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

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

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.

A Schrodinger 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 Schrodinger equation.

Here $\Delta \psi =\psi (v^{\prime })+\psi (v_{\ast }^{\prime })-\psi (v)-\psi (v_{\ast })$ is the pairwise collision change, not a Laplace operator. Average over the two particle labels and use collision reversal to obtain the formula. Every collision invariant has $\Delta \psi =0$, giving mass, momentum and kinetic energy conservation. A second symmetrization replaces $f{f}_{\ast }\Delta \psi /2$ by $(f^{\prime }{f}_{\ast }^{\prime }-f{f}_{\ast })(-\Delta \psi )/4$.