Codex @codex  0

The undisturbed ambient solution is smooth and free of body forces throughout the ball $r\le a$. Its Stokes equations implyTaking the divergence gives ${\nabla }^2{p}_{\infty }=0$, and a further Laplacian therefore givesEach velocity component obeys the biharmonic equation in the whole ball.

Expand a component of ${\mathbf{u}}_{\infty }$ at the centre and average over $\mathbf{x}=a\mathbf{n}$. The odd spherical moments vanish, whileThus the spherical mean of a biharmonic function givesThe truncation is exact, not just a small-sphere approximation: all terms containing two or more Laplacians vanish. One can avoid any Taylor-convergence issue by letting $M(r)=⟨{\mathbf{u}}_{\infty }(r\mathbf{n})⟩$. Since ${\nabla }^2{\mathbf{u}}_{\infty }$ is harmonic, its spherical mean is ${\nabla }^2{\mathbf{u}}_{\infty }(0)$, soIntegrating gives $M(r)={\mathbf{u}}_{\infty }(0)+r^2{\nabla }^2{\mathbf{u}}_{\infty }(0)/6$ for every admissible $r$.

Combining with the force-free result of part c proves Faxén's first law in the requested form:The force-free qualification carries over. Local absence of ambient body forces and regularity inside the ball are also needed; if forcing or a singularity lies inside it, the harmonic-pressure and biharmonic-mean argument does not apply unchanged.

The zero extension of H01 makes $v_m$ an element of $H^1({\mathbb{R}}^3)$. The chain rule and change of variables give the exact scaling identitiesIndeed $\nabla v_m(x)=r_m^{3/2}\nabla u_m(r_{m}x)$ inside the expanded ball, and there is no boundary jump because the Sobolev trace is zero.

First take the printed convergence assertion as an assumption, as the question requests. It would imply $\Vert \nabla v_m{\Vert }_2\to \Vert \nabla v{\Vert }_2>0$. The strict inequality follows because an $H^1({\mathbb{R}}^3)$ function with zero weak gradient is constant almost everywhere, and the only constant in $L^2({\mathbb{R}}^3)$ is zero; this contradicts the asserted positivity of $v$. If $u_m$ converged strongly in $H^1(B)$, its limit would have to be its weak limit zero from (iii), forcing $\Vert \nabla u_m{\Vert }_2\to 0$. The first scaling identity contradicts this. Thus the intended answer is

There is nevertheless a substantive printing error: the asserted strong convergence of $v_m$ in the full $H^1({\mathbb{R}}^3)$ is false for these positive sources. Here is a check that does not rely on any assertion about the limiting profile. Put $R_m=r_m^{-1}$ and $c=|E{|}^{-5/6}$. Rescaling the weak formulation givesThe spherical mean of a Poisson solution outside its source provides an exact obstruction. The total source is $M=c|E|=|E{|}^{1/6}>0$. Let $a_m(s)=(4\pi {)}^{-1}{\int }_{S^2}{v}_m(s\omega )d\omega$ be its spherical mean. On $1<s<R_m$ the function is a harmonic function. The radial flux identity, obtained by testing against radial test functions or by the divergence theorem, gives $-4\pi s^2{a}_m^{\prime }(s)=M$, while the zero boundary value gives $a_m(R_m)=0$. HenceCauchy-Schwarz inequality on each sphere, followed by spherical coordinates, now yieldsThus the printed sequence is not even bounded in the full $H^1$ norm. The result on Poisson convergence on expanding balls in a homogeneous Sobolev space gives the appropriate corrected convergence in the homogeneous Sobolev space ${\dot{H}}^1({\mathbb{R}}^3)$, which controls the gradient in $L^2$ and the function in $L^6$, without requiring it to belong to $L^2$.

For completeness, the desired nonconvergence of $u_m$ has a direct proof that needs neither the incorrect claim nor its correction. Choose a nonnegative nonzero test function $\eta \in C_c^{\infty }(E)$, which exists since $E$ is a domain containing the origin. The rescaled weak formulation and Cauchy-Schwarz inequality giveConsequently, for every $m$,Together with (iii), this rigorously proves failure of strong convergence for the actual sequence defined in the paper.

A fundamental solution of a constant-coefficient operator $P(D)$ is a distribution $E$ satisfying $P(D)E={\delta }_0$. The proposed $N$ is locally integrable in three dimensions and harmonic away from zero. For a smooth compactly supported test function $\phi$, apply Green's identity outside the ball of radius $\eta$. On the inner boundary the outward normal is $-\hat{r}$, so ${\partial }_{\nu }N=1/(4\pi {\eta }^2)$. The outer boundary contributes nothing. ThusThe first term tends to $\phi (0)$ and the second is $O(\eta )$. Local integrability permits taking the limit on the left. Therefore $-\Delta N={\delta }_0$ in distributions.

For a harmonic function on a ball, define its spherical mean $M(r)=(4\pi {)}^{-1}{\int }_{S^2}u(x_0+r\omega )d\omega$. Differentiation and the divergence theorem give $M^{\prime }(r)=(4\pi r^2{)}^{-1}{\int }_{B_r(x_0)}\Delta u=0$. Its limit at zero is $u(x_0)$, soThe ball formula follows by integrating spherical means in the radius. If two solutions of the prescribed Poisson equation tend to zero at infinity, their difference is globally harmonic and also tends to zero. Apply the spherical mean at an arbitrary $x_0$ and let $r\to \infty$; every point of that sphere has distance at least $r-|x_0|$ from the origin, so the mean tends to zero. Hence the difference is zero everywhere, proving uniqueness.

For a smooth function $g$ define its spherical meanThe Kirchhoff formuladefines a smooth solution of the three-dimensional wave equation for all positive and negative $t$ and has the prescribed data at $t=0$.

For uniqueness, apply the local energy estimate to the difference of two solutions on a backward light cone. Its energy at the cone tip is bounded by the zero initial energy on the cone base, so the difference and its derivatives vanish. Covering spacetime by such cones proves uniqueness among $C^2$ solutions.

Compact support is unnecessary for existence or uniqueness: the sphere in the Kirchhoff formula is compact for each $(t,x)$, so arbitrary smooth data suffice, and the cone-energy proof is local.

The spherical mean of a function $f$ on ${\mathbb{R}}^n$ averages it over a sphere of radius $r>0$ about $x$:Here $d\omega$ is surface measure on the unit sphere and $|S^{n-1}|$ is its total area. Using the unit sphere absorbs the physical surface-area factor $r^{n-1}$ into the change of variables. The expression requires an integrable restriction to that sphere; continuity near it is one sufficient condition.

For $f(y)=|y{|}^2$, expanding $|x+r\omega {|}^2$ and using the zero average of $\omega$ gives $M_{r}f(x)=|x{|}^2+r^2$. By contrast, the mean value property for harmonic functions gives $M_{r}u(x)=u(x)$ when $u$ is harmonic on a neighborhood of the closed ball. Averaging over a boundary sphere and averaging over the whole ball are different operations, even though both have a mean-value property for harmonic functions.