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Introduction to Linear Algebra with Mathematica

Laplace Equation


Let’s look at Laplace’s equation in 2D, using Cartesian coordinates:
\[ \Delta\,f = \nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0 , \qquad (x,y) \in \Omega . \]
We introduce new variables
\[ \begin{split} \eta &= x + {\bf j}\,y , \\ \xi &= x - {\bf j}\,y ; \end{split} \qquad \begin{split} x &= \frac{1}{2} \left( \eta + \xi \right) , \\ y &= \frac{1}{2{\bf j}} \left( \eta - \xi \right) , \end{split} \]
where j (also denoted by ⅉ) is the imaginary unit on the complex plane ℂ, so j² = −1. Then the chain rule gives
\[ \frac{\partial}{\partial x} = \frac{\partial}{\partial \eta} + \frac{\partial}{\partial \xi} , \qquad \frac{\partial}{\partial y} = {\bf j} \left( \frac{\partial}{\partial \eta} - \frac{\partial}{\partial \xi} \right) , \]
and the Laplace equation becomes
\[ \Delta \,f= \nabla^2 f = 4\,\frac{\partial}{\partial \eta} \, \frac{\partial f}{\partial \xi} = 0 . \]
Hence, its solution follows:
\[ f = p(\eta ) + q(\xi ) = p \left( x + {\bf j}\,y \right) + q \left( x - {\bf j}\,y \right) , \]
where p and q are differentiable complex functions; and assuming we wanted a real solution to the original (real) PDE, we have an additional constraint that the sum of the two functions must have no imaginary part. We can formalise this in more standard notation: if we use the (x, y) plane to represent the complex plane in the usual way, we introduce the complex variable z = x + jy. Then its complex conjugate is \( \displaystyle \quad \oveline{z} = z^{\ast} = x − {\bf j}\,y \quad \) and the solution we have just found is
\[ f = p(z ) + q(\overline{z} ) = p \left( z \right) + q \left( z^{\ast} \right) , \]
Note that in may other areas of science such as physics and engineering, complex conjugate number is denoted with asterisk rather than overline as in mathematics. Since function p(η) does not depend on ξ, we have
\[ \frac{\partial p}{\partial \xi} = \frac{\partial p}{\partial \overline{z}} = 0, \]
and using the chain rule, we obtain
\[ \frac{1}{2}\,\frac{\partial p}{\partial x} - \frac{1}{2{\bf j}}\, \frac{\partial p}{\partial y} = 0 \qquad \Longrightarrow \qquad \frac{\partial p}{\partial x} = - {\bf j}\, \frac{\partial p}{\partial y} . \]
If we separate the real part amd the imaginary part of function p(z), then
\[ p(z) = u(x, y) + {\bf j}\, v(x, y) , \]
where u and v are real-valued functions, and we get
\[ \frac{\partial u}{\partial x} + {\bf j}\, \frac{\partial v}{\partial x} = - {\bf j}\, \frac{\partial u}{\partial y} + \frac{\partial v}{\partial y} . \]
This complex equation is equivalent to the pair of real equations:
\[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} , \qquad \frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y} , \]
known as the Cauchy-Riemann equations, and are satisfied by the real and imaginary parts of any differentiable function of a complex variable z = x + ⅉy = x + jy. In fact in a given domain, u and v (continuously differentiable) satisfy the Cauchy-Riemann equations if and only if p is an analytic function of z. For proof, see Chapter 1 of APMA Fundamentals. Recall f(z) is analytic ≡ holomorphic within a domain Ω if, in every circle |zz₁| < ε lying in Ω, f can be represented as a power series in z.

