Past exam of the theoretical mechanics course of the University of Oxford / 2008 / A / 2

Let $\delta =R{e}^{-1/2}$, stretch $Y=y/\delta$ and write the original transverse velocity as $v=\delta V$. Continuity then gives $u_x+V_Y=0$. In the streamwise momentum equation $u_{yy}/Re=u_{YY}$ is order one, while $u_{xx}/Re$ is negligible. The transverse equation gives $p_Y=0$ to leading order, so pressure is supplied by the outer Euler flow. Its streamwise balance is $p_x=-U{U}^{\prime }$. ConsequentlyFor a stationary impermeable plate, $\boxed{u(x,0)=V(x,0)=0}$, with matching $\boxed{u(x,Y)\to U(x)}$ as $Y\to \infty$. The transverse velocity must match the outer displacement flow; it is generally wrong to impose $V\to 0$ independently at infinity.

At the leading edge the wall no-slip data meet the external stream abruptly, and the boundary-layer approximation is nonuniform. For a constant external speed, the thickness is $O(\sqrt{x/Re})$, whose ratio to $x$ ceases to be small when the local Reynolds number is order one. A leading-edge region retaining neglected derivatives is needed there. An adverse pressure gradient can also cause separation farther downstream; a large global Reynolds number alone does not prevent it.

Use $u={\psi }_Y$, $V=-{\psi }_x$, and put $\eta =Y/x^{\gamma }$, $\psi =x^{\beta }f(\eta )$. ThenMatching $u\sim x^m$ requires $\beta -\gamma =m$. Equating the powers in streamwise inertia $x^{2m-1}$ and transverse diffusion $x^{m-2\gamma }$ givesThe two convective terms combine to $x^{2m-1}[m(f^{\prime }{)}^2-\beta f{f}^{\prime \prime }]$, while $U{U}^{\prime }=m{x}^{2m-1}$. Hence the Falkner-Skan equation in this scaling isThe additive stream-function constant is fixed by $f(0)=0$; when $\beta \ne 0$ this also encodes no penetration. For $\beta =0$ it is a harmless choice of additive constant choice. Existence of an attached physical branch for a given $m$ is a separate issue from deriving the similarity reduction.

On an attached branch with $w=f^{\prime }>0$ away from the wall, use $f$ itself as the independent variable. The chain rule gives $f^{\prime \prime }=w{w}_f$ and $f^{(3)}=w(w_f^2+w{w}_{ff})$. ThereforeIf the wall shear $s=f^{\prime \prime }(0)>0$, then $f\sim s{\eta }^2/2$, so $w(f)\sim \sqrt{2sf}$. Thus $w_f$ is singular at zero while $w{w}_f\to s$ is finite. Imposing a finite arbitrary $w_f(0)$ would remove the usual attached branch. This is the velocity as dependent variable in the Falkner-Skan equation reduction.