Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\Omega$ is a bounded domain in $\mathbb R^n$ with smooth boundary. Consider the Dirichlet problem:$$-\Delta u = (\lambda - \log u)u ~~~{\rm on}~~~ \Omega~~~ {\rm and}~~~u=0 ~~~{\rm on}~~\partial {\Omega}$$ where the solution $u\in {C^\infty }(\bar \Omega )$ and $u>0$ on $\Omega$ and $\lambda$ is a positive constant.

My question is: what conditions do we need for the existence and uniqueness of the solution $u$ ?

share|improve this question
    
Why do yout think that a solution of this problem is $C^\infty(\overline{\Omega})$? –  Tomás May 14 '13 at 12:25

1 Answer 1

Define $$ f(t) = \left\{ \begin{array}{rl} -t\log{t} &\mbox{ if $t\in[0,1]$} \\ 0 &\mbox{ otherwise} \end{array} \right. $$

Note that $f$ is continuous. COnsider the problem

$$\tag{1} \left\{ \begin{array}{rl} -\Delta u-\lambda u=f(u) &\mbox{ in $\Omega$} \\ u=0 &\mbox{ in $\partial\Omega$} \end{array} \right. $$

1 - Existence

I - There is some methods avaliable to find a solution to $(1)$. For example, we can use the fact that $f$ is continuous with compact support, to prove that the energy functional $I:H_0^1\to\mathbb{R}$ associated with $(1)$ is (under some additional restriction) weakly lower semi continuous and coercive. Remember that if $F(t)=\int_0^t f(s)ds$, then $$Iu=\frac{1}{2}\int_\Omega |\nabla u|^2-\frac{\lambda}{2}\int_\Omega u^2-\int_\Omega F(u)$$

The additional restriction necessary here is $\lambda<\lambda_1$ (to guarantee coercivity).

II - (This is just a guess - I dont have verified the calculations): Also, you can prove existence, by noting that if $\lambda$ is not a eigenvalue of $-\Delta$, then the operator $-\Delta-\lambda I$ is invertible, and you can use arguments similar to this one.

Remark 1: The original function $t\log{t}$ is not monotone, so I dont know if this problem has uniqueness.

Remark 2: The function $f$ being continuous, we can conclude that $u\in C^1(\overline{\Omega})$.

Now, let's find conditions such that the solution $u$ of $(1)$ satisfies $0<u\leq 1$. Because $f\geq 0$ and there is a open set where $f>0$, you can conclude by the maximum principle that $u>0$. On the other hand, by regularity theory, you have that the solution $u$ of $(1)$ (See Breziz chapter 9) satisfies $(1)$ almost everywhere, i.e. $u$ is a Strong solution. If we suppose that the set where $u>1$ is non0empty, then we would have $-\Delta u-\lambda u=0$ almost everywhere in this set. If $\lambda$ is not a eingevalue, then this is a absurd, which implies that $u\geq 1$.

From the last paragraph, we can conclude that in fact $u$ is a solution to the original problem and satisfies $0<u\leq 1$. Hence, by using the first method proposed, we can conclude that if $\lambda<\lambda_1$, then you problem has a solution $u\in C^{2,\alpha}(\overline{\Omega})$ satisfying $0<u\leq 1$. I dont know if we can get more regularity, because the functions $-t\log{t}$ is only Holder continuous. A argument of bootstrap can be applied in the interior of $\Omega$, but in the boudary I dont know, because $-t\log{t}$ is not even Lipschitz for $t=0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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