Problem 8 - Advanced Engineering Mathematics

Question

Referring to Example 1 in the text we have \\[R(r)=c_{1} r^{n}+c_{2} r^{-(n-1)} \quad \text { and } \quad \Theta(\theta)=P_{n}(\cos \theta)\\] since we expect \(u(r, \theta)\) to be bounded as \(r \rightarrow \infty,\) we define \(c_{1}=0 .\) Also \(\Theta(\pi / 2)=0\) 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}=(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 is \(u(r, \theta) = \sum_{n=0}^{\infty} A_{2n+1} r^{-2(n+1)} P_{2n+1}(\cos \theta)\) where \(A_{2n+1}\) are specified by integration.

Step 1: Understanding the Given Formulas

We begin by identifying two functions: \(R(r) = c_{1}r^{n} + c_{2}r^{-(n-1)}\) and \(\Theta(\theta) = P_{n}(\cos \theta)\). These describe the radial and angular parts, respectively, in a polar coordinate system.

Step 2: Determine Boundedness Condition at Infinity

Since the solution \(u(r, \theta)\) needs to remain bounded as \(r \to \infty\), we set \(c_{1} = 0\). This simplifies \(R(r)\) to only contain \(r^{-(n-1)}\) terms.

Step 3: Analyze Angular Dependency Condition

Given the condition \(\Theta(\pi / 2) = 0\), it implies that the function \(P_{n}(\cos \theta)\) vanishes for \(\theta = \pi/2\). This typically occurs for odd \(n\), suggesting \(n\) must be odd.

Step 4: Formulate the General Solution for u(r, θ)

The conditions lead to the series sum: \(u(r, \theta) = \sum_{n=0}^{\infty} A_{2n+1} r^{-2(n+1)} P_{2n+1}(\cos \theta)\), where terms are in terms of odd \(n\).

Step 5: Match Boundary Conditions using Coefficients

The boundary condition \(u(c, \theta) = f(\theta)\) leads to \(f(\theta) = \sum_{n=0}^{\infty} A_{2n+1} c^{-2(n+1)} P_{2n+1}(\cos \theta)\).

Step 6: Derive Coefficient Expression

The coefficients \(A_{2n+1}\) can be determined by the orthogonality condition: \(A_{2 n+1} c^{-2(n+1)} = (4n+3) \int_{0}^{\pi/2} f(\theta) \sin \theta P_{2n+1}(\cos \theta) \, d\theta\). Simplifying gives: \(A_{2n+1} = (4n+3) c^{2(n+1)} \int_{0}^{\pi/2} f(\theta) \sin \theta P_{2n+1}(\cos \theta) \, d\theta\).

Key Concepts

Legendre Polynomials

Legendre Polynomials are a set of polynomials that are solutions to Legendre's differential equation. They are quite important in physics and engineering, particularly in problems involving spherical coordinates. These polynomials are denoted by \( P_n(x) \), where \( n \) denotes the

• degree of the polynomial and
• \( x \) is usually \( \, \cos(\theta) \) in many applications involving spherical symmetry.

Legendre polynomials have an interesting property—they can be used to expand functions through a series in which they act as basis functions. This means they are particularly useful for solving boundary value problems where conditions are stipulated around a central axis like a sphere.

To exemplify, in the given exercise involving polar coordinates, the angular part is defined as \( \Theta(\theta)=P_n(\cos \theta) \). The choice of Legendre polynomials ensures that our function is well-suited to the boundary conditions we may later impose, due to the orthogonality property they possess.

Boundary Conditions

In the context of solving differential equations, boundary conditions refer to the specified values or behaviors of the solution at the boundary of the domain. Having proper boundary conditions is crucial for finding the unique solution to a problem.

In our exercise, two boundary conditions are particularly noteworthy:

• As \( r \to \infty \), the solution \( u(r, \theta) \) must remain bounded. To meet this condition, we set \( c_{1} = 0 \) to eliminate terms that grow unbounded.
• The condition \( \Theta(\pi/2) = 0 \) suggests specific angular restrictions. For Legendre polynomials, this condition implies that \( n \) is odd because \( P_n(\cos \theta) \) is zero for odd \( n \) when \( \theta = \pi/2 \).

These boundary conditions guide our approach by determining the form of the solution, ensuring it aligns with the physical context described by the problem.

Orthogonality Condition

The orthogonality condition is a vital mathematical feature that simplifies many problems involving series expansions of functions. Specifically, for Legendre polynomials, the orthogonality property states that the integral of the product of two different Legendre polynomials over the interval

• \([ -1, 1 ]\) is zero unless the two polynomials are the same.

This trait allows us to determine coefficients in a series expansion uniquely, such as in the expansion of a function in terms of Legendre polynomials.

In our exercise, the orthogonality condition helps to solve the given equation by allowing us to find \( A_{2n+1} \). Specifically, the coefficients are computed using the integral:\[A_{2 n+1} c^{-2(n+1)} = (4n+3) \int_{0}^{\pi/2} f(\theta) \sin \theta P_{2n+1}(\cos \theta) \, d\theta\]This relationship leverages the orthogonal nature of \( P_{2n+1}(\cos \theta) \), isolating each coefficient and simplifying the solution process. Such techniques are widely used in solving and analyzing problems with symmetry or periodicity.