= Solution
More surfactant is encountered on the side toward which the far-field concentration $\Gamma$ increases. Adsorption lowers the <surface tension> there, so the resulting <Marangoni stress> drives interfacial flow toward the cleaner, higher-tension side. The reaction force propels the bubble along the concentration gradient; this is <chemophoresis>.
The bulk concentration obeys the <advection-diffusion equation>
$$
\frac{\partial\Gamma}{\partial t}
+\mathbf u\mathbin\cdot\nabla\Gamma
=D\nabla^2\Gamma.
$$
At $r=a$,
$$
D\mathbf n\mathbin\cdot\nabla\Gamma=-k(C-b\Gamma)
$$
equates the outward bulk diffusive flux to minus the net adsorption rate: $C>b\Gamma$ favors desorption into the bulk, while $C<b\Gamma$ favors adsorption onto the interface.
With advection neglected, $\Gamma$ is <harmonic function>[harmonic]. Rotational symmetry about $\mathbf G$ and the imposed far-field gradient select the dipolar form
$$
\Gamma=\Gamma_0+\mathbf G\mathbin\cdot\mathbf x
+B\frac{\mathbf G\mathbin\cdot\mathbf x}{r^3}.
$$
When $kab\ll D$, the boundary flux is smaller than the characteristic diffusive flux, so the leading boundary condition is $\partial_r\Gamma=0$ at $r=a$. This gives $B=a^3/2$, and hence
$$
\boxed{\Gamma(a,\mathbf n)=\Gamma_0+\frac32a\mathbf G\mathbin\cdot\mathbf n}.
$$
For an interface with unit normal directed from the bubble into the exterior liquid, the <interfacial stress balance with variable surface tension> may be written
$$
(\boldsymbol\sigma-\boldsymbol\sigma^{\rm in})\mathbin\cdot\mathbf n
=\gamma\kappa\mathbf n-\nabla_s\gamma,
$$
with signs tied to the stated curvature convention. Put $C'=A\mathbf G\mathbin\cdot\mathbf n$ and $\gamma=\gamma_0-\gamma_1C'$. The constant part of the normal traction is balanced by the uniform bubble pressure. Since $\kappa=2/a$ and
$$
\nabla_s(\mathbf G\mathbin\cdot\mathbf n)
=\frac1a(I-\mathbf n\mathbf n)\mathbin\cdot\mathbf G,
$$
the remaining exterior traction is
$$
\boxed{
\boldsymbol\sigma\mathbin\cdot\mathbf n
=\frac{\gamma_1A}{a}
\left\{\mathbf G-3(\mathbf G\mathbin\cdot\mathbf n)\mathbf n\right\}}.
$$
Its resultant vanishes because
$$
\int_{S_a}\mathbf n\mathbf n\,dS
=\frac{4\pi a^2}{3}I,
$$
and therefore the integrals of the two terms cancel. This is required because the bubble and its interfacial stresses exert no external body force on the combined bubble–fluid system.
The traction is a first spherical harmonic, so the decaying, force-free <Stokes flow> has no <Stokeslet> and is generated by the indicated <Papkovich–Neuber representation>. Comparing the supplied traction
$$
\frac{3\beta}{a}
\{\mathbf G-3(\mathbf G\mathbin\cdot\mathbf n)\mathbf n\}
$$
with the capillary traction gives
$$
\boxed{\beta=\frac{\gamma_1A}{3}}.
$$
In the convention for these potentials, the normal velocity at $r=a$ is $(\beta/\mu)\mathbf G\mathbin\cdot\mathbf n$. The <kinematic boundary condition> for a translating sphere is $\mathbf u\mathbin\cdot\mathbf n=\mathbf U\mathbin\cdot\mathbf n$, so
$$
\boxed{\mathbf U=\frac{\gamma_1A}{3\mu}\mathbf G}.
$$
Linearize the surface transport equation about $C=C_0$ by writing $C=C_0+C'$ and neglecting products of small perturbations. The tangential velocity relative to the translating bubble obtained from the same potential is
$$
\mathbf u_s=-\frac{\gamma_1A}{2\mu}
(I-\mathbf n\mathbf n)\mathbin\cdot\mathbf G.
$$
Using the identities supplied in the question,
$$
\nabla_s\mathbin\cdot\mathbf u_s
=\frac{\gamma_1A}{\mu a}\mathbf G\mathbin\cdot\mathbf n,
\qquad
\nabla_s^2C'=-\frac{2A}{a^2}\mathbf G\mathbin\cdot\mathbf n.
$$
The bulk result gives $\Gamma'=\frac32a\mathbf G\mathbin\cdot\mathbf n$ on the surface. The linearized equation
$$
C_0\nabla_s\mathbin\cdot\mathbf u_s
=D_s\nabla_s^2C'-k(C'-b\Gamma')
$$
then yields
$$
A\left(k+\frac{2D_s}{a^2}
+\frac{\gamma_1C_0}{\mu a}\right)
=\frac32kab.
$$
Thus
$$
\boxed{
A=\frac{3kab/2}
{k+2D_s/a^2+\gamma_1C_0/(\mu a)}}.
$$
Increasing $k$ strengthens exchange with the imposed bulk gradient, so $A$ and $U$ increase toward a saturation value. Increasing $D_s$ smooths surface-concentration differences and decreases $U$. Increasing $C_0$ strengthens advective redistribution of the background surfactant; the resulting feedback opposes the imposed dipole, so $U$ decreases.
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