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Since $q=-\kappa \nabla T$ and $\nabla \cdot q=H$, the divergence theorem givesSpherical symmetry gives $(r^2\kappa T^{\prime }{)}^{\prime }=-r^{2}H$. Regularity at zero and decay at infinity yieldFor $\alpha \le -1$, the exterior solution cannot tend to zero: it is logarithmic at $-1$ and grows below it.

On the exterior domain, the energy-minimizing solution with boundary value one is $\varphi =r^{-\alpha -1}$. Applying the Dirichlet principle to any admissible $w$ gives

The identity follows immediately from the product rule:If ${\varphi }_1,{\varphi }_2$ have the same boundary data, put $\psi ={\varphi }_1-{\varphi }_2$. Integrating the identity with $\varphi =\psi$ givesbecause the volume equation and boundary term vanish. Positivity of $\kappa$ makes $\psi$ constant, and its boundary value makes it zero, proving uniqueness.

For any $w=\varphi +\psi$ with $\psi =0$ on the boundary, expansion givesbecause the cross term vanishes by the same integration by parts. This proves the inequality and the Dirichlet principle.

Given $Z_i=n$, the probability of no friendship among people sharing day $i$ is $2^{-n(n-1)/2}$. The day counts are independent, so

By Poisson thinning, allocating a Poisson$(365)$ population independently and uniformly among $365$ days gives independent Poisson$(1)$ counts. Hence

Conditional on $N=n$, there are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairs, and each is both friendly and birthday-matched with probability $1/(2\cdot 365)$. Since $\mathbb{E}[N(N-1)]=365^2$ for a Poisson variable,

The system matrix has eigenvalues $2$ and $3$, with eigenvectors in the directions $y=-x$ and $y=-x/2$, respectively. Both eigenvalues are positive, so the origin is an unstable node. The two eigendirections are straight trajectories directed away from the origin. Every other trajectory also moves outward, tangent near the origin in backward time to the faster eigendirection and asymptotic in forward time to the eigenvalue-$3$ direction.

For $t<0$ both inputs vanish. Eliminating $y$ gives $x^{\prime \prime }+4x=0$, and the data at $t=-\pi$ giveFor all $t$, differentiating the first equation and using the second givesThus $x$ is continuous at zero while $x^{\prime }$ jumps by one. For $t>0$ the resulting solution isSince $\cos 2t=1-2{\sin }^{2}t$, this iswhereThus $a=1/2$, $b=1$, and the sketch of $q$ is a downward unit step at $t=\pi$.

For $x>0$, division by $x$ givesAn integrating factor is $e^{-x^2}/x$, soEvery choice exceptgrows like $x{e}^{x^2}$. The bounded solution selected by a terminal condition is thereforeIntegration by parts, or the Gaussian-tail asymptotic, givesIt starts from $0$ at $x=0$, remains negative, and decreases monotonically toward the horizontal asymptote $-1/2$.

LetThe integrating factor identity $(Iy{)}^{\prime }=If$ givesIf $y_1,y_2$ are distinct solutions, their difference is a nonzero homogeneous solution, so every solution is

Let $E_{\ell }$ be the event that the block beginning at position $\ell$, $1\le \ell \le k+1$, is a spalindrome. Each has probability $(10{)}_k/10^{2k}$. If two starting positions differ by $d<k$, the first half of the later block contains the middle digit of the earlier palindrome twice, contradicting distinctness. Thus the only possible overlap is $E_1\cap E_{k+1}$.

That intersection consists of words $A{A}^{\mathrm{rev}}A$, where $A$ has $k$ distinct digits, and hence has probability $(10{)}_k/10^{3k}$. This overlap structure of digit spalindromes makes inclusion--exclusion give

The first $k$ digits must be distinct and may be chosen inways; the remaining digits are then forced in reverse order. Thus, for $k\le 10$,and the probability is zero for $k>10$.

For any $t>0$, the exponential Markov bound and part (b) giveThe minimum occurs at $t=v$, so the Gaussian tail bound is

Completing the square gives