Falkner-Skan equation

For $U=A{x}^m$, $\psi =\sqrt{\nu xU}f(\eta )$ giveswith $f(0)=f^{\prime }(0)=0$ and $f^{\prime }(\infty )=1$.

This equation and its boundary conditions determine the Falkner-Skan boundary layer.

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.

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.