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Rabinowitz proves (using the Mountain Pass Theorem) that for a bounded smooth domain $\Omega \in \mathbb{R}^n$, and $f(x,\xi)\in C(\bar{\Omega}\times \mathbb{R},\mathbb{R})$ satisyfing the growth condition $$f(x,\xi) \leq A + B|\xi|^s,\ \ \text{where}\ \ s+1<\frac{2n}{n-2}$$ then there is a weak solution $u \in H_0^1(\Omega)$ to the problem $$ -\Delta u = f(x,u), \ x\in \Omega$$ $$ u = 0, \ \ x \in \partial \Omega$$ He then makes the remark that if in addition $f$ is locally Lipschitz, the solution is classical. The proof is made by referencing a paper by Agman, here. Unfortunately, this was made before $\LaTeX$ so it is quite difficult to read, as in it actually hurts my eyes. So, I thought I would try to use some arguments presented in Gilbarg-Trudinger to solve a particular case that came up in Evans presentation for the mountain pass theorem. Let $f = |u|^{s-1}u$, where $s$ as is above. Here, maybe we need this paper, maybe we don't.

My argument goes like this : Since $u \in H_0^1$, by the Sobolev embedding theorem, we have $u \in L^{2^{*}} = L^{\frac{2n}{n-2}}$, and so $f \in L^{\frac{2^{*}}{s}}$. Using the global $L^p$ estimates (GT thm 9.13) $u \in W^{2,\frac{2^{*}}{s}}$. Now use the Sobloev theorem again to get a better estimate for $u$. Continue this process a finite number of times until we hit the critical value for the Sobolev theorem, in which case we get $u \in C^{0,\alpha_1}$ for some $\alpha_1$ (and thus $f \in C^{0,\alpha_2}$ for some $\alpha_2$). Now use the Dirichlet theory for elliptic operators (GT 6.11) to conclude that the solution is $C^{2,\alpha_2}$, by uniqueness.

Problems :

I know this argument can't be correct for at least two reasons :

  1. The theory for the Dirichlet problem says the solution is unique, and it can be proved that there are at least 2 solutions if $f$ is locally Lipschitz continuous (Rabinowitz : Minimax Methods p 11)
  2. The $L^p$ theory used in the "proof" above requires we know apriori that $u$ is a strong solution, ie $u\in W^{2,p}$. The normal regularity result that I would use to get this, GT-theorem 8.12, requires that $f \in L^2$. Our $f$ seems to always be in $L^r$, $r<2$.

Any help you could provide would be appreciated!


Edit : I have found the result in a book "Qualitative Analysis of Nonlinear Elliptic Partial Differential Equations", but I am still having trouble understanding the proof. At least it has some of the general ideas that I had. For convenience, the argument from the book is given below, noting they use different letters for exponents (they consider the problem $-\Delta u = u^p$)

We know until now that $u\in H_0^1(\Omega)\subset L^{2^{*}}(\Omega)$.In a general framework, assuming that $u \in L^q$, it follows that $u^p \in L^{\frac{q}{p}}$, that is, by Schauder regularity and Sobolev embeddings, $u \in W^{2,\frac{q}{p}} \subset L^s$, where $\frac{1}{s} = \frac{p}{q} − \frac{2}{N}$. So, assuming that $q_1 > \frac{(p−1)N}{2}$, we have $u \in L^{q_2}$, where $\frac{1}{q_2} = \frac{p}{q_1} − \frac{2}{N}$. In particular, $q_2 > q_1$. Let $(q_n)$ be the increasing sequence we may construct in this manner and set $q_{\infty} = \lim_{n\rightarrow \infty}{q_n}$. Assuming, by contradiction, that $q_n < \frac{Np}{2}$, we obtain, passing at the limit as $n \rightarrow \infty$, that $q_{\infty} = \frac{N(p − 1)}{2} < q_1$, contradiction. This shows that there exists $r > \frac{N}{2}$ such that $u \in L^r(\Omega)$ which implies $u \in W^{2,r}(\Omega) \subset L^{\infty}(\Omega)$. Therefore, $u \in W^{2,r}(\Omega) \subset C^k(\Omega)$, where $k$ denotes the integer part of $2 − \frac{N}{r}$. Now, by Holder continuity, $u \in C^2(\Omega)$.


I still don't understand how we get $u \in W^{2,\frac{q}{p}}$. Also, the last statement by Holder continuity, $u \in C^2(\Omega)$. Why is this?

