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Let $(M,g)$ be a closed, Riemannian manifold of dimension greater than two. Let $u$ a positive solution of the equation

$\Delta u - c u = -du^\frac{n+2}{n-2}$,

where $\Delta = -div\nabla$ and $c$ and $d$ are positive constants. I've read that a consequence of the maximum principle is that $u$ will be the unique non-trivial solution but I can't find a proof anywhere. Does someone know how to prove this assertion?

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What do you mean that $u$ is unique and non-trivial? Do you mean that there is unique non-zero solution (since there is always a trivial one and uniqueness of $u$ implies the triviality of a solution) – Ilya Jun 28 '11 at 18:11
Can you say where you've read that? Maximum principles usually require the coefficient in front of the zero'th order term to be the same sign as the Laplacian (as an operator). In your case you defined $\triangle$ as a positive operator with a $-c$, so maybe it will be easier if I have the context. – Willie Wong Jun 28 '11 at 19:01
Gortaur: Like you guessed, a non-trivial solution is one that is not identical to zero in this context. – Q-the-curvature Jun 28 '11 at 20:02
Willie Wong: I've encountered this result in a few papers about the Yamabe problem. It is stated as a fact in a paper by Richard Schoen (I forgot the title, but it has something to do with counting the number of constant scalar curvature metrics in a conformal class.) The specific statement is that there is one smooth positive solution to the Yamabe equation when the Yamabe constant is negative. – Q-the-curvature Jun 28 '11 at 20:09
up vote 0 down vote accepted

I figured it out. Actually the method is not so much the maximum principle as it is a type of maximum principle. One just writes: $\Delta u = -d u^{\frac{n+2}{n-2}} + c u$ and checks what constraints on $u$ must be satisfied at a min (where $\Delta u < 0$) and at a max (where $\Delta u < 0$). The inequalities that you get rule out the possibility of a positive solution that is not constant. It still is possible, though, for a solution to exist that is negative somewhere.

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