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The radial Poisson equation with constant source one has $\psi =r^2/6+a+b/r$. The boundary conditions giveSubtracting gives $b=1$, then $a=-1/6$. HenceIts radial Laplacian is one and both endpoint values are one. The prescribed values ensure continuity with the adjacent region solutions, but do not impose derivative continuity.

For a real-valued function on a bounded regular region, apply the divergence theorem to $\varphi \nabla \varphi$. The product rule yields $\nabla \cdot (\varphi \nabla \varphi )=|\nabla \varphi {|}^2+\varphi {\nabla }^2\varphi$. The prescribed zero boundary values and Laplace equation therefore implyThe continuous nonnegative integrand must vanish everywhere. Hence $\varphi$ is constant on each connected component, and its boundary value makes each constant zero. This proves zero-boundary harmonic uniqueness. If complex-valued functions are allowed, apply the same argument separately to their real and imaginary parts. If ${\psi }_1,{\psi }_2$ solve the same Poisson equation and Dirichlet boundary data, their difference is harmonic with zero boundary value, so is zero. Thus there is at most one solution. An exterior unbounded region would require an additional condition at infinity; the preceding bounded-domain proof does not assert uniqueness without one.

For the radial calculation use Cartesian coordinates with $r=(x_i{x}_i{)}^{1/2}>0$. The chain rule gives ${\partial }_{i}r=x_i/r$ andDifferentiate again and sum the three coordinates:This derives the radial Laplacian directly in Cartesian coordinates. For ${\nabla }^2\psi =c$, set $u=r\psi$, so $u^{\prime \prime }=cr$. Integrating twice gives $u=c{r}^3/6+ar+b$, and consequentlyIf the solution must extend boundedly through the origin, $b=0$. The remaining polynomial is smooth there.

A martingale describes an observable after its predictable drift has been removed. For a Markov chain the drift is the change produced by its transition operator; for Brownian motion it is one half of the Laplacian. Requiring the resulting processes to be martingales for a sufficiently rich class of observables determines the transition law itself. The same identities, combined with optional stopping, determine hitting probabilities, expected exit times and recurrence or transience. The duplicated Q5 paragraph in the PDF is a repeated printing of this one essay request.

For a finite or countable-state Markov chain with transition matrix $P$, define $(Pf)(x)={\sum }_{y}P(x,y)f(y)$ for a bounded test function $f$. The Markov property says $\mathbb{E}[f(X_{n+1})\mid {\mathcal{F}}_n]=(Pf)(X_n)$. Consequentlyis a martingale, since each increment has conditional expected value zero. Conversely, if these processes are martingales for every bounded $f$, taking the conditional expectation of an increment recovers that identity. With $f=1_{\{y\}}$ it gives $\mathbb{P}(X_{n+1}=y\mid {\mathcal{F}}_n)=P(X_n,y)$ for every state $y$. Thus, given the initial law, these identities characterize the Markov chain, not merely a few of its moments. This is the discrete-time Markov chain martingale characterization.

The drift operator is $P-I$. If $(P-I)h=0$ away from a target set, $h(X_{n\wedge T})$ is a stopped martingale, where $T$ is the first hit of that set. Under boundedness or suitable uniform integrability, the optional stopping theorem expresses $h$ in terms of boundary hitting probabilities. Similarly, if $(P-I)v=-1$ away from the target and $v=0$ on it, then $v(X_{n\wedge T})+(n\wedge T)$ is a martingale, giving $v(x)={\mathbb{E}}_{x}T$ when the terminal limit is justified.

For a concrete calculation, take a simple symmetric random walk on $\{0,1,\dots ,N\}$ stopped at $T$, its first hit of $0$ or $N$. In the interior, $h(i)=i$ has zero drift, while $v(i)=i(N-i)$ satisfiesThe stopped identity gives ${\mathbb{E}}_i(T\wedge n)=v(i)-{\mathbb{E}}_{i}v(X_{T\wedge n})\le v(i)$, proving $T$ finite with finite mean. Bounded convergence in both stopped identities now yields the gambler's ruin answersThis illustrates why the integrability justification is part of using a martingale, rather than an automatic consequence of writing an equation.

For a finite-state continuous-time Markov chain with generator matrix $Q=(q_{ij})$, the corresponding martingale problem for a continuous-time Markov chain usesThese are martingales. Conversely, suppose an adapted càdlàg finite-state process satisfies these identities. Let $P_t=e^{tQ}$, fix $T$, and set $g(s,i)=(P_{T-s}f)(i)$. Then ${\partial }_{s}g+Qg=0$. Apply deterministic integration by parts to the finite collection of indicator martingales to obtain the time-dependent Dynkin formula: $g(s,X_s)-g(0,X_0)-{\int }_0^s({\partial }_{r}g+Qg)(r,X_r)dr$ is a martingale. ThereforeThis recovers the Markov property and the entire semigroup, proving the characterization. On countable spaces one uses nonexplosion and localization to justify the analogous unbounded-rate formulas.

