Consider a closed Riemannian manifold $(M,g)$ and let $u \in C^{2,\alpha}(M)$ be a positive function on $M$.

I am interested on the existence of solution for the following problem: given a continuous function $\psi$ on $M$, when does there exists unique $v \in C^{2,\alpha}(M)$ such that $$\Delta v + F(u,\nabla u)v + G(u,\nabla u)g(\nabla u,\nabla v) = \psi$$ where $F$ and $G$ are smooth functions on their parameters and $u\in C^{2,\alpha}(M)$ is fixed.

This should not be a very difficult problem for one reason: this PDE is nothing more than a linear elliptic second order PDE, so it has a well consolidated theory.

The problem is that in general, to find a solution for this problem one needs to ensure that $\psi$ is in the orthogonal complement of the dual of the linear and elliptic operator $$P_u : v \mapsto \left(\Delta(\cdot) + F(u,\nabla u)\cdot + G(u,\nabla u)g(\nabla u,\nabla\cdot)\right)v,$$ and in particular, this seems very difficult to compute since the expressions for $F$ and $G$ are quite complicated.

My question is: does there exists a way of ensuring the existence for solution to my problem in an easy way, i.e, without looking for the kernel of $P^*_u$?

I have a particular guess that this can be done, how so?


You are dealing with a linear, second-order elliptic linear partial differential operator (LPDO) $P_u$ with $C^{2,\alpha}$ coefficients (you have to take $\alpha\in(0,1)$) on the left hand side of the equation, whose principal part is the Laplacian $\Delta$ on $(M,g)$. The standard approach to existence of solutions to such an equation when the right hand side $\psi\in C^\alpha$ ($\psi$ just continuous will not work, see below) is by combining a priori estimates for $P_u$ in Hölder spaces with an abstract method called the method of continuity - the idea is to interpolate between $\Delta$ and $P_u$ by setting $P_0=\Delta$, $P_1=P_u$ and $P_t=(1-t)P_0+tP_1$ ($t\in[0,1]$). The basis of such a method is the following theorem (see e.g. Theorem 5.2, pp. 75 of the book by D. Gilbarg and N. S. Trudinger, Elliptic Partial Differential Equations of the Second Order, Springer-Verlag, 1983):

Theorem: Let $P_0,P_1:B\rightarrow V$ be bounded linear maps from the Banach space $B$ into the normed vector space $V$, and $P_t$ as above. If there is $C>0$ independent of $t$ such that $$\|v\|_B\leq C\|P_t v\|_V$$ for all $t\in[0,1]$, then $P_1$ is surjective if and only if $P_0$ is (injectivity of both operators is obvious from the above estimate).

In other words, the solvability of your equation will come from the solvability of the Poisson equation on $(M,g)$ with the same right hand side (this, on its turn, is standard). The above estimate is the a priori estimate we need - it usually comes from the so-called Schauder estimates in Hölder spaces (see e.g. the book by Gilbarg and Trudinger cited above in the case $(M,g)$ is a compact domain in $\mathbb{R}^n$ with smooth boundary).

The problem with assuming $\psi$ being just continuous is that there may be not even $C^2$ solutions $v$ to $\Delta v=\psi$ (let alone $C^{2,\alpha}$) in this case. Problem 4.9 (a), page 71 of Gilbarg-Trudinger, loc. cit. provides a (counter)example of a continuous, compactly supported function $f$ in $\mathbb{R}^n$, $n\geq 2$, such that $\Delta v=f$ fails to have a $C^2$ solution $v$ in any given open neighborhood of the origin.

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    $\begingroup$ Pedro, thank you very much for such a complete answer. I did not make the calculations yet, but I am confident this will solve my problem, since although $F,G$ are a little bit complicated, I think I can handle it. About $\psi$ being $C^0,$ in my case it is smooth indeed, but I had no idea about the existence of counter examples when it is just continuous. Thank you. $\endgroup$ – L.F. Cavenaghi Nov 19 '19 at 1:38
  • $\begingroup$ @WillieWong, thank you for your comment, indeed Pedro's answer seems to be applicable on my case, but I do appreciate your comment. $\endgroup$ – L.F. Cavenaghi Nov 19 '19 at 1:39
  • $\begingroup$ In that case I withdraw my comment. Glad to hear the method can be applied. $\endgroup$ – Willie Wong Nov 19 '19 at 14:06

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