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The Dirichlet boundary condition gives $B=-a^2/6$, so the unique radial solution isIts radial Laplacian is one, it is smooth at the origin, and its boundary value is zero. The uniqueness result in part (i) shows that this is also the only solution among all classical functions on the ball, not merely among spherically symmetric ones.

For a spherically symmetric function in three dimensions, the radial Laplacian is ${\nabla }^{2}u=r^{-2}(r^2{u}^{\prime }{)}^{\prime }$. On $0<r\le a$, integrate $(r^2{u}^{\prime }{)}^{\prime }=r^2$ twice to obtainThe region includes the origin. A classical solution must be regular there, so $A=0$: the $1/r$ term is singular and is not a solution of the stated full-ball problem. Hence the complete family on the ball isThe boundary cases below use this regular family. Allowing punctured-ball solutions would change the problem and the compatibility calculation.

Let ${\tau }_r=\inf \{t\ge 0:|B_t|=r\}$, ${\tau }_R=\inf \{t\ge 0:|B_t|=R\}$, and $\tau ={\tau }_r\wedge {\tau }_R$. The function $u(y)=|y{|}^{-1}$ is harmonic away from the origin: its radial Laplacian in dimension three isThe Itô formula therefore makes $u(B_{t\wedge \tau })$ a local martingale, bounded between $1/R$ and $1/r$, hence a true martingale. The annulus exit time is finite almost surely. For instance, exit from the containing open ball is finite because from every point in the open ball a sufficiently large Gaussian vector increment has a uniform positive probability of leaving it in unit time; iterate the geometric tail bound from a uniform escape probability.

The optional stopping theorem at $t\wedge \tau$, followed by the dominated convergence theorem, gives $\mathbb{E}u(B_{\tau })=u(x)$. By continuity, there is no radial overshoot and the two exit spheres cannot be hit simultaneously. Writing $p={\mathbb{P}}_x({\tau }_r<{\tau }_R)$ givesAll harmonicity and boundedness statements apply only in the annulus, which stays away from the singularity at zero.

The data are rotationally symmetric, so seek a radial solution. The radial Laplacian in two dimensions givesTwice integrating,Regularity at the center forces $C=0$, and $u(1)=1$ gives $D=0$. Thus

For a radial function in three dimensions, the radial Laplacian givesHere regularity at zero selectsA further integration giveswhose apparent singularity is removable, with limit $3+C$ at zero. Since the nonconstant term has value $4e-8$ at $r=1$, the boundary condition $\varphi (1)=3$ gives $C=11-4e$. Therefore

For a radial function $u=u(r)$ on ${\mathbb{R}}^d$, the radial Laplacian isThe problem therefore becomesThe equation has the scaling symmetryChoosing $\lambda =u(0{)}^{-(p-1)/2}$ normalizes any positive solution to $u_{\lambda }(0)=1$.