Let $(M,g)$ be a closed (compact, $\partial M=\varnothing$) smooth connected Riemannian manifold with Laplace–Beltrami operator $\Delta_g$. As it is well-known, every harmonic function ($u\in \ker \Delta_g$) on $M$ is constant (e.g. Hodge Isomorphism Theorem).

Question: What about biharmonic functions ($u\in \ker \Delta_g^2$)? In particular, can we say that every (everywhere defined, at least continuous) (distributionally) biharmonic function on $M$ is constant?

A few remarks: of course, the answer is negative on non-compact manifolds (think polynomials on $\mathbb R^n$), as well as on bounded domains with appropriate boundary conditions. (See e.g. this question).

Equivalently, we may look for smooth solutions to the Poisson equation $\Delta_g u\equiv 1$. By multiplying on both sides by $u$ and integrating w.r.t. the Riemannian volume $\mu_g$ we get: $$\int u \Delta_g u d\!\mu_g=\int |du|^2 d\!\mu_g=\int u d\!\mu_g$$ If $u$ has mean $0$, we immediately get that it is constant. If otherwise, we may assume that $\int u d\!\mu_g=1$, in which case the same holds for $|du|^2$. In particular if $u$ is a solution to the eikonal equation $|du|^2\equiv\mu_g(M)^{-1}$, then it is biharmonic. If we allow $du$ to have a singularity at some point $x_0$ in $M$, then this solution should be (?) the Green kernel for the bi-Laplacian pinned at $x_0$. If however $du\equiv 1$ everywhere, I do not see why a solution should exist.


1 Answer 1


There is no solution to the Poisson equation $\Delta u =1$ on a closed Riemannian manifold. Just use (weak) maximum principle: at the maximum $x_0$ of $u$ one has $\Delta _g u(x_0) \le 0$.

The same argument shows that $\Delta u = c$ has a solution only when $c = 0$, and in this case $u$ is harmonic and thus constant. So $\Delta^2 u = 0$ also implies that $u$ is constant.

  • $\begingroup$ (To self:) Seems it was not so difficult in the end... thank you very much! $\endgroup$
    – AlephBeth
    Nov 20, 2020 at 9:48

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