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The PDF specifies decay as $y\to \infty$; the local TeX drops this limit from its displayed boundary data. Take $a=4\pi$, so the conformal map is $w=\sin (z/4)$. Its components areFor $-2\pi <x<2\pi$ and $y>0$, $\eta >0$. The bottom edge maps monotonically to $(-1,1)$, and the two vertical sides map to the real rays $(-\infty ,-1)$ and $(1,\infty )$. The derivative $\cos (z/4)/4$ never vanishes in the interior. Injectivity follows from the identities for equal sines: the alternatives $z_1/4=z_2/4+2\pi n$ and $z_1/4=\pi -z_2/4+2\pi n$ give only the identical point within this strip. Surjectivity can be seen directly: for fixed $\xi$ and $v=y/4>0$, solve $\sin (x/4)=\xi /\cosh v$; thenOn $v>\max (0,\mathrm{arcosh}|\xi |)$, with the lower endpoint read as $0$ when $|\xi |\le 1$, the right side increases strictly from zero to infinity. Thus every point of the upper half-plane has a unique preimage.

By conformal invariance of harmonicity, write the transformed harmonic function as $\Phi (\xi ,\eta )$. Its real-axis boundary data areThe inverse sine on $(-1,1)$ is its real principal branch. For bounded integrable data, take the Fourier transform convention $\hat{\Phi }(k,\eta )={\int }_{\mathbb{R}}\Phi (\xi ,\eta )e^{-ik\xi }d\xi$. The Laplace equation becomes ${\hat{\Phi }}_{\eta \eta }-k^2\hat{\Phi }=0$, with the bounded decaying solution $\hat{\Phi }=\hat{F}(k)e^{-|k|\eta }$. Inverting the Fourier transform givesConsequently the Poisson kernel for the upper half-plane yieldsThe decay is uniform across the strip: $|w{|}^2={\sin }^2(x/4)+{\sinh }^2(y/4)\ge {\sinh }^2(y/4)$ and, for $|w|>1$, $|\Phi |\le \Vert F{\Vert }_{L^1}|w|/(\pi (|w|-1{)}^2)\to 0$. This verifies the printed decay condition. Boundary limits give the prescribed values at continuity points away from the corners.

For the given sine data,Values assigned at $s=\pm 1$ do not alter the integral. There is no solution continuous on the entire closed half-strip for these data: the bottom edge has corner limits $\pm 1$, but the side edges have corner limits zero. The formula is the bounded Dirichlet problem solution with the specified limits away from those two corners. This is the usual interpretation of the printed problem; boundedness excludes additional boundary-singular harmonic functions, while decay alone without a regularity or boundedness condition does not establish uniqueness.

The Möbius transformationmaps the unit disc conformally onto the upper half-plane. On the upper semicircle of its boundary, $w$ approaches the positive real axis; on the lower semicircle it approaches the negative real axis. If $0<\arg w<\pi$, the harmonic functionhas the required boundary limits.

For $z=x+iy$,Since the imaginary part is positive inside the disc,Therefore the solution of the Dirichlet problem isThe exceptional boundary endpoints are precisely the jump points of the prescribed data.

Fix $B=B(x_0,r)⋐\Omega$. Solvability of the Dirichlet problem on a ball gives a unique harmonic function $h\in C^2(B)\cap C^0(\overline{B})$ with $h=u$ on $\partial B$. Both $u$ and $h$ have the mean value property for harmonic functions, so $v=u-h$ has it as well and vanishes on $\partial B$.

If $v$ were not zero, compactness of $\overline{B}$ would give either a positive maximum or a negative minimum in the interior. Part 1(i) would make $v$ constant, contradicting its zero boundary values. Hence $u=h$ on $B$. Every point lies in such a ball, so

If $u_1,u_2$ solve the same Dirichlet problem, their difference $v=u_1-u_2$ satisfiesPart 2(ii) applied to $v$ gives $v\le 0$, and applied to $-v$ gives $v\ge 0$. Hence $v=0$ and the solution is unique.

For $w\in X_{\delta }$, hypothesis (2) gives $F(w)\in C^{0,\alpha }(\overline{\Omega })$. Thus $f-F(w)$ is Hölder continuous and the classical Dirichlet problem for the Poisson equation has a unique solution $u=T(w)\in C^{2,\alpha }(\overline{\Omega })$. The global Schauder estimate givesThe maximum estimate for the Poisson problem also givesThese estimates prove that $T$ is well defined and give the requested norm control.

The required boundary values on the real axis are zero to the left of the origin and one to the right. The bounded harmonic function with those values is the upper-half-plane harmonic measure of the positive half-axis,Indeed, $\arg z$ is harmonic in the upper half-plane, and the displayed function tends to $1$ on the positive half-axis and to $0$ on the negative half-axis. The uniqueness of bounded solutions of the Dirichlet problem identifies it with the Brownian exit probability.

The boundary identity gives a numerical route to shape recovery from a Herglotz boundary identity. First recover an admissible density $g$ from measured far-field patterns, enforcing the supplied Herglotz pairing with an obstacle far field for available incident directions:Then evaluate its Herglotz wave function throughout a search region and find a smooth positive radial profile satisfyingThe same $h(\theta )$ must work for every azimuth $\varphi$, expressing the assumed axial symmetry. In practice expand $g$ and $h$ in basis functions, enforce these equations at collocation points, and minimize the complex residual together with the far-field-data residual. The exclusion of an interior Dirichlet eigenvalue ensures uniqueness of the interior Dirichlet problem for the trial boundary data. Use regularization of an inverse problem to control measurement error and unstable density components; neither uniqueness nor numerical stability follows merely from writing the residual equations.

