Problem 6 - Advanced Engineering Mathematics

Question

Referring to Example 1 in the text we have \\[R(r)=c_{1} r^{n} \quad \text { and } \quad \Theta(\theta)=P_{n}(\cos \theta)\\] Now \(\Theta(\pi / 2)=0\) implies that \(n\) is odd, so \\[u(r, \theta)=\sum_{n=0}^{\infty} A_{2 n+1} r^{2 n+1} P_{2 n+1}(\cos \theta)\\] From \\[u(c, \theta)=f(\theta)=\sum_{n=0}^{\infty} A_{2 n+1} c^{2 n+1} P_{2 n+1}(\cos \theta)\\] we see that \\[A_{2 n+1} c^{2 n+1}=(4 n+3) \int_{0}^{\pi / 2} f(\theta) \sin \theta P_{2 n+1}(\cos \theta) d \theta\\] Thus \\[u(r, \theta)=\sum_{n=0}^{\infty} A_{2 n+1} r^{2 n+1} P_{2 n+1}(\cos \theta)\\] where \\[A_{2 n+1}=\frac{4 n+3}{c^{2 n+1}} \int_{0}^{\pi / 2} f(\theta) \sin \theta P_{2 n+1}(\cos \theta) d \theta\\]

Step-by-Step Solution

Answer

The solution involves representing \(u(r,\theta)\) as a series with coefficients derived from boundary conditions and integrating terms involving \(f(\theta)\).

Step 1: Understanding the Problem

We are given a function in terms of its components \(R(r)\) and \(\Theta(\theta)\). We need to combine them to express \(u(r, \theta)\) as an infinite sum of terms involving \(r\), \(\theta\), and associated Legendre polynomials \(P_{n}(\cos \theta)\).

Step 2: Applying the Condition

The condition \(\Theta(\pi/2) = 0\) implies that only odd \(n\) values are considered, leading us to express \(u(r, \theta)\) with terms involving odd indices: \(2n+1\). Thus, \(u(r,\theta) = \sum_{n=0}^{\infty} A_{2n+1} r^{2n+1} P_{2n+1}(\cos \theta)\).

Step 3: Using the Boundary Condition

The boundary condition \(u(c, \theta) = f(\theta)\) specifies that when \(r = c\), \(u\) equals a given boundary function \(f(\theta)\). It can be represented as a sum involving \(A_{2n+1}c^{2n+1}\).

Step 4: Deriving the Coefficients Expression

By equating the boundary condition to the series expansion, we solve for the coefficients \(A_{2n+1}\). It shows that \(A_{2n+1}c^{2n+1} = (4n+3) \int_{0}^{\pi/2} f(\theta) \sin \theta P_{2n+1}(\cos \theta) d\theta\).

Step 5: Solving for Coefficients

Isolating \(A_{2n+1}\) gives us the final form of coefficients: \(A_{2n+1} = \frac{4n+3}{c^{2n+1}} \int_{0}^{\pi/2} f(\theta) \sin \theta P_{2n+1}(\cos \theta) d\theta\).

Key Concepts

Boundary Conditions in Differential Equations

Boundary conditions are essential tools in solving differential equations. They provide additional information that helps pin down the exact solution to a problem. Imagine being given a recipe without instructions on how long to bake a dish; the boundary conditions are like these crucial instructions, allowing you to "bake" your solution correctly.

In the context of this exercise, we're dealing with a function defined in terms of radius \(r\) and angle \(\theta\). We know from the given problem that at \(r = c\), the solution must become some boundary function \(f(\theta)\). This condition strongly influences how we adjust the coefficients in our infinite series solution.

In essence, boundary conditions ensure our mathematical solutions fit the physical, geometrical, or practical constraints posed by the problem. They help us focus on the realistic solutions out of many potential mathematical solutions.

Infinite Series Solutions

Infinite series are incredibly useful in mathematics, acting as sums of endless sequences to represent functions or solve equations. When we use an infinite series to solve differential equations, we are essentially breaking down complex functions into simpler, digestible parts. This can make otherwise unsolvable problems approachable.

In this exercise, the solution to the differential equation is constructed as an infinite series involving Legendre polynomials, which are special functions often used because of their nice mathematical properties. They have the advantage of being able to represent complicated functions accurately through addition.

Here, only terms where \(n\) is odd are included in the sum, derived from the condition \(\Theta(\pi / 2) = 0\). The series then is a sum of terms like \(A_{2n+1} r^{2n+1} P_{2n+1}(\cos \theta)\), meaning it's comprised only of odd polynomial terms. The constants \(A_{2n+1}\) in front of these terms remind us of how each part of our solution is weighted.

Orthogonality of Functions

Orthogonality is a helpful concept borrowed from geometry. Just as perpendicular vectors have no component of one in the other, orthogonal functions are distinct in a similar sense. This comes with a powerful property: their integrals over a specific interval vanish when multiplied by each other, unless they are the same function.

In this exercise, the orthogonality property of Legendre polynomials plays a key role. It ensures that when we manipulate the coefficients \(A_{2n+1}\), the integration process leading to these coefficients involves a product of the function \(f(\theta)\) with the Legendre polynomial. The result is a straightforward computation due to these orthogonal characteristics.

The orthogonal nature simplifies solving and finding these coefficients because it separates out different "modes" of solution without interference, allowing a clean mathematical separation where each coefficient equals a simple expression involving integrals.