= Solution
The <material derivative> follows <surfactant> carried by the moving interface. The term $-C\nabla_s\cdot u_s$ is dilution or concentration by tangential area expansion or contraction. The term $-C(u\cdot n)\nabla_s\cdot n$ is the additional area change produced by normal motion of a curved interface. The term $D_s\nabla_s^2C$ represents <surface diffusion>, while $-k(C-C_0)$ relaxes the concentration towards a reservoir value through exchange. All these terms refer to concentration per <surface area>, as in <surfactant transport with exchange relaxation>.
In the steady bubble frame, the spherical interface has $u\cdot n=0$. Write $C=C_0+C'$ with uniform $C_0$ and regard both $u$ and $C'$ as first order in the rising speed. Advection of $C'$ and its product with surface dilation are second order. Hence the linear steady equation is
$$
\boxed{D_s\nabla_s^2C'-kC'=C_0\nabla_s\cdot u_s.}
$$
Spherical symmetry and linearity leave only the tangential projection of the imposed translation vector: $u_s=A I_sU$. An azimuthal term $n\times U$ would change sign under reflection and is excluded for this non-chiral rising bubble. The degree-one forcing does not excite other spherical-harmonic degrees in the linear problem.
On $r=a$, differentiating $n=x/r$ gives
$$
\boxed{(\nabla_s n)_{ij}=\frac{\delta_{ij}-n_in_j}{a}=\frac{(I_s)_{ij}}a,\qquad \nabla_s\cdot n=\frac2a.}
$$
Each component of $n$ is a degree-one <spherical harmonic>, so
$$
\boxed{\nabla_s^2n=-\frac{2n}{a^2}.}
$$
One may also obtain the coefficient from $n\cdot n=1$: its <surface Laplacian> gives $n\cdot\nabla_s^2n=-|\nabla_sn|^2=-2/a^2$, while rotational symmetry makes the vector Laplacian parallel to $n$. Expanding $I_sU=U-n(U\cdot n)$ then yields
$$
\boxed{\nabla_s\cdot(I_sU)=-\frac2aU\cdot n.}
$$
The candidate $C'=B U\cdot n$ therefore solves the linear transport equation if
$$
-\left(k+\frac{2D_s}{a^2}\right)B=-\frac{2AC_0}{a},\qquad
\boxed{B=\frac{2AC_0a}{ka^2+2D_s}.}
$$
Its spherical mean is zero. This is the <dipolar surfactant distribution>. Assuming $ka^2+2D_s>0$, its consistency condition is
$$
\boxed{\frac{\max|C'|}{C_0}=\frac{2|A||U|a}{ka^2+2D_s}\ll1.}
$$
Diffusion or exchange must restore uniform concentration faster than the interfacial flow redistributes it: this holds with a small surface <Péclet number>, fast exchange, or sufficiently immobilized surface <velocity>. These same conditions justify neglecting the perturbation's advection. Small <capillary number> and sufficiently small buoyancy-induced shape distortion are separate requirements for a spherical interface.
Take $+$ to mean the outer liquid, $-$ the bubble interior and $n$ outward from the bubble. The general <interfacial stress balance with variable surface tension> is
$$
\boxed{(\sigma^+-\sigma^-)n=\gamma\kappa n-\nabla_s\gamma,\qquad \kappa=\nabla_s\cdot n.}
$$
For $\gamma=\gamma_0-\gamma_1C'$, its tangential part is $-\nabla_s\gamma=\gamma_1 B I_sU/a$. Thus the requested form is
$$
\boxed{[I_s\sigma n]^+_-=\frac{6\mu A\lambda}{a}I_sU,\qquad
\lambda=\frac{\gamma_1C_0a}{3\mu(ka^2+2D_s)}.}
$$
This positive parameter measures the strength of the restoring <Marangoni stress> relative to viscous <stress>, including the relaxation of concentration by diffusion and exchange.
