One has for a static charge distribution, ρ(r), that the electric field, E = ∇ϕ, satisfies \[ \nabla\cdot{\bf E} = \rho /\epsilon_0 , \] where ε₀ is the permittivity of free space, a fundamental constant in physics, with a value of 8.854 × 10^-12 C²N^-1m⁻², ρ (rho) represents the volume charge density.

In regions devoid of charge, this equation yields the Laplace equation ∇²ϕ = 0.

Solutions to Laplace’s equation in 2D wedge domains have powerful applications in electrostatics and fluid flow, especially for modeling singularities, boundary layers, and field behavior near corners or edges. These problems arise naturally in engineering, physics, and applied mathematics.

Electrostatics Applications:

1. Electric Field Near Conducting Corners. A wedge domain models the region near the corner of a conducting surface. Solving Laplace’s equation with Dirichlet conditions (fixed potential) or Neumann conditions (specified flux) reveals how the electric potential behaves near edges. For wedge angles α > π, the solution exhibits singular behavior, mimicking field intensification at sharp tips — crucial for understanding corona discharge and dielectric breakdown.

2. Capacitor Edge Effects In parallel plate capacitors with finite edges, the field near the edge resembles a wedge. The wedge solution helps quantify fringing fields, which affect capacitance and energy storage.

3. Electrostatic Shielding and Conformal Mapping Wedge domains are used in conformal mapping techniques to design electrostatic shields and solve problems involving complex geometries. Mapping wedge domains to simpler ones (e.g., half-plane) allows analytic solutions for charge distributions and potential contours.

Electrostatics Applications

1. Field Intensification at Conducting Corners

• A wedge models the region near a sharp conducting corner.
• Solving Laplace’s equation with Dirichlet conditions (fixed potential) shows how the electric field behaves near the tip.
• If the wedge angle α > π, the solution becomes singular near the origin: \[ u(r, \theta) \sim r^{\pi/\alpha} \sin\left(\frac{\pi \theta}{\alpha}\right) \]
• The electric field magnitude \( \displaystyle \quad |\nabla u| \sim r^{\pi/\alpha - 1} \quad \) blows up as r → 0.

2. Fringing Fields in Capacitors

• Near the edge of a parallel plate capacitor, the field lines bend outward.
• The wedge solution helps quantify fringing effects, which influence capacitance and energy storage.
• The asymptotic behavior near the origin (or corner) is critical for electrostatics: it determines field strength and energy density

Sources and Further Reading:

Laplace Equation: Theory and Applications
Laplace Equation in Physics and Engineering
Analytical Solution via Separation of Variables

Example 1:    Let us consider the complex potential \( \displaystyle \quad F(z) = \frac{k}{2\pi}\,\ln \frac{z -a}{z - b} , \quad \) where k takes two values: k = q ∈ ℝ and k = −ⅉq.

With z = x + ⅉy, \begin{align*} \ln \frac{z-a}{z-b} &= \ln \sqrt{(x-a)^2 + y^2} - \ln \sqrt{(x-b)^2 + y^2} \\ & \quad + {\bf j} \mbox{arctan} \frac{y}{x-a} - {\bf j} \mbox{arctan} \frac{y}{x-b} . \end{align*}

For k = q. we have \begin{align*} \psi (x, y) &= \frac{q}{2\pi} \left[ \ln \sqrt{(x-a)^2 + y^2} - \ln \sqrt{(x-b)^2 + y^2} \right] = c_1 , \\ \phi (x, y) &= \frac{q}{2\pi} \left[ \mbox{arctan} \frac{y}{x-a} - \mbox{arctan} \frac{y}{x-b} \right] = c_2 . \end{align*} The potential lines are circles and the streamlines are circular arcs as shown in Figure below. These correspond to a source at z = 𝑎 and a sink at z = b. One can also view these as the electric field lines and equipotentials for an electric dipole consisting of two point charges of opposite sign at the points z = 𝑎 and z = b.

The equations for the curves are found from \begin{align*} \left( x - a \right)^2 + y^2 &= C_1 \left[ (x-b)^2 + y^2 \right] , \\ \left( x - a \right) (x -b) + y^2 &= C_2 y \left( a -b \right) , \end{align*} here these can be rewritten, respectively, in the more suggestive forms \begin{align*} \left( x - \frac{a - C_1 b}{1 - C_1} \right)^2 + y^2 &= \frac{C_1 (a-b)^2}{(1- C_1 )^2} , \\ \left( x - \frac{a+b}{2} \right)^2 + \left( y - \frac{C_2 (a-b)}{2} \right)^2 &= \left( 1 + C_2^2 \right) \left( \frac{a-b}{2} \right)^2 . \end{align*} Note that the first family of curves are the potential curves and the second give the streamlines.

In the case that k = −ⅉq, we have \begin{align*} F(z) &= - \frac{{\bf j}q}{2\pi}\, \ln \frac{z-a}{z-b} \\ &= - \frac{{\bf j}q}{2\pi} \left[ \ln \sqrt{(x-a)^2 + y^2} - \ln \sqrt{(x-b)^2 + y^2} \right] \\ & \quad + \frac{q}{2\pi} \left[ \mbox{arctan} \frac{y}{x-a} - \mbox{arctan} \frac{y}{x-b} \right] . \end{align*} So, the roles of the streamlines and potential lines are reversed and the corresponding plots give a flow for a pair of vortices as shown in the following figure.

The streamlines are found using the identity \[ \mbox{arctan}\alpha - \mbox{arctan}\beta = \mbox{arctan}\frac{\alpha - \beta}{1 + \alpha\beta} . \]    ■

Similarly, we can derive Laplace’s equation for an incompressible, ∇·v = 0, irrotational, ∇×v = 0, fluid flow. From well-known vector identities, we know that ∇ × ∇ϕ = 0 for a scalar function ϕ. Therefore, we can introduce a velocity potential, ϕ, such that v = ∇ϕ. Thus, ∇·v = 0 implies ∇²ϕ = 0. So, the velocity potential satisfies Laplace’s equation.

Fluid flow is probably the simplest and most interesting application of complex variable techniques for solving Laplace’s equation. So, we will spend some time discussing how conformal mappings have been used to study two-dimensional ideal fluid flow, leading to the study of airfoil design.

The study of fluid flow and conformal mappings dates back to Euler, Riemann, and many others. The method was further elaborated upon by physicists like Lord Rayleigh (1877) and applications to airfoil theory we presented in papers by Kutta (1902) and Joukowski (1906) on later to be improved upon by others.

Fluid Flow Applications

1. Stagnation Points and Corner Flow:
In potential flow theory, Laplace’s equation governs the velocity potential. Wedge domains model flow near sharp corners, such as in ducts, nozzles, or around obstacles. Solutions reveal stagnation zones, vortex formation, and pressure singularities.

2. Flow Past Slits and Cracks:
In 2D, flow past a slit or crack is modeled by a wedge with angle α = 2π. The solution describes streamlines and velocity fields, useful in aerodynamics, hydraulics, and microfluidics.

