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.

The question is like this: Let $ B(r)=\{x\in R^3| |x|<r\}$

$\Delta u+u=f$, in $B(r)$

$u=0$ on $\partial B(r)$

prove that there exists $\epsilon>0$ s.t. the equation has a unique weak solution $u\in H^1_0(B(r))$ for each $f\in L^2(B(r))$ for all $0<r<\epsilon$?

I want to use the First existence theorem here, and I think the Poincare inequality corresponding to the ball will be helpful (Evans 2ND edition P 291 ), but I cannot figure out the argument. Please help!

share|improve this question
2  
If I didn't mess up the calculations for every $v\in C_c^\infty(B_r)$ we have $$ \int_{B_r} |v|^2 \leq (2r)^2 \int_{B_r} |\nabla v|^2$$ from which you can obtain coercivity of the bilinear form associated with your problem for $r$ small enough. Then just use Lax-Milgram. –  Jose27 Jan 5 '13 at 21:52
    
Great! Thank you very much! –  Siming HE Jan 5 '13 at 23:33
add comment

1 Answer 1

up vote 5 down vote accepted

This is the Euler-Lagrange equation for the functional $$F(u)=\int_{B(r)} \left(|\nabla u|^2-\frac12 (u+f)^2\right).$$ If we had $+$ there, the existence and uniqueness for any $f\in L^2$ would be immediate. With the minus sign it's not even obvious that $F$ is bounded below. But when $r$ is sufficiently small, it is bounded because $\int_{B(r)} |\nabla u|^2 \ge Cr^{-2}\int u^2 $ for any $u\in W^{1,2}_0(B(r))$ -- indeed, the Poincaré inequality is helpful.

Next item, sequential weak lower semicontinuity: it holds because the gradient part is convex and $u^2_n$ converge strongly in $L^2$ whenever $u_n$ converge weakly in $W^{1,2}_0(B(r))$. This takes care of existence.

I leave uniqueness for you to do, since it is not much different from the argument for the lower bound of $F(u)$. The point is that when $f= 0$, you can't do better than $u=0$.

share|improve this answer
    
Thank you! That's very helpful! –  Siming HE Jan 5 '13 at 23:41
add comment

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.