0
$\begingroup$

I'm working on the following problem:

Let $\Omega$ be a bounded domain in $\mathbb{R}^n$ and let $H$ be a closed linear subspace of $L^2(\Omega)$. Let $\gamma : \mathbb R \to \mathbb R$ be a continuous increasing function satisfying $|\gamma(t)| \leq A + B|t|$ for some constants $A,B$. Let $F \in L^2(\Omega)$.

Show that there exists a unique $u \in H$ such that $u + \gamma \circ u -F \in H^{\perp} $.

(Hint: solve the variational inequality $(u + \gamma \circ u -F,v-u) \geq 0 \quad \forall v \in H $.)

Question: if $u \in H$ is a solution of the inequality, why does this imply $u + \gamma \circ u -F \in H^{\perp}$?

$\endgroup$
1
+50
$\begingroup$

We assume that $u \in H$ is such that $(u + \gamma \circ u -F,v-u) \geq 0$ for every $v \in H$. In particular, this means that for $w \in H$ we can take $v = u \pm w$ to obtain $(u + \gamma \circ u -F,\pm w) \geq 0$ which in turn implies that $(u + \gamma \circ u -F, w) = 0$. Since $w$ was arbitrary, this shows that $u + \gamma \circ u -F \in H^\perp$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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