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Use the pressure scale $\rho {\hat{U}}^2$, and denote the scaled transverse velocity by $v$. The steady incompressible flow equations under the prescribed coordinate scaling areThe no-slip boundary condition is $u=v=0$ at $y=1$ and at the lower wall $y=ϵF(x)$, taking $F=0$ upstream. Far upstream, $u\to U_0=6y(1-y)$, $v\to 0$, and $p_x\to -12\lambda /R$. These are the dimensional Navier-Stokes equations divided by the streamwise and transverse inertial scales, respectively; the transverse equation's ${\lambda }^{-2}$ is essential.

A consistent nonlinear wall-layer regime can be chosen asNear a wall the oncoming plane Poiseuille flow is linear in distance, so a layer of thickness $ϵ$ has $u=O(ϵ)$ and $v=O({ϵ}^2)$. Its inertia is $O({ϵ}^2)$; its transverse viscous term is $(\lambda /R)O({ϵ}^{-1})=O({ϵ}^2)$, explaining the distinguished balance. The resulting pressure perturbation is $O({ϵ}^2)$. In the core the induced transverse pressure variation is $O(ϵ/{\lambda }^2)$, negligible compared with that pressure perturbation when ${\lambda }^2ϵ\gg 1$, equivalently $\lambda \gg R^{1/7}$. Also $\lambda \ll R$ makes both the indentation and the viscous boundary layer thin. Streamwise viscous derivatives are smaller still. Thus the core perturbation is inviscid at first order and pressure is common to the two wall layers at their leading order.

Write $u=U_0+ϵu_1+\cdots$ and $v=ϵv_1+\cdots$. The core pressure gradient only enters at order ${ϵ}^2$, so first-order continuity equation and streamwise momentum giveEliminating $u_{1x}$ shows $(v_1/U_0{)}_y=0$. Let that ratio be $-A^{\prime }(x)$; integration in $x$, with the unperturbed upstream condition, then givesThis displacement form automatically has zero first-order change in total core flux, because ${\int }_0^1{U}_0^{\prime }dy=0$.

For the lower boundary layer, usewhere $p_0^{\prime }=-12{ϵ}^3$ is the smaller upstream pressure gradient. The extra term in $v$ removes the velocity induced by the moving coordinate surface. The chain rule gives ${\partial }_x{|}_y={\partial }_x{|}_z-F^{\prime }{\partial }_z$ and ${\partial }_y={ϵ}^{-1}{\partial }_z$; the wall-slope terms cancel from both continuity equation and convection. At leading order the Prandtl boundary-layer equation isMatching to the core near the lower wall gives $U\sim 6[z+F+A]$, so $H_{\mathrm{lower}}=F+A$.

For the upper boundary layer, set $y=1-ϵz$, $u=ϵU$, $v=-{ϵ}^{2}V$. The same equations, wall data and common $P(x)$ result, but the core matching gives $U\sim 6[z-A]$, so $H_{\mathrm{upper}}=-A$. The two pressure-driven wall-layer problems have the same upstream shear, wall conditions and pressure forcing. Under the question's uniqueness assumption their solutions coincide, and their displacement limits must coincide. Thus the symmetric viscous interaction in an indented Poiseuille channel gives

For $F=b{x}^{1/3}$, one has $A=-(b/2)x^{1/3}$ and $H=(b/2)x^{1/3}$. Matching $U\sim 6[z+H(x)]$ in a similarity solution $U=x^{\beta }{G}^{\prime }(z{x}^{-\gamma })$ requires $\beta =\gamma =1/3$. Equating the powers in convection $x^{2\beta -1}$, viscosity $x^{\beta -2\gamma }$ and pressure gradient $x^{\sigma -1}$ gives $\sigma =2/3$. HenceUse the stream function $\psi =x^{2/3}G(\eta )$, with $\eta =z{x}^{-1/3}$. ThenDirect differentiation gives $U{U}_x+V{U}_z=x^{-1/3}[(G^{\prime }{)}^2/3-2G{G}^{\prime \prime }/3]$, whereas $-P^{\prime }+U_{zz}=x^{-1/3}[-2\bar{p}/3+G^{\prime \prime \prime }]$. The one-third-power similarity of an indented channel boundary layer is therefore governed byThe original PDF has $(G^{\prime }{)}^2$ in this equation; the converted TeX loses the derivative and gives $G^2$. Existence remains conditional on a value of $\bar{p}$ for which this boundary value problem has a solution, as stipulated in the question. The similarity describes the stated thin-indentation regime on $x>0$.