= Solution
In the <surfactant transport with exchange relaxation>, $DC/Dt$ is the <material derivative> following an interfacial material element. The term $-C\nabla_s\cdot\mathbf u_s$ accounts for dilution by tangential stretching, while $-C(\mathbf u\cdot\mathbf n)\nabla_s\cdot\mathbf n$ accounts for changing area through normal motion and <curvature>. The <surface diffusion> term redistributes <surfactant> down concentration gradients. The exchange term relaxes its concentration towards $C_0$, through adsorption or desorption from a surrounding reservoir, on time scale $k^{-1}$.
In the bubble frame the spherical interface is stationary and impermeable. A steady <material derivative> is $\mathbf u_s\cdot\nabla_s C'$. For a linear response in the rising speed this and $C'\nabla_s\cdot\mathbf u_s$ are second order, whereas stretching of $C_0$ is first order. Thus
$$
\boxed{(D_s\Delta_s-k)C'=C_0\nabla_s\cdot\mathbf u_s.}
$$
Rotational symmetry, linear response to $\mathbf U$, and zero normal velocity allow only $\mathbf u_s=A P\mathbf U$, where $P=I-\mathbf n\mathbf n$ is the <surface tangent projector>. This is the dipolar, axisymmetric sector; axisymmetry alone, without the linear-response assumption, would allow other angular dependence.
On a sphere $\mathbf n=\mathbf x/r$ and $\partial_j n_i=(\delta_{ij}-n_i n_j)/r$. Applying the <surface tangent projector> to the derivative gives the <sphere surface derivative identities>
$$
\boxed{\nabla_s\mathbf n=P/a,\quad\nabla_s\cdot\mathbf n=2/a,\quad\Delta_s\mathbf n=-2\mathbf n/a^2,\quad\nabla_s\cdot(P\mathbf U)=-2(\mathbf U\cdot\mathbf n)/a.}
$$
For example, taking a second <surface divergence> of the first identity yields the third. Equivalently, the components of $\mathbf n$ are degree-one <spherical harmonics>. Hence a <dipolar surfactant distribution> solves the linear equation:
$$
\boxed{C'=B\mathbf U\cdot\mathbf n,\qquad B=\frac{2A C_0 a}{ka^2+2D_s}.}
$$
The required smallness condition is \b[$2|A|Ua/(ka^2+2D_s)\ll1$]. <Surface diffusion> or exchange must smooth the concentration faster than it is redistributed by the actual interfacial velocity. We require $k\geq0$, $D_s\geq0$ and $ka^2+2D_s>0$; with neither process there is no such steady linear balance at nonzero $A$. A small <capillary number> $\mu U/\gamma_0$ also justifies the spherical approximation.
Take the normal from the inner bubble to the outer liquid, define $[\sigma]_-^+=\sigma_+-\sigma_-$, and set $\kappa=\nabla_s\cdot\mathbf n$. The <interfacial stress balance with variable surface tension> is
$$
\boxed{[\boldsymbol\sigma]_-^+\mathbf n=\gamma\kappa\mathbf n-\nabla_s\gamma.}
$$
Because $\nabla_s C'=B P\mathbf U/a$ and $\gamma=\gamma_0-\gamma_1C'$, its tangential part becomes
$$
\boxed{P[\boldsymbol\sigma]_-^+\mathbf n=\frac{6\mu A\lambda}{a}P\mathbf U,\qquad\lambda=\frac{\gamma_1 C_0 a}{3\mu(ka^2+2D_s)}.}
$$
This is a <Marangoni stress>: surface-tension gradients oppose the clean-bubble circulation.