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Just a comment on the title (since unfortunately I cannot help with the PDEs): You probably want to add something about PDEs and the problem at hand to the title. I think you'd be more likely to get hits that way. –  mck Feb 2 '13 at 2:40
    
Comment has been noted. Thanks. –  Euler....IS_ALIVE Feb 2 '13 at 6:59

1 Answer 1

up vote 3 down vote accepted
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The theory for the Dirichlet problem says the solution is unique, and it can be proved that there are at least 2 solutions if $f$ is locally Lipschitz continuous.

No contradiction here. The theory to which you refer (GT Chapter 6) is for the Poisson equation $\Delta u=g$ with given $g$. The equation $\Delta u=f(u)$ may well have multiple solutions, since the right-hand side is different for different $u$.

The normal regularity result that I would use to get this, GT-theorem 8.12, requires that $f\in L^2$.

Indeed, GT do not prove a version of Theorem 8.12 for $1<p<\infty$. But it is true: if $u\in H_0^1(\Omega)$ and $\Delta u=g\in L^p(\Omega)$, then $u\in W^{2,p}(\Omega)$. The proof goes like this:

  1. extend $u$ and $g$ by zero to $\mathbb R^n$.
  2. Let $v=\mathcal{N}g$, the Newtonian potential of $g$. GT consider it in Chapter 4, but give only Hölder estimates. Sobolev estimates are based on the theory of singular integrals: $D_{ij}v$ is obtained from $g$ via Riesz transforms which are bounded on $L^p$ for $1<p<\infty$. See Theorem 3.2 here, which gives $v\in W^{2,p}(\mathbb R^n)$.
  3. Both $u$ and $v$ vanish at infinity, and $u-v$ is weakly harmonic, hence harmonic, hence identically zero.

Also, the last statement by Holder continuity, $u\in C^2(\Omega)$. Why is this?

That book is strangely written. Taking "the integer part of $2-N/r$" is not helpful because this part may well be equal to zero. The point is that the Morrey-Sobolev embedding gives $u\in C^{\alpha}(\overline \Omega)$ for some $\alpha>0$, which implies $g:=u^p$ is in $C^\alpha$. By Corollary 4.14 of GT, $\Delta u=g$ has a solution in $C^{2,\alpha}(\overline \Omega)$. Since the classical solution is also a weak solution, and the weak solution is unique, we conclude that $u \in C^{2,\alpha}(\overline \Omega)$.

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Thank you for your response. Indeed, result 2 is proved in GT. Can you expand a bit more please? I'm very surprised that this theorem isn't in GT, since you need it for all the estimates. How can you conclude that $u-v$ is weakly harmonic? We only know that $u$ is in $W^{1,p}$. So, $\int{(u-v)\Delta \phi}$ = $\int{D(u-v)\cdot D\phi}$. To get this integral is $0$, I presume we need to be able to integrate by parts again. Also, how do you know the weak solution is unique? My first thought was Lax-Milgram, but this requires $L^2$ again. –  Euler....IS_ALIVE Feb 4 '13 at 6:53
    
@Euler... I was thinking of $\Delta u$ and $\Delta v$ as distributional laplacian (the distribution happens to be the same $L^p$ function). In the distributional sense $\Delta(u-v)=0$, which by Weyl's lemma means that $u-v$ is harmonic. –  user53153 Feb 4 '13 at 15:40
    
@Euler... There is at most one $u\in H^1_0(\Omega)$ that satisfies $\Delta u=g$ in the weak sense; this follows from Corollary 8.2 in GT. –  user53153 Feb 4 '13 at 15:54
    
This is a good enough explanation for the bounty... although I wonder; Should we always think of the Laplacian as the distributional Laplacian? –  Euler....IS_ALIVE Feb 5 '13 at 4:47
    
@Euler... You don't have to, although I can't think of an instance when it can hurt (as long as Laplacian is a linear operator). Distributions are a convenient common denominator for function spaces. In this situation there is $u\in H_0^1$ which satisfies $\Delta u=g$ in the weak sense, and $v\in W^{2,p}$ which satisfies $\Delta v=g$ in the strong sense. To save myself the headache of thinking of function spaces, I say: whatever, they both have Laplacian $g$ in the sense $\int (u\text{ or }v)\Delta \phi=\int g \phi$ for Schwartz functions $\phi$. This is all Weyl's lemma needs, anyway. –  user53153 Feb 5 '13 at 4:57

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