Chapter 12: Problem 1

By a method similar to that we used to show that the \(P_{l^{\prime}}\) s are an orthogonal set of functions on \((-1,1),\) show that the solutions of \(y_{n}^{\prime \prime}=-n^{2} y_{n}\) are an orthogonal set on \((-\pi, \pi) .\) Hint: You should know what functions the solutions \(y_{n}\) are; do not use the functions themselves, but you may use their values and the values of their derivatives at \(-\pi\) and \(\pi\) to evaluate the integrated part of your equation.

Short Answer

Expert verified

The solutions \(y_n = A \cos(nx) + B \sin(nx)\) to \(y_n'' = -n^2 y_n\) are orthogonal on \((-\pi, \pi)\). The integration by parts and boundary conditions ensure the orthogonality for different \(n\) values.

Step by step solution

- Identify Solutions

First, identify that the equation \(y_n'' = -n^2 y_n\) is a second-order differential equation with sinusoidal solutions. The general solutions are \(y_n = A \cos(nx) + B \sin(nx)\).

- Apply Boundary Conditions

Consider the interval \((-\pi, \pi)\). We'll use that the solutions and their derivatives are periodic with period \(2\pi\). Specifically, for any function \(y_n\), we have \(y_n(-\pi) = y_n(\pi)\) and \(y_n'(-\pi) = y_n'(\pi)\).

- Orthogonality Condition

We need to show that \(\int_{-\pi}^{\pi} y_m y_n \, dx = 0\) for \(m eq n\). This relies on integrating by parts and utilizing boundary conditions.

- Integrate by Parts

Consider the integral \(I = \int_{-\pi}^{\pi} y_n y_m \, dx\). Use integration by parts on \(u=y_n\) and \(dv=y_m dx\). Thus, \(I = \left. y_n y_m \right|_{-\pi}^{\pi} - \int_{-\pi}^{\pi} y_n' y_m' \, dx\).

- Evaluate Boundary Terms

The boundary terms \(\left. y_n y_m \right|_{-\pi}^{\pi} = y_n(\pi) y_m(\pi) - y_n(-\pi) y_m(-\pi)\) are zero because of the periodicity constraints: \(y_n(\pi) = y_n(-\pi)\) and \(y_m(\pi) = y_m(-\pi)\).

- Inner Integral Simplification

Now the integral simplifies to \(-\int_{-\pi}^{\pi} y_n' y_m' \, dx\). Again integrate by parts on \(y_n'\) and \(y_m'\) and repeat the boundary condition arguments to show that the integral vanishes if \(m eq n\).

- Conclude Orthogonality

Since the cross terms cancel and the integral is zero if \(m eq n\), thus the solutions \(y_n'' = -n^2 y_n\) are orthogonal on \((-\pi, \pi)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second-order Differential Equations

Second-order differential equations are equations that involve the second derivative of a function. In general, they can be written as: \[ y'' + p(x)y' + q(x)y = r(x) \]In this example, our equation is: \[ y_n'' = -n^2 y_n \]This is a simple form where the second derivative of the function is proportional to the function itself, scaled by \(-n^2\). In the context of this exercise, the solutions are sinusoidal functions, specifically:\[ y_n = A \cos(nx) + B \sin(nx) \]Here, \( A \) and \( B \) are constants that depend on initial or boundary conditions.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and is given by the formula:\[ \int u \, dv = uv - \int v \, du \]In our problem, we use integration by parts to simplify the integral that checks for orthogonality. We let \( u = y_n \) and \( dv = y_m \, dx \). This gives us:\[ I = \int_{-\pi}^{\pi} y_n y_m \, dx = \left. y_n y_m \, \right|_{-\pi}^{\pi} - \int_{-\pi}^{\pi} y_n' y_m' \, dx \]The term \( \left. y_n y_m \, \right|_{-\pi}^{\pi} \) vanishes due to the periodic boundary conditions, leaving us with an integral involving the derivatives of \( y_n \) and \( y_m \).

Boundary Conditions

Boundary conditions are conditions imposed on the solutions of a differential equation at the boundaries of the domain. In our case, the boundary conditions are:\[ y_n(-\pi) = y_n(\pi) \]\[ y_n'(-\pi) = y_n'(\pi) \]These conditions come from the periodic nature of sinusoidal functions. To determine orthogonality, we evaluate the boundary terms after performing integration by parts. Because of the periodic nature of \( y_n \) and \( y_m \), these boundary terms cancel out, making the integral much simpler.

Orthogonality

Orthogonality in the context of functions means that the inner product (or integral) of the functions over a given interval is zero. For functions \( y_n \) and \( y_m \), this is written as:\[ \int_{-\pi}^{\pi} y_n y_m \, dx = 0 \] if \( n eq m \).In our exercise, we use integration by parts and the boundary conditions to show that the integral is zero when \( n eq m \). This establishes that the solutions to the differential equation are orthogonal over the interval \( (-\pi, \pi) \).