Write down the function $\psi(u, v)$ that satisfies

$$\frac{\partial^{2} \psi}{\partial u^{2}}+\frac{\partial^{2} \psi}{\partial v^{2}}=0, \quad \psi\left(-\frac{1}{2}, v\right)=-1, \quad \psi\left(\frac{1}{2}, v\right)=1$$

The circular arcs $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ in the complex $z$-plane are defined by

$$|z+1|=1, z \neq 0 \text { and }|z-1|=1, z \neq 0,$$

respectively. You may assume without proof that the mapping from the complex $z$-plane to the complex $\zeta$-plane defined by

$$\zeta=\frac{1}{z}$$

takes $\mathcal{C}_{1}$ to the line $u=-\frac{1}{2}$ and $\mathcal{C}_{2}$ to the line $u=\frac{1}{2}$, where $\zeta=u+i v$, and that the region $\mathcal{D}$ in the $z$-plane exterior to both the circles $|z+1|=1$ and $|z-1|=1$ maps to the region in the $\zeta$-plane given by $-\frac{1}{2}<u<\frac{1}{2}$.

Use the above mapping to solve the problem

$$\nabla^{2} \phi=0 \quad \text { in } \mathcal{D}, \quad \phi=-1 \text { on } \mathcal{C}_{1} \text { and } \phi=1 \text { on } \mathcal{C}_{2}$$