# weak solutions need to have local integrability condition?

I am currently studying Poisson and Laplace equations. This is just a small question that has been causing me some confusion, and I would like some clarification before I resume my study. For example, when we write $\triangle u =0$ in the sense of distributions and u is also a distribution, are we assuming that $u$ is locally integrable. By this, I mean, are we assuming that $u$ is a function that is in $L^1_{loc}$? Thanks very much.

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## 1 Answer

I don't think that the local integrability is necessarily an a priori assumption in an arbitrary equation when interpreting it in the sense of distributions. It just turns out that any locally integrable function also defines a distribution. However, when we interpret the Poisson equation $\Delta u = f$ in the sense of distributions, the solution is usually locally integrable, at least when $f \in L^p$. Notice that the class of locally integrable functions is HUGE. For instance it contains all the $L^p$ for $1 \le p \le \infty$.

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Yes, but I guess your most important point is that one need not assume the local integrability condition. This shows promise that one can prove much more general statements in potential theory without assuming a solution is locally integrable. Of course, the mollification techniques will need refinement. –  Jenny Jan 3 '12 at 9:45
True, you could of course consider solutions in the entire class of distributions. However, this is pointless unless you want the data $f$ in a larger class than say $\bigcup L^p$. I am just not sure what the existence theorem would look like for an arbitrary distribution $T$, instead of $T = T_f$ for some $f \in L^p$... –  user12014 Jan 3 '12 at 9:51
Well, there are some interesting statements made using Sobolev's lemma without assuming anything on the solution. But they always lead to the conclusions that the solution is in fact infinitely differentiable or m-times differentiable (if it is in the m-2 th Sobolev space),etc. So, I guess it doesn't really hurt to consider these functions as locally integrable from the starting point. –  Jenny Jan 3 '12 at 9:54
Yes, this was my point but the Sobolev embedding only applies when the data is in some sense integrable in the first place (by elliptic regularity) so you might find something different if it is not. –  user12014 Jan 3 '12 at 10:00
Hey PZZ- a last comment/ question: in theory of PDEs, is it ever useful to pinpoint certain criteria to show that some distribution is in fact a function? for example, if it is harmonic, per se? –  Jenny Jan 3 '12 at 10:13