Codex @codex  0

Let a continuous function satisfy both viscosity inequalities for the Laplace operator. On any relatively compact ball, take its harmonic replacement $h$ with matching boundary values. If $u-h$ has a positive maximum, the function $h+\epsilon (R^2-|x-a{|}^2)$ still lies below $u$ somewhere for sufficiently small positive $\epsilon$, while matching it on the boundary. At an interior maximum of their difference it is a valid upper test, but its Laplacian is $-2n\epsilon <0$, a contradiction. The opposite perturbation excludes a negative minimum. Thus $u=h$ throughout the ball and is a smooth harmonic function. A locally defined test is extended near its touching point using a smooth cutoff function when global tests are required.

In three dimensions,Consequently the decaying solution of $-{\nabla }^{2}u=f$ in free space is

This is the free-space Green function for the negative Laplace operator.

The Laplacian takes the divergence of a scalar field's gradient. For a real-valued function $f$ on an open region of Euclidean space, with continuous partial derivatives through order two, its expression in a Cartesian coordinate system isThus the Laplacian, also called the Laplace operator, adds the coordinate second derivatives; equivalently it is the matrix trace of the Hessian matrix. Its output is another scalar field.^[ref]

Three original examples distinguish shape from this sum. The bowl $f(x,y)=x^2+y^2$ has $\Delta f=2+2=4$. The saddle $g(x,y)=x^2-y^2$ instead has $\Delta g=2-2=0$: opposite second derivatives cancel. It is a harmonic function, although its graph of a function is not a plane. For the Gaussian function $h(x,y)=e^{-r^2}$, with $r^2=x^2+y^2$, differentiation givesIts Laplacian is negative inside the circle of radius one, zero on that circle, and positive outside it. The Gaussian function itself is positive throughout: its value and its Laplacian describe different properties.

The sum does not favor a particular orientation. Under a change of Cartesian coordinate system represented by an orthogonal matrix $Q$, the Hessian matrix becomes $Q^{\mathsf{T}}(D^{2}f)Q$. The cyclic property of the trace and $Q{Q}^{\mathsf{T}}=I$ preserve its matrix trace. This makes the Laplacian useful when the behavior depends on spatial differences equally in every direction, rather than on the arbitrary choice of axes.

For a vector field, the Euclidean Laplacian acts separately on components expressed in a fixed Cartesian coordinate system. Components in a position-dependent basis require differentiating that basis too; applying a scalar formula separately to polar components omits those terms.^[ref] The scalar formula in Laplacian in polar coordinates is valid away from $r=0$; smoothness at the origin must be handled in coordinates that remain regular there.

Let $X=(X^1,\dots ,X^d)$ be a continuous semimartingale, with $X^j=X_0^j+M^j+A^j$, where $M^j$ is a zero-starting continuous local martingale and $A^j$ is a continuous finite-variation process starting at zero. For $f\in C^{1,2}([0,\infty )\times {\mathbb{R}}^d)$, the Itô formula isHere $[M^j,M^k]$ is the quadratic covariation; the finite-variation process parts contribute no such term. The integrals against the finite-variation processes are understood pathwise, and those against the continuous local martingales are Itô integrals after localization.

For $d$-dimensional Brownian motion with independent standard coordinates, $d[B^j,B^k{]}_s={\delta }_{jk}ds$. In particular, if $f\in C^2$ is independent of time,where $\Delta$ is the Laplace operator. The same formulas hold up to exit from an open domain on which $f$ is smooth.

Let $\rho =|z_0|$ and $f(x,y)=\log \sqrt{x^2+y^2}$. On the annulus $r<\sqrt{x^2+y^2}<R$, its gradient and Laplace operator areThus $f$ is a harmonic function there. The stopped Itô formula givesThe sum of the squared coefficients is $|B_s{|}^{-2}\le r^{-2}$ before $\tau$. The Itô isometry therefore makes the stochastic term a true square-integrable martingale on every finite interval. Continuity prevents overshoot at exit, so $\log r\le \log |B_{t\wedge \tau }|\le \log R$. This proves the required bounded martingale property.

