Minimization problem associated to a PDE $-\Delta y + f(y) = 0$?

Question

I want to write the PDE $$-\Delta y + f(y) = 0$$ $$y|_{\partial\Omega} = 0$$ where $f$ is some smooth function from $\mathbb{R}$ to $\mathbb{R}$ as the solution of $$\min_{y \in H^1_0(\Omega)} \frac{1}{2}\int_\Omega |\nabla y|^2 + \int_\Omega F(y).$$ How can I relate $F$ to $f$?

Answer 1

Define $J(y)=\int_\Omega \frac12 |\nabla y(x)|^2 + F(y(x)) ~dx$. Let $y$ be a minimizer of $J$ on $H_0^1$, i.e., $y \in H_0^1$ and $J(y)=\min_z J(z)$. Then $y+\lambda \varphi \in H_0^1$ for all $\varphi \in C_0^\infty$ and all $\lambda \in \mathbb{R}$. Then

$$\tilde J(\lambda):=J(y+\lambda \varphi)=\int_\Omega \frac12 |\nabla(y+\lambda \varphi)|^2 + F(y+\lambda\varphi) ~dx.$$

Since $0$ is a minimzer of $\tilde J$ it yields $\tilde{J}'(0)=0$. In other words,

$$\tilde J'(0)=\int_\Omega \nabla y \cdot \nabla \varphi + F'(y)\varphi ~dx=\int_\Omega (-\Delta y+ F'(y))\varphi ~dx.$$

By the fundamental lemma of calculus of variations $-\Delta y+ F'(y)=0$. Now, compare it to your question. It gives $F'(y)=f(y)$.