Question

Solve the eigenvalue problem. $$ y^{\prime \prime}+\lambda y=0, \quad y(-\pi)=y(\pi), \quad y^{\prime}(-\pi)=y^{\prime}(\pi) $$

Short answer

Question: Determine the eigenvalues and eigenfunctions for the given ordinary differential equation $$y'' + \lambda y = 0$$ with boundary conditions $$y(-\pi)=y(\pi)$$ and $$y^{\prime}(-\pi)=y^{\prime}(\pi)$$. Answer: The eigenvalues are $$\lambda_n = n^2$$, and the corresponding eigenfunctions are $$y_n(x) = A\cos(nx) + B\sin(nx)$$, where $$n$$ is an integer.

Step-by-step solution

For the given ODE $$y'' + \lambda y = 0$$, we have a second-order homogeneous linear ODE. Depending on the sign of the eigenvalue $$\lambda$$, the general solutions will be different: 1. If $$\lambda > 0$$, let $$\lambda = k^2$$, and the general solution is: $$y(x) = A\cos(kx) + B\sin(kx)$$. 2. If $$\lambda = 0$$, the general solution is: $$y(x) = Ax + B$$. 3. If $$\lambda < 0$$, let $$\lambda = -k^2$$, and the general solution is: $$y(x) = Ae^{kx} + Be^{-kx}$$. #Step 2: Apply the boundary conditions for eigenvalues and eigenfunctions#

The given boundary conditions are $$y(-\pi)=y(\pi)$$ and $$y^{\prime}(-\pi)=y^{\prime}(\pi)$$. We'll apply these conditions to each case from step 1: 1. If $$\lambda > 0$$: $$y(-\pi) = A\cos(-k\pi) + B\sin(-k\pi) = A\cos(k\pi) - B\sin(k\pi)$$, and $$y(\pi) = A\cos(k\pi) + B\sin(k\pi)$$. Therefore, $$A\cos(k\pi) - B\sin(k\pi) = A\cos(k\pi) + B\sin(k\pi) \implies B\sin(k\pi) = 0$$. Also, we have: $$y^{\prime}(-\pi) = -A\sin(-k\pi) + B\cos(-k\pi) = -A\sin(k\pi) - B\cos(k\pi)$$, and $$y^{\prime}(\pi) = -A\sin(k\pi) + B\cos(k\pi)$$. Therefore, $$-A\sin(k\pi) - B\cos(k\pi) = -A\sin(k\pi) + B\cos(k\pi) \implies B\cos(k\pi) = 0$$. For non-trivial solutions, we need either $$B = 0$$ or $$k\pi = n\pi$$, where $$n$$ is an integer. Therefore, we have: $$k = n \implies \lambda_n = n^2$$, and the corresponding eigenfunction is $$y_n(x) = A\cos(nx) + B\sin(nx)$$. 2. If $$\lambda = 0$$: $$y(-\pi) = y(\pi) \implies -A\pi + B = A\pi + B \implies A = 0$$, and $$y^{\prime}(-\pi) = y^{\prime}(\pi) \implies B = B$$. Therefore, all solutions are trivial in this case. 3. If $$\lambda < 0$$: The solutions are in the form $$y(x) = Ae^{kx} + Be^{-kx}$$. For non-trivial solutions, we require both $$A$$ and $$B$$ to be non-zero. However, the boundary conditions imply exponential growth (or decay) at both ends, which is not possible given the periodic nature of the problem. Hence, there are no non-trivial solutions for this case. In summary, the eigenvalues are $$\lambda_n = n^2$$, and the corresponding eigenfunctions are $$y_n(x) = A\cos(nx) + B\sin(nx)$$, where $$n$$ is an integer.

Key concepts

Understanding Boundary Value Problems

Boundary value problems (BVPs) are a class of differential equations coupled with a set of additional constraints, called boundary conditions. These problems are essential in physics and engineering because they model phenomena with specified conditions at the boundaries of the domain. Unlike initial value problems, which define conditions at a single point, BVPs define conditions at multiple points.

For example, when solving the eigenvalue problem in the exercise, we look for solutions to the ordinary differential equation (ODE) that also satisfy the given boundary conditions. These conditions are that the function value and its derivative are the same at the points \( -\pi \) and \( \pi \). This periodic boundary condition gives rise to a set of allowable solutions (eigenfunctions) associated with specific characteristic values (eigenvalues).

Understanding the nature of boundary conditions is crucial because they determine the form and number of solutions. In the provided exercise, the periodic boundary conditions ensure that the solution is periodic, leading us to search for sine and cosine functions, which are inherently periodic, among possible solutions.

Homogeneous Linear ODEs

Homogeneous linear ordinary differential equations (ODEs) are differential equations of the form \( ay'' + by' + cy = 0 \), where \( a \) , \( b \) , and \( c \) are constants, and the right side of the equation is zero. These equations are called homogeneous because the absence of a non-zero term on the right side implies that if \( y(x) \) is a solution, then any constant multiple of \( y(x) \) is also a solution.

In our exercise, the ODE is given by \( y'' + \lambda y = 0 \), where \( \lambda \) is an eigenvalue. The solutions of this equation depend critically on the sign of \( \lambda \): positive, negative, or zero. Each case leads to a different general form for the solution, including trigonometric functions for positive \( \lambda \), exponential functions for negative \( \lambda \), and linear functions for \( \lambda = 0 \).

Understanding the structure of these solutions is fundamental in solving BVPs because the properties of the eigenfunctions, such as orthogonality and completeness, originate from the type of ODE and the boundary conditions applied.

Eigenvalues and Eigenfunctions

In the context of ODEs and BVPs, eigenvalues and eigenfunctions emerge when we solve problems like the one in this exercise. An eigenvalue is a special value of \( \lambda \) for which the ODE has a non-trivial (non-zero) solution that also satisfies the boundary conditions. These non-zero solutions are called eigenfunctions.

In our exercise, applying the boundary conditions to the general solutions obtained from the ODE, we find conditions for the constants involved. These conditions lead to quantification of possible values for \( k \) in the eigenfunctions, hence determining the eigenvalues \( \lambda_n = n^2 \) and the corresponding eigenfunctions \( y_n(x) = A\cos(nx) + B\sin(nx) \) where \( n \) is an integer.

The concept of eigenvalues and eigenfunctions reveals a profound insight: not every function can satisfy the boundary conditions for a given ODE, and only the special eigenfunctions corresponding to the eigenvalues fit within the constraints imposed by the problem. This is widely applicable, providing a foundation for understanding phenomena in quantum mechanics, string theory, and many more areas where BVPs are paramount.