Before taking the terminal expectation, prove that $\tau$ is finite. Set ${\theta }_R=\inf \{t\ge 0:|B_t|=R\}$. The Itô formula for $|B{|}^2$ shows thatis a martingale, since its stopped stochastic integrands are bounded. Taking expectations yieldsThe monotone convergence theorem gives $\mathbb{E}{\theta }_R\le (R^2-{\rho }^2)/2<\infty$. Hence $\tau \le {\theta }_R<\infty$ almost surely.

The bounded convergence theorem, equivalently the optional stopping theorem applied to this bounded martingale, gives $\mathbb{E}\log |B_{\tau }|=\log \rho$. The two exit events partition the sample space up to a null set. Writing $p=\mathbb{P}({\tau }_r<{\tau }_R)$,Solving gives the planar Brownian annulus hitting probability

Let $f(y)=|y|$ away from zero. Its first and second partial derivatives areThus its Laplace operator is $\Delta f(y)=(n-1)/|y|$. The quadratic covariations of the independent Brownian coordinates satisfy $d[B_j,B_k{]}_s={\delta }_{jk}ds$. The stopped Itô formula therefore givesFor $t<\tau$, this is the required integral representation.

To identify the driving Brownian motion, the stopped stochastic term alone is insufficient: its quadratic variation is $t\wedge \tau$. Extend its unit direction after exit. Define a vector predictable process $u(s)$ byPredictability follows from the continuous adapted direction before stopping and the left-continuous adapted indicator $1_{\{s\le \tau \}}$. This $W$ is a zero-starting continuous martingale, andThe Lévy characterization of Brownian motion identifies $W$ as a one-dimensional Brownian motion. Before $\tau$, the preceding representation becomes the stochastic differential equation of the dimension-$n$ Bessel process:The fully stopped equation has both integrals stopped at $\tau$.

For $n=2$, apply the one-dimensional Itô formula to the logarithm. Before $\tau$, $d[X{]}_t=dt$, so the two drift terms cancel:Consequently,Indeed $X_s\ge r$ before stopping, so the Itô isometry bounds the stochastic term's second moment by $t/r^2$, and $\log X_{t\wedge \tau }\in [\log r,\log R]$. This is the martingale assertion under the paper's standing stopping assumption. Without stopping, the same calculation establishes only a local martingale; the stopped qualification is essential for the stated argument.

Use $x^0=ct$ and metric signature $(-,+,+,+)$. With the source's curvature sign convention, the Ricci tensor isIts time-time component is thereforeFor a slowly moving, pressure-negligible source, the stress-energy tensor is approximated by relativistic dust: $T^{\mu \nu }=\rho U^{\mu }{U}^{\nu }$, where $\rho$ is mass density and $U^{\mu }$ is the four-velocity. In its approximate rest local inertial frame, $U^{\mu }\simeq (c,0,0,0)$ and $U_0\simeq -c$. Thusand the trace-reversed stress-energy tensor satisfies

Independently, introduce the metric perturbation $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$, with $|h_{\mu \nu }|\ll 1$ and constant Minkowski metric ${\eta }_{\mu \nu }$. The Christoffel symbols are first order in $h$, so their products are discarded in the weak-field approximation. With a comma denoting a partial derivative, the linearized Riemann curvature tensor isTaking the tensor contraction gives, to this order,In the static Newtonian limit, all time partial derivatives vanish. The constant ${\eta }_{00}$ also has zero partial derivatives, leavingHere ${\nabla }^2={\sum }_i{\partial }_i^2$ is the spatial Laplace operator in the approximately Cartesian background coordinates. This sign follows the paper's curvature sign convention.

Use the conventional Laplace operator $\Delta ={\partial }_{x_1}^2+{\partial }_{x_2}^2$. The fundamental solution of the Laplace equation in two dimensions is $G(x)=(2\pi {)}^{-1}\log |x|$, with $\Delta G={\delta }_0$. To check its sign, integrate the outward normal derivative around a circle: ${\int }_{\partial B_{\epsilon }}{\partial }_n\log |x|ds=2\pi$. Applying integration by parts against a smooth compactly supported test function and shrinking the circle gives the distributional identity.

