Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 36E (Fluid Dynamics II)

For steady unidirectional pressure-driven channel flow, write $G=-dp/dz>0$. The axial momentum equation and no-slip boundary condition areNear an ideal wedge, the constant forcing has an $r^2$ particular solution by scaling. Substitution gives $f^{\prime \prime }+4f=-1$, with $f(\pm \alpha )=0$. For $\cos 2\alpha \ne 0$, the symmetric particular solution isThis corner expansion of pressure-driven sector flow is a particular solution, not necessarily the dominant corner term; a homogeneous corner mode can be larger, as the requested cases below show.

Separation of the symmetric harmonic correction with zero values on the straight walls gives modes $r^{{\lambda }_n}\cos ({\lambda }_n\theta )$, whereRegularity at the vertex excludes negative powers. Orthogonality on $[-\alpha ,\alpha ]$ has norm $\alpha$. To cancel the particular solution on $r=a$, use its cosine coefficientsThe integral follows by product-to-sum and $\sin ({\lambda }_n\alpha )=(-1{)}^n$. ThusThe arc data are satisfied by the cosine expansion, and the Poisson equation by construction. At the omitted thresholds $2\alpha =\pi /2,3\pi /2$, a mode has ${\lambda }_n=2$, so the pure $r^{2}f$ ansatz is resonant. The corresponding finite-limit term in the full sector solution is $(G/\mu )(-1{)}^n{r}^2\log (a/r)\cos 2\theta /(4\alpha )$; the other modes remain regular. This explains rather than hides the division by $\cos 2\alpha$.

For $2\alpha <\pi /2$, ${\lambda }_0>2$ and ${\lambda }_1>6$. Every harmonic correction decays faster than $r^2$ at the corner, so the $r^{2}f(\theta )$ particular solution is the leading corner velocity. It is positive inside this acute wedge and the velocity gradient tends to zero at the vertex.

For $\pi /2<2\alpha <3\pi /2$, $2/3<{\lambda }_0<2$ while $2<{\lambda }_1<6$. Thus the first harmonic mode dominates the particular $r^2$ term, and the remaining modes are higher order. Its coefficient is positive. The particular solution alone can be negative here and cannot describe the physical leading flow. Once $2\alpha >\pi$, ${\lambda }_0<1$ and the leading velocity gradient behaves as $r^{{\lambda }_0-1}$, singular at the corner although the velocity itself tends to zero.

For $3\pi /2<2\alpha <2\pi$, $1/2<{\lambda }_0<2/3$ and $3/2<{\lambda }_1<2$, whereas ${\lambda }_2>5/2$. Hence both of the first two harmonic modes precede the $r^2$ particular term in the corner expansion; the first still dominates as $r\to 0$. The first-mode gradient is singular. These are reentrant wedges, and the global arc condition determines the important homogeneous coefficients.