= Solution
{c}
In a thin <boundary layer>, take streamwise <velocity> and distance to be order one, set $z=\delta\zeta$, and use <incompressibility> to set $w=\delta W$. The streamwise <convective acceleration> is order one, whereas its dominant viscous contribution after division by the <Reynolds number> is $\mathrm{Re}^{-1}\delta^{-2}u_{\zeta\zeta}$. Their <dominant balance> selects
$$
\boxed{\delta=\mathrm{Re}^{-1/2}.}
$$
With this choice, the exact rescaled <Navier-Stokes equations> are
$$
\begin{aligned}
u_x+W_\zeta&=0,\\
u u_x+W u_\zeta&=-p_x+u_{\zeta\zeta}+\delta^2u_{xx},\\
\delta^2(uW_x+WW_\zeta)&=-p_\zeta+\delta^2W_{\zeta\zeta}+\delta^4W_{xx}.
\end{aligned}
$$
Thus the leading <pressure> is independent of $\zeta$. Matching it to the <outer solution> from part (a), evaluated near $z=0$, gives $p=p_*-x^2/2$, hence $-p_x=x$. The leading <boundary-layer equation> is therefore
$$
\boxed{u_x+W_\zeta=0,\qquad
u u_x+W u_\zeta=x+u_{\zeta\zeta},\qquad \zeta<0.}
$$
The exact surface <shear stress> condition rescales to $u_\zeta+\delta^2W_x=\delta\beta x=\gamma x$. Since $W(x,0)=0$ for all $x$, its tangential <derivative> also vanishes there. The leading surface <boundary conditions> are
$$
\boxed{W(x,0)=0,\qquad u_\zeta(x,0)=\gamma x.}
$$
Matching to the <outer solution> requires
$$
u(x,\zeta)\longrightarrow x\quad(\zeta\to-\infty),
\qquad W(x,\zeta)\sim-\zeta\quad(\zeta\to-\infty).
$$
The second matching condition specifies the leading growth of $W$, allowing a constant displacement correction. Requiring $W+\zeta\to0$ would impose an extra condition not supplied by the leading <outer solution>. This is the <stress-driven stagnation-point boundary layer>.
Back to article page
