 Solution

The unit normal pointing out of the fluid at its lower boundary is $-{\mathbf{e}}_y$. Since impermeability gives $v(x,0)=0$, also $v_x(x,0)=0$, and the tangential viscous traction exerted on the fluid is $t_x=-\mu u_y(x,0)$. The cilia do positive work at rate $t_{x}u=\rho P$, soFor positive downstream velocity this requires $u_y<0$, as expected when the wall-driven flow slows away from the cilia. The wall velocity is not prescribed independently; the driving condition prescribes power.

In a laminar boundary layer, momentum diffuses over a thickness $\delta$ during the downstream advection time $x/U$, giving ${\delta }^2\sim \nu x/U$. The wall power condition gives $U^2/\delta \sim P/\nu$. With these balances the downstream Reynolds number grows and $\delta /x\propto x^{-3/5}$ tends to zero. This explains the thin layer for large $x$ within the laminar approximation. Pressure is uniform across it and has no imposed streamwise gradient because the outer fluid is at rest. The two-dimensional boundary-layer equations are

Choose the scales to satisfy the balances exactly:The stream function $\psi =U\delta f(\eta )$, $\eta =y/\delta$, enforces continuity equation and givesThus the advective term is $U^2[\frac{1}{5}{f}^{\prime 2}-\frac{3}{5}f{f}^{\prime \prime }]/x$, while $\nu u_{yy}=\nu U{f}^{\prime \prime \prime }/{\delta }^2=U^2{f}^{\prime \prime \prime }/x$. HenceImpermeability, imposed wall power and matching to resting outer fluid giveA regular decaying profile also has $f^{\prime \prime }\to 0$ and finite positive $f_{\infty }$; no arbitrary additional wall value for $f^{\prime }$ should be imposed.

The volume flux per unit span is $Q={\int }_0^{\infty }udy=U\delta f_{\infty }\propto x^{3/5}$, so continuity equation gives $v_{\infty }=-dQ/dx<0$: outer fluid is entrained toward the wall. In the physical positive decaying similarity profile, $f^{\prime }>0$ and $f^{\prime \prime }<0$. To justify monotonicity, $f^{\prime \prime }(0)<0$ by the power condition; if $f^{\prime \prime }$ first crossed zero with $f^{\prime }>0$, the equation would give $f^{\prime \prime \prime }=f^{\prime 2}/5>0$, after which $f^{\prime \prime }>0$ cannot cross back and $f^{\prime }$ could not decay to zero. Therefore $f(\eta )={\int }_0^{\eta }{f}^{\prime }(s)ds\ge \eta f^{\prime }(\eta )$, and the velocity formula gives $2\eta f^{\prime }-3f<0$ for $\eta >0$. Consequently the transverse velocity is negative throughout that physical layer, and zero at the wall. This pointwise conclusion uses the positive decaying branch; the flux argument already establishes entrainment of the exterior.