The far <velocity> $-U$ is produced by the constant harmonic vector potential $\Phi=U$. The two independent decaying degree-one modes are the force-carrying vector potential $\alpha aU/r$ and the scalar dipole $\chi=\beta a^3U\cdot\nabla(1/r)$. They are harmonic outside the bubble and have precisely the symmetry of uniform translation. The <Unscaled Papkovich–Neuber representation> gives, at arbitrary radius,
$$
u=-\left(1+\frac{\alpha a}{r}+\frac{\beta a^3}{r^3}\right)U
+\left(-\frac{\alpha a}{r}+\frac{3\beta a^3}{r^3}\right)(U\cdot n)n,
\qquad p=-\frac{2\mu\alpha a}{r^2}U\cdot n.
$$
The <velocity> at $r=a$ agrees with the printed expression. However, the original PDF's <traction> formula has a factor-of-two error. Direct differentiation with $\sigma=-pI+\mu(\nabla u+\nabla u^T)$ gives
$$
\boxed{\sigma n=\frac{6\mu}{a}\{\beta U+(\alpha-3\beta)(U\cdot n)n\},}
$$
not the printed prefactor $12\mu/a$. This is <traction of translating-sphere Papkovich–Neuber potentials>. The $\alpha$ term follows immediately from the <Stokeslet> <stress>, and the scalar-potential part gives $6\mu\beta[U-3(U\cdot n)n]/a$; neither contribution contains the extra factor two.
With this consistent <traction>, no penetration requires $\beta-\alpha=1/2$, tangential <velocity> requires $A=-(1+\alpha+\beta)$, and tangential <stress> balance requires $\beta=A\lambda$. Solving,
$$
\boxed{A=-\frac1{2(1+2\lambda)},\qquad
\beta=-\frac{\lambda}{2(1+2\lambda)},\qquad
\alpha=-\frac{1+3\lambda}{2(1+2\lambda)}.}
$$
For comparison, taking the printed $12\mu/a$ literally would instead give
$$
A_{\rm printed}=-\frac1{2(1+\lambda)},\qquad
\beta_{\rm printed}=-\frac{\lambda}{4(1+\lambda)},\qquad
\alpha_{\rm printed}=-\frac{2+3\lambda}{4(1+\lambda)}.
$$
Those solve the supplied algebraic pair but not the Newtonian <stress> calculated from the supplied potentials. The factor-six result is the physically consistent <translating surfactant-coated bubble> solution.
As $\lambda\to0$, $A\to-1/2$ and $\beta\to0$: \b[the clean bubble has a mobile interface and zero tangential <traction>]. As $\lambda\to\infty$, $A\to0$, $\alpha\to-3/4$ and $\beta\to-1/4$: \b[the interface becomes immobilized and the exterior flow approaches that past a no-slip rigid sphere]. The limiting tangential <traction> is $-3\mu I_sU/(2a)$, a useful check on the correct normalization.
The normal dynamic <traction> is
$$
n\cdot\sigma n=\frac{6\mu}{a}(\alpha-2\beta)U\cdot n
=-\frac{3\mu(1+\lambda)}{a(1+2\lambda)}U\cdot n,
$$
apart from an arbitrary constant <pressure>. In the <normal stress balance on a translating bubble>, its degree-one part is balanced by the buoyant <hydrostatic pressure> difference and by the variation $-2\gamma_1C'/a$ of normal capillary <stress>. A uniform internal gas <pressure> balances the constant Laplace-pressure term but cannot by itself balance this angular variation. If $U$ is upwards and $\Delta\rho$ is liquid minus bubble <mass density>, the degree-one balance can be written
$$
n\cdot\sigma_{\rm dyn}n+\Delta\rho g a\,n_z=-\frac{2\gamma_1 B}{a}U n_z.
$$
It fixes $U=\Delta\rho g a^2(1+2\lambda)/[3\mu(1+3\lambda)]$, tending to the clean-bubble and rigid-sphere rise speeds. A degree-one radial displacement translates a sphere rather than changing its curvature, so a supposed dipolar shape distortion cannot replace this <force> balance.
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