Past exam of the theoretical mechanics course of the University of Oxford / 2010 / A / 2

Balance streamwise convection of order one with transverse viscous diffusion $R{e}^{-1}{\partial }_y^{2}u$. This requires a thickness $R{e}^{-1/2}$. Continuity then makes the normal velocity of that same small order. SetThe normal momentum equation makes the leading pressure independent of $Y$. It equals the outer pressure, whose inviscid tangential momentum balance gives $-p_x=U_s{U}_s^{\prime }$. The leading tangential equation and continuity are consequentlyThe divergence-free field $(u,V)$ has a streamfunction with $u={\Psi }_Y$, $V=-{\Psi }_x$. Substituting yieldsThe stationary impermeable wall has $u=V=0$, so ${\Psi }_Y(x,0)=0$ and $\Psi (x,0)$ is constant, chosen to be zero. Matching to the outer slip velocity gives ${\Psi }_Y(x,\infty )=U_s(x)$.

For the proposed form,Matching the outer velocity requires $\alpha -\beta =m$. The convective and outer-pressure terms have order $x^{2m-1}$, while $u_{YY}$ has order $x^{\alpha -3\beta }$. Equality of exponents givesThe two terms containing $\eta f^{\prime }{f}^{\prime \prime }$ cancel in $u{u}_x+V{u}_Y$, leaving $x^{2m-1}[m(f^{\prime }{)}^2-\alpha f{f}^{\prime \prime }]$. Since $U_s{U}_s^{\prime }=m{A}^2{x}^{2m-1}$, the reduced equation isThe wall and matching conditions become $\boxed{f(0)=f^{\prime }(0)=0,f^{\prime }(\infty )=A}$. This establishes a consistent similarity reduction; existence of an actual profile is a separate boundary-value question for a particular choice of parameters.

At these parameters the equation becomes $g^{\prime \prime }+g^2-4=0$, where $g=f^{\prime }$. Multiply by $g^{\prime }$ and integrate:Outer matching gives $g\to -2$ and $g^{\prime }\to 0$, so $C=16/3$. No slip gives $g(0)=0$. At that point the first integral fixes $g^{\prime }(0)=\pm 4\sqrt{2/3}$, so there are exactly two smooth initial-value solutions of the second-order equation satisfying the matching condition. Put ${\eta }_0=\mathrm{arctanh}\sqrt{2/3}$. Translation invariance of the supplied profile gives them explicitly:The first decreases monotonically from zero to $-2$. The second first rises to $4$ at ${\eta }_0$, crosses zero again at $2{\eta }_0$, and then tends to $-2$. Integrating either profile from zero recovers $f$ with $f(0)=0$.

One would not expect both branches to be equally stable. The rising branch contains flow opposite to its negative outer stream and interior inflection points, typical ingredients for shear instability. The monotone branch has $g^{\prime \prime }=4-g^2>0$ at finite positive $\eta$ and no interior inflection point, so the classical inviscid inflection-point mechanism is absent there. This motivates expecting the reversed-flow branch to be unstable and the monotone branch to be the stable candidate. A definitive viscous temporal-stability result requires perturbing the time-dependent equations; the steady first integral alone is not a stability proof.