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Differentiating the axisymmetric stream function givesThe normal kinematic boundary condition fixes $C=Ua/2$, soBoth velocity components vanish at infinity, and the axisymmetric stream function is regular at the polar axes. The tangential component of the strain-rate tensor vanishes everywhere:Thus the stress-free boundary condition is satisfied. The vorticity is ${\omega }_{\varphi }=2C\sin \theta /r^2=Ua\sin \theta /r^2$, and the supplied dipolar pressure is consistent with the Stokes equation.

Use the Newtonian fluid stress tensor $\mathbf{\sigma }=-p\mathbf{I}+2\eta \mathbf{e}$. At the bubble surface,The vertical force exerted by the liquid is the integral of the vertical traction:Since the integral is $2/3$, the drag is downward and has magnitudeThe constant pressure integrates to zero force. The pressure perturbation contributes $4\pi \eta aU/3$ and the normal viscous stress contributes $8\pi \eta aU/3$; there is no tangential traction contribution.

For the free-interface qualification in part (b), the calculated normal traction at $\theta =0$ and $\theta =\pi$ differs by $6\eta U/a$. It cannot equal a single uniform gas pressure if every restoring effect is discarded. Restoring the external-minus-internal variation of hydrostatic pressure supplies a term $({\rho }_w-{\rho }_a)ga\cos \theta$. It cancels the dynamic normal traction at the terminal velocity $U=({\rho }_w-{\rho }_a)g{a}^2/(3\eta )$, consistent with a spherical clean bubble without surface tension. This distinguishes the drag calculation from the extra conditions needed for a freely rising bubble.

Take the upward direction as ${\mathbf{e}}_z$, with polar angle $\theta$ measured from it. The liquid is quiescent at infinity. On a translating clean inviscid bubble, the kinematic boundary condition is $(\mathbf{u}-U{\mathbf{e}}_z)\cdot \mathbf{n}=0$, and the stress-free boundary condition is zero tangential traction. The normal traction must balance uniform gas pressure and any constant-curvature surface-tension jump.

For a vertical Stokeslet of strength $F$, the radial velocity is $u_r=F\cos \theta /(4\pi \mu r)$. Its traction on a concentric sphere is purely normal, so zero tangential traction is automatic. The kinematic condition at $r=a$ givesThe singularity is only a device for generating the exterior solution; there is no liquid point force inside the actual bubble. The integrated dynamic traction on the bubble is $-F{\mathbf{e}}_z$, giving the clean-bubble Stokes drag. The upward buoyancy is $4\pi \rho g{a}^3/3$, hence force balance gives

The normal stress also shows that surface tension is not required for this particular spherical solution. Let the ambient hydrostatic pressure at the bubble centre be $p_c$, so on its surface $p_h=p_c-\rho ga\cos \theta$. Adding the Stokeslet's dynamic normal traction giveswhere the terminal buoyancy value of $F$ cancels the dipole. This is the spherical clean bubble without surface tension balance: an initially spherical bubble can remain spherical in this ideal steady Stokes solution even when surface tension is zero. With uniform surface tension $\gamma$, choose gas pressure $p_b=p_c+2\gamma /a$; with $\gamma =0$, $p_b=p_c$. This proves compatibility of the spherical shape, not stability of arbitrary deformations or validity when inertia matters.