In plane polar coordinates $(r, \theta)$, the orthonormal basis vectors $\mathbf{e}_{r}$ and $\mathbf{e}_{\theta}$ satisfy

$$\frac{\partial \mathbf{e}_{r}}{\partial r}=\frac{\partial \mathbf{e}_{\theta}}{\partial r}=\mathbf{0}, \quad \frac{\partial \mathbf{e}_{r}}{\partial \theta}=\mathbf{e}_{\theta}, \quad \frac{\partial \mathbf{e}_{\theta}}{\partial \theta}=-\mathbf{e}_{r}, \quad \text { and } \quad \boldsymbol{\nabla}=\mathbf{e}_{r} \frac{\partial}{\partial r}+\mathbf{e}_{\theta} \frac{1}{r} \frac{\partial}{\partial \theta}$$

Hence derive the expression $\nabla \cdot \nabla \phi=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial \phi}{\partial r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} \phi}{\partial \theta^{2}}$ for the Laplacian operator $\nabla^{2}$.

Calculate the Laplacian of $\phi(r, \theta)=\alpha r^{\beta} \cos (\gamma \theta)$, where $\alpha, \beta$ and $\gamma$ are constants. Hence find all solutions to the equation

$$\nabla^{2} \phi=0 \quad \text { in } \quad 0 \leqslant r \leqslant a, \quad \text { with } \quad \partial \phi / \partial r=\cos (2 \theta) \text { on } r=a$$

Explain briefly how you know that there are no other solutions.