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For prescribed boundary data $\phi$, the affine Sobolev spaceconsists of the $W^{1,p}$ functions whose trace equals the trace of $\phi$. It is the natural admissible set for a variational Dirichlet problem.

For the SLE angle process, the bounded Dirichlet problem on the unbounded surviving domain has data $\pi$ on the intrinsic boundary mapped to the left of the Loewner driving function and zero on the boundary mapped to its right. For a simple Loewner trace these are its left and right banks, with the negative and positive real rays respectively. The tip is a discontinuity point of the boundary data.

Given a uniquely solvable Dirichlet problem, this map sends prescribed Dirichlet boundary data to the solution's outward normal derivative. It depends on the differential operator and domain. Spectral global relations offer one way to compute it numerically from boundary traces, while a finite collocation scheme still requires a rank and conditioning check.

An open set satisfies an exterior cone condition at $z$ if a solid cone with vertex $z$ lies outside the set in some neighbourhood of $z$. The stronger global version requires the closures of the cone and the set to meet only at $z$. In two dimensions, a power barrier for an exterior cone makes such a point a regular boundary point for the Dirichlet problem. Local cone conditions suffice for local boundary regularity; a global cone is an additional geometric assumption.

To solve a linear Dirichlet problem with boundary value $g$, choose a sufficiently regular extension $G$ whose trace is $g$ and write $v=w+G$. Then $w$ has zero trace, while moving the known term $LG$ to the right-hand side turns the equation $Lv=f$ into the homogeneous-boundary problem $Lw=f-LG$.

Under the usual bounded-domain hypotheses, with piecewise smooth boundary and a sufficiently smooth vector field, the divergence theorem stateswhere $\mathbf{n}$ is outward. Apply it to $g\mathbf{F}$ and use the product rule $\nabla \cdot (g\mathbf{F})=\mathbf{F}\cdot \nabla g+g\nabla \cdot \mathbf{F}$ to obtainFor uniqueness, let $w=u_1-u_2$ for two classical solutions with the same Dirichlet boundary data. Then $\Delta w=0$ and $w=0$ on the boundary. Taking $\mathbf{F}=\nabla w$, $g=w$ gives Green's first identity in the formThe continuous nonnegative integrand must vanish, so $w$ is constant on each connected component. The zero boundary data make every such constant zero. Thus the solution of the Dirichlet problem for the Laplace equation is unique, if it exists. This establishes Uniqueness of the Dirichlet problem; existence is assumed. Boundedness, or suitable conditions at infinity, is needed: on an unbounded half-space the harmonic function $w=z$ has zero boundary values but is not identically zero.

For a simple Loewner trace, the domain mapped by $g_t$ is $D_t=\mathbb{H}\setminus \gamma [0,t]$. For a non-simple Loewner trace, use instead the unbounded component $D_t=\mathbb{H}\setminus K_t$; the map $g_t$ is not defined on swallowed bounded components. This is the necessary domain interpretation when $\kappa >4$.

The function $\mathrm{Im}\log (g_t(z)-{\xi }_t)$ is harmonic in $D_t$ because the logarithm is holomorphic on the complex upper half-plane. Let $S_t^{-}$ and $S_t^{+}$ be the intrinsic boundary sets mapped to $(-\infty ,{\xi }_t)$ and $({\xi }_t,\infty )$ respectively. The Dirichlet problem has boundary dataFor a simple Loewner trace these are the left bank of the Loewner trace together with the negative real boundary, and the right bank together with the positive real boundary. Left and right refer to the orientation from the starting point towards the tip. The tip and infinity correspond to discontinuities of the data; no unique limit is imposed there. They have zero harmonic measure.

More precisely, the bounded solution isby the Poisson kernel for the upper half-plane and conformal invariance of planar Brownian motion. This establishes the boundary values in the intrinsic sense and uniqueness among bounded solutions of the Dirichlet problem, without treating the two banks as one Euclidean boundary point.