3. Lubrication and Thin Film Flow Wedge-shaped gaps arise in lubrication theory, where fluid flows between surfaces with angular separation. • Laplace solutions help predict pressure distribution, film thickness, and load capacity.

• The asymptotic behavior near the origin (or corner) is critical:
• For fluid flow: it governs velocity gradients and shear stress
• These solutions are foundational in fracture mechanics, MEMS design, antenna modeling, and boundary layer theory

Wedge-shaped gaps arise in lubrication theory, where fluid flows between surfaces with angular separation. Laplace solutions help predict pressure distribution, film thickness, and load capacity.

Example 1:    Suppose that the 2D flow is given by \begin{align*} F(z) &= \frac{U_0 e^{-{\bf j}\alpha}}{z - z_0} \\ &= U_0 \frac{\cos\alpha + {\bf j}\sin\alpha}{(x- x_0 )^2 + (y- y_0 )^2} \left[ ( x- x_0 ) - {\bf j} \left( y - y_0 \right) \right] \\ &= \frac{U_0}{(x - x_0 )^2 + (y - y_0 )^2} \left[ (x - x_0 )\,\cos\alpha + (y- y_0 )\,\sin\alpha \right] \\ &\quad + \frac{{\bf j}\,U_0}{(x - x_0 )^2 + (y - y_0 )^2} \left[ -(y - y_0 )\,\cos\alpha + (x- x_0 )\,\sin\alpha \right] . \end{align*} The level curves become \begin{align*} \phi (x, y) &= \frac{U_0}{(x - x_0 )^2 + (y - y_0 )^2} \left[ (x - x_0 )\,\cos\alpha + (y- y_0 )\,\sin\alpha \right] = c_1 , \\ \psi (x, y) &= \frac{U_0}{(x - x_0 )^2 + (y - y_0 )^2} \left[ -(y - y_0 )\,\cos\alpha + (x- x_0 )\,\sin\alpha \right] = c_2 . \end{align*} The level curves for the stream and potential functions satisfy equations of the form \[ \beta_i \left( \Delta x^2 + \Delta y^2 \right) - \cos (\alpha + \delta_i )\,\Delta x - \sin (\alpha + \delta_i )\,\Delta y = 0 , \] where Δx = x − x₀, Δy = y − y₀, \( \displaystyle \quad \beta_i = c_i /U_0 , \quad \delta_1 = 0, \quad \) and δ₂ = π/2. These expressions can be written in the more suggestive form \[ \left( \Delta x - \gamma_i \cos (\alpha - \delta_i ) \right)^2 + \left( \Delta y - \gamma_i \sin (\alpha - \delta_i \right)^2 = \gamma_i^2 \] for γ_i = c_i/U₀/2,    i = 1, 2. Thus, the stream and potential curves are circles with varying radii (γi) and centers \( \displaystyle \quad \left( (x_0 + \gamma_i \cos (\alpha - \delta_i ) , y + \gamma_i \sin (\alpha - \delta_i ) \right) . \quad \) Examples of this family of curves is shown for α = 0 below.

The components of the velocity field for α = 0 are found from \begin{align*} \frac{{\text d}F}{{\text d}z} &= \frac{{\text d}}{{\text d}z} \left( \frac{U_0}{z - z_0} \right) = - \frac{U_0}{(z - z_0 )^2} \\ &= - U_0 \frac{\left[ (x - x_0 ) - {\bf j} (y - y_0 ) \right]^2}{\left[ (x- x_0 )^2 + (y - y_0 )^2 \right]^2} \\ &= - U_0 \frac{(x - x_0 )^2 + (y - y_0 )^2 - 2 {\bf j} (x- x_0 )(y-y_0 )}{\left[ (x- x_0 )^2 + (y - y_0 )^2 \right]^2} \\ &= - U_0 \frac{(x - x_0 )^2 + (y - y_0 )^2}{\left[ (x- x_0 )^2 + (y- y_0 )^2 \right]^2} + {\bf j} U_0 \frac{2 (x- x_0 )(y - y_0 )}{\left[ (x- x_0 )^2 + (y- y_0 )^2 \right]^2} \\ &= - \frac{U_0}{(x - x_0 )^2 + (y - y_0 )^2} + {\bf j} U_0 \frac{2 (x- x_0 )(y - y_0 )}{\left[ (x- x_0 )^2 + (y- y_0 )^2 \right]^2} . \end{align*} Thus, we have \begin{align*} u &= - \frac{U_0}{(x - x_0 )^2 + (y - y_0 )^2} , \\ v &= U_0 \frac{2 (x - x_0 )(y- y_0 )}{\left[ (x- x_0 )^2 + (y- y_0 )^2 \right]^2} . \end{align*}

Example 2:    Suppose that the flow around a cylinder is defined by \( \displaystyle \quad F(z) = U_0 \left( z + \frac{a^2}{z} \right) , \quad a, U_0 \in \mathbb{R} . \)

We have \begin{align*} F(z) &= U_0 \left( z + \frac{a^2}{z} \right) \\ &= U_0 \left( x + {\bf j}y + \frac{a^2}{x + {\bf j} y} \right) \\ &= U_0 \left( x + {\bf j}y + \frac{a^2}{x^2 + y^2} \left( x - {\bf j} y \right) \right) \\ &= U_0 x \left( 1 + \frac{a^2}{x^2 + y^2} \right) + {\bf j} U_0 y \left( 1 + \frac{a^2}{x^2 + y^2} \right) . \end{align*} The level curves become \begin{align*} \phi (x,y ) &= U_0 x \left( 1 + \frac{a^2}{x^2 + y^2} \right) = c_1 , \\ \psi (x, y) &= U_0 y \left( 1 + \frac{a^2}{x^2 + y^2} \right) = c_2 . \end{align*} Note that for the streamlines when |z| is large, then ψ ∼ U₀y, or horizontal lines. For x² + y² = 𝑎², we have ψ = 0. This behavior is shown below.

 
restart: with(plots): 
k0:=20: 
for k from 0 to k0 do 
    P[k]:=implicitplot(sin(t)*(r-1/r)*1=(k0/2-k)/20, r=1..5, 
    t=0..2*Pi, coords=polar,view=[-2..2, -1..1], axes=none, 
    grid=[150,150],color=black): 
    od: 
display({seq(P[k],k=1..k0)},scaling=constrained);

Example 3:    We consider a slightly modified flow: \( \displaystyle \quad F(z) = U_0 \left( z + \frac{a^2}{z} \right) - {\bf j}\,\frac{\Gamma}{2\pi}\,\ln \frac{r}{a} . \) The combination of the two terms gives the streamlines, \[ \psi (x, y) = U_0 y \left( 1 - \frac{a^2}{x^2 + y^2} \right) - \frac{\Gamma}{2\pi}\,\ln \frac{r}{a} , \] which are seen in Figure below. We can see interesting features in this flow including what is called a stagnation point. A stagnation point is a point where the flow speed, \( \displaystyle \quad \left\vert \frac{{\text d}F}{{\text d}z} \right\vert = 0 . \)

Let us consider a similar flow \[ F(z) = U_0 \left( z + \frac{1}{z} \right) - {\bf j}\,\ln z . \quad \) \] Since the flow speed vanishes at the stagnation points, we consider \[ \frac{{\text d}F}{{\text d}z} = 1 - \frac{1}{z^2} - \frac{\bf j}{z} = 0 . \] This equation can be rewritten as \[ z^2 - {\bf j} z -1 = 0 , \] which has two complex conjugate solutions \( \displaystyle \quad z = \frac{1}{2} \left( {\bf j} \pm \sqrt{3} \right) . \quad \) Thus, there are two stagnation points on the cylinder about which the flow is circulating. These are shown in the following figure.    ■

Its derivation can be obtained via the method of Fourier transforms or conformal mapping from the unit disk to the half-plane.

