Variational formulation of the Helmholtz equation and doubt about how to deal with coercivity. I am looking to made the variational formulation of the Helmholtz equation $$-\Delta u +k^2 u = f\quad \text{in}\quad \Omega.$$ With homogeneous Dirichlet conditions $$u=0\quad \text{on}\quad \partial\Omega.$$
The variational formulation is $$\int_\Omega \nabla u\cdot \nabla u\, dx +k^2\int_\Omega u\cdot v\, dx= \int_{\Omega}fv\, dx,\quad \forall v\in H^1_0(\Omega).$$
After, we identify the bilinear form and linear form as:
$$a(u,v)=\int_\Omega \nabla u\cdot \nabla v\, dx+k^2\int_\Omega u\cdot v \,dx$$
and $$L(v)=\int_{\Omega}fv \,dx.$$
It is relatively easy to see that the bilinear form and the linear form are continuous but now when studying the coercivity of the bilinear form, I did as follows:
$$a(u,u)=\int_\Omega |\nabla u|^2\, dx+k^2\int_\Omega |u|^2 \,dx,$$
after I took the minimum between $k^2$ and $1$, i.e, $\min\{k^2,1\}$, thus arrived at $$a(u,u)=\int_\Omega |\nabla u|^2\, dx+k^2\int_\Omega |u|^2 \,dx \geq \min\{k^2,1\}\left(\int_\Omega |\nabla u|^2\, dx+\int_\Omega |u|^2 \,dx\right)=M\|u\|^2_{H^1_0(\Omega)}$$
with $M:=\min\{k^2,1\}>0$.
I am not sure if the way I approach coercivity is correct or not. Moreover, this question is linked to the fact that in some documents it is mentioned that coercivity is satisfied when $\lambda_{\min}>k^2$ with $\lambda_{\min}$ the smallest eigenvalue of the Laplacian $-\Delta$.
 A: First of all, I believe you have made a small sign error - the Hemholtz equation is usually given as $$\tag{$\ast$}-\Delta u -k^2u=f \qquad \text{in } \Omega. $$ (Think of eigenvalues: $-\Delta u = k^2 u$). Of course you can consider the equation as you have written it, but the you will always find that it is coercive (independent of the value of $k$) via the Poincare inequality.
The bilinear form corresponding to $(\ast)$ is: $$B(u,v) = \int_\Omega \nabla u \cdot \nabla v \, dx - k^2\int_\Omega uv \, dx. $$ The proof that $B$ is continuous will be identical to what you already have. For coercivity, \begin{align*}
B(u,u) = \int_\Omega \vert \nabla u \vert^2 \, dx - k^2 \int_\Omega u^2 \, dx.
\end{align*} It is well known that the first Dirichlet eigenvalue of $-\Delta$ (which is by definition the smallest) can be written as $$\lambda_1(\Omega) = \inf_{v\in H^1_0(\Omega), \, v\neq 0} \frac{\int_\Omega \vert \nabla v \vert ^2 \, dx}{\int_\Omega v^2 \, dx}.$$ Hence, for all $u\in H^1_0(\Omega)$, $$\int_\Omega u^2 \, dx\leqslant \frac 1 {\lambda_1(\Omega)} \int_\Omega \vert \nabla u \vert ^2 \, dx$$ and so $$ B(u,u) \geqslant \bigg (1-\frac{k^2}{\lambda_1(\Omega)}\bigg )\int_\Omega \vert \nabla u\vert^2 \, dx. $$ Finally, to get the $\|\cdot\|_{H^1(\Omega)}$ norm on the right hand side, we use that $\|\nabla v\|_{L^2(\Omega)}\geqslant C \|v\|_{H^1(\Omega)}$ for all $v\in H^1_0(\Omega)$ which is true via the Poincare inequality. Thus, $$B(u,u) \geqslant C\bigg (1-\frac{k^2}{\lambda_1(\Omega)}\bigg )\|v\|_{H^1(\Omega)}, $$ so $B$ is coercive provided $\lambda_1(\Omega)>k^2$ as required.