For rich incident-direction data, the conjugated pairing constraints form an adjoint far-field integral equation for $g$. For only one fixed incidence, the displayed pairing is a single complex scalar constraint and cannot by itself determine an arbitrary density. Additional data or a joint constrained shape fit is then needed; the trace condition supplies the actual geometric information.

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.

The method of images replaces the earthed plane by an image charge $-q$ at $(-a,0,0)$. The two Coulomb potentials cancel at $x=0$, so uniqueness for the Dirichlet problem makes the resulting field the physical field in $x>0$. With $\mathbf{r}=(x,y,z)$ and ${\mathbf{e}}_x=(1,0,0)$,and

The electric multipole expansion from part (a) gives, for $r\gg a$,Thus the leading field is that of an electric dipole.

On the plane,Taking the plane normal to be $+{\mathbf{e}}_x$ and using polar coordinates,With the outward normal of the region $x>0$, the sign is reversed. This is consistent with Gauss's law: all electric flux from the real charge terminates on the grounded conductor.

The electrostatic boundary condition gives the induced surface charge densityand its integral isIn the plane $z=0$, the field lines leave the positive charge, meet the conductor normally, and are the right-half-plane portions of the field lines joining the real charge to its negative image.

Write $s=\log r$, so the two circular boundaries are $s=0$ and $s=2$. The zero angular mode must interpolate between $-1$ and $1$, giving $s-1=\log r-1$.

For the $n$th cosine mode, the radial factor has equal value $A_n$ at both boundaries. The unique harmonic function with those data isHence the solution of the annular Dirichlet problem isAt $r=1$ and $r=e^2$, the hyperbolic cosine quotient equals one, so the boundary conditions are satisfied term by term.

WriteThis is the harmonic measure of $A$ viewed from $z$, and is a bounded harmonic function outside the closed target disc. The disc is contained in $B(0,r_2)$. In the half-plane cut out by a line $L$ through the origin, the probability of reaching $B(0,r_2)$ before $L$, from a point of modulus $R$, is $O(r_2/R)$ uniformly in the direction. One sees this by mapping the half-plane outside the disc with $z\mapsto \log (z/r_2)$ to a half-strip and solving the corresponding Dirichlet problem by a sine series.

Part (b) consequently shows that the angular oscillation of $H_A$ on a large circle tends to zero, uniformly in the Borel set $A$. The same half-strip estimate in the annulus $R-|a|\le |z|\le R+|a|$ shows that shifting the center of that circle by the fixed vector $a$ changes the average of $H_A$ by $o(1)$, uniformly in $A$. Hence, for every $\epsilon >0$, all sufficiently large $R$ satisfyfor every Borel subset $A$ of the target disc.

The covariance isIndeed, the killed-walk Green function satisfiesOn the other hand, the conditional mean from part 2 makes the covariance harmonic in $x\ne y,a$. At $x=y$, the conditional variance $1/2$ gives the same equation with source $\frac{1}{2}{1}_{\{x=y\}}$. Uniqueness of the Dirichlet problem therefore identifies the covariance with $G_{\{a\}}/2$. The factor $1/2$ comes from the coefficient $1/(2d)$ in this paper's energy density.

Fix a closed ball $\overline{B_R(a)}\subset \Omega$, and let $v$ solve the Dirichlet problem for the Laplace equation in this ball with boundary data $u$. Put $w=u-v$. The function $w$ is continuous, vanishes on the boundary, and inherits the restricted spherical mean identity because $v$ has the full mean value property for harmonic functions.

Suppose $M={\max }_{\overline{B_R(a)}}w>0$. Its maximum set $E$ is a nonempty compact subset of the open ball. Choose $x\in E$ maximizing $|x-a|$. For every sufficiently small radius in the sequence attached to $x$,Equality of the average with the maximum and continuity imply that the whole sphere belongs to $E$. Its point in the direction from $a$ through $x$ lies farther from $a$ than $x$ does; if $x=a$, any point on the sphere does. Both cases contradict the choice of $x$. Hence $w\le 0$, and applying the same argument to $-w$ gives $w=0$.

Fix $v\in C^{1,\beta }(\overline{\Omega })$. Its jet map $x\mapsto (x,v(x),Dv(x))$ is $C^{0,\beta }$, while the barred coefficients are $C^{0,\alpha }$ in all their variables. Composition therefore makes$C^{0,\alpha \beta }$. The image of the jet map is compact, so strict ellipticity on the coefficient domain has a positive uniform lower bound on this image. The assumed linear Dirichlet problem theory now gives a uniquesolving ${\bar{a}}^{ij}(x,v,Dv)D_{ij}u+\bar{b}(x,v,Dv)=0$ with boundary value $\phi$. Thus $T:C^{1,\beta }\to C^{1,\beta }$ is well defined.

Suppose $v_k\to v$ in $C^{1,\beta }$ and write $u_k=T(v_k)$. The preceding uniform Schauder estimate bounds $(u_k)$ in $C^{2,\alpha \beta }$. Every subsequence therefore has a further subsequence converging in $C^{1,\beta }$ by the compact embedding of Hölder spaces. The composed coefficients converge uniformly, and lower-exponent Hölder compactness provides enough convergence of the second derivatives to pass to the linear equation. Every subsequential limit solves the problem defining $T(v)$.

Uniqueness of that linear Dirichlet problem forces every such limit to equal $T(v)$. Since every subsequence has a further subsequence with this same limit, the whole sequence converges to $T(v)$ in $C^{1,\beta }$. Thus $T$ is continuous.

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.