Recall some properties and interpretation. The kernel \[ P(x - \xi, y) = \frac{1}{\pi} \cdot \frac{y}{(x - \xi)^2 + y^2} \] is the Poisson kernel, a fundamental solution that acts like a “smeared delta function” as y → 0. As y → 0, u(x, y) → f(x) in the sense of boundary limits (pointwise or in 𝔏^p depending on f).

Example: Let’s take \( \displaystyle \quad f(x) = \frac{1}{1 + x^2}. \quad \) Then \[ u(x, y) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{y}{(x - \xi)^2 + y^2} \cdot \frac{1}{1 + \xi^2} \, d\xi \] This integral can be evaluated numerically or symbolically (e.g., via residue calculus or convolution with known transforms).

We use separation of variables in Cartesian coordinates. Assume: \[ u(x, y) = X(x) Y(y) \] Substituting into Laplace’s equation, we obtain \[ X''(x) Y(y) + X(x) Y''(y) = 0 \quad \Rightarrow \quad \frac{X''(x)}{X(x)} = -\frac{Y''(y)}{Y(y)} = -\lambda \] This gives two ODEs: \[ \begin{split} X'' + \lambda X &= 0 , \\ Y'' - \lambda Y &= 0 . \end{split} \] General Solution \[ u(x, y) = \int_0^\infty \left[ A(\lambda) \sin(\sqrt{\lambda} x) + B(\lambda) \cos(\sqrt{\lambda} x) \right] \cdot \left[ C(\lambda) e^{-\sqrt{\lambda} y} + D(\lambda) e^{\sqrt{\lambda} y} \right] \, {\text d}\lambda \] To satisfy decay and boundary conditions, we choose appropriate terms and determine coefficients via Fourier sine/cosine transforms.

Let us use the Conformal Mapping Approach. Alternatively, map the quarter-plane to the upper half-plane using: \[ w = z^2, \quad \text{where } z = x + {\bf j}\,y \in \mathbb{C} . \] We have \[ z = x + {\bf j}\,y \qquad \Longrightarrow \qquad w = x^2 - y^2 + 2{\bf j}xy . \] The boundaries become \[ \begin{split} x > 0,\ y = 0 \qquad \Longrightarrow \qquad w \in \mathbb{R}_+ , \\ x = 0, \ y > 0 \qquad \Longrightarrow \qquad w \in \mathbb{R} . \end{split} \] So the positive real and imaginary axes in z-space map to the positive and negative real axes in w-space. This maps the quarter-plane x > 0, y > 0 to the upper half-plane Im(w) > 0.

Let U(w) be harmonic in ℍ with boundary data: \[ \begin{split} U(\xi) &= f(\sqrt{\xi}) \quad \mbox{for}\quad \xi > 0 , \\ U(\xi) &= g(i\sqrt{-\xi}) \quad \mbox{for} \quad \xi < 0 . \end{split} \] where F(ξ) is the combined boundary data: \[ F(\xi) = \begin{cases} f(\sqrt{\xi}), & \xi > 0 , \\ g(i\sqrt{-\xi}), & \xi < 0 . \end{cases} \] Then apply the Poisson integral formula in w-coordinates: \[ U(w) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\text{Im}(w) \cdot F(\xi)}{(\text{Re}(w) - \xi)^2 + (\text{Im}(w))^2} \, {\text d}\xi , \] Pull Back to the Quarter-Plane Finally, define: \[ u(z) = U(z^2) . \] This gives a harmonic function in the quarter-plane that satisfies the original boundary conditions. \[ u(z) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\text{Im}(w) \cdot F(\xi)}{(\text{Re}(w) - \xi)^2 + (\text{Im}(w))^2} \, {\text d}\xi , \] where F(ξ) is the transformed boundary data.

Example 1:    Let’s take: \[ \begin{split} f(x) &= \sin(ax) , \\ g(y) &= 0 \end{split} \] Then: \[ \begin{split} F(\xi) &= \sin(a\sqrt{\xi})\quad \mbox{for} \quad \xi > 0 , \\ F(\xi) &= 0 \quad \mbox{for}\quad \xi < 0 . \end{split} \] So: \[ U(w) = \frac{1}{\pi} \int_0^{\infty} \frac{\text{Im}(w) \cdot \sin(a\sqrt{\xi})}{(\text{Re}(w) - \xi)^2 + (\text{Im}(w))^2} \, d\xi \] and the solution in the quarter-plane is: \[ u(z) = U(z^2) \] So the solution becomes: \[ u(x, y) = \sin(ax) \cdot e^{-a y} \]

This solution is constructed from the real part of the analytic function: \[ f(z) = z^2 + z^4 = r^2 e^{2{\bf j}\theta} + r^4 e^{4{\bf j}\theta} . \] Taking the real part gives: \[ u(r, \theta) = r^2 \cos(2\theta) + r^4 \cos(4\theta) . \] It satisfies Laplace’s equation in the wedge. It obeys homogeneous Neumann conditions on both boundaries (θ = 0 and θ = π/2) because the angular derivatives vanish there.    ■

We seek a harmonic function u(x, y) such that:

Δ u = ∇²u = 0   in Ω,    Laplace's equation;
u = f    on the boundary ∂ Ω (i.e., along the positive x- and y-axes),    the Dirichlet boundary conditions;
\( \displaystyle \quad u(x, y) \to 0 \mbox{ as } \sqrt{x^2 + y^2} \to \infty , \quad \) regularity conditions,
\( \displaystyle \quad \lim_{r \to 0} u(x, y) \quad \) is bounded (corner condition).

One particular solution is observed \[ u(z) = \frac{1}{2}\,\arg (z) , \qquad z \in \Omega . \] This function is harmonic, shows logarithmic growth at infinity, and satisfies the Dirichlet conditions:     u(x=+0) = 1 and u(y= +0) = 0.

If one tries to employ the separation of variables method, boundary conditions with respect to one independent variable must be homogeneous, It is convenient to use polar coordinates (r, θ). The exterior of the quarter-plane corresponds to: \[ \theta \in \left( \frac{\pi}{2}, 2\pi \right) \] We look for solutions of the form: \[ u(r, \theta) = \sum_{n=1}^\infty A_n r^{-\lambda_n} \sin\left( \lambda_n \left(\theta - \frac{\pi}{2}\right) \right) , \] where \( \displaystyle \quad \lambda_n = \frac{n\pi}{3\pi/2} = \frac{2n}{3} \quad \) The sine term ensures vanishing on the boundaries θ = π/2 and θ = 2π.

