Green's Identity and Method of Images for the Half-Space

Source: Unofficial personal solution to Cambridge Mathematical Tripos Part IB 2025, Paper 3, Question 14D (Methods).

Prove that, for scalar fields $\phi(\mathbf{x})$ and $\psi(\mathbf{x})$ in a three-dimensional domain $\mathcal{D}$ with boundary $\partial\mathcal{D}$, \[\int_{\mathcal{D}}(\phi\nabla^2\psi - \psi\nabla^2\phi)\,dV \equiv \int_{\partial\mathcal{D}} \!\left(\phi\frac{\partial\psi}{\partial n} - \psi\frac{\partial\phi}{\partial n}\right)dS,\tag{1.1}\] where $\mathbf{n}$ is the outward unit normal to $\partial\mathcal{D}$.

Let $\psi(\mathbf{x})$ satisfy \[\nabla^2\psi = \delta(\mathbf{x}-\mathbf{x}_0) \quad \text{in } z>0,\tag{1.2}\] with \[\psi = 0 \text{ on } z=0, \qquad \psi\to 0 \text{ as } |\mathbf{x}|\to\infty,\tag{1.3}\] in Cartesian coordinates $\mathbf{x}=(x,y,z)$. Use the method of images to determine $\psi(x,y,z)$.

Now use the identity above to solve the equation \[\nabla^2\phi = 0 \quad \text{in } z>0, \qquad \phi\to 0 \text{ as } |\mathbf{x}|\to\infty,\tag{1.4}\] with $\phi=1$ on $z=0$, $x^2+y^2<1$, while $\phi=0$ on $z=0$, $x^2+y^2>1$, in terms of a surface integral. Find the closed-form solution for $\phi(0,0,z)$.

Write the integrand as a divergence: \[\phi\nabla^2\psi - \psi\nabla^2\phi = \nabla\cdot(\phi\nabla\psi) - \nabla\phi\cdot\nabla\psi - \nabla\cdot(\psi\nabla\phi) + \nabla\psi\cdot\nabla\phi = \nabla\cdot(\phi\nabla\psi - \psi\nabla\phi).\tag{1.5}\] Applying the divergence theorem immediately gives the result.

The free-space Green's function satisfying $\nabla^2 G = \delta(\mathbf{x}-\mathbf{x}_0)$ with $G\to 0$ at infinity is $G = -1/4\pi|\mathbf{x}-\mathbf{x}_0|$. To impose $\psi=0$ on $z=0$, place an image of opposite sign at the reflected point $\bar{\mathbf{x}}_0 = (x_0,y_0,-z_0)$: \[\psi(\mathbf{x}) = -\frac{1}{4\pi}\!\left( \frac{1}{|\mathbf{x}-\mathbf{x}_0|} - \frac{1}{|\mathbf{x}-\bar{\mathbf{x}}_0|} \right).\tag{1.6}\] On $z=0$, $|\mathbf{x}-\mathbf{x}_0|=|\mathbf{x}-\bar{\mathbf{x}}_0|$ so $\psi=0$.

Apply Green's identity to $\phi$ (harmonic) and $\psi$ ($\nabla^2\psi = \delta(\mathbf{x}-\mathbf{x}_0)$) over $z>0$: \[\phi(\mathbf{x}_0) = \int_{\partial\mathcal{D}} \!\left(\phi\frac{\partial\psi}{\partial n} - \psi\frac{\partial\phi}{\partial n}\right)dS.\tag{1.7}\] The boundary has two parts: the plane $z=0$ and a sphere at infinity. The sphere at infinity contributes nothing since $\phi,\psi\to 0$. On $z=0$, $\psi=0$ so the $\psi\partial\phi/\partial n$ term vanishes. The outward normal from $z>0$ at $z=0$ is $\hat{\mathbf{n}}=-\hat{\mathbf{z}}$, so $\partial\psi/\partial n = -\partial\psi/\partial z\big|_{z=0}$. Computing (setting $x_0=y_0=0$ for $\phi(0,0,z_0)$): \[\frac{\partial\psi}{\partial z}\bigg|_{z=0} = -\frac{1}{4\pi}\!\left[ -\frac{z-z_0}{|\mathbf{x}-\mathbf{x}_0|^3} +\frac{z+z_0}{|\mathbf{x}-\bar{\mathbf{x}}_0|^3} \right]_{z=0} = -\frac{z_0}{2\pi(x^2+y^2+z_0^2)^{3/2}},\tag{1.8}\] so $\partial\psi/\partial n\big|_{z=0} = z_0/2\pi(x^2+y^2+z_0^2)^{3/2} > 0$. Since $\phi=1$ on $x^2+y^2<1$ and zero otherwise: \[\phi(0,0,z_0) = \int_{x^2+y^2<1,\;z=0} \frac{z_0}{2\pi(x^2+y^2+z_0^2)^{3/2}}\,dS.\tag{1.9}\]

Converting to polar coordinates $r^2=x^2+y^2$: \[\begin{aligned}\phi(0,0,z_0) &= \int_0^{2\pi}d\theta\int_0^1 \frac{z_0}{2\pi(r^2+z_0^2)^{3/2}}\,r\,dr = z_0\int_0^1\frac{r\,dr}{(r^2+z_0^2)^{3/2}} \notag\\ &= z_0\left[-\frac{1}{\sqrt{r^2+z_0^2}}\right]_0^1 = z_0\!\left(\frac{1}{z_0}-\frac{1}{\sqrt{1+z_0^2}}\right).\end{aligned}\tag{1.10}\] Hence: \[\boxed{\phi(0,0,z_0) = 1 - \frac{z_0}{\sqrt{1+z_0^2}}.}\tag{1.11}\] As $z_0\to 0^+$, $\phi\to 1$ (matches the BC on the unit disk). As $z_0\to\infty$, $\phi\to 0$ (matches the far-field BC).