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A hyperbola is a conic section with two unbounded branches. In suitable coordinates its equation is $x^2/a^2-y^2/b^2=1$ with $a,b>0$. The unit hyperbola is parametrized by $x=\cosh t$, $y=\sinh t$ on its right branch, relating it to the hyperbolic functions.

The boundary ray has $y/x=3/5$, and its intersection with the hyperbola is $(5/4,3/4)$. Under the given change of variables, $x^2-y^2=r^2$ and $y/x=\tanh \theta$. Thus the region becomesThe hyperbolic functions satisfy ${\cosh }^2\theta -{\sinh }^2\theta =1$, so the Jacobian determinant isIt is positive in the interior. The coordinate degeneracy at $r=0$ is a boundary set of area zero and does not affect the change of variables formula. ThereforeHere $\cosh (\log 2)=5/4$. The PDF specifies two line segments and one hyperbolic arc; the duplicated $y=0$ line in the TeX is not an extra boundary condition.

The wall removes the far-field decay condition and instead imposes ${\psi }_{1x}(d)=0$, ${\psi }_{1y}(d)=U_1$. The force-free condition eliminates any first-order mean shear; since mean sheet tangential velocity is zero, $U_1=0$. Thus ${\psi }_1=f_d(y)\sin \theta$, whereStart with the hyperbolic function form $f_d=(a+by)\cosh y+(c+ey)\sinh y$. The lower conditions give $a=1$, $c=-b$. Defining $\Delta ={\sinh }^{2}d-d^2$, the wall conditions giveConsequently the first-order flow for Taylor-sheet swimming next to a rigid wall isFor $d>0$, $\sinh d>d$ ensures a nonzero denominator; as $d\to \infty$, this reduces to the unbounded first-order field on every fixed height interval.