I need to find conditions on $\alpha$ such that the bilinear, symmetric form $a(u,v) = \int \nabla u \nabla v + \alpha \int u v$ is coercive in order to use Lax-Milgram over the Sobolev space $H_{0}^{1}([0,1])$.
We have $a(u, u) = ||\nabla u||_{L_{2}}^{2} + \alpha ||u||_{L_{2}}^{2}$, and part of the hint includes using Lagrange multiplier method to deduce an appropriate range of $\alpha$ for coercivity:
My attempt was that, up to rescaling, I can suppose $||u||_{L_{2}}^{2} = 1$, and try to find $\min \int |\nabla u|^{2}$ under this condition. Then, the Lagrange multipliers formulation assures that there is a $C \in \mathbb{R}$ such that: $\frac{d}{du} [\int (\nabla u)^{2}] + C \frac{d}{du} [\int (u^{2}) - 1] = 0$. But, this does not help me achieve any inequality to prove the coercivity of $a(u, u)$ and hints towards a wrong formulation.
However, in the solution I'm provided, it is asserted that there exists an $\alpha$ such that $2 \int \nabla u \nabla h - 2 \alpha \int u h = 0 $ for all $h \in H_{0}^{1}([0,1])$, and hence $ -2 \Delta u - 2 \alpha u = 0$ a.e. such that $u = \sum\limits_{k=1}^{\infty} \sin(k \pi x) $. From thereon, one can find approriate inequalities for the problem.
I am confused as to how one uses Lagrange multipliers to reach the formulation $2 \int \nabla u \nabla h - 2 \int u h = 0 $ or solve the problem at large.