Use the <Unscaled Papkovich–Neuber representation>, $\mathbf u=\nabla(\mathbf x\cdot\boldsymbol\Phi+\chi)-2\boldsymbol\Phi$ and $p=2\mu\nabla\cdot\boldsymbol\Phi$. The constant vector potential gives $\mathbf u\to-\mathbf U$. In the decaying, linear axisymmetric sector, a vector monopole and scalar dipole provide the two constants needed at the sphere; other equivalent potential choices amount to a representation freedom. The PDF potential contains the factor $a$ omitted in the TeX aid. With that factor restored, the exterior velocity is
$$
\mathbf u=-\mathbf U-\alpha a\left(\frac{\mathbf U}{r}+\frac{(\mathbf U\cdot\mathbf x)\mathbf x}{r^3}\right)+\beta a^3\left(-\frac{\mathbf U}{r^3}+\frac{3(\mathbf U\cdot\mathbf x)\mathbf x}{r^5}\right),\qquad p=-\frac{2\mu\alpha a\,\mathbf U\cdot\mathbf x}{r^3}.
$$
At the interface, no penetration gives $\beta-\alpha=1/2$, and the tangential velocity gives $A=-(1+\alpha+\beta)$. \b[Using the traction coefficient printed in equation (3)], the tangential stress condition gives $2\beta=A\lambda$. The requested algebraic answers are therefore
$$
\boxed{A_{\rm printed}=-\frac1{2(1+\lambda)},\quad\alpha_{\rm printed}=-\frac{2+3\lambda}{4(1+\lambda)},\quad\beta_{\rm printed}=-\frac{\lambda}{4(1+\lambda)}.}
$$
\b[There is a factor-of-two error in the PDF traction.] Direct differentiation of the preceding velocity with the <Newtonian fluid stress tensor> gives
$$
\boldsymbol\sigma\mathbf n=\frac{6\mu}{a}\left[\beta\mathbf U+(\alpha-3\beta)(\mathbf U\cdot\mathbf n)\mathbf n\right],
$$
rather than its printed coefficient $12\mu/a$. In particular the tangential <traction> is $6\mu\beta P\mathbf U/a$. Keeping the same definition of $\lambda$, the physically consistent <translating surfactant-coated bubble> instead has $\beta=A\lambda$ and
$$
\boxed{A=-\frac1{2(1+2\lambda)},\quad\alpha=-\frac{1+3\lambda}{2(1+2\lambda)},\quad\beta=-\frac{\lambda}{2(1+2\lambda)}.}
$$
The two sets must not be silently conflated. As a useful check, the no-slip limit has the usual total <Stokes flow> drag $6\pi\mu aU$ with the corrected <traction>; the printed <traction> would double it.
Both versions give $A=-1/2$, $\alpha=-1/2$, $\beta=0$ when $\lambda\to0$: the interface is tangentially stress-free and circulates as a clean inviscid bubble. For $\lambda\to\infty$, $A\to0$, $\alpha\to-3/4$, $\beta\to-1/4$: the interface is immobile in the bubble frame and behaves as a no-slip sphere. The <Marangoni stress> remains finite because $\lambda A$ has a nonzero limit. The concentration smallness condition still has to hold; large $\lambda$ is not by itself a complete linearization criterion.
The normal component is balanced by the inner gas pressure and the normal capillary stress $\gamma\kappa$, including the variation of $\gamma$, together with the <hydrostatic pressure> difference responsible for buoyancy. A spatially uniform inner pressure alone cannot balance the dipolar <traction>. To make the <normal stress balance on a translating bubble> explicit, let $\Delta\rho_b$ be the outer-minus-inner <mass density>. After removing the equilibrium Laplace pressure, the corrected outer normal <traction> has dynamic part $-3\mu(1+\lambda)(\mathbf U\cdot\mathbf n)/[a(1+2\lambda)]$. Its degree-one balance is
$$
\Delta\rho_bga\cos\vartheta-\frac{3\mu(1+\lambda)}{a(1+2\lambda)}U\cos\vartheta=-\frac{2\gamma_1 B}{a}U\cos\vartheta.
$$
Thus, if a terminal speed is desired, \b[$U=\Delta\rho_bga^2(1+2\lambda)/[3\mu(1+3\lambda)]$], interpolating between the clean-bubble and rigid-sphere speeds. A degree-one shape displacement is merely a translation of the sphere, so the balance determines its rise speed, rather than a dipolar shape distortion.
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