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In the upper half-plane, a line source of outward volume flux $Q$ at height $a$ above an impermeable wall has an equal image source below it. On the wall $u=Qx/[\pi (x^2+a^2)]$ and the normal velocity vanishes. The steady Bernoulli equation gives $p-p_{\infty }=-\rho Q^2{x}^2/[2{\pi }^2(x^2+a^2{)}^2]$. Sincethe ambient-relative force per unit length pulls the wall toward the source with the displayed magnitude. The torque about the source's projection onto the wall is zero in the symmetric Cauchy principal value sense. Individual one-sided moments diverge logarithmically. This planar result differs from the three-dimensional hydrodynamic attraction of a plane wall to a point source.

Define source strength $m$ here as the volume flux through a small sphere around the source. For incompressible potential flow, $\mathbf{u}=\nabla \varphi$ and the governing conditions areEquivalently the Laplace equation holds away from the source, with local behavior $\varphi \sim -m/(4\pi |\mathbf{x}-{\mathbf{x}}_s|)$. The boundary condition expresses impermeability. The method of images uses an equal source below the plane:The image singularity lies outside the fluid, the two vertical derivatives cancel on the wall, and the source has the specified outward flux.

Let $s=\sqrt{x^2+y^2}$. On the wall $u_z=0$ and the tangential radial velocity is $u_s=ms/[2\pi (s^2+a^2{)}^{3/2}]$. With ambient pressure $p_{\infty }$, the steady Bernoulli equation in this irrotational flow givesThe fluid pressure is reduced everywhere on the wall except at its axis and in the infinite-distance limit. The finite hydrodynamic force is the excess over the uniform ambient-pressure force. Integrating its magnitude over the plane yieldsPutting $u=s^2$ gives the last integral as $\frac{1}{2}{\int }_0^{\infty }u(u+a^2{)}^{-3}du=1/(4a^2)$. HenceRelative to ambient pressure on the other side, the boundary is attracted toward the source. This is the hydrodynamic attraction of a plane wall to a point source.

There is an alternative source-strength convention: if $m$ denotes the coefficient in $u_r=m/r^2$ rather than total volume flux, replace the flux above by $4\pi m$. Then $\varphi =-m(1/r_{+}+1/r_{-})$, $p-p_{\infty }=-2\rho m^2{s}^2/(s^2+a^2{)}^3$, and $F=\pi \rho m^2/a^2$. The physical result is identical once the normalization is fixed.