Question

Find the solution \(u(r, \theta)\) of Laplace's equation in the circular sector \(0

Short answer

#tag_title# Short Answer#tag_content# The solution of Laplace's equation in the circular sector with the given boundary conditions is $$ u(r, \theta) = \sum_{n=1}^{\infty} r^n b_n \sin(\frac{n\pi \theta}{\alpha}) $$ where \(b_n\) are coefficients obtained from the Fourier series of the function \(f(\theta)\) representing the boundary condition at \(r=a\).

Step-by-step solution

Write the Laplace's Equation in Polar Coordinates

The Laplace's equation in polar coordinates is $$\nabla^2 u = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial u}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0$$

Use Separation of Variables

Assume a solution in the form \(u(r, \theta) = R(r)\Theta(\theta)\). Substitute this into the Laplace's equation, then divide by \(R(r)\Theta(\theta)\) to separate the variables: $$\frac{1}{R(r)}\frac{\partial}{\partial r}\left(r \frac{\partial R(r)}{\partial r}\right) + \frac{1}{\Theta(\theta)r^2}\frac{\partial^2 \Theta(\theta)}{\partial \theta^2} = 0$$

Separate into Two ODEs

This equation can be separated into two ordinary differential equations (ODEs) by setting each term equal to a constant: $$ \frac{1}{R(r)}\frac{\partial}{\partial r}\left(r \frac{\partial R(r)}{\partial r}\right) = -\lambda $$ $$ \frac{1}{\Theta(\theta)}\frac{\partial^2 \Theta(\theta)}{\partial \theta^2} = \lambda r^2 $$ Where \(\lambda\) is a constant.

Solve the ODEs

Solve the first ODE for \(R(r)\): $$ \frac{\partial}{\partial r}\left(r \frac{\partial R(r)}{\partial r}\right) = -\lambda R(r) $$ This leads to the following solution: $$ R(r) = A r^\nu + B r^{-\nu} $$ Where \(A\) and \(B\) are constants, and \(\nu= \sqrt{\lambda}\). Solve the second ODE for \(\Theta(\theta)\): $$ \frac{\partial^2 \Theta(\theta)}{\partial \theta^2} = \lambda \Theta(\theta) $$ This has the general solution: $$ \Theta(\theta) = C \cos(\nu \theta) + D \sin(\nu \theta) $$ Where \(C\) and \(D\) are constants.

Apply Boundary Conditions and Fourier Series

Apply the given boundary conditions: $$ u(r, 0) = R(r)\Theta(0) = 0 \\ u(r, \alpha) = R(r)\Theta(\alpha) = 0 \\ u(a, \theta) = R(a)\Theta(\theta) = f(\theta) $$ From the boundary condition \(u(r, 0)\) and \(u(r, \alpha)\), we have either \(\Theta(0) = 0\) or \(\Theta(\alpha) = 0\). From this, we can choose the sine function for \(\Theta(\theta)\), which would satisfy both conditions: $$ \Theta(\theta) = B \sin(\nu \theta) $$ To satisfy the last boundary condition, we can express \(f(\theta)\) as a sine (Fourier) series of the form: $$ f(\theta) = \sum_{n=1}^{\infty} b_n \sin(\frac{n\pi \theta}{\alpha}) $$ Now, we can solve for the general solution by multiplying \(R(r)\) and \(\Theta(\theta)\) and finding coefficients to match the Fourier series: $$ u(r, \theta) = \sum_{n=1}^{\infty} r^n b_n \sin(\frac{n\pi \theta}{\alpha}) $$ This is the solution for \(u(r, \theta)\) that satisfies Laplace's equation and the given boundary conditions.

Key concepts

Polar Coordinates

Polar coordinates offer a two-dimensional coordinate system where each point on a plane is identified by a distance from a reference point and an angle from a reference direction. This system is crucial in solving problems with symmetrical properties, especially those involving circular or spherical shapes.

In the context of Laplace's equation, using polar coordinates simplifies the problem as it aligns with the geometry of a circular sector. Instead of working with Cartesian coordinates (x, y) which would require complex transformations for circular boundaries, we use the more natural (r, \theta) system. Here, 'r' represents the radius or distance from the pole (origin) and '\theta' is the angle formed with the polar axis (typically the positive x-axis in Cartesian coordinates).

Transformation to Polar Coordinates

In the given exercise, the Laplace equation is expressed in polar coordinates as:
\[\begin{equation}abla^2 u = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial u}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0\end{equation}\]This accounts for the curvature of the space in which the function 'u' exists, thus is essential for solving the problem in a circular sector.

Separation of Variables

Separation of variables is a method used to solve partial differential equations (PDEs), such as Laplace's equation, by reducing them to a set of ordinary differential equations (ODEs). This method assumes that the solution to the PDE can be written as the product of functions, each depending on a single variable.

In our problem, the solution is assumed to be in the form of
\[\begin{equation}u(r, \theta) = R(r)\Theta(\theta)\end{equation}\]which is then substituted into the transformed Laplace's equation. By dividing through by the product
\[\begin{equation}R(r)\Theta(\theta)\end{equation}\]we effectively \'separate\' the variables r and \theta.

Deriving Ordinary Differential Equations (ODEs)

The separation leads to two ODEs by setting each part of the equation equal to a constant
\[\begin{equation}\lambda\end{equation}\], effectively uncoupling the variables:
\[\begin{equation}\frac{1}{R(r)}\frac{\partial}{\partial r}\left(r \frac{\partial R(r)}{\partial r}\right) = -\lambda\end{equation}\]and
\[\begin{equation}\frac{1}{\Theta(\theta)}\frac{\partial^2 \Theta(\theta)}{\partial \theta^2} = \lambda r^2\end{equation}\]These ODEs can then be solved independently, which greatly simplifies the process.

Fourier Series

The Fourier series is a powerful tool used to represent a periodic function as an infinite sum of sine and cosine functions. It breaks down any periodic signal into its constituent frequencies, allowing for the analysis and reconstruction of the signal from these basic components.

In the context of boundary value problems, the Fourier series can provide a convenient way to express complex boundary conditions. For the given problem, the boundary function
\[\begin{equation}f(\theta)\end{equation}\]is assumed to be periodic within the interval
\[\begin{equation}\Theta\end{equation}\], and can therefore be represented as a Fourier sine series:
\[\begin{equation}f(\theta) = \sum_{n=1}^{\infty} b_n \sin(\frac{n\pi \theta}{\alpha})\end{equation}\]where
\[\begin{equation}b_n\end{equation}\]are the Fourier coefficients. By satisfying the boundary conditions using the Fourier series, we can match the coefficients to the series to solve for the terms of the solution.

Applying Fourier Series to Boundary Conditions

The general solution for our problem involves matching the found ODE solutions with the Fourier series to satisfy the boundary conditions, culminating in a sum that represents the final solution:
\[\begin{equation}u(r, \theta) = \sum_{n=1}^{\infty} r^n b_n \sin(\frac{n\pi \theta}{\alpha})\end{equation}\]Each term in this series corresponds to a specific mode of the solution that resonates with the geometry and conditions of the problem.