The Laplace equation in polar coordinates (r, θ) can be written as

\[ \Delta\,f = \nabla^2 f = \frac{1}{r}\,\frac{\partial}{\partial r} \left( r\,\frac{\partial f}{\partial r} \right) + \frac{1}{r^2} \,\frac{\partial^2 f}{\partial \theta^2} = 0 . \]
It was shown in the previous section that the Laplace equation has a periodic with period 2π solution in whole plane except the origin (more precisely, in any circle 0 < |z| < R of arbitrary radius R) of the form:
\[ f(r, \theta ) = A + B\,\ln r + \sum_{n\ge 1} \left( a_n \cos (n\theta) + b_n \sin (n\theta)\right) \left( c_n r^n + d_n r^{-n} \right) \]
In complex polar coordinates \( \displaystyle \quad z = r\, e^{{\bf j}\theta} , \quad \) this solution has the follwoing form:
\[ f(r, \theta ) = \mbox{Real} \left\{ A + B\,\ln r + \sum_{n\ge 1} \left[ c_n \left( a_n - {\bf j} b_n \right) z^n + d_n \left( a_n + {\bf j} b_n \right) z^{-n} \right] \right\} . \]
This means that our solution is the real part of a function of z only:
\[ f(r, \theta ) = \mbox{Real} \left\{ g(z) \right\} , \]
where g(z) is a holomorphic function in any simply connected domain that does not include the origin; if B = 0 it is analytic everywhere except the origin, and if additionally dₙ = 0, it is analytic everywhere. We have shown that the real solution to Laplace’s equation we had found is the real part of an analytic function of z = x + iy in our domain; we can show the converse very quickly from the Cauchy-Riemann equations. Consider an analytic function

A function f: ℝ → ℝ is called conformal if it preserves angles.

However, what does it mean to “preserve angles”? In the Euclidean norm, the angle between two vectors is defined by their dot product. Nevertheless, most analytic maps are nonlinear, and so will not map vectors to vectors since they will typically map straight lines to curves. However, if we interpret “angle” to mean the angle between two curves†, then it makes sense of the conformality requirement. Thus, in order to realize complex functions as conformal maps, we first need to understand their effect on curves. In general, a curve C ∈ C in the complex plane is parametrized by a complex-valued function

\[ z(t) = x(t) + {\bf j} y(t), \qquad a ≤ t ≤ b, \]
that depends on a real parameter t. Note that there is no essential difference between a complex curve and a real plane curve; we have merely switched from vector notation x(t) = (x(t), y(t)) to complex notation z(t) = x(t) + ⅉy(t). All the usual vectorial curve terminology — closed, simple (non-self intersecting), piecewise smooth, etc. — is employed without modification. In particular, the tangent vector to the curve at the point z(t) = x(t) + ⅉy(t) can be identified with the complex number \( \displaystyle \quad \dot{z}(t) = \dot{x}(t) + {\bf j}\dot{y}(t). Smoothness of the curve is guaranteed by the requirement that \( \displaystyle \quad \dot{z}(t) \ne 0. \)

When we interpret the curve as the trajectory of a particle in the complex plane, so that z(t) is the position of the particle at time t, the tangent \( \displaystyle \quad \dot{z}(t) \quad \) represents its instantaneous velocity. The modulus of the tangent, \( \displaystyle \quad |\dot{z}(t)| = \sqrt{\dot{x}^2 (t) m+ \dot{y}^2 (t)} , \quad \) indicates the particle’s speed, while its phase ph\( \displaystyle \dot{z}(t) \quad \) measures the direction of motion, as prescribed by the angle that the curve makes with the horizontal.

The (signed) angle between between two curves is defined as the angle between their tangents at the point of intersection, \( \displaystyle \quad z = z_1 (t_1 ) + z_2 (t_2 ) . \quad \) If the curve C₁ is an angle \( \displaystyle \theta_1 = \mbox{ph}\dot{z}_1 (t_1 ) \quad \) while the curve C₂ is at angle \( \displaystyle \theta_2 = \mbox{ph}\dot{z}_2 (t_2 ) , \quad \) then the angle θ between C₁ and C₂ at z is their difference

\[ \theta = \theta_2 - \theta_1 = \mbox{ph}\dot{z}_1 - \mbox{ph}\dot{z}_2 = \mbox{ph}\left( \frac{\dot{z}_2}{\dot{z}_1} \right) . dot{z}_2 - \]

Plane problems


If f(z) is holomorphic (analytic) in a domain Ω ⊂ ℂ, then both its real part Re(f(z)) and imaginary part Im(f(z)) are harmonic functions in Ω, that is, they satisfy the Laplace equation
\[ \Delta\,u = \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 , \qquad (x,y) \in \Omega . \]
This means every analytic function yields two solutions to Laplace’s equation, and conversely, every harmonic function locally corresponds to the real or imaginary part of some analytic function.

