= Solution
For the given harmonic potentials, $\nabla^2(\boldsymbol\Phi\cdot\mathbf x)=2\nabla\cdot\boldsymbol\Phi$. Taking the <divergence> of the velocity representation therefore gives $\nabla\cdot\mathbf u=0$. Taking its <Laplacian> gives $\nabla^2\mathbf u=-2\nabla(\nabla\cdot\boldsymbol\Phi)$, so the <Stokes equations> hold with
$$
\boxed{p=-2\mu\nabla\cdot\boldsymbol\Phi+p_\infty.}
$$
This is the <Papkovich–Neuber representation> in the sign and normalization of this problem. The potentials used in other conventions can differ by an overall factor and sign; the <velocity> and <pressure> must be scaled together.
\b[(a) Point force.] Put the singularity at the origin, $r=|\mathbf x|$, and choose $\boldsymbol\Phi=\mathbf F/(8\pi\mu r)$, $\chi=0$. The resulting <Stokeslet> is
$$
\boxed{\mathbf u=\frac1{8\pi\mu}\left(\frac{\mathbf F}{r}+\frac{(\mathbf F\cdot\mathbf x)\mathbf x}{r^3}\right),\qquad
p-p_\infty=\frac{\mathbf F\cdot\mathbf x}{4\pi r^3}.}
$$
The harmonic calculation applies away from the force. Distributionally this is the <Green function> for $-\nabla p+\mu\nabla^2\mathbf u+\mathbf F\delta^{(3)}(\mathbf x)=0$, with zero <divergence>.
\b[(b) Point source.] Set $\boldsymbol\Phi=0$, $\chi=-Q/(4\pi r)$. The <three-dimensional point source> field is
$$
\boxed{\mathbf u=\frac{Q\mathbf x}{4\pi r^3},\qquad p=p_\infty.}
$$
Its outward <volume flux> across every enclosing <sphere> is $Q$. The <divergence> is zero away from the source and equals $Q\delta^{(3)}(\mathbf x)$ distributionally, so the singular point represents volume injection rather than an incompressible material point.
\b[(c) Stresslet.] Let $G_{ij}(\mathbf x)=(\delta_{ij}/r+x_ix_j/r^3)/(8\pi\mu)$ and use a symmetric trace-free <stresslet tensor> $S$ with the convention $u_i=-S_{jk}\partial_kG_{ij}$. Differentiating the <Stokeslet> and using symmetry and zero trace gives
$$
\boxed{u_i=\frac{3x_i(\mathbf x\cdot S\mathbf x)}{8\pi\mu r^5},\qquad
p-p_\infty=\frac{3\mathbf x\cdot S\mathbf x}{4\pi r^5}.}
$$
For an axial <force-dipole flow> with axis $\mathbf n$, write $S=s(\mathbf n\mathbf n-I/3)$. Then the radial factor is proportional to $s[3(\mathbf n\cdot\widehat{\mathbf x})^2-1]$. Specifying the derivative convention is essential for interpreting the signs of image strengths.
\b[(d) Source dipole.] A positive source and negative source separated by a vector $\boldsymbol\ell$ have moment $\mathbf M=Q\boldsymbol\ell$. Their dipole limit is minus the directional derivative of the unit source field. Hence the <source dipole> has
$$
\boxed{\mathbf u=\frac1{4\pi}\left[\frac{3(\mathbf M\cdot\mathbf x)\mathbf x}{r^5}-\frac{\mathbf M}{r^3}\right],\qquad
\chi=-\frac{\mathbf M\cdot\mathbf x}{4\pi r^3},\qquad p=p_\infty.}
$$
It carries no net <volume flux>, and unlike a <stresslet> it is an irrotational <potential flow>.
Now set $\mathbf r=\mathbf x-d\mathbf n$ and $\mathbf R=\mathbf x+d\mathbf n$, with $d>0$. Define the unscaled kernels $K(\mathbf R)=\mathbf n/R+(\mathbf n\cdot\mathbf R)\mathbf R/R^3$, $H(\mathbf R)=\mathbf R/R^3$, and let $\partial_n=\mathbf n\cdot\nabla_{\mathbf R}$. The <no-slip image system of a normal Stokeslet> is
$$
\boxed{\mathbf u=\frac{F}{8\pi\mu}\{K(\mathbf r)-K(\mathbf R)+2d\partial_nK(\mathbf R)-2d^2\partial_nH(\mathbf R)\}.}
$$
The reflected <Stokeslet> has force $-F\mathbf n$. With the conventions above, the remaining image strengths are
$$
\boxed{S=-2Fd(\mathbf n\mathbf n-I/3),\qquad \mathbf M=\frac{Fd^2}{\mu}\mathbf n.}
$$
The isotropic part of $S$ produces zero <velocity>, because $\partial_jG_{ij}=0$. Its inclusion simply puts the tensor in standard trace-free form. The <pressure> follows by superposition:
$$
p-p_\infty=\frac F{4\pi}\left[\frac{\mathbf n\cdot\mathbf r}{r^3}-\frac{\mathbf n\cdot\mathbf R}{R^3}+2d\left(\frac1{R^3}-\frac{3(\mathbf n\cdot\mathbf R)^2}{R^5}\right)\right].
$$
To verify the <no-slip boundary condition> explicitly, write $\mathbf x=\boldsymbol\rho$ on the plane, with $\boldsymbol\rho\cdot\mathbf n=0$, and let $b^2=\rho^2+d^2$. Then
$$
\begin{aligned}
K(\mathbf r)-K(\mathbf R)&=-2d\boldsymbol\rho/b^3,\\
\partial_nK(\mathbf R)&=\mathbf R(1/b^3-3d^2/b^5),\\
\partial_nH(\mathbf R)&=\mathbf n/b^3-3d\mathbf R/b^5.
\end{aligned}
$$
The $b^{-5}$ terms cancel, while the remaining image contribution is $2d(\mathbf R-d\mathbf n)/b^3=2d\boldsymbol\rho/b^3$. Thus both normal and tangential <velocity> vanish. All image singularities lie outside the fluid, so the physical half-space contains only the prescribed force.
For the far field, expand about the wall origin. The pair of opposite forces starts with $-2d\partial_nK$; the image <stresslet> cancels this entire $r^{-2}$ term. The first surviving field is
$$
\mathbf u\sim\frac{Fd^2}{4\pi\mu}(\partial_n^2K-\partial_nH).
$$
Writing $z=\mathbf x\cdot\mathbf n$ and now $r=|\mathbf x|$, this is
$$
\mathbf u\sim\frac{3Fd^2}{4\pi\mu}\left[-\frac{2z\mathbf x}{r^5}-\frac{z^2\mathbf n}{r^5}+\frac{5z^3\mathbf x}{r^7}\right].
$$
Therefore \b[the generic fixed-angle far-field velocity decays as $r^{-3}$], rather than the $r^{-1}$ free-space <Stokeslet> or an uncancelled $r^{-2}$ <stresslet>. Along the wall it is exactly zero; at fixed height and large horizontal distance the leading tangential field instead scales as $\rho^{-4}$, consistent with the angular zeros of the displayed far field.
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