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Reflect the source across the equatorial plane and subtract the two ball kernels. This gives the Dirichlet Green function inside a hemisphere $G_{\mathrm{hemi}}(\mathbf{x};{\mathbf{x}}_0)=G_B(\mathbf{x};{\mathbf{x}}_0)-G_B(\mathbf{x};{\mathbf{x}}_0^{-})$. It is zero on the spherical boundary because each ball kernel is zero there, and on the plane because reflection makes their values equal.

For a point charge $e$, Poisson's equation in SI units is $\Delta V=-e\delta (\mathbf{x}-{\mathbf{x}}_0)/{\epsilon }_0$, so $V=-e{G}_{\mathrm{hemi}}/{\epsilon }_0$. Writing ${\mathbf{x}}_0^{\ast -}$ for the reflected inverse point givesThe construction uses the real charge, its negative reflected image, a negative inversion image of magnitude $ea/r_0$, and its positive reflected image. Reflection and inversion commute. All three images lie outside the open upper half-ball, and the prescribed zero boundary values determine the electrostatic potential uniquely.

Let $\overline{{\mathbf{r}}^{\prime }}=(x^{\prime },y^{\prime },-z^{\prime })$ be reflection of the interior source across the equatorial plane. Using the ball Green's function from part (c), formBoth terms vanish on the unit sphere. On the flat disk $z=0$, distances to reflected points coincide, and the same is true of their inversion images, so the difference is zero there as well. Only ${\mathbf{r}}^{\prime }$ is inside the upper half-ball; the reflected source is in the lower half-ball, and both inversion images lie outside the full ball. Thus there is exactly one interior delta source, of the required strength.

Explicitly, with $s=|{\mathbf{r}}^{\prime }|$ and stars denoting inversion in the unit sphere,This is the Dirichlet Green function inside a hemisphere, constructed from four image-source terms.

Reflect the physical source across $z=0$, writing $\overline{{\mathbf{r}}^{\prime }}=(x^{\prime },y^{\prime },-z^{\prime })$. Its radius is still $s=|{\mathbf{r}}^{\prime }|$, and its spherical image is $\overline{{\mathbf{r}}^{\ast }}$. The Dirichlet Green function inside a hemisphere isExplicitly this four-source method of images isBoth $G_0$ terms vanish on the spherical cap. On the flat equatorial disc, reflection equates the two physical-source distances and also the two image-source distances, so both pairs cancel. The reflected physical source lies in the lower half-ball and the two inverted images lie outside the sphere. Consequently the only source in the upper half-ball is ${\mathbf{r}}^{\prime }$, with the correct coefficient:The Dirichlet uniqueness for the Poisson equation then identifies this as the required Green function.

Let $\overline{{\mathbf{r}}^{\prime }}=(x^{\prime },y^{\prime },-z^{\prime })$ be the reflection of the source in the equatorial plane. For a source in the upper half-ball, the Dirichlet Green function inside a hemisphere isBoth terms vanish on the spherical boundary. On the flat boundary $z=0$, the two source distances are equal, and the distances to their spherical image points are also equal, so the difference vanishes there too. The reflected physical source is in the lower half-ball and both inverted image points are outside the sphere. Therefore the only source singularity in the desired region is the original one, with coefficient $+1$ in $\Delta G$. This combines reflection with the spherical method of images.