Question

In each exercise, a circular disk of radius \(\rho\) is given, as well as a function \(f(\theta)\) defined on the boundary of the disk. Solve the corresponding Dirichlet problem, $$ \begin{array}{lll} r^{-1}\left(r u_{r}(r, \theta)\right)_{r}+r^{-2} u_{\theta \theta}(r, \theta)=0, & 0

Short answer

Question: Find the solution to the Dirichlet problem with the given partial differential equation and boundary condition in polar coordinates. PDE: \(r^{-1}(r u_{r}(r, \theta))_{r} + r^{-2} u_{\theta \theta}(r, \theta) = 0\), for \(0 < r < 1\) and \(0 \leq \theta \leq 2\pi\) Boundary condition: \(u(1, \theta) = f(\theta)\) for \(0 \leq \theta < 2\pi\) Function on the boundary: For \(0 \leq \theta \leq \pi\), \(f(\theta) = \sin\theta\). For \(\pi < \theta < 2\pi\), \(f(\theta) = 0\). Solution: The solution of the Dirichlet problem for the given PDE and boundary condition is given by the expression: \(u(r,\theta) = \sum_{n=1}^{\infty}\frac{1-(-1)^n}{n}(C_n r^n + D_n r^{-n})\sin(n\theta)\).

Step-by-step solution

Identify the given information

We are given a Dirichlet problem with the following information: 1. The partial differential equation: \(r^{-1}(r u_{r}(r, \theta))_{r} + r^{-2} u_{\theta \theta}(r, \theta) = 0\) for \(0 < r < 1\) and \(0 \leq \theta \leq 2\pi\) 2. The boundary condition: \(u(1, \theta) = f(\theta)\) for \(0 \leq \theta < 2\pi\) 3. The function \(f(\theta)\) on the boundary: For \(0 \leq \theta \leq \pi\), \(f(\theta) = \sin\theta\) For \(\pi < \theta < 2\pi\), \(f(\theta) = 0\)

Solve the PDE using separation of variables

To solve the partial differential equation using separation of variables, we assume the solution is in the form \(u(r, \theta) = R(r)Θ(\theta)\), where \(R(r)\) is the radial part and \(Θ(\theta)\) is the angular part. Substitute the assumed solution into the PDE and divide by the product \(R(r)Θ(\theta)\) to separate variables: \(\frac{r^{-1}(r R'(r))'Θ(\theta) + r^{-2} R(r)Θ''(\theta)}{R(r)Θ(\theta)} = 0\) \(r(r R'(r))'Θ(\theta) + R(r)Θ''(\theta) = 0\) Now, we can separate the equation for each variable: \(\frac{r(r R'(r))'}{R(r)} = -\frac{Θ''(\theta)}{Θ(\theta)} = \lambda\) Here, \(\lambda\) is the separation constant. This gives us two ordinary differential equations (ODEs) for each variable.

Solve the ODEs

The two ODEs from the previous step are as follows: 1. \(r(r R'(r))' - \lambda R(r) = 0\) 2. \(Θ''(\theta) + \lambda Θ(\theta) = 0\) Solve the angular ODE first. For \(\lambda = 0\), \(Θ(\theta) = A\theta + B\). Since it is a periodic function, for a solution to exist, we need \(A = 0\). Thus, \(Θ_0(\theta) = B\). For \(\lambda > 0\), the solution takes the form: \(\Theta(\theta) = A_n \cos(n\theta) + B_n \sin(n\theta)\) for \(n = 1, 2, 3, ...\) Now, solve the radial ODE for each \(\lambda\). For \(\lambda = 0\), the general solution is \(R(r) = C + Dr\ln(r)\). For \(\lambda > 0\), the general solution is \(R(r) = C_n r^n + D_n r^{-n}\). So, we have a general solution \(u(r, \theta) = R(r)Θ(\theta)\).