You are probably familiar with the logarithm function f(z) = Ln(z), z ∈ ℂ. This analytic functions consists of infinite many holomorphic functions, called branches, defined not on whole complex plane, but on some part of it, usually obtained by deleting semi-infinite slit. Let ln(z) be one its branch. Then

  • \( \displaystyle \quad \Re(f(z)) = \mbox{Re}(f(z)) = \ln |z| \quad \) is harmonic → radial potential,
  • \( \displaystyle \quad Im(f(z)) = \mbox{Im}(f(z)) = \arg(z) \quad \) is harmonic → angular potential.
These harmonic functions solve Laplace’s equation in polar coordinates and model, for instance, the potential around a point charge or vortex.    
Example 1: Let ϕ(r) be the electric potential (also called the electric field potential, potential drop, the electrostatic potential), which is the difference in electric potential energy per unit of electric charge between two points in a static electric field. More precisely, electric potential is the amount of work needed to move a test charge from a reference point to a specific point in a static electric field, normalized to a unit of charge. The test charge used is small enough that disturbance to the field-producing charges is unnoticeable, and its motion across the field is supposed to proceed with negligible acceleration, so as to avoid the test charge acquiring kinetic energy or producing radiation.

RJB
Figure 1: electric potential around two oppositely charged conducting spheres

One has for a static charge distribution, ρ(r), that the electric field, E = ∇ϕ, satisfies \[ \nabla\cdot{\bf E} = \rho /\epsilon_0 , \] where

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

   ■
End of Example 1
Modeling of irrotational, incompressible flows is simplified because the velocity field, v, can be expressed as the gradient of a scalar velocity potential, ϕ (i.e., v = ∇ϕ), and the continuity equation for an incompressible fluid becomes ∇·ϕ = 0. As a result, flow potentials are harmonic functions.

The velocity potential is not uniquely defined since one can add to it an arbitrary function of time without affecting the relevant physical quantity. The non-uniqueness is usually removed by suitably selecting appropriate initial or boundary conditions.    

Example 2:

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.

Mathematical and Engineering Insight:
  • 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.    ■

End of Example 2
   
Example 3: We consider the upper half-plane: \[ \mathbb{H} = \{ (x, y) \in \mathbb{R}^2 \mid y > 0 \} \] and seek a harmonic function u(x, y) satisfying Laplace’s equation: \[ \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \quad \text{in } \mathbb{H} \] and the Dirichlet boundary condition on the real axis: \[ u(x, 0) = f(x), \quad x \in \mathbb{R} \] We also need to impose the decay condition: \[ u(x, y) \to 0 \quad \text{as } x,y \to \infty \] The classical solution is given by the Poisson kernel for the upper half-plane: \[ u(x, y) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{y \, f(\xi)}{(x - \xi)^2 + y^2} \, d\xi \] This integral represents the harmonic extension of f(x) into the upper half-plane.

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 𝔏p 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).

Integrate[y/((x - t)^2 + y^2)/(1 + t^2), {t, -Infinity, Infinity}]
(\[Pi] y (x^2 (1 + Sqrt[1/y^2]) - (-1 + Sqrt[1/y^2]) (-1 + y^2)))/((1 + x^2)^2 + 2 (-1 + x^2) y^2 + y^4)
\[ u(x,y) = \frac{x^2 \left( y+1 \right) - (1-y) \left( y^2 -1 \right)}{(x^2 +1)^2 + 2 \left( x^2 -1 \right) y^2 + y^4} . \]    ■
End of Example 3
   
Example 4: We consider the quarter-plane domain: \[ \Omega = \{ (x, y) \in \mathbb{R}^2 \mid x > 0, y > 0 \} \] and seek a function u(x, y) such that Laplace’s equation holds: \[ \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \quad \text{in } \Omega . \] Dirichlet boundary conditions are prescribed: \[ u(x, +0) = f(x), \quad u(+0, y) = g(y) . \] Decay condition (optional): \[ u(x, y) \to 0 \quad \text{as}\quad x, y \to \infty \] This boundary value problem can be solved using Method of Reflection and Separation of Variables.

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: 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: \[ u(x, y) = \sin(ax) \cdot e^{-a y} \]

RJB
This function satisfies Laplace’s equation and the boundary conditions.

