Perfect Harmony

Suppose that $\phi$ , $\psi$ are elements of $\mathcal{F}$ . Suppose moreover that $\nabla^2\phi = 0$ . By Green's Theorem $\begin{aligned} 0 & = \; \oint_{\partial V}(\phi-\psi)\nabla \phi \cdot d\mathbf{S} \; = \; \int_V \nabla \cdot\big((\phi-\psi)\nabla \phi\big)\,d\tau \; = \; \int_V\left((\phi-\psi)\nabla^2\phi + \nabla(\phi-\psi) \cdot \nabla \phi\right)d\tau \\ & = \; \int_V \left(\big|\nabla \phi\big|^2 - \nabla \phi \cdot \nabla \psi\right)\,d\tau \end{aligned}$ Here the surface integral $\int_{\partial V}$ is the sum of the surface integral over the sphere centred at the origin with radius $2$ (with outward-pointing normal) and the surface integral over the sphere centred at the origin with radius $1$ (and inward pointing normal). This means that $\int_V \nabla \phi \cdot \nabla \psi \; = \; \int_V \big|\nabla \phi\big|^2\,d\tau$ and hence $\int_V\big|\nabla \phi - \nabla \psi\big|^2\,d\tau \; = \; \int_V\left(\big|\nabla \phi\big|^2 + \big|\nabla \psi\big|^2 - 2\nabla \phi \cdot \nabla \psi\right)\,d\tau \; = \; \int_V \big|\nabla \psi\big|^2\,d\tau - \int_V \big|\nabla \phi\big|^2\,d\tau$ so that $I(\psi) \; = \; \int_V \big|\nabla \psi\big|^2\,d\tau \; \ge \; \int_V \big|\nabla \phi\big|^2\,d\tau \; = \; I(\phi)$ Thus the smallest value of $I(\psi)$ over all $\psi \in \mathcal{F}$ is obtained by choosing $\psi$ to be harmonic. We can also use this argument to show that $\mathcal{F}$ contains at most one harmonic function.Thus we need to find the harmonic function $\phi$ for $1 \le r \le 2$ such that $\phi = 3$ when $r=1$ and $\phi = 2$ when $r=2$ . In other words, we want $\phi \; = \; 1 + \frac{2}{r}$ so that $\nabla \phi = -2r^{-3}\mathbf{r}$ , and hence $I(\phi) \; = \; \int_V \big|\nabla \phi\big|^2\,d\tau \; = \; 4\pi\int_1^2 4r^{-4} \,r^2\,dr \; = \; 8\pi$ making the answer $\boxed{8}$ .

By symmetry, I assume that $\psi$ depends on $r$ alone, so that the integral becomes $\int_V (\psi'(r))^2dV$ $=4\pi\int_{1}^{2}(\psi'(r))^2r^2dr$ . Calculus of variations gives $\psi''(r)r+2\psi'(r)=0$ for the minimal integral so $\psi(r)=1+\frac{2}{r}$ for the given boundary conditions. The integral now comes out to be $\boxed{8}\pi$ .

A little knowledge of physics gives the solution in just one line. Think of $\psi$ as the electrostatic potential. Then $I(\psi)$ is simply the energy of the associated electrostatic field enclosed between the spherical shells at $r=1$ and $r=2$ . It is well-known that Coulomb's potential minimizes this energy , when the shells are held at fixed potentials. Hence, the optimal $\psi$ in the spherical coordinates is of the form $\psi^*(r)= \frac{A}{r}+b.$ Supplying the boundary conditions $\psi^*(1)=3, \psi^*(2)=2$ , the optimal $\psi$ turns out to be $\psi^*(r)= \frac{2}{r}+1.$ Hence, $I(\psi^*)= \int_{1}^{2} \bigg(-\frac{d\psi^*(r)}{dr}\bigg)^2 4\pi r^2 dr=\int_{1}^{2} \big(\frac{2}{r^2}\big)^2 4\pi r^2 dr= 8\pi.$