Question

In Exercises 23-26 solve the eigenvalue problem. $$ y^{\prime \prime}+\lambda y=0, \quad y(0)=0, \quad \int_{0}^{L} y(x) d x=0 $$

Short answer

Based on the given differential equation and boundary conditions, identify the eigenvalues and corresponding eigenfunctions. Eigenvalues: \(\lambda_n = \frac{n^2\pi^2}{L^2}\), where \(n = 1, 2, 3, \dots\) Eigenfunctions: \(y_n(x) = A\sin\Big(\frac{n\pi x}{L}\Big)\), where \(n = 1, 2, 3, \dots\)

Step-by-step solution

Identify the General Form of the Differential Equation

First, note that the given equation is a second-order linear homogeneous differential equation: $$ y^{\prime \prime}+\lambda y=0 $$ This equation can have three types of general solutions depending on the value of the eigenvalue, \(\lambda\): real and positive, real and negative, or zero.

Solve the Differential Equation According to the Type of Eigenvalue

Let's solve the given equation for each type of eigenvalue: 1. Real and positive eigenvalue \((\lambda > 0)\): Let \(\lambda = k^2\), where \(k > 0\). Then, the differential equation becomes: $$ y^{\prime \prime} + k^2y = 0 $$ The general solution of this equation is: $$ y(x) = A\sin(kx) + B\cos(kx) $$ 2. Real and negative eigenvalue \((\lambda < 0)\): Let \(\lambda = -k^2\), where \(k > 0\). Then, the differential equation becomes: $$ y^{\prime \prime} - k^2y = 0 $$ The general solution of this equation is: $$ y(x) = A\sinh(kx) + B\cosh(kx) $$ 3. Zero eigenvalue \((\lambda = 0)\): The differential equation becomes: $$ y^{\prime \prime} = 0 $$ The general solution of this equation is: $$ y(x) = Ax + B $$

Apply the Boundary Condition \(y(0) = 0\)

Applying the boundary condition \(y(0) = 0\) to the above general solutions, we get: 1. Real and positive eigenvalue: $$ y(0) = 0 = A\sin(0) + B\cos(0) \Rightarrow B = 0 $$ Thus, the solution becomes: $$ y(x) = A\sin(kx) $$ 2. Real and negative eigenvalue: $$ y(0) = 0 = A\sinh(0) + B\cosh(0) \Rightarrow B = 0 $$ Thus, the solution becomes: $$ y(x) = A\sinh(kx) $$ 3. Zero eigenvalue: $$ y(0) = 0 = A(0) + B \Rightarrow B = 0 $$ Thus, the solution becomes: $$ y(x) = Ax $$

Apply the Boundary Condition \(\int_{0}^{L} y(x) d x=0\) and Find the Eigenvalues and Eigenfunctions

Now, we will apply the boundary condition \(\int_{0}^{L} y(x) d x=0\) for each type of eigenvalue: 1. Real and positive eigenvalue: $$ \int_{0}^{L} A\sin(kx) dx = 0 $$ Solving for the integral, we get: $$ A\Big[-\frac{1}{k}\cos(kx)\Big]_0^L = 0 $$ For this equation to hold true, we need the eigenvalue to be \(k_n = \frac{n\pi}{L}\), where \(n = 1, 2, 3, \dots\). Thus, the eigenvalues and corresponding eigenfunctions will be: $$ \lambda_n = k_n^2 = \frac{n^2\pi^2}{L^2}, \quad y_n(x) = A\sin\Big(\frac{n\pi x}{L}\Big) $$ 2. Real and negative eigenvalue: As \(\sinh(kx)\) can't fulfill the given boundary condition \(\int_{0}^{L} y(x) d x=0\), there are no eigenvalues and eigenfunctions for this case. 3. Zero eigenvalue: $$ \int_{0}^{L} Ax dx = 0 $$ Solving for the integral, we get: $$ A\Big[\frac{1}{2}x^2\Big]_0^L = 0 $$ For this equation to hold true, we need \(A = 0\). This leads to the trivial solution \(y(x) = 0\). As we are not considering the trivial solution, there are no eigenvalues and eigenfunctions for this case.