Interpret the printed coordinate notation as the square corners $1+i$, $1-i$, $-1-i$, $-1+i$. This listed order is clockwise, contrary to the counterclockwise convention in (ii). Keep the printed first side directed from top to bottom. Then $m_1=1$, $h_1=-i$, and $z(s)=1-is$. Put $g_1(s)=u(1,-s)$ and $q_1(s)=u_x(1,-s)$, where $q_1$ is the outward normal derivative on the right side. The pullback formula, without any orientation shortcut, givesIndeed $h_1{u}_z-{\bar{h}}_1{u}_{\bar{z}}=-i(u_z+u_{\bar{z}})=-i{u}_x$, and $i\beta (\lambda h_1+{\bar{h}}_1/\lambda )=\beta (\lambda -{\lambda }^{-1})$. If one reverses the side to match the counterclockwise convention, its parameter is $1+is$ and its integrand is $-W_1(-s,\lambda )$. Reversing every side multiplies the global relation by minus one, leaving its zero value unchanged.

The unheaded numerical reconstruction request. The four unknown Neumann boundary conditions are four functions on the sides, not four scalar values. The polygonal global relation is linear in their outward normal derivatives, and its remaining terms depend only on the prescribed Dirichlet boundary data.

Choose a finite approximation on each side, for example an expansion of $q_j(s)$ in Legendre polynomials or piecewise polynomials. Substitute those expansions into the consistently oriented global relation. At chosen nonzero spectral parameters for a linear boundary value problem, integrate the exponential kernels against each basis function to assemble a complex linear system; the known right-hand side is obtained by integrating the given Dirichlet boundary data. Use enough independent samples to resolve all side coefficients, and preferably oversample. The conjugate global relations for the modified Helmholtz equation provide useful companion tests; for complex data use the corresponding independent adjoint relation rather than assuming the traces real.

Solve the scaled system by least-squares solution using a stable factorization such as a singular value decomposition. Sampling directions should probe all sides, and exponential row scaling avoids overflow and poor conditioning. Refine the side approximation and spectral samples until the recovered traces and unused global relation residuals stabilize. Corner incompatibilities or limited corner regularity call for mesh refinement or enriched endpoint basis functions. This realizes a numerical Dirichlet-to-Neumann map without first discretizing the whole interior.

The underlying Dirichlet problem is uniquely solvable in the usual trace class for $\Delta u-4{\beta }^{2}u=0$ with $\beta >0$: the homogeneous problem has ${\int }_{\Omega }(|\nabla u{|}^2+4{\beta }^2|u{|}^2)dA=0$. This supports the boundary reconstruction, although uniqueness of the continuous problem alone does not guarantee that an arbitrary finite set of spectral samples is well conditioned.

For the usual Dirichlet problem on a bounded regular volume, suppose ${\varphi }_1,{\varphi }_2$ both satisfy the stated data. Their difference $u={\varphi }_1-{\varphi }_2$ is a harmonic function, with ${\nabla }^{2}u=0$ in $V$ and $u=0$ on $\partial V$. In Green's first identity, choose both functions equal to $u$:The nonnegative continuous integrand forces $\nabla u=0$. Thus $u$ is constant on each connected component, and its zero boundary values force each constant to vanish. This proves uniqueness for the Dirichlet problem.

If the boundary value is a constant $c$, the constant function $c$ is itself a harmonic function with those data. Uniqueness therefore yieldsBoundedness, or an appropriate condition at infinity replacing it, is essential. On the upper half-space both $u=0$ and $u=z$ are harmonic functions with zero data on its boundary plane. They show why unrestricted uniqueness must not be asserted for an unbounded volume.

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 $\Delta ={\sum }_j{\partial }_j^2$.

In the bounded-domain setting of the weak maximum principle, let $u_1,u_2$ be two classical solutions with the same forcing and boundary values. Their difference $z=u_1-u_2$ satisfies $Lz=0$ and $z=0$ on the boundary. Apply the weak maximum principle first to $z$ and then to $-z$:Thus $u_1=u_2$, proving uniqueness of the Dirichlet problem. The same proof works on an unbounded domain if the difference and its negative satisfy the required condition at infinity. As literally printed without either restriction, uniqueness need not hold: on $\{x_n>0\}$ the two harmonic functions $0$ and $x_n$ have the same zero boundary values.