# Biharmonic functions on closed manifolds?

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.

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.