Do we have Pohozaev's identity on compact manifolds without boundary？

Question

Recently I got to know about Pohozaev's identity, and I calculated several examples. The basic idea is multiplying $x \cdot \nabla u$ on both sides of the equation, but I noticed that all the materials I read are all about domains with boundary, so I wonder if we still have the Pohozaev identity on compact manifolds without boundary？ If we do, what should be multiplied on both sides of the equation? I will be very glad if you could show me some examples!

Answer 1

To answer this question, it is better to understand Pohozaev's identity using the heuristic argument given in

Berestycki, Henri; Lions, Pierre-Louis, Nonlinear scalar field equations. I: Existence of a ground state, Arch. Ration. Mech. Anal. 82, 313-345 (1983). ZBL0533.35029.

I reproduce their argument here (paraphrased):

Suppose we are interested in studying the critical points of the functional $\int_{\mathbb{R}^n} |\nabla u|^2 ~dx + \int_{\mathbb{R}^n} G(u) ~dx$. Call the first term $K[u]$ and the second term $P[u]$ (for kinetic and potential energies). Consider the one parameter family of scaling transformations $u_\lambda(x) := u(\lambda x)$. Then it is easy to compute that $K[u_\lambda] = \lambda^{2-n} K[u]$ and $P[u_\lambda] = \lambda^{-n} P[u]$. In order for $u$ to be a critical point, we must have that $$\label{1}\tag{$*$} \frac{d}{d\lambda} ( K[u_\lambda] + P[u_\lambda] ) \Big|_{\lambda = 1} = 0$$ Inserting the explicit form of the energies as constant times powers of $\lambda$, we find we must have. $$ (2-n) K[u] + (-n) P[u] = 0 $$

The connection of this heuristic argument to the multiplier $x\cdot \nabla$ is that the vector field corresponding to $x\cdot \nabla$ is the generator of the scaling transformation $x \mapsto \lambda x$. (More precisely, let $\phi: \mathbb{R}\times\mathbb{R}^n\to\mathbb{R}^n$ be such $\phi(\lambda,x) = \lambda x$, then $\partial_{\lambda}\phi(1,\cdot)$ is exactly the vector field $x\cdot \nabla$.) So multiplying by $x\cdot \nabla$ is a way to "make precise/rigorous" the $\frac{d}{d\lambda}$ heuristic argument above.

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Let's dissect the heuristic argument a bit more. Suppose now that $u$ lives instead on a Riemannian manifold $\Omega$ with some metric $g$. At the heart of the Pohozaev identity is a one parameter family of smooth maps $ \phi:I\times \Omega\to\Omega$, where $I$ is an interval in $\mathbb{R}$ containing $1$. We can then define $u_\lambda(x) = u\circ \phi(\lambda,x)$. Then formally a critical point of $K[u] + P[u]$ will need to satisfy \eqref{1}.

Now let $v$ be the vector field defined on $\Omega$ from $\partial_\lambda\phi(1,x)$. Then it is not too hard to see that if $P[u] = \int_\Omega G(u) ~\mathrm{dvol}_g$, then $$ \frac{d}{d\lambda} P[u_\lambda] \Big|_{\lambda = 1} = -\int_{\Omega} G(u) \mathrm{div}(v) ~\mathrm{dvol}_g $$

To implement the Pohozaev argument exactly, you'd need $\mathrm{div}(v)$ to be a constant (so the right hand side can be expressed as a constant times $P[u]$). But this is not possible on a closed manifold: by Stokes theorem $\int \mathrm{div}(v) ~\mathrm{dvol}_g$ for any vector field $v$ on a closed manifold is $0$, so the only possibility is $\mathrm{div}(v) \equiv 0$.

(On an open manifold this can be avoided by having $\mathrm{div}(v)$ not integrable against $\mathrm{dvol}_g$, voiding the Stokes theorem argument.)

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This is not to say similar lines of argument cannot be useful. In fact, equation \eqref{1} will give you a weighted integral identity concerning solutions of the corresponding PDE. But they will not be in as nice a form as the Pohozaev one.