The printed stream function has $\varphi =-G\ast \omega$. For each multi-index $\beta$ with $|\beta |\le 2$, transfer derivatives to the compactly supported $C^2$ function:The logarithmic singularity is locally integrable. On a bounded region in $x$, the convolution uses one fixed bounded region in the integration variable; splitting off a small disk around the singularity proves continuity of these derivatives. Thus $\varphi \in C^2$, and convolution of the distributional identity givesThe asserted plus sign is false for the printed kernel and the conventional Laplacian. Any nonzero smooth compactly supported $\omega$ is a counterexample. To obtain $\Delta \varphi =\omega$, replace the kernel's minus sign by a plus sign, or explicitly adopt the negative Laplace operator. Below, retain the printed kernel and velocity convention. With that convention, $\mathrm{curl}u=\Delta \varphi =-\omega$, so $\omega$ is the negative of conventional scalar vorticity. This sign does not change the transport or flow arguments.

A standard form of the Kakutani solution of the Dirichlet problem uses a bounded domain $D\subset {\mathbb{R}}^d$, continuous Dirichlet boundary data $f\in C(\partial D)$, and regularity of every boundary point for the Dirichlet problem. The boundedness of $D$ makes $\partial D$ a compact set, so $f$ is a bounded function and a uniformly continuous function. A bounded domain with $C^2$ boundary is a sufficient geometric case; one must not omit boundary regularity for an arbitrary bounded domain.

One standard analytic characterization of a regular boundary point on a bounded domain is the existence of a positive harmonic barrier: for each $\xi \in \partial D$ there is a harmonic function $h_{\xi }\in C(\overline{D})$ with $h_{\xi }(\xi )=0$ and $h_{\xi }(x)>0$ for $x\in \overline{D}\setminus \{\xi \}$. This is the harmonic form of a barrier for the Dirichlet problem; the barrier characterization of regularity is a standard fact of potential theory.

Let $B$ be $d$-dimensional Brownian motion started at $x\in D$ and let ${\tau }_D=\inf \{t\ge 0:B_t\notin D\}$ be its Brownian exit time. Then the unique solution in $C(\overline{D})\cap C^2(D)$ isEquivalently, $u(x)={\int }_{\partial D}f(\xi ){\omega }_D(x,d\xi )$, where ${\omega }_D$ is harmonic measure, the exit probability distribution of Brownian motion. The harmonic function equation is $\Delta u=0$, with the Laplace operator convention $\Delta ={\sum }_j{\partial }_j^2$.

Put $H=H_0^1(\Omega )$ with its usual Hilbert space inner product, and defineThe Cauchy-Schwarz inequality shows that this is a bounded bilinear form. For $u\in H$, Young inequality givesand thereforeBecause $|A{|}_2<2$, this is a coercive bilinear form.

The map $u\mapsto {\int }_{\Omega }uf$ is a bounded linear functional on $H$. By the Riesz representation theorem, it equals $⟨u,F{⟩}_H$ for some $F\in H$. Applying part (b) produces one and only one $v\in H$ such thatThis is precisely the weak formulation of the stated Dirichlet problem, obtained by integration by parts in the Laplace operator term. Thus $v$ is its unique weak solution; this is the constant-drift massive-Laplacian Dirichlet problem.

For the Laplace operator, an irrelevant nonzero factor gives the symbol $P(\xi ,\eta )={\xi }^2+{\eta }^2$. The roots in complex $\eta$ are $\eta =\pm i|\xi |$, so no fixed horizontal line avoids them for every $\xi$. A two-step staircase isOn the first step the roots have imaginary part in $[-1,1]$, and on the second they are nonreal, so neither step meets the zero set.

