# How to show the weak solution to this equation $\Delta u+u=f$ is unique?

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!

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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

## 1 Answer

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 $u=0$.

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Thank you! That's very helpful! – Siming HE Jan 5 '13 at 23:41