Past exam of the theoretical mechanics course of the University of Oxford / 2009 / A / 2

Let $\delta =R{e}^{-1/2}$ and use the inner variables $\boxed{Y=y/\delta =R{e}^{1/2}y,V=v/\delta =R{e}^{1/2}v}$. Continuity balances $u_x$ with $V_Y$. In the tangential momentum balance, advection is order one while the normal viscous term is $R{e}^{-1}{\delta }^{-2}{u}_{YY}=u_{YY}$. Streamwise viscosity is smaller by $R{e}^{-1}$. The normal momentum balance forces the leading pressure to be independent of $Y$. ThusOn the stationary impermeable plate, no-slip boundary condition and no-penetration boundary condition give $u=V=0$ at $Y=0$, $x>0$. Matching to the uniform outer stream gives $u\to 1$ as $Y\to \infty$, and the leading-edge/upstream data are $u\to 1$ as $x\downarrow 0$ at fixed $Y>0$. There is no requirement that the scaled transverse velocity $V$ vanish at infinity; its physical value $v=\delta V$ matches a smaller outer correction. The outer streamwise Euler equation with constant speed one gives $p_x=0$, and $p_Y=0$ carries this pressure gradient throughout the layer.

The Von Mises boundary-layer transformation uses $\Psi$ as transverse coordinate. With $u=U(x,\Psi )$ and ${\Psi }_x=-V$, ${\Psi }_Y=u=U$, the chain rule givesThe streamwise advection terms reduce to $u{u}_x+V{u}_Y=U{U}_x$, so, for $Y>0$ where $U>0$, division by $U$ givesContinuity was already enforced by the stream function. Substitute $U=g(\eta )$ with $\eta =\Psi /\sqrt{x}$. Then $U_x=-\eta g^{\prime }/(2x)$ and ${\partial }_{\Psi }(U{U}_{\Psi })=(g{g}^{\prime }{)}^{\prime }/x$, yieldingThe wall coordinate is zero because $\Psi$ was set to zero there, and the far condition matches unit external speed. This is the transformed Blasius similarity problem. One must not additionally prescribe a finite $g^{\prime }(0)$: the coordinate change degenerates at the no-slip wall, and the regular physical shear corresponds to $g\propto \sqrt{\eta }$ near zero. This similarity reduction is obtained for the attached monotone boundary layer assumed in the question.

Existence of a positive normalized profile can also be seen directly. Put $q=g{g}^{\prime }$ and choose $q(0)=a>0$. In $g$ coordinates the regular initial-value system is $dq/dg=-\eta /2$, $d\eta /dg=g/q$, with $\eta (0)=0$. It starts as $g\sim \sqrt{2a\eta }$. For finite positive $\eta$, $q^{\prime }/q=-\eta /(2g)$ keeps $q$ positive and $q\le a$, while $(g^2{)}^{\prime }=2q$ gives $g\le \sqrt{2a\eta }$. Consequently $q^{\prime }/q\le -\sqrt{\eta }/(2\sqrt{2a})$, so $q$ is integrable at infinity. It follows that $g$ increases to a finite positive limit $M$. The rescaled profile $\tilde{g}(\eta )=M^{-1}g(\sqrt{M}\eta )$ satisfies the same equation and has limit one. This supplies the requested similarity solution with both boundary conditions, rather than only its formal differential equation.