MHB Is Every Solution to a Certain PDE on a Closed Manifold Constant?

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Here is this week's POTW:

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Let $M$ be a closed, connected Riemannian manifold. Prove that every $C^\infty(M;\Bbb R)$-solution of the PDE

$$f\Delta f = -\alpha |\nabla f|^2\quad (\alpha\in \Bbb R)$$ is constant.

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No one answered this week's problem. You can read my solution below.

Spoiler

Let $f$ be a solution of the PDE in $C^\infty(M;\Bbb R)$. Suppose $\alpha = 1$. Then

$\displaystyle \quad \Delta(f^2) = 2f\Delta f + 2|\nabla f|^2 = 0$.

Therefore $f^2$ is harmonic on $M$, and thus $f$ is constant.

Now suppose $\alpha \neq 1$. Then

$\displaystyle \text{div}(f\nabla f) = \frac{\Delta(f^2)}{2} = f\nabla f + |\nabla f|^2 = (1 - \alpha)|\nabla f|^2$.

Since $M$ is closed, the divergence theorem gives

$0 = \displaystyle \int_M \text{div}(f\nabla f)\, dV = (1 - \alpha) \int_M |\nabla f|^2\, dV$.

As $\alpha \neq 1$, it follows that

$\displaystyle \int_M |\nabla f|^2 \, dV = 0$

and thus $|\nabla f| = 0$. Hence $f$ is constant.