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 $\psi_1=f_d(y)\sin\theta$, where
$$
(\partial_y^2-1)^2f_d=0,\qquad f_d(0)=1,\quad f_d'(0)=0,\quad f_d(d)=f_d'(d)=0.
$$
Start with the <hyperbolic function> form $f_d=(a+by)\cosh y+(c+ey)\sinh y$. The lower conditions give $a=1$, $c=-b$. Defining $\Delta=\sinh^2d-d^2$, the wall conditions give
$$
b=\frac{d+\sinh d\cosh d}{\Delta},\qquad e=-\frac{\sinh^2d}{\Delta}.
$$
Consequently the first-order flow for <Taylor-sheet swimming next to a rigid wall> is
$$
\boxed{\psi_1=\left[\cosh y+\frac{d+\sinh d\cosh d}{\Delta}(y\cosh y-\sinh y)-\frac{\sinh^2d}{\Delta}y\sinh y\right]\sin\theta.}
$$
For $d>0$, $\sinh d>d$ ensures a nonzero denominator; as $d\to\infty$, this reduces to the unbounded first-order field on every fixed height interval.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 334 / 1 / e / Solution

{c}

For a transverse <Taylor swimming sheet> below a flat <no-slip boundary condition> at height $d$, the first-order amplitude $f(y)$ satisfies $(\partial_y^2-1)^2f=0$, $f(0)=1$, $f'(0)=f(d)=f'(d)=0$. Writing $\Delta=\sinh^2d-d^2$ gives
$$
f(y)=\cosh y+\frac{d+\sinh d\cosh d}{\Delta}(y\cosh y-\sinh y)-\frac{\sinh^2d}{\Delta}y\sinh y.
$$
The <mean boundary velocity determines Taylor-sheet swimming speed>, yielding
$$
\overline U_2=-\frac12f''(0)=\frac{\sinh^2d+d^2}{2(\sinh^2d-d^2)}.
$$
This is greater than the unbounded value $1/2$ for every $d>0$. In a narrow gap it scales as $3/d^2$, with the small-amplitude calculation requiring $\epsilon\ll d$ as well as $\epsilon\ll1$. https://arxiv.org/html/2410.02278v1[Taylor's swimming sheet near a soft boundary] recovers this rigid-wall limit while studying how compliance changes propulsion.

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