For $d$-dimensional Brownian motion, the Itô formula shows that, for $f\in C_c^{\infty }({\mathbb{R}}^d)$,is a true martingale, since its bounded derivatives make the integral square-integrable on each bounded horizon. This is the martingale problem for Brownian motion with generator $L=\frac{1}{2}\Delta$.

Here is why it also characterizes Brownian motion. Suppose a continuous adapted process $X$, with $X_0=x$, satisfies that martingale problem. Use smooth cutoff test functions equal to $x_i$ and $x_i{x}_j$ inside successively larger balls, and stop before leaving those balls. The coordinate tests show that $X^i-x_i$ are continuous local martingales; the product tests show that $X^i{X}^j-x_i{x}_j-{\delta }_{ij}t$ are local martingales. The Itô product rule consequently givesthe difference between the two proposed compensators is a continuous finite-variation local martingale, hence constant and zero at time zero. For each $u\in {\mathbb{R}}^d$, another application of the Itô formula gives the exponential local martingaleIts modulus is the deterministic quantity $e^{|u{|}^{2}t/2}$, so it is a true martingale on every bounded time interval. Taking its conditional expectation and dividing by its known value at time $s$ givesThe conditional characteristic function identifies a centered multivariate normal distribution of covariance $(t-s)I$, independent of the past. Together with continuity, this is the defining law of Brownian motion. This proves the Lévy characterization of multidimensional Brownian motion in the form needed for the generator characterization.

The analytic use of these martingales parallels the chain case. A harmonic function $h$ satisfies $\Delta h=0$, so it yields a stopped local martingale until the process leaves its domain of harmonicity. Bounded stopped observables permit optional stopping and solve boundary hitting problems. A solution of $\frac{1}{2}\Delta v=-1$ with zero boundary values gives an expected exit time. For example, in the ball of radius $R$, $v(x)=(R^2-|x{|}^2)/d$ gives ${\mathbb{E}}_x{T}_R=(R^2-|x{|}^2)/d$; stopping $|B_t{|}^2-dt$ first at $T_R\wedge n$ proves finiteness and justifies the limiting calculation. The Brownian local time construction in Q4 is another example: the predictable compensators of approximations to $|B_t|$ recover its increasing occupation term.

We now prove the required three-dimensional transience of Brownian motion in dimension at least three, in the strong sense $|B_t|\to \infty$ almost surely. Start first at a point $x$ with $r<|x|<R$, and let $T$ be the first exit from this annulus. The radial Laplacian in dimension three givesThus $|B_{t\wedge T}{|}^{-1}$ is a bounded martingale. The exit time is finite: stopping $|B_t{|}^2-3t$ gives $3\mathbb{E}(t\wedge T)\le R^2-|x{|}^2$. At exit, continuity puts $|B_T|$ at either $r$ or $R$. The optional stopping theorem therefore yieldsand hence the Brownian sphere-hitting probability in dimension threeAs $R\to \infty$, the events $\{{\tau }_r<{\tau }_R\}$ increase to $\{{\tau }_r<\infty \}$: a continuous path is bounded on the finite interval preceding any finite inner hit. This argument assumes no transience. ConsequentlyTo pass from a positive escape probability to almost-sure eventual escape, take Brownian motion started at zero. Let ${\sigma }_R$ be its first hit of the sphere of radius $R$. Stopping the squared-norm martingale proves $\mathbb{E}{\sigma }_R=R^2/3<\infty$. At ${\sigma }_R$, the Strong Markov property and the preceding hitting formula giveThe Strong Markov property here follows from independent Brownian increments by approximating a stopping time from above with dyadic-valued stopping times and using path continuity. Choose $R_m=2^{m}r$. The exit times ${\sigma }_{R_m}$ increase to infinity, since a continuous path is bounded on every compact time interval. The events of a return to the closed ball of radius $r$ after ${\sigma }_{R_m}$ decrease and have probabilities $2^{-m}$. Their intersection has probability zero. Thus almost surely there is some finite exit time after which the path never again enters that ball. Apply this simultaneously to all positive integer $r$. This proves the last visit to a bounded ball for three-dimensional Brownian motion conclusion and therefore

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 $u_{\lambda }(0)=1$.

For $x>0$, differentiation of $u=y/x$ givesThus the radial Laplacian equation reduces toBoundedness of $u$ at zero implies $y(0)=0$. Conversely, a regular reduced solution with $y(0)=0$ gives finite $u(0)=y^{\prime }(0)$.

