Exterior sphere Stokes solution, force and decay-class uniqueness

For any radial scalar functions $A,B$, direct Cartesian differentiation with ${\mathrm{\partial }}_{i}r=x_i\mathrm{/}r$ and ${\mathrm{\partial }}_{i}s=U_i$ gives $\mathrm{div}(AU+Bsx)=(s\mathrm{/}r)(A^{\mathrm{\prime }}+r^2{B}^{\mathrm{\prime }}+4rB)$ and $\mathrm{\Delta }(AU+Bsx)=(A^{\mathrm{\prime }\mathrm{\prime }}+2A^{\mathrm{\prime }}\mathrm{/}r+2B)U+(B^{\mathrm{\prime }\mathrm{\prime }}+6B^{\mathrm{\prime }}\mathrm{/}r)sx$. For the stated coefficients, $A^{\mathrm{\prime }}=3a\mathrm{/}(4r^2)+3a^3\mathrm{/}(4r^4)$, $A^{\mathrm{\prime }\mathrm{\prime }}=-3a\mathrm{/}(2r^3)-3a^3\mathrm{/}{r}^5$, $B^{\mathrm{\prime }}=9a\mathrm{/}(4r^4)-15a^3\mathrm{/}(4r^6)$, and $B^{\mathrm{\prime }\mathrm{\prime }}=-9a\mathrm{/}{r}^5+45a^3\mathrm{/}(2r^7)$. Substitution yields $A^{\mathrm{\prime }}+r^2{B}^{\mathrm{\prime }}+4rB=0$ and $\mathrm{\Delta }u=-3aU\mathrm{/}(2r^3)+9asx\mathrm{/}(2r^5)$. Also $\mathrm{\nabla }(s\mathrm{/}{r}^3)=U\mathrm{/}{r}^3-3sx\mathrm{/}{r}^5$, so $\mathrm{\nabla }p=\mu \mathrm{\Delta }u$ and $\mathrm{div}u=0$.

At $r=a$, $A(a)=B(a)=0$, giving $u=0$. As $r\to \mathrm{\infty }$, $A(r)-1=O(r^{-1})$ and $\mathrm{\mid }B(r)sx\mathrm{\mid }=O(r^{-1})$ uniformly because $\mathrm{\mid }s\mathrm{\mid }\le \mathrm{\mid }U\mathrm{\mid }r$; hence $u-U=O(r^{-1})$. Differentiating the displayed radial expression gives $\mathrm{\nabla }u=O(r^{-2})$, and $\mathrm{\mid }p-p_{\mathrm{\infty }}\mathrm{\mid }\le (3\mu a\mathrm{\mid }U\mathrm{\mid })\mathrm{/}(2r^2)$; these also establish the required uniform far-field limits.

Let $(\tilde{u},\tilde{p})$ be any other pair in the stated class and put $w=\tilde{u}-u$, $q=\tilde{p}-p$. Subtracting the equations gives $\mathrm{div}w=0$, $\mu \mathrm{\Delta }w=\mathrm{\nabla }q$, $w=0$ on $S_a$, and $w=O(r^{-1})$, $\mathrm{\nabla }w=O(r^{-2})$, $q=O(r^{-2})$ uniformly. For each $R>a$, apply [F2] on the bounded smooth annulus ${\mathrm{\Omega }}_{a,R}=\{a<r<R\}$ to $qw$ and to ${\sum }_i{w}_i\mathrm{\nabla }{w}_i$. The product rule and $\mathrm{div}w=0$ give ${\int }_{{\mathrm{\Omega }}_{a,R}}w\cdot \mathrm{\nabla }q={\int }_{\mathrm{\partial }{\mathrm{\Omega }}_{a,R}}qw\cdot n_{R}dS$ and ${\int }_{{\mathrm{\Omega }}_{a,R}}w\cdot \mathrm{\Delta }w={\int }_{\mathrm{\partial }{\mathrm{\Omega }}_{a,R}}w\cdot {\mathrm{\partial }}_{n_R}wdS-{\int }_{{\mathrm{\Omega }}_{a,R}}\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}^{2}dx$. Substitution into $\mu \mathrm{\Delta }w=\mathrm{\nabla }q$ yields $\mu {\int }_{{\mathrm{\Omega }}_{a,R}}\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}^{2}dx={\int }_{r=R}(\mu w\cdot {\mathrm{\partial }}_{r}w-qw\cdot n)dS$, since both inner-boundary terms vanish by $w=0$.

Write $U_t=U-(U\cdot n)n$ on $S_a$. There $A^{\mathrm{\prime }}(a)=3\mathrm{/}(2a)$, $B(a)=0$, and $B^{\mathrm{\prime }}(a)=-3\mathrm{/}(2a^3)$, so the outward radial derivative is ${\mathrm{\partial }}_{n}u=(3\mathrm{/}(2a))U_t$. Since the boundary trace is the zero vector, every tangential derivative there vanishes; thus $Du=(3\mathrm{/}(2a))U_t\otimes n$, $(Du{)}^{\mathsf{T}}n=0$, and $Dun=(3\mathrm{/}(2a))U_t$. Further, $p{\mathrm{\mid }}_{S_a}=p_{\mathrm{\infty }}-(3\mu \mathrm{/}(2a))(U\cdot n)$. It follows that $\sigma n=-p_{\mathrm{\infty }}n+(3\mu \mathrm{/}(2a))U$.

For $\mathrm{\Phi }(\theta ,\phi )=a(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$, $0\le \theta \le \pi$, $0\le \phi \le 2\pi$, direct differentiation gives tangent Gram coefficients $E=a^2$, $F=0$, $G=a^2{\sin }^2\theta$. By [F3], the surface density is $\sqrt{EG-F^2}=a^2\sin \theta$ (also $\mathrm{\mid }{\mathrm{\Phi }}_{\theta }\times {\mathrm{\Phi }}_{\phi }\mathrm{\mid }$). The spherical chart is regular in the parameter interior, with the poles and seam on its content-zero parameter boundary. By [F4] those boundary values contribute zero to the integral, so $\mathrm{area}(S_a)=a^2{\int }_0^{2\pi }{\int }_0^{\pi }\sin \theta d\theta d\phi =4\pi a^2$. The parameter change $(\theta ,\phi )\mapsto (\pi -\theta ,\phi +\pi <mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2\pi )$ has absolute Jacobian one on each half of the longitude interval, preserves the density since $\sin (\pi -\theta )=\sin \theta$, and induces the antipodal map. Thus the antipodal map $n\mapsto -n$ preserves spherical area and reverses $n$, so ${\int }_{S_a}ndS=0$. Integrating the expression in step 3.1 and using $\mathrm{area}(S_a)=4\pi a^2$ gives ${\int }_{S_a}\sigma ndS=(3\mu \mathrm{/}(2a))4\pi a^{2}U=6\pi \mu aU$.

On $r=R$ the integrand in step 2.2 is $O(R^{-3})$ and the area is $4\pi R^2$, so its integral tends to zero as $R\to \mathrm{\infty }$. The left side is nonnegative and nondecreasing in $R$; since it equals a quantity tending to zero, it is zero for every $R>a$. Continuity then gives $\mathrm{\nabla }w=0$ throughout the connected exterior, so $w$ is constant there; its far-field limit is zero, hence $w=0$. The equation now gives $\mathrm{\nabla }q=0$, so $q$ is constant, and its zero far-field limit gives $q=0$.