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

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.