For technical reasons, it is preferable to consider quarter IV of the plane instead of the first quadrant; Ω = {(x, y) : x < 0 or y > 0}. The exterior of the fourth quadrant is a wedge of angle 3π/2, bounded by the negative y-axis and positive x-axis.

The boundary value problems in wedge shape domains is best treated with the Conformal Mapping. This technique involves a change independent variable so that the particular domain is mapped into a simpler domain (e.g., upper half-plane or unit disk). In our case of exterior domain to quadrant IV, the transformation \[ w = z^{2/3} , \qquad z = x + {\bf j}\,y \in \mathbb{C} , \] maps the exterior quarter-plane into a half-plane, where the Dirichlet problem is easier to solve using the Poisson integral formula. This map flattens the exterior of the fourth quadrant onto the upper half-plane because the map w = z^{α maps a wedge of angle β to a wedge of angle α β. The boundary rays arg(z) = 0 and arg(z) = 3&pi'/2 map to the real axis in w-space. The interior of the wedge maps to Im(w) > 0.}

Let ϕ(w) be harmonic in the upper half of complex plane ℂ and satisfy: \[ \phi(x) = g(x) \quad \text{on } \mathbb{R} . \] Then: \[ \phi(w) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\operatorname{Im}(w) \cdot g(s)}{(s - \operatorname{Re}(w))^2 + (\operatorname{Im}(w))^2} \, {\text d}s . \]

Now we pull back to the original domain with transformation \[ z = w^{3/2} . \] Then \[ u(z) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\operatorname{Im}(z^{2/3}) \cdot g(s)}{(s - \operatorname{Re}(z^{2/3}))^2 + (\operatorname{Im}(z^{2/3}))^2} \, {\text d}s . \] This is the explicit solution to the Dirichlet problem in the exterior of the fourth quadrant, provided the boundary data f(z) satisfies: \[ f(z) = g(z^{2/3}) \quad \text{on } \partial \Omega \]

For nonhomogeneous Laplace's equation, it is important to know the corresponding Green function. It is a function G(x, y, ξ, η) satisfying the following conditions:

• Δ G = −δ(x − ξ, y − η)    in the exterior of Q. Here δ(·,·) is the Dirac delta function.
• G = 0 on the boundaries x = 0 and x = 0.

Then the Green's function becomes: \[ G(x, y; \xi, \eta) = G_0(x, y; \xi, \eta) - G_0(x, y; -\xi, \eta) - G_0(x, y; \xi, -\eta) + G_0(x, y; -\xi, -\eta) \] This function satisfies the Dirichlet condition on both axes and is valid in the exterior of the quarter-plane. This construction mirrors the classical Sommerfeld method for wedges, adapted to the quarter-plane.

The quarter-plane problem is a classic in potential theory, electrostatics, and elasticity, especially in crack and edge singularity analysis. Techniques go back to Sommerfeld, Muskhelishvili, and Sneddon, who used complex variable methods to handle such domains. The decay condition at infinity is crucial for uniqueness and selects the decreasing radial solutions.

Example 4E.1:    Let’s take: \[ f(z) = 1 \quad \text{on } \partial \Omega \] Then in the upper half-plane, the boundary data becomes: \[ g(s) = 1 \quad \text{for } s \in \mathbb{R} \] The Poisson integral gives: \[ \phi(w) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\operatorname{Im}(w)}{(s - \operatorname{Re}(w))^2 + (\operatorname{Im}(w))^2} \,{\text d}s = 1 \] So the solution is: \[ u(z) = \phi(z^{2/3}) = 1 \] This makes sense: the harmonic extension of constant boundary data is constant.

Example 4E.2:    Let’s take linear boundary data \[ f(z) = \begin{cases} 0 & \text{on } x > 0, y = 0 \\ 1 & \text{on } x = 0, y < 0 \end{cases} \] This is a step function on the boundary: 0 on the positive real axis, 1 on the negative imaginary axis.

First, we map boundary to real axis \[ \begin{split} &\mbox{ On } x > 0, y = 0: z = x \quad \Rightarrow \quad w = x^{2/3} \in \mathbb{R}^+ \\ &\mbox{ On } x = 0, y < 0: z = {\bf j} y, y < 0 \quad \Rightarrow \quad z = r e^{-i\pi/2} \quad \Rightarrow \quad w = r^{2/3} e^{-i\pi/3} \in \mathbb{R}^- \end{split} \] So the boundary data in w-space becomes: \[ g(s) = \begin{cases} 1 & s < 0 \\ 0 & s > 0 \end{cases} \] This is the Heaviside step function H(-s).

Poisson Integral in upper half-plane \[ \phi(w) = \frac{1}{\pi} \int_{-\infty}^{0} \frac{\operatorname{Im}(w)}{(s - \operatorname{Re}(w))^2 + (\operatorname{Im}(w))^2} \, ds \] This evaluates to: \[ \phi(w) = \frac{1}{\pi} \tan^{-1} \left( \frac{\operatorname{Im}(w)}{\operatorname{Re}(w)} \right) + \frac{1}{2} \] Pull Back to original domain \[ u(z) = \phi(z^{2/3}) = \frac{1}{\pi} \tan^{-1} \left( \frac{\operatorname{Im}(z^{2/3})}{\operatorname{Re}(z^{2/3})} \right) + \frac{1}{2} \] This is a smooth harmonic function in the exterior of the fourth quadrant, interpolating between 0 and 1 on the boundary. It satisfies the Dirichlet boundary conditions:

• u = 0 on x > 0, y = 0,
• u = 1 on x = 0, y < 0.

Near the corner of the fourth quadrant (i.e., near the origin), the behavior of the harmonic solution u(z) reveals the singular structure induced by the reentrant angle of &frac32;.

Lets ana;yze the asymptotic behavior of function u(z near the origin. Let \( \displaystyle \quad z = r\, e^{{\bf j}\theta}, \quad \) with r → 0⁺ and θ ∈ (0, 3π/2). Then: \[ z^{2/3} = r^{2/3} e^{{\bf j} \cdot \frac{2}{3} \theta} . \]; So: u(z) = \frac{1}{\pi} \cdot \frac{2}{3} \theta + \frac{1}{2} = \frac{2\theta}{3\pi} + \frac{1}{2} . This shows that near the origin, the solution behaves like a linear function of the angle θ, with a discontinuity across the branch cut (from θ = 0 to θ = 3π/2.

• The solution has a logarithmic-type singularity at the origin, typical of wedge domains.
• The angular derivative of u is discontinuous across the boundary rays.
• The gradient blows up near the corner, reflecting the stress concentration in analogous elasticity problems.