Apply the boundary condition and write the Fourier series

Applying the boundary condition \(u(1, \theta) = f(\theta)\), we get: \(R(1)Θ(\theta) = f(\theta)\) Since \(R(1) \neq 0\), we can write \(f(\theta)\) as a Fourier series: \(f(\theta) = Σ[a_n \cos(n\theta) + b_n \sin(n\theta)]\) for \(n=0,1,2,...\) Now, we need to find the Fourier coefficients \(a_n\) and \(b_n\). We are given the function \(f(\theta)\) as follows: For \(0 \leq \theta \leq \pi\), \(f(\theta) = \sin\theta\) For \(\pi < \theta < 2\pi\), \(f(\theta) = 0\) All even terms vanish due to the odd symmetry of the function, so \(a_n = 0\) for all n. To find \(b_n\), we use the following formula: \(b_n = \frac{1}{\pi}\int_{0}^{2\pi}f(\theta)\sin(n\theta)d\theta\) For \(0 \leq \theta \leq \pi\), \(b_n = 2\int_{0}^{\pi}\sin(\theta)\sin(n\theta)d\theta\) Solve the integral to find: \(b_n = \frac{1-(-1)^n}{n}\) for \(n=1,2,3,...\) Now, we can write the final solution as: \(u(r,\theta) = \sum_{n=1}^{\infty}\frac{1-(-1)^n}{n}(C_n r^n + D_n r^{-n})\sin(n\theta)\)

Key concepts

Boundary Value Problem

Boundary value problems (BVPs) are a class of mathematical problems in which one seeks a solution to a differential equation subject to certain fixed conditions, known as boundary conditions. In the context of partial differential equations (PDEs), these conditions are typically specified on the boundaries of the domain where the PDE is defined. The Dirichlet problem presented in the exercise is a classic example of a BVP, where the solution is required to take on prescribed values along the boundary of a circular disk. Understanding BVPs is crucial because they appear across various physical and engineering contexts, such as heat conduction, electrostatics, and fluid flow, where the conditions at the boundaries are known, but the interior values must be determined.

In the case of the exercise, the boundary condition is given by the function f(\(\theta\)), which assigns values to the solution at the edge of the disk. Finding a function that satisfies both the PDE and boundary condition can be quite complex. However, methods such as separation of variables are instrumental in simplifying and solving these problems, transforming the PDE into more manageable ordinary differential equations (ODEs).

Partial Differential Equations

Partial differential equations (PDEs) encompass a vast area of mathematics that deals with functions of several variables and their partial derivatives. PDEs form the cornerstone of mathematical physics and modeling, as they describe how quantities like temperature, pressure, displacement, and electromagnetic fields change over space and time. The PDE presented in the exercise is typical of those that arise in circular or spherical domains, reflecting the radially symmetric nature of the problem.

The equation r^{-1}(r u_{r}(r, \(\theta\)))_{r} + r^{-2} u_{\(\theta\theta\)}(r, \(\theta\)) = 0 is known as the Laplace equation in polar coordinates. It serves as a foundation to many physical situations where there is no change over time, and conditions are in steady state. To solve such an equation, one usually employs techniques such as separation of variables to reduce the PDE to much simpler ODEs that can be solved independently. This equation is particularly important in potential theory and has broad implications across various fields in science and engineering.

Separation of Variables

Separation of variables is an elegant mathematical technique often employed to solve PDEs. The method involves assuming that the solution can be written as a product of functions, each depending on only a single coordinate. For instance, in the exercise, the solution u(r, \(\theta\)) is assumed to be of the form R(r)Θ(\(\theta\)), with R being a radial function and Θ being an angular function.

By substituting this product into the original PDE and rearranging, one arrives at separate ODEs for each variable. The beauty of this approach lies in its reduction of multidimensional problems into one-dimensional problems. Each ODE addresses a particular aspect of the problem and can be solved using well-established techniques for ODEs. Especially in problems involving symmetry, like those with circular or spherical domains, separation of variables is particularly powerful and leads to solutions that reflect the symmetry of the problem.

Fourier Series

Fourier series are a powerful tool used to represent periodic functions as infinite sums of sines and cosines. These series are hinged on Fourier's remarkable insight that almost any function can be broken down into a superposition of harmonics. In our exercise, once we have separated the variables, we are left with the task of expressing the boundary values — provided by the function f(\(\theta\)) — as a Fourier series to find the coefficients that will be used in the final solution.

The Fourier series expansion is particularly suitable for solving the Dirichlet problem as it perfectly encapsulates the periodic nature of the boundary values in circular domains. In the context of this exercise, it provides a systematic way of determining the angular part of the solution. Solving for the Fourier coefficients involves integrating the product of the function and a sine or cosine function, over one period. These coefficients tell us how much of each harmonic is present in our boundary function and hence, in our final solution. By considering the symmetry and specific form of f(\(\theta\)), we significantly simplify the process of finding these coefficients, allowing us to write down a complete solution in terms of a Fourier series.