Some nonlinear PDE problem

Question

I'm facing the following problem: Let $ q < p_n < p $ such that $ p_n \to p $ as $ n \to + \infty, $ where $ 1 < q < p $ are constants. Let $ \Omega $ be a bounded domain and $ f \in L^{q'}( \Omega) $ where $ q' = \frac{q}{q-1} $ the conjugate of $ q. $ Let $ u_n \in W_0^{1,p_n}(\Omega) $ be such that $$ \int_{\Omega} \left| \nabla u_n\right|^{p_n-2} \nabla u_n \nabla v dx = \int_{\Omega} f v dx,\ \forall\ v \in W^{1,p_n}( \Omega). $$ Clearly, there exists $ c > 0 $ such that $ \left| \nabla u_n\right|_{L^{p_n}( \Omega)} \leq c,\ \forall\ n. $ By consequence, $ (u_n)_n $ is bounded in $ W_0^{1,q} (\Omega) $ and there exists $ u \in W_0^{1,q}( \Omega) $ such that $ u_n $ is weakly convergent to $ u $ in $ W_0^{1,q}( \Omega). $ We can also show that $$ \int_{\Omega} \left| \nabla u\right|^{p} \leq \liminf_{n \to + \infty} \int_{\Omega} \left| \nabla u_n\right|^{p_n} dx. $$ Thus, $ u \in W_0^{1,p}(\Omega). $ The problem is that I want to show that $$ \int_{\Omega} \left| \nabla u\right|^{p_n-2} \nabla u \nabla (u_n -u) dx \to 0. $$ In fact, we do not have that $ u_n $ is weakly convergent to $ u $ in $ W_0^{1,p}( \Omega) $ but this fact is needed as i see to prove our result. I'm blocked. Any help is welcome.

Answer 1

[This is an attempt to prove something that I feel is the motivation of the question: to prove that $u$ satisfies a nonlinear equation. I do not see how to prove the claimed convergence.]

Let $v\in C_c^\infty(\Omega)$. Then because of monotonicity of $\nabla u \mapsto |\nabla u|^{p_n-2}\nabla u$ we get $$ \int_\Omega f(u_n-v) = \int_\Omega |\nabla u_n|^{p_n-2}\nabla u_n \nabla (u_n-v) \ge \int_\Omega |\nabla v|^{p_n-2}\nabla v \nabla (u_n-v). $$ Here we can pass to the limit to get $$ \int_\Omega f(u-v) \ge \int_\Omega |\nabla v|^{p-2}\nabla v \nabla (u-v). $$ By density, this is true for all $v\in W^{1,p}_0(\Omega)$. Now set $v:=u+tw$, $w\in W^{1,p}_0(\Omega)$, $t>0$. Dividing the resulting inequality by $t$ yields $$ \int_\Omega f(-w) \ge \int_\Omega |\nabla (u+tw)|^{p-2}\nabla (u+tw) \nabla (-w). $$ Passing to the limit $t\searrow0$ gives $$ \int_\Omega f(-w) \ge \int_\Omega |\nabla u|^{p-2}\nabla u \nabla (-w). $$ But $w$ was arbitrary, so we find $$ \int_\Omega fw = \int_\Omega |\nabla u|^{p-2}\nabla u \nabla w \quad \forall w\in W^{1,p}_0(\Omega). $$ Using the convergence $\int_\Omega fu_n \to \int_\Omega f u$, we can prove $$ \int_\Omega |\nabla u_n|^{p_n} \to \int_\Omega |\nabla u|^p. $$