= Crack pressure in elliptic coordinates
{title2=$p=\frac{\mu Ua}{k}\left[-\cosh\xi+\frac\alpha{1+\alpha}e^{-\xi}\right]\cos\eta,\quad\alpha=\frac{H_0^3}{24ka}$}
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 $F'(0)=\alpha F(0)$ for the cosine mode, while uniform far flow gives $F\sim-e^\xi/2$. Solving $F''=F$ 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.
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