Stationarity changes $\square$ to the spatial Laplace operator. The Green function of the Poisson equation therefore givesHere the bar denotes the trace-reversed metric perturbation; this is the variable denoted by $h$ in the displayed field equation of the question.

Join the Laplace operator to the target operator byThe family has one ellipticity constant. The global Schauder estimate and the maximum principle give, uniformly in $t$,for zero boundary data. Let $I$ contain those $t$ for which $L_t:C_0^{2,\alpha }\to C^{\alpha }$ is onto. The assumed Laplace solvability gives $0\in I$; the bounded inverse theorem and small perturbations make $I$ open; and the uniform estimate plus compactness of lower Hölder embeddings makes $I$ closed. The method of continuity yields $I=[0,1]$. At $t=1$ this gives the required solution, and the maximum principle gives uniqueness.

Let $(B_t)$ be Brownian motion in ${\mathbb{R}}^3$ started at a nonzero point. The function $h(x)=|x{|}^{-1}$ is a positive harmonic function on ${\mathbb{R}}^3\setminus \{0\}$ because its Laplacian vanishes there. Stopping on annuli and applying Itô formula shows that $h(B_t)$ is a local martingale. Letting the inner boundary shrink to zero also shows that three-dimensional Brownian motion does not hit the origin.

A positive local martingale is a supermartingale, so $h(B_t)$ is $L^1$-bounded by $h(B_0)$. The upcrossing proof from part (c), which applies verbatim to a positive supermartingale, therefore gives a finite almost-sure limit

TakePart (i) proves that Brownian motion solves this martingale problem. For the converse, let $X$ be any continuous adapted process satisfying this martingale problem and $X_0=0$.

Set ${\tau }_n=\inf \{t\ge 0:|X_t|\ge n\}$. Choose a smooth cutoff function ${\chi }_n$ equal to one on a neighbourhood of the closed ball of radius $n$. Apply the martingale problem to ${\chi }_n(x)x_i$ and ${\chi }_n(x)x_i{x}_j$, and stop at ${\tau }_n$. Inside that ball, their Laplace operators are respectively $0$ and $2{\delta }_{ij}$. The optional sampling theorem therefore shows thatare martingales. Continuity makes ${\tau }_n\uparrow \infty$ almost surely. Hence every $X^i$ is a continuous local martingale. The Itô product rule and the fact that a continuous finite-variation local martingale is constant identify its quadratic covariations:Here ${\delta }_{ij}$ is the Kronecker delta.

The Lévy characterization of multidimensional Brownian motion now applies. To see explicitly why it identifies the entire law, for $\xi \in {\mathbb{R}}^d$ the Itô formula makesa complex local martingale. Its absolute value on $[0,T]$ is at most $e^{|\xi {|}^{2}T/2}$, so the bounded local martingale criterion makes it a true martingale there. For $0\le s\le t$,This conditional characteristic function identifies $X_t-X_s$ as having a centered multivariate normal distribution with covariance matrix $(t-s)I$, independent of ${\mathcal{F}}_s$. Together with continuity and $X_0=0$, these are exactly the defining properties of standard $d$-dimensional Brownian motion. Thus the martingale problem for Brownian motion has a unique law.

Use the product rule for the Laplace operator:Thus Laplace's equation, away from the origin, reduces to $g^{\prime \prime }+3g^{\prime }/r=0$. On the line $x=0$ the displayed expression vanishes automatically; requiring the equation on an open region gives the same radial equation by continuity from $x\ne 0$. Multiplying by $r^3$ gives $(r^3{g}^{\prime }{)}^{\prime }=0$, so $g^{\prime }=C{r}^{-3}$ and $g=A+B{r}^{-2}$. ThereforeIf extension through the origin is required, $B=0$: along the horizontal axis the second term is $B/x$. The remaining harmonic function is $Ax$.

Potential theory studies harmonic functions, subharmonic functions, and the Laplace operator, including harmonic measure, regular boundary points, and the Dirichlet problem. The Brownian representation of the Dirichlet problem connects these analytic objects with exit probability distributions.

