Past exam of the viscous flow course of the University of Oxford / 2023 / 2 / c

At zero surface shear stress, the solution connected to part (a) isIndeed, it has $f_0^{\prime }=1$, $f_0^{\prime \prime }=f_0^{\prime \prime \prime }=0$ and $f_0(0)=0$. Insert the regular perturbation $f=\zeta +\gamma f_1+O({\gamma }^2)$ into the nonlinear ordinary differential equation. The terms areThe coefficient of $\gamma$ therefore givesExpanding the surface boundary conditions and the far-field matching condition givesThe printed $+\infty$ limit is a sign error. It is present in the native PDF on paper page 3, not introduced by transcription. The fluid occupies $z<0$, the inner variable is $\zeta =z/\delta <0$, and part (b)(iii) correctly matches at $-\infty$. We retain that physical domain throughout. A change to the positive coordinate $\eta =-\zeta$ would also change the differential equation and surface derivative signs; it cannot be made by changing only the far-field sign.

The inequality ${\delta }^2\ll \gamma$ ensures that effects discarded at order ${\delta }^2$ in the boundary-layer scaling are smaller than the retained order-$\gamma$ perturbation. The inequality $\gamma \ll 1$ justifies this regular perturbation. The two conditions control different approximations.

Set $g=f_1^{\prime }$. The boundary value problem becomesFor $y_1=1+{\zeta }^2$, direct substitution gives $y_1^{\prime \prime }+\zeta y_1^{\prime }-2y_1=2+2{\zeta }^2-2(1+{\zeta }^2)=0$. This growing polynomial solution is useful for reduction of order.

Write $g=y_{1}v$. Substitution into the linear ordinary differential equation cancels the terms proportional to $v$ and givesAn integrating factor for this first-order equation for $v^{\prime }$ yieldsThe far-field boundary condition selects the convergent integral from $-\infty$:Any additional multiple of $1+{\zeta }^2$ would grow at $-\infty$. The chosen expression tends to zero: for $r=-\zeta \ge 1$,Here $(1+t^2{)}^{-2}\le (1+r^2{)}^{-2}$ and ${\int }_r^{\infty }{e}^{-t^2/2}dt\le r^{-1}{\int }_r^{\infty }t{e}^{-t^2/2}dt$. Thus the far-field boundary condition is verified without a cancellation of large terms.

At $\zeta =0$, differentiating gives $g^{\prime }(0)=C$, so $C=1$. Using the antiderivative supplied in the paper gives the explicit answerHere $\mathrm{erf}$ is the error function. Finally,recovers the unique correction satisfying $f_1(0)=0$. The decay bound also shows that $f_1$ approaches a finite constant at $-\infty$, as allowed by the normal velocity matching. This is the weak surface-shear perturbation of stagnation flow.

If the printed $+\infty$ condition is instead imposed literally on a positive half-line, reduction of order givesIt satisfies $g_{+}^{\prime }(0)=1$ but grows as $\zeta \to -\infty$, so it cannot describe the fluid domain specified in this paper. This makes the consequence of the printed sign error explicit.

At the interface, the tangential velocity is $u(x,0)=x{f}^{\prime }(0)$ and the normal velocity remains zero. The preceding error function expression givesHence, within the leading boundary layer model,Allowing the formally discarded finite-Reynolds number terms gives an additional order-${\delta }^2$ error, smaller than the order-$\gamma$ correction under the assumed ordering. The coefficient is $\sqrt{\pi /8}\approx 0.626657$.

Thus positive surface shear increases the outward surface speed linearly: for $x>0$ it increases $u$, and for $x<0$ it makes $u$ more negative. The stagnation point remains at $x=0$. A negative coefficient would result from using the printed $+\infty$ condition literally, and would describe the different half-line problem identified in part (ii).