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For outer velocity $U_s=-2/x$, the Falkner-Skan equation reduces for the scaled velocity $g=f^{\prime }$ to $g^{\prime \prime }+g^2-4=0$. Multiplication by $g^{\prime }$ gives $\frac{1}{2}(g^{\prime }{)}^2+g^3/3-4g=16/3$ for profiles tending to $-2$ at infinity. No slip gives $g(0)=0$, so $g^{\prime }(0)=\pm 4\sqrt{2/3}$. The two smooth solutions are the displayed translates with ${\eta }_0=\mathrm{arctanh}\sqrt{2/3}$. One decreases monotonically to the outer velocity; the other rises to $4$ before descending, producing an internal reversed-flow region. The latter has inflection points, motivating concern about shear instability, but the steady profile equation alone does not determine temporal stability.

A Falkner-Skan boundary layer is a similarity boundary-layer flow driven by an outer velocity proportional to a power of distance along the wall. Its streamfunction reduction gives the Falkner-Skan equation, with no-slip wall and outer-velocity matching conditions. It generalizes the Blasius flat-plate boundary layer.

Boundary-layer theory assumes steady incompressible high-Reynolds-number flow, $\delta /x\ll 1$, weak streamwise viscous diffusion, nearly constant pressure across the layer, and outer pressure gradient $-{\rho }^{-1}{p}_x=U{U}^{\prime }$.

For $\varphi =-A{r}^k\cos (k\theta )/k$,At $\theta =\pm \pi /k$ the normal velocity vanishes; along the upper wall $U(x)=A{x}^{k-1}$. Boundary-layer scaling givesSet $m=k-1$. The Falkner-Skan equation becomesso $\alpha =k-1$, $\beta =-k/2$, $\gamma =1-k$, with
$f(0)=f^{\prime }(0)=0$ and $f^{\prime }(\infty )=1$. Finallyrequires $x\gg (\nu /A{)}^{1/k}$; thus $a=1/k$ and the tip region violates the approximation.

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.

For the Falkner-Skan equation $f^{(3)}+\beta f{f}^{\prime \prime }+m(1-(f^{\prime }{)}^2)=0$, an attached branch has $f^{\prime }>0$ away from the wall. It is then possible to replace the similarity coordinate by $f$ itself. Define $w(f)=f^{\prime }$. The chain rule gives $f^{\prime \prime }=w{w}_f$ and $f^{(3)}=w(w_f^2+w{w}_{ff})$, proving the displayed second-order equation. The wall and outer conditions become $w(0)=0$ and $w(\infty )=1$ when the stream function is set to zero at the wall.

The reduction is singular at the wall: if $s=f^{\prime \prime }(0)>0$, then $f\sim s{\eta }^2/2$ and $w\sim \sqrt{2sf}$. Hence $w_f$ diverges while $w{w}_f\to s$ remains finite. A finite-slope condition on $w$ at zero would incorrectly exclude the ordinary positive-wall-shear branch. At flow reversal or separation, the invertibility of the coordinate change must be checked again.