$.

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  • $\begingroup$ Thank you so much for this. Could you give me a reference for this too? $\endgroup$ – Vera Jul 15 '16 at 4:03
  • $\begingroup$ @Vera : For what part? $\endgroup$ – DisintegratingByParts Jul 15 '16 at 4:13
  • $\begingroup$ Since I've never seen this before, it would be good for me to have some book writing about the operator $e^{s \Delta}$ and its properties. So basically everything... $\endgroup$ – Vera Jul 15 '16 at 4:21

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