= Solution
Take positive $x$ in the snail's direction of travel, and let $y$ measure height above the rigid plane. The wave moves at laboratory <velocity> $U-V$. In the wave frame the plane's horizontal <velocity> is $V-U$ and the foot's is $V$. The foot is stationary as a shape, so its material <velocity> is tangent to $y=h(x)$: to leading <lubrication theory> order its components are $(V,Vh_x)$. Apply the <no-slip boundary condition> to both surfaces.
The leading <Stokes flow> equations are $p_y=0$ and $\mu u_{yy}=p_x$. Their solution is the local <Couette-Poiseuille flow>
$$
\boxed{u(x,y)=V-U+\frac{Uy}{h(x)}+\frac{p_x}{2\mu}y(y-h).}
$$
The normal-velocity condition on each steady surface and <incompressibility> make the wave-frame flux independent of $x$. Integrating the <velocity> gives
$$
q=h\left(V-\frac U2\right)-\frac{h^3p_x}{12\mu}.
$$
Put $C=V-U/2$. Periodicity of <pressure> gives $0=\int_0^Lp_xdx=12\mu(CI_2-qI_3)$, and hence
$$
\boxed{q=C\frac{I_2}{I_3},\qquad
p_x=12\mu\left(\frac C{h^2}-\frac q{h^3}\right),\qquad
I_j=\int_0^Lh^{-j}dx.}
$$
The shear traction exerted by the fluid on the plane, positive in the $x$ direction, is
$$
\boxed{\tau_0=\mu u_y(x,0)=\frac{\mu U}{h}-\frac h2p_x
=\frac{\mu(4U-6V)}h+\frac{6\mu q}{h^2}.}
$$
The snail is force-free horizontally: its gait consists of internal motion and it has no externally applied horizontal drive. Uniform locomotion therefore requires zero net horizontal fluid <force> on the foot per wavelength. The stresses on opposite periodic fluid faces cancel, so horizontal momentum balance also makes the net <force> on the plane zero. Thus
$$
0=\int_0^L\tau_0dx=\mu(4U-6V)I_1+6\mu qI_2.
$$
Substitution of $q$ proves the <lubrication model of a crawling snail>:
$$
\boxed{U=\frac{6(1-\alpha)}{4-3\alpha}V,\qquad
\alpha=\frac{I_2^2}{I_1I_3}.}
$$
The <Cauchy-Schwarz inequality> applied to $h^{-1/2}$ and $h^{-3/2}$ gives $0<\alpha\leq1$, with equality exactly for a uniform gap. Thus the denominator is positive, and a nonuniform waveform gives forward motion; the flat gap gives $U=0$.
For the sinusoidal shape, the given integrals yield
$$
I_1=\frac{L}{h_0\sqrt{1-A^2}},\qquad
I_2=\frac{L}{h_0^2(1-A^2)^{3/2}},\qquad
I_3=\frac{L(1+A^2/2)}{h_0^3(1-A^2)^{5/2}}.
$$
Therefore $\alpha=1/(1+A^2/2)$ and
$$
\boxed{U=\frac{3A^2}{1+2A^2}V.}
$$
For small amplitude, $U\sim3A^2V$: the leading locomotion is quadratic, and there is no propulsion from a flat foot. As $A\to1^-$, $U\to V$, so the wave becomes nearly stationary in the laboratory frame while the snail advances relative to it. However, the minimum gap tends to zero; finite speed in this limit does not imply finite dissipation, and $A=1$ itself is excluded by the noncontact model.
For the general waveform, use the mechanical work identity for <viscous dissipation>. <Pressure> does no boundary work because the boundary motion is tangential in the wave frame. The plane's <velocity> is constant, and its total <force> is zero, so it contributes zero net work. The upper boundary supplies, at leading lubrication order,
$$
\Phi=\mu V\int_0^Lu_y(x,h)dx
=V\left(\mu UI_1+\frac12\int_0^Lhp_xdx\right).
$$
The zero plane-force relation is $\mu UI_1=\tfrac12\int hp_xdx$. Consequently
$$
\boxed{\Phi=2\mu UVI_1.}
$$
As a direct check, integrating the squared shear rate gives $\Phi=\int_0^L[\mu U^2/h+h^3p_x^2/(12\mu)]dx=\mu I_1[U^2+12C^2(1-\alpha)]$, which reduces to the same result using the force-free speed relation. Zero horizontal <force> on the foot does not mean zero gait work: the tangential material <velocity> along its sloping surface includes vertical motion.
For prescribed speed $U>0$ and sinusoidal shape, eliminate $V=U(1+2A^2)/(3A^2)$ to obtain
$$
\Phi(A)=\frac{2\mu LU^2}{3h_0}\frac{1+2A^2}{A^2\sqrt{1-A^2}}.
$$
Writing $y=A^2$, the logarithmic derivative of the amplitude-dependent factor is
$$
\frac2{1+2y}-\frac1y+\frac1{2(1-y)}=0
\quad\Longleftrightarrow\quad2y^2+3y-2=0.
$$
The only root in $0<y<1$ is $y=1/2$. The dissipation diverges at both endpoints, so this is the global minimum and gives the <optimal sinusoidal waveform for viscous snail locomotion>:
$$
\boxed{A_{\rm opt}=\frac1{\sqrt2},\qquad V_{\rm opt}=\frac43U,\qquad
\Phi_{\min}=\frac{8\sqrt2}{3}\frac{\mu LU^2}{h_0}.}
$$
The magically driven flat glider has simple <Couette flow> $u=Uy/h_0$ in the laboratory frame, and dissipation $\Phi_{\rm glide}=\mu LU^2/h_0$ per wavelength. Hence
$$
\boxed{\Phi_{\min}/\Phi_{\rm glide}=8\sqrt2/3.}
$$
The comparison is at the same $U$, mean thickness $h_0$ and wavelength $L$; the flat glider is externally propelled, whereas the snail must power its internal waveform.
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