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Let $M$ be a compact Riemannian manifold. Let $u\in M\times[0,T)\to\mathbb R$ be a smooth function. Let $\Delta$ be the Laplace operator on $M$. Suppose there is a constant $C$ such that $$\Delta u-\frac{\partial u}{\partial t}\geq Cu$$ Prove that if $u(x,0)\leq0$, then $u\leq0$ always holds.

I've proved the more well-known version of the maximum principle, i.e. under the same assumptions, if $$\Delta u-\frac{\partial u}{\partial t}\geq 0$$ then $u(x,t)$ attains its maximum only if $t=0$.

I suppose I need to do a transform on $u$ and apply the second version. Is this the right direction? If so, how should I do that?

trisct
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1 Answers1

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Let $w=e^{Ct}u$. Then $w$ satisfies $$\Delta w- w_t \geq 0.$$ Now apply the maximum principle for the above equation to get $w\leq 0$ which implies $u \leq 0$.

Mike Williams
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