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.    ■

End of Example 4
   
Example 4E:    ■
End of Example 4E
   
Example 5:    ■
End of Example 5
   
Example 6: We consider the Laplace equation in polar coordinates: \[ \Delta u = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial u}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0 \] in a wedge domain: \[ \Omega = \{ (r, \theta) \mid 0 < r < \infty,\ 0 < \theta < \alpha \} \] with the Dirichlet boundary conditions: \[ u(r, +0) = f_1(r), \quad u(r, \alpha -0) = f_2(r) . \] We use the fact that the real part of any analytic function w(z) is harmonic. Let u(x, y) be harmonic in the wedge. Then there exists a harmonic conjugate v(x, y) such that \[ (z) = u(x, y) + {\bf j}\, v(x, y) \] 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 = reⅉθ be the complex variable with ⅉ being the imaginary unit so ⅉ² = −1. Some times, we will denote this unit vector of complex plane ℂ as j; note that mathematicians prefer to use Euler's notation i for this unit. The wedge domain corresponds to: \[ \arg(z) \in (0, \alpha) . \] We map this wedge to the upper half-plane using: \[ w = z^{\pi/\alpha} . \] This maps the wedge arg(z) ∈ (0, α) to arg(w) ∈ (0, π), i.e., 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(zπ/α)exp(ⅉπθ/α). Let U(w) be harmonic in the upper half-plane with boundary values prescribed on the real axis. The Dirichlet problem in the upper half-plane can be solved using the Poisson integral formula: \[ U(x + {\bf j}y) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{y\, g(t)}{(x - t)^2 + y^2} {\text d}t , \] where g(t) is the boundary data on the real axis and j is the imaginary unit (also denoted by ⅉ), so j² = −1.

To use this formular, we need to express the boundary data f₁(r) and f₂(r) in terms of the real axis of the w-plane. That is, we define: \[ g(w) = \begin{cases} f_1(r), & \text{if } w = r^{\pi/\alpha} \text{ for } \theta = 0 \\ f_2(r), & \text{if } w = r^{\pi/\alpha} e^{i\pi} \text{ for } \theta = \alpha \end{cases} . \] Finally, we pull back the solution. Once U(w) is found, the solution in the original wedge domain is: \[ u(z) = \Re[U(z^{\pi/\alpha}) . \] This 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^{\pi/\alpha}, which linearizes the angular domain

Connection to separation of variables:    Both methods, Analytic Functions and Eigenfunction Expansions, solve the same PDE — Laplace’s equation — but they approach it from different angles:

  • Complex variable method: uses conformal mapping and analytic functions, exploiting the fact that harmonic functions are real parts of holomorphic functions.
  • Separation of variables: uses orthogonal eigenfunction expansions in polar coordinates, typically involving sine and cosine functions in θ and powers of r.
Despite their differences, they are intimately connected. In polar coordinates, Laplace’s equation becomes: \[ \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0 \] Assume a solution of the form \( \displaystyle \quad u(r, \theta) = R(r)\Theta(\theta). \quad \) Plugging into the PDE and separating variables gives: \[ r^2 \frac{R''}{R} + r \frac{R'}{R} = -\frac{\Theta''}{\Theta} = \lambda \] This leads to two ODEs: Angular part: \( \displaystyle \quad \Theta'' + \lambda \Theta = 0 , \)
Radial part: \( \displaystyle \quad r^2 R'' + r R' - \lambda R = 0. \)

The general solution is: \[ u(r, \theta) = \sum_{n=1}^\infty A_n r^{n\pi/\alpha} \sin\left( \frac{n\pi \theta}{\alpha} \right) . \] This function satisfies the Dirichlet conditions u(r, 0) = 0 and u(r, α) = 0, and the coefficients Aₙ are determined by expanding the boundary data on θ = θ₀ in a sine series.

Connection to Complex Variables: Now we compare this expression to the solution obtained with the aid of the complex variable method. The conformal map \( \displaystyle \quad w = z^{\pi/\alpha} \quad \) transforms the wedge to the upper half-plane. In the upper half-plane, harmonic functions can be represented as real parts of analytic functions. The general form of an analytic function in the upper half-plane (or wedge) is: \[ f(z) = \sum_{n=1}^\infty a_n z^{n\pi/\alpha} \] \[ u(r, \theta) = \sum_{n=1}^\infty a_n r^{n\pi/\alpha} \sin\left( \frac{n\pi \theta}{\alpha} \right) \] This matches the separated solution exactly — the powers r^{n\pi/\alpha} and angular modes \sin(n\pi\theta/\alpha) or \cos(n\pi\theta/\alpha) arise naturally in both approaches.

The general solution is: \[ u(r, \theta) = \sum_{n=1}^\infty A_n r^{n\pi/\alpha} \sin\left( \frac{n\pi \theta}{\alpha} \right) \] This function satisfies the Dirichlet conditions u(r, 0) = 0 and u(r, α) = 0, and the coefficients Aₙ are determined by expanding the boundary data on θ = &theta:₀ in a sine series.

