Question

(a) Find the solution \(u(r, \theta)\) of Laplace's equation in the semicircular region \(r

Short answer

#tag_title# Answer #tag_content# For the given boundary condition \(f(\theta)=\theta(\pi-\theta)\) and \(a=2\), the solution to the Laplace's equation in the semicircular region is given by: $$ u(r, \theta) = -\frac{4}{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^n-1}{n^2}r^n\cos(n\theta) $$ To visualize the solution, plot the function using software like Mathematica, MATLAB or free tools like Desmos or Python libraries (NumPy and Matplotlib) for various values of \(r\) and \(\theta\).

Step-by-step solution

Setting up the equation and boundary conditions

We are given Laplace's equation in polar coordinates: $$ \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} = 0 $$ with boundary conditions: $$ u(r, 0)=0, \quad u(r, \pi)=0, \quad 0 \leq r<a \\ u(a, \theta)=f(\theta), \quad 0 \leq \theta \leq \pi $$

Separation of variables

Assume \(u(r, \theta) = R(r) \Theta (\theta)\). We will now separate variables by plugging this ansatz into Laplace's equation and then dividing by \(R\Theta\): $$ \frac{1}{rR}\frac{d}{dr}\left(r\frac{dR}{dr}\right) + \frac{1}{r^2\Theta}\frac{d^2\Theta}{d\theta^2} = 0 $$ We can now rearrange the equation so that each variable (r or θ) is separated: $$ \frac{r}{R}\frac{d}{dr}\left(r\frac{dR}{dr}\right) + \frac{1}{\Theta}\frac{d^2\Theta}{d\theta^2} = 0 $$

Solving for Θ

Since the left side of the equation contains only r terms and the right side contains θ terms, each side must be equal to a constant. Let's call this constant -m^2: $$ \frac{1}{\Theta}\frac{d^2\Theta}{d\theta^2} = -m^2 $$ This is an ordinary second-order differential equation with solution: $$ \Theta(\theta) = A\cos(m\theta) + B\sin(m\theta) $$

Applying boundary conditions to Θ

Apply the boundary conditions \(u(r,0)=0\) and \(u(r,\pi)=0\). We get: $$ \Theta(0) = A\cos(0) + B\sin(0) = 0 \\ \Theta(\pi) = A\cos(m\pi) + B\sin(m\pi) = 0 $$ Since \(B=0\), we have \(A\cos(m\pi)=0\). Thus, \(m=n\) (an integer), and: $$ \Theta_n(\theta) = A_n\cos(n\theta), \quad n = 1,2,3,\dots $$

Solving for R

From the separated equation for r, we have: $$ \frac{r}{R}\frac{d}{dr}\left(r\frac{dR}{dr}\right) = m^2 $$ This is an Euler's equation and can be solved easily. The solution is given by: $$ R(r) = C_1 r^n + C_2 r^{-n} $$

Overall solution and applying boundary conditions

The product of the two solutions gives the overall solution: $$ u_n(r,\theta) = (C_1 r^n + C_2 r^{-n}) \, A_n \cos(n\theta) $$ The last boundary condition implies that the solution must be bounded, implying \(C_2=0\). Thus: $$ u_n(r,\theta) = A_n r^n\cos(n\theta) $$

Applying Fourier series expansion

We can now expand the given boundary function \(f(\theta)\) in a Fourier series: $$ f(\theta) = \sum_{n=1}^{\infty} a_n \cos(n\theta) $$ From the orthogonality of the cosine functions, we get \(a_n = A_n a^n\). So, we can write the complete solution as: $$ u(r,\theta) = \sum_{n=1}^{\infty} a_n r^n \cos(n\theta) $$

Finding the solution for \(f(\theta)=\theta(\pi-\theta)\)

We are given \(f(\theta)=\theta(\pi-\theta)\). Using the Fourier cosine series formula, we have: $$ a_n = \frac{2}{\pi}\int_0^{\pi}\theta(\pi - \theta) \cos(n\theta) \, d\theta = -\frac{4}{n^2\pi^2}\left((-1)^n-1\right) $$ Therefore, the solution is given by: $$ u(r, \theta) = -\frac{4}{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^n-1}{n^2}r^n\cos(n\theta) $$

Plotting the solution when \(a=2\)

We can plot the above solution for different values of \(r\) and \(\theta\) when \(a=2\). The solution has the following terms: $$ u(r, \theta) = -\frac{4}{\pi^2}\left(\frac{1}{1^2}(2)^1\cos(1\theta) + \frac{1}{2^2}(2)^2\cos(2\theta) +\dots\right) $$ Plot this function for \(u\) vs \(r\), \(u\) vs \(\theta\), \(u\) vs both \(r\) and \(\theta\), and create a contour plot. To do this, you can use a software like Mathematica or MATLAB or free tools like Desmos or Python libraries (NumPy and Matplotlib).

Key concepts

Separation of Variables

Separation of variables is a mathematical method used for solving partial differential equations, like Laplace's equation, by breaking the equation into simpler, separate variables that can each be solved independently. In the context of our problem where we are looking for a function u(r, θ), this approach simplifies the process of finding a solution by assuming that the function is the product of two separate functions: one of r (radial component) and one of θ (angular component).

Using this assumption, we get u(r, θ) = R(r)Θ(θ), and plugging this into Laplace's equation allows us to rearrange terms so that all the r terms are on one side and all the θ terms are on the other. Since the two sides are independent of each other, they must each equal some constant. Thus, we can separately solve the resulting ordinary differential equations for R(r) and Θ(θ).

Fourier Series

Fourier series is a way to represent a function as an infinite sum of sine and cosine terms. When solving boundary value problems, like the one we are tackling with Laplace's equation, Fourier series can be particularly helpful to match boundary conditions given in terms of functions.

In this case, we have the boundary condition u(a, θ) = f(θ) for the semicircle's edge. We can express f(θ) as a Fourier series with coefficients that represent the projection of f(θ) onto the orthogonal functions cos(nθ), thereby finding the necessary a_n to satisfy this boundary condition. Through the superposition principle, we sum these solutions and form a convergent series that represents our solution within the specified domain.

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 and an angle from a reference direction. In our problem, we use polar coordinates (r, θ) due to the circular symmetry of the domain, which is a semicircle.

This approach greatly simplifies the process, especially for Laplace's equation, as it naturally fits problems with circular symmetry. The original Laplace's equation, which is usually expressed in Cartesian coordinates, needs to be transformed into polar form to properly utilize the symmetry and therefore ease the solving process with separation of variables.

Euler's Differential Equation

Euler's differential equation is a type of ordinary differential equation where the function's largest derivative is multiplied by the independent variable. When separated during the separation of variables process, the equation in terms of radial component r can be recognized as an Euler differential equation.

Assuming a solution of the form R(r) = C₁rⁿ, we can substitute this into the Euler's equation and determine the values that satisfy the equation and the boundary conditions. In the context of Laplace's equation and our specific problem, solving this yields the general form for the radial part of the function.

Orthogonality of Functions

Orthogonality of functions is a concept utilized in Fourier series and similar expansions, where functions are orthogonal if their inner product over a specific interval is zero. This property allows us to solve for the coefficients in a Fourier series expansion by taking advantage of the fact that each sine or cosine term in the series will only 'interact' with itself when the boundary conditions are applied.

It simplifies the computation of coefficients in boundary value problems, since we can integrate the product of the given boundary condition with each orthogonal function over the domain, and because of the orthogonality, only the integrals with the same frequency term will remain. Orthogonal functions create a sort of 'coordinate system' for function spaces, allowing us to project the boundary condition onto this space and find the solution in terms of its orthogonal function components.