2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.