= Solution
At the sheet, the second-order conditions are $\psi_{2x}(0)=0$ and $\psi_{2y}(0)=-f_d''(0)\sin^2\theta$. At the wall, $\psi_{2x}(d)=0$ and $\psi_{2y}(d)=\overline U_2$. The general solution has a polynomial mean mode and a second harmonic with exponentials $e^{\pm2y}$, as in the unbounded calculation.
Let $\overline u_2(y)=\langle\psi_{2y}\rangle$. The averaged <Stokes equation>, with periodic pressure and no imposed mean pressure gradient, gives $\overline u_2''=0$. The total horizontal hydrodynamic force on the sheet is opposite to the mean shear force on the flat wall. Since the swimmer is <force-free>, $\overline u_2'(d)=0$, making the mean velocity constant. Matching its values at the sheet and wall therefore gives
$$
\overline U_2=-\frac12 f_d''(0).
$$
The previous part gives $f_d''(0)=1-2\sinh^2d/\Delta=-(\sinh^2d+d^2)/\Delta$, hence
$$
\boxed{\overline U_2=\frac{\sinh^2d+d^2}{2(\sinh^2d-d^2)}.}
$$
This determines the speed without solving the second-harmonic flow. The force constraint matters: prescribing a translating sheet without imposing zero net force would allow a mean shear and would not determine a unique swimming speed.
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