Example 4E.3:    Let us consider the Dirichlet problem problem with the boundary data be: \[ f(z) = \begin{cases} 1, & z \in \text{segment on } y = 0, \text{ say } 0 < x < a, \\ 0, & \text{elsewhere on } \partial \Omega \end{cases} \] This is a finite step function on the positive real axis. We use the conformal mapping \[ w = z^{2/3} = e^{(2/3)\ln z} . \] This transformation maps the wedge of angle 3π/2 (i.e., the exterior of the fourth quadrant) to the upper half-plane.

• The boundary rays arg(z) = 0 and arg(z) = 3π/2 map to the real axis.
• The segment 0 < x < 𝑎 on the real axis maps to 0 < w < 𝑎^2/3.

Pull back to original domain with substitution w = z^2/3, then the solution in thegiven Dirichlet problem becomes \[ u(z) = \frac{1}{\pi} \left[ \tan^{-1} \left( \frac{a^{2/3} - \operatorname{Re}(z^{2/3})}{\operatorname{Im}(z^{2/3})} \right) - \tan^{-1} \left( \frac{-\operatorname{Re}(z^{2/3})}{\operatorname{Im}(z^{2/3})} \right) . \right] . \] Setting \[ z = r\,e^{{\bf j}\theta} \quad \Longrightarrow \quad z^{2/3} = r^{2/3} e^{(2{\bf j}/3) \theta} = r^{2/3} \left[ \cos \left( \frac{2\theta}{3} \right) + {\bf j}\,\sin \left( \frac{2\theta}{3} \right) \right] , \] we get \[ u(r, \theta ) = \frac{1}{\pi} \left[ \mbox{arctan} \left( \frac{a^{2/3} - r^{2/3}\cos\frac{2\theta}{3} }{r^{2/3} \sin \frac{2\theta}{3} } \right) + \mbox{arctan} \left( \frac{\cos \frac{2\theta}{3}}{\sin \frac{2\theta}{3}} \right) . \right] . \]

This is the explicit harmonic solution in the exterior of the fourth quadrant with a finite step boundary condition. The solution is smooth in the domain, despite the jump on the boundary. Near the corner (origin), the solution exhibits angular singularity typical of wedge domains. The gradient is concentrated near the support of the step, modeling localized potential or heat input.    ■

Let z = x + ⅉy ∈ ℂ with ⅉ² = −1. We define a complex potential Φ(z) such that: \[ \phi(x, y) = \text{Re}[\Phi(z)] . \] Φ(z) is analytic in the domain excluding the crack. A canonical solution for a semi-infinite slit is: \[ \Phi(z) = \sqrt{z} = z^{1/2} . \] This function is analytic in the complex plane cut along the negative real axis. The square root introduces a branch cut, modeling the discontinuity across the crack.

Real part: \( \displaystyle \quad \phi(x, y) = \text{Re}[\sqrt{x + {\bf j}\,y}] . \quad \) This function has a singularity at the tip of the crack z = 0, the gradient ∇ ϕ becomes singular, reflecting stress concentration or field intensification.

A semi-infinite slit in a conducting plane can represent a line of discontinuity in potential or surface charge. Model: The potential \( \displaystyle \quad \phi(x, y) = \text{Re}[\sqrt{z}] \quad \) solves Laplace’s equation in the plane with a branch cut along the negative real axis. Interpretation: The slit behaves like a line of charge or a discontinuity in the electric field. The square-root singularity at the tip models field intensification.

2. Edge Effects and Field Enhancement Near the tip of the slit (crack), the electric field \( \displaystyle \quad \vec{E} = -\nabla \phi \quad \) becomes singular. This models edge effects in capacitors, electrodes, or sharp conductors where charge accumulates and the field intensifies.

3. Conformal mapping to slit domains. Conformal maps (e.g., \( \displaystyle \quad w = \sqrt{z}) \quad \) transform simple domains (upper half-plane) to slit domains.

This approach is useful for solving problems with boundary conditions on slits, such as: Potential prescribed on one side of the slit; or zero normal field across the slit (ideal insulator)

4. Green’s Function in Slit Domains: The Green’s function for Laplace’s equation in a slit domain can be constructed using the method of images or conformal mapping. This enables computation of potential due to point charges near slits or cracks.

5. Electrostatic Analogy in Fracture Mechanics The same mathematical structure governs stress fields near crack tips in elasticity and electric fields near slit tips in electrostatics. The stress intensity factor in fracture mechanics has an electrostatic analogue: the field intensity near a sharp edge.

Example: Potential near a slit. Let the slit lie along x < 0, y = 0. Then: \[ \phi(x, y) = \text{Re}[\sqrt{x + {\bf j}y}] = \sqrt{r} \cos\left(\frac{\theta}{2}\right) , \] where (r, θ) are polar coordinates centered at the tip. This shows that the singularity is: \( \displaystyle \quad \phi \sim \sqrt{r} \quad \) near the tip Field: \( \displaystyle \quad |\vec{E}| \sim \frac{1}{\sqrt{r}}, \quad \) diverging as r → 0.    ▣

Let ϕ(x, y) satisfy Laplace’s equation: \[ \nabla^2 \phi = 0 \quad \text{in } \mathbb{R}^2 \setminus \text{(semi-infinite slit)} , \] with the slit along x < 0, y = 0, and Dirichlet boundary conditions: \[ \phi(x, 0^+) = f(x), \quad \phi(x, 0^-) = g(x), \quad x < 0 . \] The goal is to find ϕ(x, y) in the upper and lower half-planes, consistent with these boundary values.

Classical Solution Strategy.

1. Complex Variable Formulation Let z = x + ⅉy ∈ ℂ with ⅉ² = −1, and seek a harmonic function ϕ\phi(x, y) = Re[\Φ(z)], where Φ(z) is analytic in ℂ ∖ (-∞, 0].

2. Conformal Mapping. Use the mapping: \[ w = \sqrt{z} \] which maps the slit domain to the right half-plane. The slit becomes the imaginary axis in the w-plane.

3. Poisson Integral in Half-Plane In the w-plane, solve Laplace’s equation using the Poisson integral for the right half-plane: \[ \Phi(w) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\text{Re}(w) \cdot h(t)}{(t - \text{Im}(w))^2 + \text{Re}(w)^2} \, {\text d}t , \] where h(t) is the transformed boundary data from f(x) and g(x).

4. Transform back to the z-plane using z = w². The solution \( \displaystyle \quad \phi(x, y) = \text{Re}[\Phi(\sqrt{z})] \quad \) satisfies Laplace’s equation and the Dirichlet conditions on the slit.

Special Case: Symmetric Slit Conditions. If f(x) = g(x), the slit has continuous potential across it. Then the solution simplifies to: \[ \phi(x, y) = \text{Re} \left[ \frac{1}{\pi} \int_{-\infty}^{0} \frac{\sqrt{z} \cdot f(t)}{(t - \text{Im}(\sqrt{z}))^2 + \text{Re}(\sqrt{z})^2} \, {\text d}t \right] \]

Historical Context. Muskhelishvili developed integral representations for such problems in elasticity, which directly apply to electrostatics. Sneddon used Fourier transforms and dual integral equations to handle general boundary conditions on cracks. Sommerfeld pioneered the use of branch cuts and Riemann surfaces for diffraction and potential problems.

u(x, 0) = 0 on x ≥ 1; u(x, 0) = 1 on x ≤ −1.

