I was working on the following question and got stuck (Chap 2, #10 Kinderleher and Stampaccia)

Let $\Omega$ be a bounded domain with a smooth boundary $\partial \Omega$ and suppose given $f \in H^{-1}(\Omega)$ and $g \in L^2(\Omega)$. For $\lambda \in \mathbb{R}$ study the existence and uniqueness of solutions of the variational inequality:

$$u \in K : \int_\Omega[u_{x_i}(v-u)_{x_i} + \lambda(v-u)]\,dx \geq \langle f,v-u\rangle + \int_{\partial \Omega} g(v-u)\,dx \quad \forall v \in K$$

Where $K = \{v \in H^1(\Omega):v \geq 0\text{ on }\partial \Omega\}$. (Here I've used summation convention and partial derivatives are to be understood in the weak sense)

I think I have a lot of the pieces:

  1. The LHS, written as $a(u,v) = \int_\Omega[u_{x_i}(v)_{x_i} + \lambda(vu)]\,dx$ is coercive on $H^1(\Omega)$ when $\lambda >0$
  2. K is convex in $H^1$

So I think there will exist a weak solution to the Neumann Problem: $$-\Delta u + \lambda u = f \\ \frac{\partial u}{\partial n}=g$$ where $g \geq 0$.

I'm having trouble dealing with $f \in H^{-1}$ though. This $H^{-1}$ is dual to $H^1_0$, so the dual pairing on the RHS of the inequality only works when $v-u \in H^1_0$ But taking two elements arbitrary $v,u \in K$ doesn't necessarily give that $u-v$ is going to zero at boundary. This seems inconsistent and I'm having trouble resolving it.


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