Question

In each exercise, solve the Dirichlet problem for the annulus having a given inner radius \(b\), given outer radius \(a\), and given boundary values \(u(b, \theta)=g(\theta)\) and \(u(a, \theta)=f(\theta)\). $$b=1, a=2, u(1, \theta)=0, u(2, \theta)=1+\cos \theta, \quad 0 \leq \theta \leq 2 \pi$$

Short answer

Question: Solve the Dirichlet problem in the annulus with inner radius 1 and outer radius 2, where the boundary conditions are given by u(1,θ) = 0 for 0≤θ≤2π and u(2,θ) = 1+cos(θ) for 0≤θ≤2π. Answer: The harmonic function u(r,θ) that solves the Dirichlet problem in the given annulus with boundary conditions is: $$u(r, \theta) = \frac{1}{4}(r^{-1}+r) \cos\theta +\frac{1}{2}$$

Step-by-step solution

General Solution of Laplace's Equation in the Annulus

Consider the general solution of Laplace's equation in polar coordinates: $$u(r, \theta) = \sum_{n=-\infty}^{\infty}\left[ A_n r^n + B_n r^{-n} \right] e^{in\theta}$$ $$A_n, B_n \in \mathbb{C}$$ Since we're working in an annular region (b=1 and a=2), we expect u(r,θ) to be single-valued, and hence the coefficients B_n must not contain any negative parts. Thus, we can write the solution as $$u(r, \theta) = \sum_{n=-\infty}^{\infty} A_n r^n e^{in\theta}$$

Applying the Boundary Conditions

Let's apply the given boundary conditions to find the coefficients A_n. Firstly, for the inner boundary (r=1, 0≤θ≤2π), we have $$0=u(1,\theta) = \sum_{n=-\infty}^{\infty} A_n e^{in\theta}$$ Now, let's apply the outer boundary condition (r=2, 0≤θ≤2π): $$u(2,\theta) = 1 + \cos\theta = \sum_{n=-\infty}^{\infty} A_n 2^n e^{in\theta}$$

Finding the Fourier Coefficients

We can find the Fourier coefficients A_n by using the inverse Fourier formula: $$A_n = \frac{1}{2\pi} \int_{0}^{2\pi} u(2,\theta) e^{-in\theta} d\theta$$ For n=-1, 0, and 1 we get: $$A_{-1}=\frac{1}{4}, A_0=\frac{1}{2}, A_1=\frac{1}{4}$$ All other A_n's are zero.

Constructing the Solution

Now, we can write the solution of the Dirichlet problem as a Fourier series that satisfies the boundary conditions: $$u(r, \theta) = \frac{1}{4}(r^{-1}+r) e^{-i\theta} +\frac{1}{2}+\frac{1}{4}(r^{-1}+r) e^{i\theta}$$ Thus, the harmonic function u(r,θ) that solves the Dirichlet problem in the given annulus with boundary conditions is: $$u(r, \theta) = \frac{1}{4}(r^{-1}+r) \cos\theta +\frac{1}{2}$$

Key concepts

Laplace's Equation

Laplace's equation is a second-order partial differential equation named after Pierre-Simon Laplace. It is given by the equation \( abla^2 u = 0 \), where \( abla^2 \) is the Laplacian operator, which represents the sum of the second partial derivatives with respect to each spatial variable. In two dimensions, using Cartesian coordinates (x, y), the Laplacian is \( abla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \).

When solving problems in a circular region, such as the annulus in our exercise, it's more convenient to use polar coordinates (r, \( \theta \) ), in which the Laplace's equation becomes \( abla^2 u = \frac{1}{r}\frac{\partial}{\partial r}\left( r\frac{\partial u}{\partial r} \right) + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} = 0 \).

For a Dirichlet problem, we're finding a function u that satisfies Laplace's equation within a region (the annulus in this case) and takes on specified values at the boundary of the region. The fact that Laplace's equation is linear allows us to find solutions by superposition, meaning we can add solutions together to find a new solution—this is a key feature in constructing solutions using Fourier series.

Fourier Series

A Fourier series is a way to represent a function as an infinite sum of sines and cosines. In the context of the Dirichlet problem, we use Fourier series to represent the solution to Laplace's equation along the boundaries. Mathematically, any periodic function \( f(\theta) \) can be written as \[ f(\theta) = \sum_{n=-\infty}^{\infty}c_n e^{in\theta} \] where \( c_n \) are complex Fourier coefficients determined by the function's values.

In the given problem, we apply the Fourier series approach to deduce the coefficients that satisfy the boundary conditions. These coefficients are then used to construct the function within the annulus. Fourier series is a powerful tool because it transforms complicated or unknown functions into a series of simpler trigonometric functions, which are easier to analyze and compute.

Boundary Conditions

Boundary conditions are constraints necessary to uniquely determine a solution to a differential equation. There are several types of boundary conditions, and in the context of the Dirichlet problem, we are given specific values that the function must take on the boundary of the domain.

For our annulus problem, the inner and outer radii serve as boundaries where our solution, \( u(r,\theta) \) must match given functions (which, in our problem, are \( u(b,\theta) = g(\theta) \) and \( u(a,\theta) = f(\theta) \)).

Applying the boundary conditions allows us to solve for the unknown coefficients in our general solution. This bridges the gap between the abstract form of Laplace's equation and the concrete problem we seek to resolve. Ensuring the function conforms to these conditions at the annulus's edges is what makes the solution relevant to the physical or geometrical context of the problem.

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system where each point on a plane is determined by a distance from a reference point (the origin) and an angle from a reference direction. The coordinates are usually written as \( (r, \theta) \), where \( r \) is the radius or the distance to the point, and \( \theta \) is the angle measured in radians from the reference direction.

In the study of problems involving circular or annular regions, like our Dirichlet problem for an annulus, polar coordinates simplify the equations. This is because the symmetry of these problems corresponds naturally to the radial and angular coordinates of the polar system.

Using polar coordinates, the solution takes advantage of the periodic nature of the angle \( \theta \) and the natural geometry of the problem. As evidenced by the problem at hand, polar coordinates make it easier to apply Fourier series and solve for the coefficients relevant to the boundary conditions.