Minimizers of an Energy Functional

[Euge]

Let $U$ be a bounded open subset of $\mathbb{R}^n$. Given a continuous function $\phi : \overline{U} \to \mathbb{R}$, show that any real-valued function $u$ of class $C^2(\overline{U})$ such that $\Delta u = \phi$ in $U$ and $u|_{\partial U} = 0$ is a minimizer of the energy functional $$\mathscr{E}(u) = \int_U d^n x\, \left(\frac{1}{2}|\nabla u(x)|^2 + \phi(x) u(x)\right)$$ over the class of functions $\Sigma = \{v\in C^2(\overline{U}) : v|_{\partial U} = 0\}$.

 

[julian]

Spoiler

First, as

$$
\Delta u(x) = \phi (x)
$$

we have

$$
0 = \int_U d^nx \left( - \Delta u(x) + \phi (x) \right) \eta (x)
$$

for an arbitrary function $\eta (x)$. If we take $\eta \in \Sigma$, integrating by parts gives

$$
0 = \int_U d^nx \left( \nabla u(x) \cdot \nabla \eta (x)+ \phi (x) \eta (x) \right)
$$

Take an arbitrary $v \in \Sigma$. We can write down an obvious inequality

$$
0 \leq \left( \| \nabla u \| - \| \nabla v \| \right)^2
$$

where $\| \nabla u \| := \left( \sum_{i=1}^n (\partial_i u)^2 \right)^{1/2}$ ($\| \nabla u \|^2 \equiv | \nabla u |^2$). Which implies

$$
0 \leq \frac{1}{2} | \nabla u |^2 + \frac{1}{2} | \nabla v |^2 - \| \nabla u \| \| \nabla v \|
$$

or

$$
\| \nabla u \| \| \nabla v \| \leq \frac{1}{2} | \nabla u |^2 + \frac{1}{2} | \nabla v |^2 .
$$

As $\nabla u \cdot \nabla v \leq | \nabla u \cdot \nabla v | \leq \| \nabla u \| \| \nabla v \|$ (Cauchy-Schwarz) we have

$$
\nabla u \cdot \nabla v \leq \frac{1}{2} | \nabla u |^2 + \frac{1}{2} | \nabla v |^2 .
$$

From which we have

$$
\int_U d^nx \left( \nabla u \cdot \nabla v + \phi v - \frac{1}{2} | \nabla u |^2 \right) \leq \int_U d^nx \left( \frac{1}{2} | \nabla v |^2 + \phi v \right)
$$

Using $\int_U d^nx \left( \nabla u \cdot \nabla \eta+ \phi \eta \right) = 0$ in this integral gives

\begin{align*}
\int_U d^nx \left( \nabla u \cdot \nabla [\eta + v] + \phi [\eta + v] - \frac{1}{2} | \nabla u |^2 \right)
\leq \int_U d^nx \left( \frac{1}{2} | \nabla v |^2 + \phi v \right)
\end{align*}

Putting $\eta (x) = u(x) - v(x)$ we have,

$$
\int_U d^nx \left( \frac{1}{2} | \nabla u (x) |^2 + \phi (x) u (x) \right) \leq \int_U d^nx \left( \frac{1}{2} | \nabla v (x) |^2 + \phi (x) v (x) \right)
$$

which is the desired result. Obviously, this choice of $\eta (x)$ means that $\eta \in \Sigma$.