For $u \in H^{1}_0({\Omega})$ ($\Omega$ is a domain open and bounded in $R^n$). Let $0 < \lambda < \lambda_1$ ($\lambda_1$ is the first eigen value of the laplacean) and a fixed $f \in L^{2}(\Omega)$ define
$$ \varphi(u) = \displaystyle\int_{\Omega} \frac{1}{2}|\nabla u|^2 + \lambda (\cos u -1) - fu \ dx$$
I am doing trying to show that $\varphi$ is coercive, that is $\varphi(u) \rightarrow \infty$ if $|| u||_{H^{1}_{0}} \rightarrow \infty$. If i have this then a exercise that i am trying to do is done. But i am not seeing how to do this ... Someone can give me a help to show this ?
Maybe the Poincare inequality helps $|| u||^{2}_{H^{1}_{0}} \geq \lambda_1 || u||^{2}_{L^2}$
Yes, the Poincaré is the place to begin. Don't use up all of the good stuff, though. Stash an $\epsilon $ of it away: $$\int_\Omega |\nabla u|^2 \ge \epsilon \int_\Omega |\nabla u|^2 + (1-\epsilon)\lambda_1 \int_\Omega u^2$$ where $\epsilon $ is small enough so that $(1-\epsilon)\lambda_1\ge \lambda$. Then estimate $$1-\cos u = 2\sin^2 \frac{u}{2} \le \frac{u^2}{2}$$ thus obtaining $$ \varphi(u) \ge \frac{\epsilon }2 \int_\Omega |\nabla u|^2 - \int_{\Omega} fu $$ Here $fu$ contributes $O(\|u\|_{L^2})$ (linear term), which is dominated by the square of the Sobolev norm when the latter is large.