Final Eigenvalues and Eigenfunctions

The eigenvalues and corresponding eigenfunctions for the given differential equation and boundary conditions are as follows: Eigenvalues: $$ \lambda_n = \frac{n^2\pi^2}{L^2}, \quad n = 1, 2, 3, \dots $$ Eigenfunctions: $$ y_n(x) = A\sin\Big(\frac{n\pi x}{L}\Big), \quad n = 1, 2, 3, \dots $$

Key concepts

Second-Order Linear Differential Equation

A second-order linear differential equation is an equation involving the second derivative of a function. It is of the form \( y'' + p(x)y' + q(x)y = r(x) \). In our problem, the equation is simpler: \( y'' + \lambda y = 0 \). This simplicity is because there are no terms involving the first derivative \( y' \) or a known function \( r(x) \).
The equation is called 'linear' because it can be expressed as a linear combination of the function \( y \), its derivatives, and some coefficients, leading to solutions that can often be added together to form new solutions. This characteristic is fundamental in problems involving differential equations as it allows superposition of solutions.
Understanding the form of second-order linear differential equations is essential as it helps identify possible solution methods, particularly looking at how the eigenvalue \( \lambda \) affects the nature of the solutions.

Boundary Conditions

Boundary conditions are additional constraints required to solve differential equations uniquely. They specify the values the solution must satisfy at certain points, making the problem well-defined. For our problem, the boundary conditions given are \( y(0)=0 \) and \( \int_{0}^{L} y(x) \, dx = 0 \).
The first condition, \( y(0)=0 \), means that at \( x=0 \), the value of the function \( y(x) \) is zero. This is a Dirichlet boundary condition which specifies the value of the solution at a specific point.
The second condition, \( \int_{0}^{L} y(x) \, dx = 0 \), is an integral condition. It requires that the total area under the curve of \( y(x) \) from \( x=0 \) to \( x=L \) is zero. This type of boundary condition ensures the function satisfies certain physical or geometrical constraints over an interval.
Applying these boundary conditions is crucial as they significantly influence which solutions of the differential equation are acceptable.

Eigenvalues and Eigenfunctions

In the context of differential equations, eigenvalues and eigenfunctions are special values and functions that arise when solving problems with boundary conditions. They basically determine the 'shape' and 'size' of the solution functions. The eigenvalue problem involves solving \( y'' + \lambda y = 0 \) subject to given boundary conditions.
For our problem, testing whether \( \lambda \) is positive, negative, or zero plays a key role. Each of these conditions alters the form of the general solution:

• For \( \lambda > 0 \), the solutions involve trigonometric functions, leading to a particular set of eigenvalues and eigenfunctions.
• For \( \lambda < 0 \), the solutions involve hyperbolic functions, which in this case don't satisfy the given conditions.
• For \( \lambda = 0 \), the solution is a linear function, which again doesn't satisfy the non-trivial integral boundary condition.

The acceptable solutions are determined by the condition \( \int_{0}^{L} y(x) \, dx = 0 \), which leads to discrete eigenvalues \( \lambda_n = \frac{n^2\pi^2}{L^2} \) corresponding to the eigenfunctions \( y_n(x) = A\sin\left(\frac{n\pi x}{L}\right) \). This set of solutions is not just one arbitrary solution, but an infinite sequence indexed by \( n \).

Solution of Differential Equations

Solving differential equations involves finding all possible functions that satisfy the given equation and boundary conditions. In this exercise, solving \( y'' + \lambda y = 0 \) with the given boundary conditions leads us to approach the equation differently based on the nature of \( \lambda \).
First, you identify the types of solutions available depending on whether \( \lambda \) is positive, negative, or zero. You then apply the boundary conditions to these forms:

• The Dirichlet condition \( y(0)=0 \) simplifies the solution by removing constants that don't contribute to satisfying \( y(0)=0 \).
• The integral boundary condition \( \int_0^L y(x) \, dx = 0 \) helps to further refine the solutions, eliminating potential solutions that do not meet this condition.

By going through these steps, we found that only certain values of \( \lambda \) (eigenvalues) and corresponding solutions (eigenfunctions) satisfy both conditions. This method results in a comprehensive solution reflecting the nature and constraints of the differential equation.