Let $S$ be a bounded region of $\mathbb{R}^{2}$ and $\partial S$ be its boundary. Let $u$ be the unique solution to Laplace's equation in $S$, subject to the boundary condition $u=f$ on $\partial S$, where $f$ is a specified function. Let $w$ be any smooth function with $w=f$ on $\partial S$. By writing $w=u+\delta$, or otherwise, show that

$$\int_{S}|\nabla w|^{2} \mathrm{~d} A \geqslant \int_{S}|\nabla u|^{2} \mathrm{~d} A$$

Let $S$ be the unit disc in $\mathbb{R}^{2}$. By considering functions of the form $g(r) \cos \theta$ on both sides of $(*)$, where $r$ and $\theta$ are polar coordinates, deduce that

$$\int_{0}^{1}\left(r\left(\frac{\mathrm{d} g}{\mathrm{~d} r}\right)^{2}+\frac{g^{2}}{r}\right) \mathrm{d} r \geqslant 1$$

for any differentiable function $g(r)$ satisfying $g(1)=1$ and for which the integral converges at $r=0$.

$$\left[\nabla f(r, \theta)=\left(\frac{\partial f}{\partial r}, \frac{1}{r} \frac{\partial f}{\partial \theta}\right), \quad \nabla^{2} f(r, \theta)=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial f}{\partial r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} f}{\partial \theta^{2}} \cdot\right]$$