At leading order, $X_1=Vt$, $dX_1/dt=V$, $Y_1=0$, and the much smaller motion of sphere 2 may be neglected when evaluating the separation:
$$
\mathbf R\simeq(-X_1,Y_0,0),
\qquad
R=(X_1^2+Y_0^2)^{1/2}.
$$
The $y$ component of the leading $a/R$ term in the mobility from part b is
$$
\frac{dY_2}{dt}
=\frac{3aV}{4R}\frac{R_xR_y}{R^2}
=-\frac{3aV}{4}
\frac{X_1Y_0}{(X_1^2+Y_0^2)^{3/2}}.
$$
Thus
$$
\boxed{
\frac{dY_2/dt}{dX_1/dt}
=-\frac{3a}{4}
\frac{X_1Y_0}{(X_1^2+Y_0^2)^{3/2}}}.
$$
Integrating from the initial position $X_1=0$ to infinity gives the <hydrodynamic displacement of a force-free sphere>
$$
\lim_{t\to\infty}[Y_2(t)-Y_0]
=-\frac{3a}{4}
\int_0^\infty
\frac{XY_0}{(X^2+Y_0^2)^{3/2}}\,dX
=\boxed{-\frac{3a}{4}}.
$$

The leading horizontal velocity is
$$
\frac{dX_2}{dt}
=\frac{3aV}{4R}
\left(1+\frac{X_1^2}{R^2}\right)
\sim\frac{3aV}{2X_1}.
$$
Consequently $X_2\sim(3a/2)\log(X_1/Y_0)$ and
$$
\boxed{X_2(t)\longrightarrow+\infty}
$$
logarithmically. The sphere is carried arbitrarily far downstream even though its transverse displacement approaches a finite limit.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 329 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-329/1/c/solution