There is an important qualification to the reduction in the printed question: bounded $u$ at infinity requires $y=O(x)$, not bounded $y$. The example in part (ii) actually has $y\sim -x$, so it cannot be the bounded solution described in part (a). The differential equation and origin condition reduce as asserted, and the same Green's function can still be used with the correct growth condition. This is the radial reduction and growth at infinity distinction.

For the real decaying Bessel form used here, take $k>0$. The original PDF has $K_m(r)$ as the last factor in its auxiliary differential equation, although both differentiated factors have argument $kr$; that last argument must also be $kr$ for the stated Modified Bessel differential equation.

Put $B(r)=K_m(kr)$. The corrected identity is $B^{\prime \prime }=-B^{\prime }/r+(k^2+m^2/r^2)B$. Since the field is radial, the radial Laplacian givesAway from the point source this must vanish. For $k>0$, a constant ratio of $r{B}^{\prime }$ to $B$ would make $B$ a pure power, incompatible with the $k^2{r}^2$ term of its equation. Therefore both displayed coefficients must vanish:The two orders give the same Modified Bessel function of the second kind: $K_{-1/2}(z)=K_{1/2}(z)=\sqrt{\pi /(2z)}{e}^{-z}$.^[ref]

The source describes $k$ merely as real, but the following displayed real square-root formula implicitly needs $k>0$. For the operator with arbitrary real $k$, its decaying fundamental solution instead uses $\kappa =|k|$; $k=0$ is obtained by a limit. The unmodified real $A{K}_m(kr)$ ansatz is not a real Bessel formula for all negative $k$.

For the rotationally invariant free-space fundamental solution of the Laplace equation, put $s=\sqrt{(x-x^{\prime }{)}^2+(y-y^{\prime }{)}^2}$. Away from the source, the radial Laplacian gives $(s{G}^{\prime }{)}^{\prime }=0$, hence $G=A\log s^2+C$. Integrate its Poisson equation over a disk of radius $r$ about the source. The divergence theorem gives $2\pi r(2A/r)=-1/{ϵ}_0$, soA fixed reference length can be inserted inside the logarithm and absorbed into $C$. The resulting electric field decays like $1/s$.

Strictly, the stated pointwise decay as $y\to \pm \infty$ alone does not justify “most general”: one can also add the harmonic function $\mathrm{Re}{e}^{(x+iy{)}^2}$, whose gradient has that decay for every fixed $x$. This is the same growth issue as a zero-boundary harmonic function with pointwise vertical decay. The displayed answer is the usual isotropic free-space choice; requiring the field to vanish uniformly in every direction at spatial infinity removes these extra harmonic additions. An additive constant remains undetermined.

For $r>0$, the substitution $y=ru$ removes the first derivative in the three-dimensional radial Laplacian, since $u^{\prime \prime }+2u^{\prime }/r=y^{\prime \prime }/r$. Bounded $u$ at zero implies $y(0)=0$. Bounded $u$ at infinity means $y=O(r)$; it does not require bounded $y$. For example, constant forcing $g=1$ has the bounded particular solution $u=-1$, whose reduced function is $y=-r$. A Dirichlet half-line Green function for d2 minus 1 can still represent this solution when its integral converges, but the resulting reduced function has the growth demanded by the forcing.

Let $\kappa >0$ be an inverse length. In three spatial dimensions, seek the Green's function that decays at infinity and satisfieswhere ${\nabla }^2$ is the Laplacian and ${\delta }^{(3)}$ the three-dimensional Dirac delta function. The point source and the differential operator have spherical symmetry. For $r>0$, the radial Laplacian givesWriting $u(r)=r{G}_{\kappa }(r)$ reduces this to $u^{\prime \prime }-{\kappa }^{2}u=0$. Decay at infinity removes the growing exponential function, leaving $G_{\kappa }(r)=C{e}^{-\kappa r}/r$.

The Dirac delta function fixes the coefficient. Integrating over a ball $B_{ϵ}$ and using the divergence theorem givesThe first term tends to $4\pi C$, while the second is $O({ϵ}^2)$. Hence $C=1/(4\pi )$ andThis solves the source equation in the sense of distributions, including at the origin. Merely verifying the equation for $r>0$ would leave the normalization undetermined. The limit $\kappa \to 0$ at fixed $r>0$ is $1/(4\pi r)$, the Green function of the Laplacian for the convention $-{\nabla }^2$.

For an attractive scalar toy interaction, absorb constants into $A>0$ and setHere $A$ has units of energy times length, and negative $F_r$ points inward. At equal $A$, comparison with the unscreened model $V_0=-A/r$ givesThus the potential energy and force have the same exponential scale but different suppression factors. At $x=1$, these ratios are approximately $0.368$ and $0.736$; at $x=3$, they are $0.0498$ and $0.199$. Finite range means rapid decay, not an abrupt cutoff.