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Let $u \in C^0([0,\infty);L^1(M)) \cap W^{1,1}_{\text{loc}}((0,\infty);L^1(M))$ with $u(t) \in H^1(M)$ for a.e. $t$ be the solution of the porous medium equation $\dot u = \Delta (u^m)$ on a compact (closed) smooth Riemannian manifold $M$, i.e., it satisfies $$\int_M u_0\varphi(0) =-\int_0^T\int_M u(t)\varphi'(t) +\int_0^T\int_M \nabla u^m(t) \cdot \nabla \varphi(t) $$ for all test functions $\varphi \in C^1([0,T]\times M)$ that vanish at $t=T$.

Suppose $u_0 \in L^\infty(M)$. It is the claimed in this work on top of page 10 that the $L^p$ norm decreases in time: $$\lVert u(t) \rVert_{L^p(M)} \leq \lVert u_0 \rVert_{L^p(M)}$$ for all $t$ and all $p$.

Formally this can be seen by: $$\frac{d}{dt}\int_M |u(t)|^p=p\int_M|u(t)|^{p-2}u(t)u_t(t)=p\int_M \nabla(|u(t)|^{p-2}u(t))\cdot \nabla(u^m(t))\\ =...=-pm(p-1)\int_M |u(t)|^{p+m-3}|\nabla u(t)|^2 \leq 0.$$ Note that the weak formulation is used in the second equality.

Obviously this calculation is formal because $u$ is not that smooth. How do I make it rigorous? Edit: let us assume $u(t) \in L^\infty(M)$ if necessary.

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  • $\begingroup$ "It is claimed..." where is it claimed? $\endgroup$
    – Yemon Choi
    Commented Jul 3, 2014 at 20:04
  • $\begingroup$ I edited to include the source uam.es/personal_pas/mbonfort/papers/04p-BG05-JFA.pdf. $\endgroup$
    – TomJoseph
    Commented Jul 3, 2014 at 20:39
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    $\begingroup$ For non-negative solutions you can use approximation with standard $L^1$-contractivity. For signed solutions I have no idea, I don't think the argument is so straightforward... $\endgroup$ Commented Jul 4, 2014 at 8:45
  • $\begingroup$ Concerning the assumptions, don’t you have an assumption of the kind $u \in L^2 ((0, \infty); H^1 (M))$ so that the last term in your weak formulation is well-defined. $\endgroup$ Commented Aug 27, 2014 at 7:34

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