The domain looks superficially suitable for separation of variables in Cartesian coordinates, but the boundary conditions are not suitable: we would need u(x, 0) to be prescribed for all x for separation to work.

Instead we use the map \( \displaystyle \quad w(z) = z + \sqrt{z^2 -1} . \quad \) Note that the square root means this map is not analytic over the whole plane; we need a branch cut at each of z = 1, z = −1. Given the domain, we are trying to transform, it makes sense to put the branch cuts on y = 0 (or z real) and |x| ≥ 1 (or |z| ≥ 1).

The point z = 0 maps to \( \displaystyle \quad w = \sqrt{(-1)} \quad \) and we can choose which of the possible values we take for the sign of the square root here: we choose w(0) = ⅉ. This choice, with the positioning of the branch cuts, determines w(z) everywhere in our domain. So when z = x + ⅉε and x > 1, both roots are positive; when z = x and |x| < 1, the root at z = 1 has argument ⅉ and the other is positive; when z = x − ⅉε with x < −1, both roots have argument ⅉ so the product is negative, and so on:

In particular, \[ \begin{split} &w(-1) = -1, \\ &w(x) = x + {\bf j} \sqrt{1 - x^2} , \\ &w(x+{\bf j}\varepsilon ) = x - \sqrt{x^2 -1} , \\ &w(x-{\bf j}\varepsilon ) = x + \sqrt{x^2 -1} . \end{split} \] Thus, the branch cuts in the z-plane map onto the real line in the w-plane: the left-hand cut maps to (top side) w < −1 and (bottom side) −1 < w < 0, and the right-hand cut maps to (top side) w > 1 and (bottom side) 0 < w < 1. In the w-plane we now need to solve Laplace’s equation for a new function v with

v(x, 0) = 1 on x ≤ 0     v(x, 0) = 0 on x ≥ 0.

This new problem is suitable for separation of variables in polar coordinates: the boundary conditions in terms of r and θ are

v(r, 0) = 0     v(r, π) = 1.

Note that our domain now does not encircle the origin, so we must revisit our separable solution. We look for the form v = R(r) Θ(θ) and derive the coupled ODEs \[ \frac{r^2 R′′(r) + r\,R′(r)}{R(r)} = \lambda , \qquad \frac{\Theta′′(θ)}{\Theta (\theta )} = -\lambda . \] In the three cases λ < 0, λ > 0 and λ = 0 respectively these yield: \begin{align*} v &= (A_µ \exp [µθ] + B_µ \exp [−µθ])(C_µ \cos [µ ln r] + D_µ \sin [µ ln r]) \\ v &= \left( a_λ \cos [λθ] + b_λ \sin [λθ]\right)\left( c_λr^λ + d_λ r*{−λ} \right) , \\ v &= (α + β ln r)\left( γ + δθ\right). \end{align*} Applying the boundary condition Θ(θ = 0) = 0 gives the three basis functions \begin{align*} v &= \sinh [µθ]\left( C_µ \cos [µ ln r] + D_µ \sin [µ ln r]\right) . \\ v &= \sin [λθ]\left( c_λ r^{λ} + d_λ r^{−λ} \right) , \\ v &= θ(α + β ln r), \end{align*} and the condition that v must be well-behaved (bounded) at r = 0 (since the origin is in our domain) fixes further: \[ v = \alpha\theta + \sum_{\lambda} c_{\lambda} r^{\lambda} \sin \lambda\theta . \] The final boundary condition v(r, π) = 1 gives \[ 1 = \alpha\pi + \sum_{\lambda} c_{\lambda} r^{\lambda} \sin \lambda\pi , \] which is satisfied with α = 1/π and c_λ = 0. Hence, we have found \[ v(r, \theta ) = \frac{\theta}{\pi} . \] In order to convert this to a solution to our original problem, we first need to find the analytic function of which it is the real part. In this case, the function is straightforward: \[ v(r, \theta ) = \frac{\theta}{\pi} = - \frac{1}{\pi}\,\mbox{Re}\left[ {\bf j} \left[ \ln r + {\bf j}\theta \right] \right] = - \frac{1}{\pi}\,\mbox{Re}\left[ {\bf j} \ln w \right] , \] so the analytic function we need is \[ g(w ) = -\frac{{\bf j}\,ln w}{\pi} . \] Finally we need to convert back to the original variables: \] f (z) = g \circ w(z) = − \frac{\bf j}{\pi} \ln \left\{ z + \sqrt{z^2 -1} \right\} \] and the solution we need is the real part of this \begin{align*} u(x,y) &= \mbox{Re} \left( -\frac{\bf j}{\pi}\,\ln \left\{ z + \sqrt{z^2 -1} \right\} \right) \\ &= \frac{1}{\pi}\,\mbox{Im} \left( \ln \left\{ z + \sqrt{z^2 -1} \right\} \right) . \end{align*} In particular, on the “missing line” y = 0, −1 ≤ x ≤ 1, we have \[ u(x, 0) = \frac{1}{\pi}\,\mbox{Arg} \left( \left\{ x + {\bf j}\sqrt{1 - x^2} \right\} \right) = \frac{1}{\pi}\,\mbox{arctan} \frac{\sqrt{1 - x^2}}{x} . \]    ■

• the Dirichlet boundary condition specifies on the crack: u = f(x).
• The Neumann boundary condition specifies the normal derivative on the slit: ∂u/ ∂n = g(x) on the slit
• Or mixed conditions of different types (seee Duffy's book).

Let z = x + ⅉy ∈ ℂ with ⅉ² = −1. The key idea is to find a complex potential w(z) = ϕ(x, y) + ⅉψ(x, y) such that:
w(z) is analytic in ℂ ∖ Γ, so ϕ = Re(w) is harmonic and satisfies the boundary conditions,

We find a Canonical (conformal) mapping: \[ \zeta = \sqrt{z^2 - a^2} . \] This analytic function maps the complex plane with a slit [-𝑎, 𝑎] onto the complex plane cut along the negative real axis. The square root introduces a branch cut along the slit, which is essential for capturing the singular behavior near crack tips.

Example: Constant Displacement on the slit. Suppose we want u = 1 on the slit [-𝑎, 𝑎], and u → 0 at infinity. Then the solution becomes: \[ u(x, y) = \text{Re} \left[ \frac{1}{\pi} \cos^{-1} \left( \frac{z}{a} \right) \right] . \] This function is harmonic in ℂ ∖ [-𝑎, 𝑎], equals 1 on the slit, and decays at infinity.    ▣

Singularity at Crack Tips. Near the crack tips z = ± 𝑎, the solution behaves like: \[ u(x, 0) \sim \sqrt{a - x} \quad \text{as } x \to a^- . \] This square-root singularity is fundamental in fracture mechanics, representing stress concentration near crack tips.