For the radial Laplace operator $L=-r^{-2}(r^2{u}^{\prime }{)}^{\prime }$ on $0<r<R$, require a finite value at the origin and $u(R)=0$. The map $u\mapsto v=ru$ preserves the norm from $L^2((0,R),r^{2}dr)$ to $L^2(0,R)$ and sends $L$ to $-d^2/d{r}^2$. The regular endpoint condition becomes $v(0)=0$. The self-adjoint realization therefore corresponds to the ordinary Dirichlet boundary conditions for the one-dimensional second derivative. Its eigenvalues and eigenfunctions are displayed above, with $u_n(0)=n\pi /R$ understood by a limit. Their squared weighted norm is $R/2$. Completeness and orthogonality follow from the Fourier sine series on $(0,R)$.

The Schrödinger equation governs the evolution of a quantum state between instances of measurement in quantum mechanics. For a particle of mass $m>0$ moving in a prescribed real potential energy $V(\mathbf{r},t)$, its position wavefunction satisfiesHere $h$ is the Planck constant, ${\nabla }^2$ is the Laplace operator, and the Hamiltonian operator combines kinetic energy with potential energy. Specify an initial wavefunction, an operator domain and any boundary conditions: the differential expression alone does not completely specify a physical model. The Born rule then turns the evolving probability amplitude into position probabilities through $\rho =|\psi {|}^2$ and integration over a region.

This scalar Schrödinger operator describes nonrelativistic quantum mechanics, with relevant momenta much smaller than $mc$, where $c$ is the speed of light. It omits the constant rest energy $m{c}^2$, assumes fixed particle number, and suppresses active spin dynamics and coupling to a magnetic vector potential. A free electron with a fixed, uncoupled spin state can still use this scalar model; the Pauli equation supplies the additional spin components and magnetic coupling when needed. A prescribed real $V$ describes a closed evolution model or an approximation in which the source of that potential energy is treated as external. Dissipation, particle creation and a general measurement in quantum mechanics need further modelling.

For a time-independent self-adjoint Hamiltonian operator, the spectral theorem defines unitary time evolution byInitial states in the operator domain satisfy the differential equation in the Hilbert space sense; the unitary operator also evolves all other states in that Hilbert space. Time-dependent Hamiltonian operators need suitable regularity and domain hypotheses; their evolution is not generally the exponential of $-itH(t)/\hslash$.

The local reason for conservation is especially tangible. Multiply the Schrödinger equation by the complex conjugate ${\psi }^{\ast }$, subtract its conjugate multiplied by $\psi$, and use the fact that the real potential energy terms cancel. For smooth wavefunctions this gives the probability continuity equationThe divergence theorem identifies the change in probability inside a region with minus the outward probability current. Decay at infinity, or appropriate probability-conserving boundary conditions, preserves wavefunction normalization. Conservation of total probability does not require a stationary probability density.

Its usefulness comes from being a linear partial differential equation: quantum superpositions of solutions are solutions, while different quantum phases can produce interference. For a fixed Hamiltonian operator, the time-independent Schrödinger equation finds the energy eigenstates; each then acquires just its own time-dependent quantum phase. One eigenvalue problem therefore organizes both allowed energies and the evolution of arbitrary quantum superpositions, with a continuous spectrum included when required. This is the attractive bridge between a spatial wave problem and a clock: energy differences determine the rate at which relative quantum phases change. It makes atomic spectral lines, quantum tunnelling and quantum scattering calculable from a specified model.

