Smooth Dirichlet Energy and Laplace Operator

Let $M \subset \mathbb{R}^2$ be a domain with smooth boundary $\partial M$ and outpointing normal vector field $N$. For a smooth function $f \in C^{\infty}(M,\mathbb{R})$ the gradient vector field $\mbox{grad} \, f :M \rightarrow \mathbb{R}^2$ is defined as :
\[ \mbox{grad} \, f := \left( \begin{array}{c} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{array} \right)  = \left( f’\right)^t. \]
For a vector field $v:M \rightarrow \mathbb{R}^2$ the divergence is given by:
\[\mbox{div} \, v := \frac{\partial v_1}{\partial x} +\frac{\partial v_2}{\partial y}.\]
Taking this together we can define the laplace operator $\Delta : C^{\infty}(M,\mathbb{R})\rightarrow C^{\infty}(M,\mathbb{R})$:
\[\Delta \,f := \mbox{div} \, \mbox{grad}\, f = \frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}.\]
The laplace operator plays an important role in physics and geometry. A typical problem from physics is the following called Dirichlet boundary problem: Given $g \in C^{\infty}(\partial M,\mathbb{R})$ find $f \in C^{\infty}(M,\mathbb{R})$ such that:
\[\left\{ \begin{array}{cc} \Delta \, f = 0 \\  \left. f\right|_{\partial M} = g.\end{array} \right.\]
If $f$ is a temperature distribution on $M$ and $g$ is the prescribed temperature on the boundary, then $f$ has to solve the heat equation: $\dot{f} = \Delta \, f.$ For a stationary temperature distribution there holds $\dot{f}=0$ and we obtain the above Dirichlet boundary problem.

1) Given $g \in C^{\infty}(\partial M,\mathbb{R})$ and $d \in C^{\infty}(M,\mathbb{R})$, then there is a unique $f \in C^{\infty}(M,\mathbb{R})$ such that: \[\left\{ \begin{array}{cc} \Delta \, f = d \\ \left. f\right|_{\partial M} = g.\end{array} \right.\]
2) If $d=0$ and $f$ is a solution of the corresponding Dirichlet boundary problem, then $f$ minimizes the Dirichlet energy \[ E_D(f):=\frac{1}{2}\int_M \left|\mbox{grad}\,f\right|^2,\] among all the functions $h \in C^{\infty}(M,\mathbb{R})$ with $\left. h\right|_{\partial M} = g.$

1) The proof of the existence of $f$ is hard and can be found in some good books about PDE’s. We only proof the uniqueness of $f$.
Suppose there exists $f,\tilde{f} \in C^{\infty}(M,\mathbb{R})$ with: \( \Delta \, f =  \Delta \,\tilde{f} =0\) and \(\left. f\right|_{\partial M} =\left. \tilde{f}\right|_{\partial M} =g\), then $u:= \tilde{f} -f$ solves:
\[\left\{ \begin{array}{cc} \Delta \, u = 0 \\\left. u\right|_{\partial M} = 0.\end{array} \right.\]Now we consider the vector field $X:= u\, \mbox{grad} \,u$. For the divergence of $X$ we get:
\[ \mbox{div}\,X = \mbox{div} (u\, \mbox{grad} \,u) =\langle \mbox{grad}\,u, \mbox{grad} \,u\rangle\ + u\, \mbox{div}\, \mbox{grad} \, u =  \langle \mbox{grad}\,u, \mbox{grad} \,u\rangle\ + u \Delta \,u.\]
Using the fact that $X$ vanishes on the boundary and applying the divergence theorem we obtain:
\[0=\int_{\partial M}\langle X, N\rangle = \int_M \mbox{div}\,X = \int_M \langle \mbox{grad}\,u, \mbox{grad} \,u\rangle\ + u \Delta \,u = \int_M \left|\mbox{grad}\,u\right|^2.\]
Therefore, we have  $\mbox{grad} \,u=0$ on the whole of $M$ and $u$ is locally constant. Due to the fact that, $\left. u\right|_{\partial M} = 0$ $u$ is identically zero and therefore $f$ is unique.