References:

• Muskhelishvili, Some Basic Problems of the Mathematical Theory of Elasticity
• Sneddon, Mixed Boundary Value Problems in Potential Theory
• Analytic Solutions to Laplace’s Equation in 2D (Exeter Notes).

For the Dirichlet boundary value problem, we seek a harmonic function u(x, y) such that:
u = f(x) on the slit (Dirichlet condition);
u → 0 as |z| → ∞;
u is harmonic in ℂ ∖ Γ.

We use the fact that the real or imaginary part of any analytic function w(z) is a harmonic function. Let u(x, y) be a harmonic function in the wedge Ω. Then there exists a harmonic conjugate v(x, y) such that \[ f (z) = u(x, y) + {\bf j}\, v(x, y) , \qquad z = x + {\bf j}\,y \in \mathbb{C} , \] is an analytic function in the wedge domain. The key idea is to map the wedge domain to a simpler domain (like the upper half-plane) using a conformal map, solve the problem there, and then pull back the solution.

Let z = x + ⅉy = r e^ⅉθ be a complex variable with ⅉ (also denoted by j) being the imaginary unit so ⅉ² = −1. Note that mathematicians prefer to use Euler's notation i for this unit. The wedge domain corresponds to \[ \arg(z) \in (0, \alpha) . \] To simplify the geometry, we map the wedge Ω to the upper half-plane H = { w ∈ ℂ : Im(w) > 0 } using \[ w = z^{\pi/\alpha} = e^{(\pi /\alpha) \ln z} . \tag{6.1} \] The branches of an analytic power function are the single-valued functions that result from a multivalued power function by selecting a specific "sheet" or range of the argument. Each branch is defined over a domain that excludes a branch cut, which is a curve where the function is discontinuous. For a power function z^c with a non-integer exponent c, the branches can be obtained by choosing a specific value for the logarithm Lnz = ln|z| + ⅉ(arg(z) + 2πk), where k is an integer.

This power transformation (6.1) is conformal (analytic and invertible) in the wedge domain. It maps

• The ray θ = 0 to the positive real axis;
• the ray θ = α to the negative real axis;
• the interior of the wedge to the upper half-plane.

Now we solve in the upper half-plane problem. In the w-plane, we seek an analytic function U(w) such that u = ReU(z^π/α)exp(ⅉπθ/α), and such that \[ U(w) = g(w)\ \text{on}\ \mathbb{R} , \] where g(w) is the transformed boundary data: \begin{align*} g(w) &= f(z(w)) = f(w^{\alpha/\pi}) \\ &= \begin{cases} f_1(r), & \text{if } w = r^{\pi/\alpha} \text{ for } \theta = 0 \\ f_2(r), & \text{if } w = r^{\pi/\alpha} e^{{\bf j}\pi} = - r^{\pi/\alpha} \text{ for } \theta = \alpha . \end{cases} \end{align*} The solution in the upper half-plane H is given by the Poisson integral: \[ U(x + {\bf j}y) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{y\, g(t)}{(x - t)^2 + y^2} {\text d}t . \tag{6.2} \] Finally, we pull back the solution in the original wedge domain by applying the inverse conformal transformation \[ u(z) = U(z^{\pi/\alpha}) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\operatorname{Im}(z^{\pi/\alpha})\, f(t)}{(\operatorname{Re}(z^{\pi/\alpha}) - t)^2 + (\operatorname{Im}(z^{\pi/\alpha}))^2} {\text d}t . \tag{6.3} \] This formula gives an explicit integral representation of the harmonic function in the wedge domain. \[ u(z) = \Re\left[ U\left( z^{\pi/\alpha} \right) \right] = \mbox{Re} \left[ U\left( z^{\pi/\alpha} \right) \right] . \] This elegant method dates back to Riemann and Schwarz, and was popularized in the context of electrostatics and fluid flow. The wedge geometry is particularly amenable to complex variable techniques due to the power mapping z^π/α, which linearizes the angular domain. For wedges with angle α = π/n, n ∈ ℤ, the mapping becomes w = z^ₙ, simplifying computations. If the wedge is exterior (i.e., α > π), branch cut and multivaluedness must be handled carefully.

Example 6.1:    Let us consider the wedge of angle α = π, with boundary data u(r, 0) = 0, u(r, π) = 1. Since this Dirichlet problem is linear, we can always assume that that the boundary condition at one ray is homogeneous because the solution to this problem is a sum of two solution with one boundary data zero.

Since the wedge \( \displaystyle \quad \Omega = \{ z = r e^{{\bf j}\theta} \mid r > 0,\ 0 < \theta < \pi \} \quad \) is already the upper half-plane, we don't need to use the conformal transformation. So we can apply formula (6.3). The harmonic function with these boundary conditions becomes \[ u(x + iy) = \frac{1}{\pi} \left[ 1 - \tan^{-1}\left( \frac{x}{y} \right) \right] = \frac{1}{\pi}\, \arg(z) . \]

This function u(x, y) is bounded in a neighbohood of the origin, but it has no limit as |z| → 0.

• At the origin z = 0, arg(z) is undefined.
• The function u(z) has a jump discontinuity across the negative real axis.
• The gradient ∇ u ∼ 1/r near the origin, so the energy integral diverges: \[ \int_{W} |\nabla u|^2\, dx\,{\text d}y = \infty . \]

Now we shift the boundary data and consider the Dirichlet problem:

This finction satisfies the corner condition: \( \displaystyle \quad \lim_{|z|\to 0} u(x,y) = 0 . \quad \)

Example 6.2:    Let us consider the wedge domain with angle α = π/2 (see Example 6.4 above). The Dirichlet boundary conditions are as foolows:

• θ = 0 ⟶ arg(w) = 0,
• θ = π/2 ⟶ arg(w) = π.

• u(z) = 0 on arg(w) = 0,
• u(z) = ln(r) = ½ ln(|w|) on arg(w) = π.

Example 6.3:    Let us consider a wedge of arbitrary angle α; with boundary conditions \[ u(r, 0) = 0,\quad u(r, \alpha) = 1, \] the corresponding Dirichlet problem has the explicit solution: \[ u(r, \theta) = \frac{1}{\pi} \arg(z^{\pi/\alpha}) = \frac{\theta}{\alpha} . \] This is a harmonic function and it satisfies the boundary conditions, but: \[ \nabla u = \frac{1}{\alpha r} \hat{\theta} . \] So \( \displaystyle \quad |\nabla u|^2 = \frac{1}{\alpha^2 r^2} . \quad \) The energy integral diverges near the origin: \[ \int_0^\epsilon \int_0^\alpha |\nabla u|^2 r\, d\theta\, dr = \infty . \] This is a non-integrable singularity — the harmonic function is smooth away from the origin but has infinite energy due to the angular mismatch.

Example 6.4:    We consider a wedge domain of angle π/4: \[ \Omega = \left\{ z = r\,e^{{\bf j}\theta} \in \mathbb{C}\ :\ 0 \lt; \yjeya < \\pi /4 \right\} \] Let us consider the wedge domain with angle α = π/4

• θ = 0 ⟶ arg(w) = 0,
• θ = π/4 ⟶ arg(w) = π.