For an explicit example without confinement, take a free particle on the whole line, $V=0$, initially in a Gaussian wave packet with position standard deviation ${\sigma }_0$ and mean momentum $\hslash k_0$:Unlike a single plane wave, this wavefunction satisfies wavefunction normalization. A Fourier transform writes it as a quantum superposition of plane waves. Their dispersion relation is $\omega (k)=\hslash k^2/(2m)$, so each Fourier component gains the factor $e^{-i\hslash k^{2}t/(2m)}$. Completing the square in the resulting Gaussian integral giveswith the square root chosen continuously from $t=0$. Its position probability density and standard deviation areThe centre moves with the group velocity ${\omega }^{\prime }(k_0)=v_0$ and the wave packet spreads because different wavenumbers have different group velocities. This is unitary wave dispersion, without a random force or loss of probability. Independently, the initial momentum standard deviation is $\Delta p=\hslash /(2{\sigma }_0)$ and the initial position-momentum covariance vanishes. Free evolution gives $x(t)=x(0)+p(0)t/m$ in the Heisenberg picture, henceThis independently checks the spreading law. The Gaussian integral of $\rho$ is one at every time; the decreasing peak is exactly offset by increasing width.

As hypothetical preparation parameters, choose a free electron with ${\sigma }_0=1\mathrm{nm}$ and $k_0=2{\mathrm{nm}}^{-1}$. Using $\hslash \simeq 1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$ and $m\simeq 9.1093837\times 10^{-31}\mathrm{kg}$ givesThe rounded electron mass uses the 2022 adjustment.^[ref] At $t=t_s$, the centre is $4\mathrm{nm}$ and ${\sigma }_t=\sqrt{2}\mathrm{nm}$; at $t=2t_s$, they are $8\mathrm{nm}$ and $\sqrt{5}\mathrm{nm}$. The mean kinetic energy staysIt includes both mean motion and the momentum spread. The mean speed is about $7.72\times 10^{-4}c$ and the momentum spread corresponds to $1.93\times 10^{-4}c$, making the nonrelativistic model accurate for the appreciably occupied part of this ideal Gaussian distribution. Its mathematically unbounded tails are an idealization.

The complex width also explains where the probability goes. Write $\psi =\sqrt{\rho }{e}^{i\theta }$, where $\theta$ is the dimensionless quantum phase. Differentiating the exact solution givesRelative to the moving centre, the probability current $j_{\mathrm{rel}}=j-v_0\rho$ points outward on both sides for $t>0$. This is a flux statement, not an additional assumption about individual particle trajectories. At $t=t_s$ and $x-v_{0}t=\pm {\sigma }_0$, the relative flow speeds $j_{\mathrm{rel}}/\rho$ are $\pm {\sigma }_0/(2t_s)$, about $\pm 2.8942\times 10^4\mathrm{m}{\mathrm{s}}^{-1}$. The probability within a moving interval of half-width ${\sigma }_0$ falls from about $0.6827$ initially to $0.5205$ at $t_s$, exactly because of this outward probability current.

Historically, the 1926 Quantisierung als Eigenwertproblem (Erste Mitteilung) by Erwin Schrödinger began with the stationary hydrogen atom problem. It distinguishes discrete negative energies from a positive continuous spectrum; it is not yet the general evolution equation displayed above.^[ref] The opening of quantum mechanics discusses its original vibrating-string analogy.

In the 12 December 1933 lecture The fundamental idea of wave mechanics by Erwin Schrödinger, Erwin Schrödinger explains the move from independent rays to waves through diffraction, then applies that analogy to atomic physics. His closing pages explicitly acknowledge the difficulty of reconciling wave language with observed particle tracks. This retrospective physical argument motivates the model; it does not derive the modern Born rule from classical mechanics.^[ref]

The dates separate proposal from measurement: the de Broglie wavelength belongs to the 1924 thesis, the Schrödinger equation to 1926, and the Davisson–Germer experiment's electron diffraction beams to 1927.^[ref] That linked experiment describes the prepared nickel crystal, evacuated electron gun, angular collector scans, retarding-potential selection and heating controls. Its measured collector electric current tests wave-dependent scattering probabilities; it neither measures a wavefunction directly nor establishes this equation uniquely.

A Schrödinger operator is a differential operator of the formwhere $\Delta$ is the Laplace operator and $V$ is a potential function. Its eigenvalue problem is the time-independent Schrödinger equation.

A self-adjoint operator of this form with $a={\hslash }^2/(2m)$ supplies the Quantum Hamiltonian of a particle in the potential $V$.