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Consider the heat equation $u_t=\Delta u$ with Neumann boundary condition in a bounded domain $\Omega$.

Is this true to say:

$$\|u(. , t)-v(. , t)\|_p\leq \|u(. , 0)-v(. , 0)\|_p$$ where $u$ and $v$ are two solutions of the heat equation in $W^{2,p}$.

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Two questions: (1) Are you interested in all values of $p$ or just $p > 1$?; (2) Is $\|\cdot\|_p$ the $L^p$ norm or the $H^p$ norm? Since the equation is linear, it suffices to consider v = 0. If you care about $L^p$ norms, the maximum principle gives the result for $p = \infty$ while $\|f \ast g\|_1 \le \|f\|_1 \|g\|_1$ gives the result for $p = 1$ (together with the fact that the Green's function has unit $L^1$ norm constant in time). Interpolation now gives the result for $1 < p < \infty$. I might be missing something of course. –  Aaron Hoffman Oct 19 '11 at 0:24
    
Can we simply use the Young's inequality for convolution and say: $\|K\ast u_0\|_p\leq \|u_0\|_p\|K\|_1$ where $K$ is the kernel with $\|K\|_1=1$ or any constant? –  user18626 Oct 21 '11 at 2:47
    
no, we can't. because in general the heat semigroup is not given by a convolution (although this is true if $\Omega=R^n$). –  Delio M. Dec 12 '12 at 21:24
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2 Answers

up vote 1 down vote accepted

yes, with some regularity on the boundary.

Theorem 3.2.9 p. 90 of E.B. Davies book, Heat Kernels and Spectral Theory gives Gaussian bounds for the heat kernel of an elliptic operator with Neumann boundary conditions. These bounds imply that the heat flow preserves L^p.

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Actually, much more is true: you even have "ultracontractivity", meaning that the solution is immediately in $L^\infty$ (not clear a priori, unless you are in dimension 1 where you can use the Sobolev embedding) and moreover you can estimate the $L^\infty$-norm of $u(t)$ by the $L^1$-norm of $u(0)$. –  Delio M. Dec 20 '12 at 16:11
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A very naive answer:

Assume that the initial data is positive (the same should be true dealing with absolute values...) and take p>2 (p=2 follows the same idea). Multiply the equation by $pu^{p-1}$ and integrate. One gets $$ p\int_\Omega u^{p-1}u_t dx=\frac{d}{dt}\int_\Omega u^p dx=p\int_\Omega \Delta u u^{p-1}dx=-p\int_\Omega \nabla u\cdot((p-1)u^{p-2}\nabla u)\leq0, $$ where in the last equality we use Green formula an homogeneous Neumann BC. Due to the linearity of the equation one gets the same result for the difference of two solutions.

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