Question

Solve the eigenvalue problem. $$ y^{\prime \prime}+\lambda y=0, \quad y(-2)=y(2), \quad y^{\prime}(-2)=y^{\prime}(2) $$

Short answer

What are the eigenvalues and eigenfunctions for the given periodic boundary value problem: $$ y''+\lambda y = 0, \quad y(-2)=y(2), \quad y'(-2)=y'(2) $$ Answer: The eigenvalues are $$\lambda_n = -(2n+1)^2 \frac{\pi^2}{16}$$, where n is a non-negative integer. The corresponding eigenfunctions are given by $$y_n(x) = A_n\cosh((2n+1)\frac{\pi x}{4}) + B_n\sinh((2n+1)\frac{\pi x}{4})$$, where A_n and B_n are arbitrary constants.

Step-by-step solution

Identify the cases for the eigenvalue λ

The differential equation is given by: $$y^{\prime \prime}+\lambda y=0$$ We can have three cases for λ: 1. λ > 0: In this case, the equation becomes $$y^{\prime \prime} + \lambda y=0$$ with λ = k², k > 0. 2. λ = 0. 3. λ < 0: In this case, the equation becomes $$y^{\prime \prime} - |\lambda| y=0$$ with λ = -k², k > 0.

Solve the differential equation for each case

1. λ > 0: $$y^{\prime \prime} + k^2 y=0$$ The general solution is: $$y(x) = A\cos(kx) + B\sin(kx)$$ 2. λ = 0: $$y^{\prime \prime} = 0$$ The general solution is: $$y(x) = Ax + B$$ 3. λ < 0: $$y^{\prime \prime} - k^2 y=0$$ The general solution is: $$y(x) = A\cosh(kx) + B\sinh(kx)$$

Apply the boundary conditions

Boundary conditions: $$y(-2)=y(2), \quad y^{\prime}(-2)=y^{\prime}(2)$$ 1. λ > 0: Using the general solution and its derivative, we get: $$A\cos(-2k) + B\sin(-2k) = A\cos(2k) + B\sin(2k)$$ $$-2Ak\sin(-2k) + 2Bk\cos(-2k) = -2Ak\sin(2k) + 2Bk\cos(2k)$$ Since the system of equations doesn't seem to simplify, we move on to the next case. 2. λ = 0: Using the general solution and its derivative, we get: $$-2A+B = 2A+B$$ $$0 = 4A$$ This implies A = 0, which gives us $$y(x) = B$$ as trivial eigenfunction. Since we want non-trivial solutions, we discard this case. 3. λ < 0: Using the general solution and its derivative, we get: $$A\cosh(-2k) + B\sinh(-2k) = A\cosh(2k) + B\sinh(2k)$$ $$-2kA\sinh(-2k) + 2kB\cosh(-2k) = 2kA\sinh(2k) + 2kB\cosh(2k)$$ Cancelling the terms with similar hyperbolic functions yields: $$B\sinh(-2k) = B\sinh(2k)$$ $$A\sinh(-2k) = -A\sinh(2k)$$ Since sinh is an odd function, and we don't want trivial solutions with A = 0 and B = 0, we must have an integer n such that $$2k = (2n+1)\frac{\pi}{2}$$.

Finding the eigenvalues and eigenfunctions

The eigenvalues are given by $$\lambda_n = -(2n+1)^2 \frac{\pi^2}{16}$$. The corresponding eigenfunctions are given by $$y_n(x) = A_n\cosh((2n+1)\frac{\pi x}{4}) + B_n\sinh((2n+1)\frac{\pi x}{4})$$. In conclusion, the eigenvalues are $$\lambda_n = -(2n+1)^2 \frac{\pi^2}{16}$$, where n is a non-negative integer, and the corresponding eigenfunctions are given by $$y_n(x) = A_n\cosh((2n+1)\frac{\pi x}{4}) + B_n\sinh((2n+1)\frac{\pi x}{4})$$, where A_n and B_n are arbitrary constants.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They describe the relationship between a function and its rate of change. In this context, the differential equation we deal with is \(y'' + \lambda y = 0\). This represents a second-order linear differential equation, where the solution \(y\) depends on the independent variable and the unknown parameter \(\lambda\), known as the eigenvalue.
This type of differential equation often arises in various physical contexts, such as vibrations and wave phenomena. The solution approaches differ depending on the sign of \(\lambda\):

• For \(\lambda > 0\), the solutions involve trigonometric functions.
• For \(\lambda = 0\), the equation becomes simpler, involving only linear functions of the form \(Ax + B\).
• For \(\lambda < 0\), the solutions are expressed in terms of hyperbolic functions.

This classification helps us target appropriate methods to solve the equation.

Boundary Conditions

Boundary conditions are essential in solving differential equations as they specify additional constraints that the solution must satisfy. In the given problem, the boundary conditions are:

• \(y(-2) = y(2)\)
• \(y'(-2) = y'(2)\)

These conditions imply periodicity or symmetry in the solution over the interval from \(-2\) to \(2\). They significantly narrow down the set of possible solutions by ensuring that the function and its derivative repeat values at these specified points. It's these conditions that lead to discrete solutions—known as eigenfunctions—and corresponding discrete eigenvalues in an eigenvalue problem. Boundary conditions can even eliminate trivial solutions (like constants) from being considered as valid solutions of our differential equation.

General Solution

The general solution of a differential equation is a formula that includes all possible solutions of the equation, typically using arbitrary constants. In the exercise, we find different general solutions depending on the sign of \(\lambda\):

• For \(\lambda > 0\), the solution is \(y(x) = A\cos(kx) + B\sin(kx)\).
• For \(\lambda = 0\), the solution simplifies to \(y(x) = Ax + B\), where \(A\) and \(B\) are constants.
• For \(\lambda < 0\), the solution adopts a hyperbolic form: \(y(x) = A\cosh(kx) + B\sinh(kx)\).

The constants \(A\) and \(B\) can be determined using the boundary conditions, albeit the presence of arbitrary constants in a general solution indicates multiple specific solutions exist under various conditions.

Hyperbolic Functions

Hyperbolic functions, similar to trigonometric functions, are defined through exponential functions and appear frequently in the context of differential equations with negative \(\lambda\). They include \(\sinh(x)\) and \(\cosh(x)\). For angles, these functions model untangled geometric relations on a hyperbola, as opposed to a circle.
For the differential equation \(y'' - k^2 y = 0\) that arises when \(\lambda < 0\), the general solution is expressed in terms of hyperbolic functions:
\[y(x) = A\cosh(kx) + B\sinh(kx)\]
These functions exhibit unique properties:

• \(\sinh(x)\) is an odd function, \(\sinh(-x) = -\sinh(x)\), akin to sine but stretched out over an infinite plane.
• \(\cosh(x)\) is an even function, \(\cosh(-x) = \cosh(x)\), similar to cosine, but representing a non-oscillatory shape.

The boundary conditions involving equalities over symmetric points will leverage these properties to isolate solutions.