Crack pressure in elliptic coordinates ($p=\frac{\mu Ua}{k}\left[-\cosh \xi +\frac{\alpha }{1+\alpha }{e}^{-\xi }\right]\cos \eta ,\alpha =\frac{H_0^3}{24ka}$)

For $h=H_0(1-x^2/a^2{)}^{1/6}$, elliptic coordinates make $h^3/\sin \eta$ constant on the upper crack face. Pressure continuity and harmonicity reduce the boundary condition to $F^{\prime }(0)=\alpha F(0)$ for the cosine mode, while uniform far flow gives $F\sim -e^{\xi }/2$. Solving $F^{\prime \prime }=F$ yields the displayed pressure. The midpoint crack flux is $2aU\alpha /(1+\alpha )$; small $\alpha$ leaves the background pressure gradient nearly unchanged, while large $\alpha$ makes the crack nearly equipotential.

Subtracting the uniform-flow pressure $-\mu Ux/k$ leaves $(\mu Ua/k)\alpha e^{-\xi }\cos \eta /(1+\alpha )$. The coordinate map has $r\sim a{e}^{\xi }/2$, so $e^{-\xi }\sim a/(2r)$, giving the displayed dipole coefficient. The one-half is essential. In the perfect-conductor limit the independent expression $p=-(\mu U/k)\mathrm{Re}\sqrt{z^2-a^2}$ has the same far-field expansion.

For dipole pressure $\delta p\sim Cx/(x^2+y^2)$, choose a rectangular control volume whose transverse size tends to infinity faster than its streamwise size. Each end face contributes $\pi C$ to $\int \delta p{n}_{x}ds$, so fixed-throughflow boundary work gives $\Delta \mathcal{D}=-2\pi UC$, the displayed result. This control-surface choice removes the integrated far-field velocity correction. A circular cutoff of the pressure-only functional gives $-\pi UC$ instead and does not implement the same throughflow normalization.