• The function u(r, θ) = (4/π)*θ is linear in angle and independent of radius.
• The surface plot shows how the potential increases with angle.
• The contour plot reveals straight radial level curves — consistent with constant angle lines.

• \( \displaystyle \quad \hat{e}_r \quad \) is the unit vector in the radial direction;
• \( \displaystyle \quad \hat{e}_\theta \quad \) is the unit vector in the angular direction (perpendicular to \( \displaystyle \quad \hat{e}_r , \quad \) pointing counterclockwise).

Let u(x, y) be harmonic in the wedge. Then there exists a harmonic conjugate v(x, y) such that \[ f(z) = u(x, y) + {\bf j}\, v(x, y) \] is analytic in the wedge domain. The key idea is to map the wedge conformally to a simpler domain (typically the upper half-plane or strip), solve the problem there, and pull back the solution.

Let \( \displaystyle \quad z = r\, e^{{\bf j}\theta} \in \mathbb{C} . \quad \) Let us define the transformation: \[ w = z^{\pi/\alpha} \] This maps the wedge 0 < θ < α to the upper half-plane Im(w) > 0. The Laplace equation is preserved under conformal maps (up to scaling), and Neumann conditions transform accordingly.    ■

We are looking for partial nontrivial solutions of Laplace's equationn \[ \nabla^2 \phi = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \phi}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 \phi}{\partial \theta^2} = 0 \] in terms of products of two functions: ϕ(r, θ) = R(r) Θ(θ). Laplace’s equation in polar coordinates will be reduced to the following ODE’s for R(r) and Θ(θ) \[ \frac{1}{r}\,\frac{\text d}{{\text d}r} \left( r\,\frac{{\text d}R(r)}{{\text d}r} \right) \Theta (\theta ) + \frac{R(r)}{r^2}\, \frac{{\text d}^2 \Theta}{{\text d}\theta^2} = 0 , \] which leads to separation of variables: \[ r\, \frac{\text d}{{\text d}r} \left( r\,\frac{{\text d}R(r)}{{\text d}r} \right) = -\frac{{\text d}^2 \Theta}{{\text d}\theta^2} = \lambda . \] So we get two ordinary differential equations \[ r^2 R'' + r\, R' - \lambda\,R = 0 , \qquad \Theta'' + \lambda\,\Theta = 0 . \] Noticing that the homogeneous Dirichlet boundary conditions on the lateral sides of the wedge imply Θ(0) = Θ(α) = 0, we will have the eigenvalue problem \[ \begin{cases} \Theta'' + \lambda\,\Theta = 0 , \\ \Theta (0) = \Theta (\alpha ) = 0 . \end{cases} \] Since λ = 0 is not an eigenvalue, this Sturm--Liouville problem has the following eigenfunctions and eigenvalues: \[ \lambda_n = \left( \frac{n\pi}{\alpha} \right)^2 , \quad \Theta_n (\theta ) = \sin \left( \frac{n\pi\theta}{\alpha} \right) , \qquad n= 1,2,3,\ldots . \] The R-equation is the Euler equation; therefore, it has a power form solution when &nambda; = λₙ: \[ R_n (r) = A_n r^{n]pi /\alpha} + B_n r^{-n]pi /\alpha} , \] with some constants Aₙ and Bₙ. The Bₙ term is not defined at the origin, which is a corner point of the wedge. Thus, this term is discarded, since we are looking for a harmonic function that is also continuous on the boundary (this can be thought of as a boundary condition at r = 0 that the solution is finite). Now using the separated solutions Rₙ(r) &Thetaₙ(θ), we can write the series solution as \[ \phi (r, \theta ) = \sum_{n\ge 1} R_n (r)\,\Theta_n (\theta ) = \sum_{n\ge 1} A_n r^{n\pi /\alpha} \sin \left( \frac{n\pi\theta}{\alpha} \right) . \] The coefficients Aₙ are determined by the boundary condition on the boundary r = 𝑎. Indeed, checking this condition gives \[ \phi (a, \theta ) = h(\theta ) = \sum_{n\ge 1} A_n a^{n\pi /\alpha} \sin \left( \frac{n\pi\theta}{\alpha} \right) , \] which implies \[ A_n a^{n\pi /\alpha} = h_n = \frac{2}{\alpha} \int_0^{\alpha} h(\theta ) \,\sin \left( \frac{n\pi\theta}{\alpha} \right) {\text d}\theta . \tag{7.1} \] Hence, \[ A_n = \frac{h_n}{a^{n\pi /\alpha}} = a^{-n\pi /\alpha} \frac{2}{\alpha} \int_0^{\alpha} h(\theta ) \,\sin \left( \frac{n\pi\theta}{\alpha} \right) {\text d}\theta , \qquad n=1,2,\ldots . \] So the solution to the Dirichlet problem in wedge domain Ω is \[ \phi (r, \theta ) = \sum_{n\ge 1} h_n \left( \frac{r}{a} \right)^{n\pi /\alpha} \sin \left( \frac{n\pi\theta}{\alpha} \right) , \tag{7.2} \] with hₙ being the Fourier sine coefficients of the Dirichlet data h(θ).

Neumann problem:   

page 157 of Carrier's book    ■

1. Carrier, G.F., Krook, M., and Pearson, C.E., Functions of a Complex Variable: Theory and Technique, Society for Industrial and Applied Mathematics, 2005.
2. Kam-Tim Chau, Analytic Methods in Geomechanics (Chapter 9), CRC Press,
3. Duffy, D. G., Mixed boundary value problems, 2008, Chapman & Hall/CRC, Boca Raton, FL.
4. England, A.H., Complex Variable Methods in Elasticity, Dover Publications, 2003. ISBN-13 ‏ : ‎ 978-0486432304
5. Hewett, D.P. and Morris, A., Diffraction by a right-angled impedance wedge: An edge source formulation, The Journal of the Accoustical Society of America, 2015, Volume 137, Issue 2,
6. Kwok, Y.K., Applied Complex Variables for Scientists and Engineers, second edition, Campridge University Press.
7. Muskhelishvil, N.I., Some basic problems of the mathematical theory of elasticity, Springer, 1977.
8. Needham, T., Visual Complex Analysis, Osford University Press,
9. Osipov, A. V., & Norris, A. N. (1999). The Malyuzhinets theory for scattering from wedge boundaries: A review. Wave Motion, 29(4), 313-340. https://doi.org/10.1016/S0165-2125(98)00042-0
10. Prusov, A.A., Complex Analysis and Special Functions with Applications, De Gruyter, 2022.
11. Prusov, I.A.., Method of Conjugation in the Theory of Plates,
12. Serov, V. and Harju, M., Complex Analysis and Special Functions: Cauchy Formula, Elliptic Functions and Laplace’s Method (De Gruyter Textbook) 1st Edition De Gruyter, 2025. ISBN-13 ‏ : ‎ 978-3111632117
13. Sneddon, I. N., Mixed Boundary Value Problems in Potential Theory (North-Holland, 1966).