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For a spherically symmetric function $\Phi (r)$, the Laplacian in spherical coordinates isIntegrating from zero to $r$ and using regularity at the origin gives

For the stated density, integration and the boundary condition $\Phi (a)=0$ giveThe two pieces and their first derivatives agree at $r=b$. If another solution existed, its difference from $\Phi$ would be harmonic with zero boundary data, so part (a) proves Uniqueness of the Dirichlet problem.

On the shell $U(b,a)$, normalize the outer radial part to obtain the harmonic functionBy the Dirichlet principle, this function minimizes the energy among all $w$ with the same boundary values. Sinceits energy isTherefore

The stationary Schrödinger equation isThe Laplacian represents the kinetic-energy operator. For a spherically symmetric function, $\Delta \psi ={\psi }^{\prime \prime }+2{\psi }^{\prime }/r$. The attractive Coulomb potential of the nucleus is $V=-q^2/(4\pi {ϵ}_{0}r)$; $E$ is the energy eigenvalue. Multiplication by $-2m/{\hslash }^2$ produces the displayed radial equation with Bohr radiusFor a normalizable bound state, write ${\kappa }^2=-2mE/{\hslash }^2$ and take $\lambda >0$. Substitution of $\psi =(A+Br)e^{-\lambda r}$ ingives, after division by the exponential, the coefficientsIf $B=0$ and $A\ne 0$, these force $\lambda =1/a$ and $\kappa =\lambda$. If $B\ne 0$, the first coefficient gives ${\kappa }^2={\lambda }^2$, the second gives $\lambda =1/(2a)$, and the third gives $B=-A/(2a)$. Up to nonzero scalar factors, the two radial bound states and their energies are consequentlyThey are linearly independent as functions and correspond to different energy eigenvalues. Strictly, they are not two solutions of one fixed-energy differential equation: each is a solution at its own energy.

For the higher-energy state, the spherical Born rule includes the radial factor $4\pi r^2$. Its normalization constant and the angular factor cancel in the ratioThe given integral formula makes the denominator $8a^5$ and the numerator $8a^5-21a^5/e$. Thus

For a spherically symmetric function, $v=ru$ reduces the radial wave equation to $v_{tt}=c^2{v}_{rr}$. A time-harmonic separated solution has radial numerator $C\sin (kr)+D\cos (kr)$, with $k=\omega /c$. Finiteness at the origin excludes $D$, and the prescribed finite origin value fixes $C$. This is a standing combination of incoming and outgoing waves. A purely outgoing point-source wave instead has a $1/r$ singularity and must be specified by its source strength, not by a finite value at the origin.