Let $K$ a compact set and $(u^j)_j$ a sequence of functions bounded uniformly such that $$\Delta u^j=f^j(u^j), \ \ \ \mbox{in} \ \ K,$$ where the sequence $(f^j)$ is bounded uniformly and equicontinuous, converging uniformly in the compact set $K$ to $\widehat f$. Can I say that exist a function $\widehat u$ such that $u^j$ converge uniformly to $\widehat u$ and $\Delta\widehat u=\widehat f(\widehat u)$??? Why??

In the Gilbarg-Trudinger (corollary $4.7$, page: $61$), we have the uniform convergence to a solution of poisson equation, but in this case, the function $f$ is the same for each $u^j$. My question is when the function $f$ changes for each $j$.


1 Answer 1


Your question differs in several noticeable ways from Corollary 4.7 in Gilbarg-Trudinger, which make the answer no. I have discussed 4 critical differences.

First, I think you need the domain $K$ to be better. Strange things can happen to even the Dirichlet problem for Laplace's equation; there are examples where harmonic functions fail to have the properties you want. The estimate in Gilbarg-Trudinger assumes that your set $K$ is compactly contained in some larger open bounded domain $\Omega$ where the equation holds.

Second, you need for the functions to converge to each other somewhere, or they could be off by any harmonic function, and you get no pointwise convergence. This is why Gilbarg-Trudinger says that there exists a subsequence which converges uniformly to a solution. For instance, we could write $$ \Delta u^j = 0$$ and consider the functions $u^j = \frac{1}{j}$ if $j$ is even, and $u^j = 1$ if $j$ is odd. Then clearly the odd subsequence converges everywhere to 1, while the even subsequence converges to 0, but there's no convergence for the sequence as a whole.

Third, Gilbarg-Trudinger assumes that the $f$ is in $C^\alpha$. This assumption can be relaxed to Dini continuous without upsetting the estimates on the Newtonian potential, but if you simply assume the source functions are merely bounded and continuous, then you cannot show that $u$ is even $C^2$ (see, e.g., problem 4.9 in Gilbarg-Trudinger). Thus this case also poses a major obstacle: if we let $\hat{f}$ be such a function and $f^j$ any sequence of continuous functions that converges uniformly to $\hat{f}$ then there is no function that satisfies $\Delta \hat{u} = \hat{f}$ in the classical sense. If you are willing to move to the realm of weak solutions, this problem can be remedied, but not otherwise.

Fourth, you have changed the character of the equation. $$\Delta u(x) = f(x)$$ is a nice linear equation, while $$\Delta u(x) = f(u(x))$$ is a considerably more complicated semilinear equation. There are some well known cases where solutions to semilinear equations are known not to exist; a famous example is $$-\Delta u = u^\frac{n+2}{n-2}$$ on the unit ball (this is a famous result of Pohozaev). So it might be better to leave the semilinear nature of the problem out of it for now.

Now, if we amend the problem to: suppose we have a sequence of uniformly bounded functions $u^j(x)$, satisfying the equation $$\Delta u^j(x) = f^j(x)$$ where the $f^j$ are uniformly bounded sequence of functions in $C^\alpha$ that converge in the $C^\alpha$ norm to $\hat{f}$ on some compact set $K$, where $K$ is compactly contained inside a larger open bounded domain $\Omega$ where these equations also hold, then we can adapt the argument of Chapter 4 in Gilbarg-Trudinger to say that there is a subsequence of the $u^j$ that must converge to a solution of $$\Delta \hat{u} = \hat{f}(x)$$ The argument goes as follows:

1) We see that the $u^j$ are uniformly bounded in the $C^{2,\alpha}$ norm, by the Holder estimates on the second derivatives given in the book. The uniform bound happens because those estimates only depend on the $C^\alpha$ norms of the $f^j$, which we have taken to be uniformly bounded.

2) Since the $u^j$ are uniformly bounded in $C^{2,\alpha}$, by compactness there exists a subsequence which converges in the $C^{2,\alpha}$ norm to a function, which we will call $u_\infty$. From now on, by $u^j$ we mean only this convergent subsequence, and it remains only to show that $$\Delta u_\infty = \hat{f}$$

3) We see this by letting $\hat{u}$ be the Newtonian potential of $\hat{f}$ and analyze the difference $$\|\Delta (\hat{u} - u_\infty)\|_{C^{\alpha}} = \lim_{j \rightarrow \infty} \|\Delta(\hat{u} - u^j)\|_{C^\alpha} = \lim_{j\rightarrow \infty} \|\hat{f} - f^j|_{C^\alpha} = 0 $$

If you're interested in asking what happens if we drop the requirement that $f$ be $C^\alpha$ in the sense of weak solutions, you can get an analogous result with convergence to a weak solution - this involves using the estimates for weak solutions of the Poisson equation in the same way we have used the estimates on the Newtonian potential here.

  • $\begingroup$ This is an excellent answer and should have more upvotes $\endgroup$ Apr 11, 2013 at 2:17

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.