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To be specific, suppose $M$ is a closed oriented manifold, $g$ is a Riemannian metric of $M$. Let $\Delta_g$ be the Laplace-Beltrami operator w.r.t. $g$.

Prove: Suppose $f\in C^\alpha(M)$ satisfies $\int_M f\, dVol_g=0$, then there exists a function $u\in C^{2,\alpha}(M)$ such that $\Delta_g u=f$ in $M$, and $u$ is unique up to plus a constant, here $0<\alpha<1$.

My attempt is that, firstly use $D(u):=\int_M(\frac{1}{2} |\nabla u|^2+fu)dVol_g$ is a convex functional with a lower bound on $W_0^{1,2}(M)$ to show that there exists a weak solution $u\in W^{1,2}(M)$, next use the $L^2$-regularity theory to show that $u\in W^{2,2}(M)$, but I don't know how to improve the regularity of $u$ further. (Actually, I can use the method to prove that if $f$ is $C^\infty$, then $u$ is also $C^\infty$, but I cannot extend this result to $C^\alpha$ case.)

Another attempt is Schauder estimate. However, in Gilbarg and Trudinger's book they assume that $u\in C^{2,\alpha}(M)$ already to get some interior derivative norm bound of $u$, while I don't know how to establish $u\in C^{2,\alpha}(M)$. They give a continuity method to ensure that, but it seems their discussion works for domains in Euclidean space, not for manifolds. Therefore, I want to split the question into coordinate charts, but I failed, because I don't know how to use the condition $\int_M f\, dVol_g=0$ and how to give boundary conditions in every coordinate charts.

Since I'm a novice in PDE, my presentation of the problem might have some errors. Please correct them by comments or answers. Also, any comments or answers are welcome.

Thanks for your help.

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2 Answers 2

up vote 2 down vote accepted

I think I've got an answer to this question.

First, use the functional $D(u)=\int_M(\frac{1}{2}|\nabla u|^2+fu)dVol_g$, we can find a weak solution $u\in W^{1,2}_0(M)$, since $f\in C^\alpha(M)$ hence in $L^2(M)$. Next we shall show that $u\in C^{2,\alpha}(M)$.

We cite theorem 6.14 from Gilbarg-Trudinger's book:

Let $L$ be strictly elliptic in a bounded domain $\Omega$, with $c\leq0$, and let $f$ and the coefficients of $L$ belong to $C^\alpha(\overline{\Omega})$. Suppose that $\Omega$ is a $C^{2,\alpha}$ domain and that $\phi\in C^{2,\alpha}(\overline\Omega)$. Then the Dirichlet problem, $$ Lu=f \textrm{ in }\Omega,\quad u=\phi\textrm{ on }\partial\Omega, $$ has a unique solution lying in $C^{2,\alpha}(\overline\Omega)$.

The proof relies on the solvability of classical Dirichlet problem of laplacian, continuity method and Schauder estimates.

Suppose $U$ is a coordinate chart s.t. $\overline{U}$ is diffeomorphic to a closed ball, then by the theorem cited above, $\Delta_g v=f$ in $U$, $v = 0$ on $\partial U$ is solvable, and $v\in C^{2,\alpha}(\overline{U})$. Therefore, $u-v$ is a weak solution of $\Delta_g (u-v)=0$. By $L^2$-regularity, we have $u-v\in C^\infty(U)$, thus $u\in C^{2,\alpha}(U)$. Since we can choose finitely many $U$ to cover $M$, finally we have $u\in C^{2,\alpha}(M)$.

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Looks good to me. –  Willie Wong Jul 26 '12 at 15:04

Insofar as Gilbarg and Trudinger, I think you are looking in the wrong chapter. What you should be looking for is a method to get improved interior regularity for weak solutions. That is, you should be looking at sections 8.9 and 8.11 of G&T.

Alternatively, the theorem you want is Theorem 4.7 in T. Aubin Nonlinear analysis on Manifolds. The method of proof is much as your first attempt: first using convexity you get a weak solution in $W^{1,2}$. Then you need to upgrade the regularity to $C^{2,\alpha}$, for which he invokes his theorem in paragraph 3.54

Let $\Omega$ be an open set of $\mathbb{R}^d$ and $A$ a linear elliptic operator of order $2m$ with $C^\infty$ coefficients. If $u$ is a distributional solution to $A(u) = f$ and $f\in C^{k,\alpha}(\Omega)$, then $u\in C^{k+2m,\alpha}(\Omega)$.

For a proof Aubin refers to Theorem 6.4.3 of C.B. Morrey's Multiple Integrals in the Calculus of Variations. You need to (of course) use more than just $L^2$ regularity theory. (Unfortunately I do not know of a more modern exposition of this theorem, so you will have to suffer through Morrey's notation a bit.)

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I've found Aubin's book, and I've seen the notation of Morrey's book through google... seems to be so complicated. Anyway, thanks for your help and reminding of G&T's right chapter~ –  Yuchen Liu Jul 20 '12 at 10:06
    
Morrey's proof goes in two steps: first you develope/use an $L^q$ regularity theory for $1 < q < \infty$. This automatically wins you $C^{1,\alpha}$ (actually even a bit more) by Sobolev embedding from $W^{2,q}$. I'll have to parse through the chapter before I can tell you how to go from $C^{1,\alpha}$ to $C^{2,\alpha}$. –  Willie Wong Jul 20 '12 at 11:22
    
Do you think my proof listed below is correct? Or give some comments? –  Yuchen Liu Jul 26 '12 at 14:53

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