 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 equationAt $r=a$,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. Rotational symmetry about $\mathbf{G}$ and the imposed far-field gradient select the dipolar formWhen $kab\ll D$, 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

For an interface with unit normal directed from the bubble into the exterior liquid, the interfacial stress balance with variable surface tension may be writtenwith signs tied to the stated curvature convention. Put $C^{\prime }=A\mathbf{G}\cdot \mathbf{n}$ and $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$. The constant part of the normal traction is balanced by the uniform bubble pressure. Since $\kappa =2/a$ andthe remaining exterior traction isIts resultant vanishes becauseand 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 tractionwith the capillary traction givesIn the convention for these potentials, the normal velocity at $r=a$ is $(\beta /\mu )\mathbf{G}\cdot \mathbf{n}$. The kinematic boundary condition for a translating sphere is $\mathbf{u}\cdot \mathbf{n}=\mathbf{U}\cdot \mathbf{n}$, so

Linearize the surface transport equation about $C=C_0$ by writing $C=C_0+C^{\prime }$ and neglecting products of small perturbations. The tangential velocity relative to the translating bubble obtained from the same potential isUsing the identities supplied in the question,The bulk result gives ${\Gamma }^{\prime }=\frac{3}{2}a\mathbf{G}\cdot \mathbf{n}$ on the surface. The linearized equationthen yieldsThus

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.