Connection to Complex Variables Now compare this to the complex variable method: The conformal map \( \displaystyle \quad w = z^{\pi/\alpha} \quad \) transforms the wedge to the upper half-plane. In the upper half-plane, harmonic functions can be represented as real parts of analytic functions. The general form of an analytic function in the upper half-plane (or wedge) is: \[ f(z) = \sum_{n=1}^\infty a_n z^{n\pi/\alpha} , \] whose real part is:

This function satisfies the Dirichlet conditions u(r, +0) = 0 and u(r, α −0) = 0, and the coefficients A_n are determined by expanding the boundary data on θ = θ_0 in a sine series. 🧠 Connection to Complex Variables Now compare this to the complex variable method: The conformal map w = z^{\pi/\alpha} transforms the wedge to the upper half-plane. In the upper half-plane, harmonic functions can be represented as real parts of analytic functions. The general form of an analytic function in the upper half-plane (or wedge) is: \[ f(z) = \sum_{n=1}^\infty a_n z^{n\pi/\alpha} \] whose real part is:

Philosophical Insight: The separation of variables builds the solution from orthogonal modes — it's spectral. The complex variable method builds the solution from analytic structure — it's geometric. But both converge to the same harmonic function, just viewed through different lenses.    ■

End of Example 6
   
Example 6N: We consider the Laplace equation in a wedge domain: \[ \Delta u = 0 \quad \text{in} \quad \Omega = \{(r,\theta) \mid 0 < r < \infty, \ 0 < \theta < \alpha\} \] with Neumann boundary conditions: \[ \frac{\partial u}{\partial {\bf n}}(r, 0) = g_0 (r), \quad \frac{\partial u}{\partial {\bf n}}(r, \alpha) = g_{\alpha} (r) \] and possibly a condition at infinity or at a circular arc r = R → ∞, and corner condition at r = 0;. Here g₀ and gα are given functions of r, and n is a unit vector directed outward the wedge domain. It is well-known that the gradient operator in polar coordinates is \[ \nabla f = \frac{\partial f}{\partial r} \, \hat{e}_r + \frac{1}{r} \frac{\partial f}{\partial \theta} \, \hat{e}_\theta , \] where:
  • \( \displaystyle \quad \hat{e}_r \quad \) 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).
Then we can rewite the Neumann boundary conditions in terms of angular derivatives: \[ \frac{\partial u}{\partial \theta}(r, +0) = -r\,g_0 (r), \quad \frac{\partial u}{\partial \theta}(r, \alpha -0) = r\,g_{\alpha} (r) , \] because the outward normal derivative is proportional to θ-direction on ray θ = α and is opposite at ray θ = 0.    ■
End of Example 6N
   
Example 6M:    ■
End of Example 6M
   
Example 6F: We are looking for a harmonic function ϕ(r, θ) that is a solution to the Dirichlet problem in the finite wedge domain: \[ \nabla^2 \phi = 0 , \qquad \phi_{|z| = a} = h , \quad \phi_{\arg z = 0} = k_0 , \quad \phi_{\arg z = \alpha} = k_{\alpha} . \] where h : [0, α] ↦ ℝ is a given function. \[ \phi (r, \theta ) = \frac{a_0}{2} + \frac{a_0}{2}\,\ln r + \sum_{k\ge 1} \left(a_k \, r^k + a_k^{\ast}\, r^{-k} \right) \cos (k\theta ) + \sum_{k\ge 1} \left(b_k\, r^k + b_k^{\ast}\, r^{-k} \right) \sin (k\theta ) \]    ■
End of Example 6F
   
Example 7: Let D be the domain in the complex plane ℂ exterior to a slit AA₂ along the real axis from x = −𝑎 to x = 𝑎
RJB
Figure 1: Slit in ℂ

page 157 of Carrier's book    ■

End of Example 7

Kolosov–Muskhelishvili formalism


Kolosov–Muskhelishvili formalism Classical method using complex potentials to solve 2D problems in plane elasticity.    
Example 4:    ■
End of Example 4

 

  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. England, A.H., Complex Variable Methods in Elasticity, Dover Publications, 2003. ISBN-13 ‏ : ‎ 978-0486432304
  4. Muskhelishvil, N.I., Some basic problems of the mathematical theory of elasticity, Springer, 1977.
  5. Needham, T., Visual Complex Analysis, Osford University Press,
  6. Prusov, A.A., Complex Analysis and Special Functions with Applications, De Gruyter, 2022.
  7. Prusov, I.A.., Method of Conjugation in the Theory of Plates,
  8. 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

 

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