2)Let $f$ be the unique solution of the Dirichlet boundary problem $\Delta \,f =0, \quad\left. f\right|_{\partial M} =g$ and $h$ an arbitrary smooth function with  $\left. h\right|_{\partial M} =g$. For $u:=h-f$ there holds $\left. u\right|_{\partial M} =0$ and we get:
\[\int_M \left|\mbox{grad},h\right|^2=\int_M \left|\mbox{grad}\,f+u\right|^2=\int_M \left|\mbox{grad}\,f\right|^2+ 2 \int_M \langle \mbox{grad}\,f ,\mbox{grad}\,u \rangle + \int_M \left|\mbox{grad}\,u\right|^2.\]
Using the usual product rule:
\[\mbox{div}(u\,\mbox{grad}\,f ) = \langle \mbox{grad}\,f ,\mbox{grad}\,u \rangle + u\, \Delta\,f, \] and the divergence theorem we obtain:
\begin{align*}2E_D(h)=\int_M \left|\mbox{grad}\,h\right|^2 &=\int_M \left|\mbox{grad}\,f\right|^2+ 2 \int_M \mbox{div}(u\,\mbox{grad}\,f ) -u\, \underbrace{\Delta\,f}_{=0}  + \int_M \underbrace{\left|\mbox{grad}\,u\right|^2}_{\geq 0}\\ &\geq \int_M \left|\mbox{grad}\,f\right|^2 + \int_{\partial M} \underbrace{u\langle \mbox{grad}\, f,N \rangle}_{=0}\\ &= \int_M \left|\mbox{grad}\,f\right|^2 = 2E_D(f).\end{align*}


Let us now consider the space of smooth functions on $M$ that vanish at the boundary: \[V:= \left\{f \in  C^{\infty}(M,\mathbb{R}) \, \big\vert \, \left. f\right|_{\partial M}=0 \right\}\]
We can equip  $V$ with two  scalar products $\langle\!\langle \cdot, \cdot \rangle\!\rangle , \, \langle\!\langle\cdot, \cdot \rangle\!\rangle_D :V\times V \rightarrow \mathbb{R},$
\begin{align*}\langle\!\langle f, g\rangle\!\rangle &:=\int_M fg, \\ \langle\!\langle f, g\rangle\!\rangle_D & := \int_M \langle \mbox{grad}\,f ,\mbox{grad}\,g \rangle.\end{align*}
To see that $\langle\!\langle\cdot, \cdot \rangle\!\rangle_D$  is really a scalar product on $V$ one has to check that it is positive definite (symmetry and bi-linearity are obvious).
\begin{align*} & 0=\langle\!\langle f, f\rangle\!\rangle_D  = \int_M \left|\mbox{grad}\,f\right|^2 \\  \Leftrightarrow & \quad \left|\mbox{grad}\,f\right| = 0  \end{align*}
Therefore, $f$ is locally constant. With $\left. f\right|_{\partial M}=0$ one has $f=0$.

Propositon: On V the laplace operator is self adjoint, i.e. $\langle\!\langle \Delta \,f, g\rangle\!\rangle = \langle\!\langle f, \Delta \, g\rangle\!\rangle$ and  there holds: \[\langle\!\langle \Delta \,f, g\rangle\!\rangle =-\langle\!\langle f, g\rangle\!\rangle_D .\]

Proof: \begin{align*} \langle\!\langle \Delta \,f, g\rangle\!\rangle & = \int_M g\, \mbox{div} \, \mbox{grad}\, f  = \int_M  \mbox{div} (g \, \mbox{grad}\, f ) \,- \langle \mbox{grad} \,f ,\mbox{grad} \,g \rangle \\ & = \underbrace{\int_{\partial M} f \langle \mbox{grad} \,g , N \rangle}_{=0}  – \int_M \langle \mbox{grad} \,f ,\mbox{grad}\,g \rangle  \\ &=\,-  \langle\!\langle f, g\rangle\!\rangle_D \\ &= \,- \langle\!\langle g, f\rangle\!\rangle_D \\ & \,\, \vdots \\& =\langle\!\langle f, \Delta \, g\rangle\!\rangle.\end{align*}


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