Question

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

Short answer

Answer: The eigenvalues are given by \(-\lambda_n = \frac{(n \pi)^2}{4}\) (for \(n = 1, 2, 3, ...\)) and the eigenfunctions are given by \(y_n(x) = C_1 (e^{\sqrt{\frac{(n \pi)^2}{4}} x} - e^{-\sqrt{\frac{(n \pi)^2}{4}} x})\) (for \(n = 1, 2, 3, ...\)).

Step-by-step solution

Identify the given differential equation and boundary conditions

We have been given the following: Differential equation: \(y^{\prime \prime}+\lambda y = 0\) Boundary conditions: \(y(0) = 0\) and \(y(2) = 0\)

Solve the differential equation

The general solution to a second-order linear differential equation of the form \(y^{\prime \prime}+\lambda y = 0\) is given by: \(y(x) = C_1 \cdot e^{r_1 x} + C_2 \cdot e^{r_2 x}\) where \(r_1\) and \(r_2\) are the roots of the characteristic equation: \(r^2 + \lambda = 0\) Solving for the roots: \(r^2 = -\lambda\) \(r_{1,2} = \pm \sqrt{-\lambda}\) The general solution is now: \(y(x) = C_1 \cdot e^{\sqrt{-\lambda} x} + C_2 \cdot e^{-\sqrt{-\lambda} x}\)

Apply the boundary conditions

We will use the given boundary conditions to find the eigenvalues (if any) and eigenfunctions. Boundary condition 1: \(y(0) = 0\) Applying the boundary condition: \(0 = C_1 e^{0} + C_2 e^{0}\) \(0 = C_1 + C_2\) Boundary condition 2: \(y(2) = 0\) Applying the boundary condition: \(0 = C_1 e^{\sqrt{-\lambda} \cdot 2} + C_2 e^{-\sqrt{-\lambda} \cdot 2}\)

Determine the eigenvalues and eigenfunctions

To find the nontrivial solutions, we need to find when \(C_1\) and \(C_2\) are not equal to zero. We can eliminate \(C_1\) by substituting \(C_1 = -C_2\): \(0 = -C_2 e^{\sqrt{-\lambda} \cdot 2} + C_2 e^{-\sqrt{-\lambda} \cdot 2}\) Divide both sides with \(C_2\): \(0 = e^{-\sqrt{-\lambda} \cdot 2} - e^{\sqrt{-\lambda} \cdot 2}\) This implies that: \(e^{\sqrt{-\lambda} \cdot 2} = e^{-\sqrt{-\lambda} \cdot 2}\) Which leads us to: \(\tan(\sqrt{-\lambda} \cdot 2) = 0\) The eigenvalues are given by: \(-\lambda_n = \frac{(n \pi)^2}{4}\) (for \(n = 1, 2, 3, ...\)) And the eigenfunctions are given by: \(y_n(x) = C_1 (e^{\sqrt{\frac{(n \pi)^2}{4}} x} - e^{-\sqrt{\frac{(n \pi)^2}{4}} x})\) (for \(n = 1, 2, 3, ...\))

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function to its derivatives. They are fundamental in fields such as physics, engineering, and mathematics because they describe how physical quantities change over time. The exercise involves a second-order linear differential equation, which is given by:

• \( y'' + \lambda y = 0 \)

This type of equation is notable because it involves the second derivative of the function \( y \).
This specific differential equation represents a harmonic oscillator problem, which is commonly seen in physics for systems like springs and pendulums. The solution to this equation involves understanding the behavior of the system depending on the parameter \( \lambda \). Differentiating further means solving for \( y \) in terms of known functions or constants.

Boundary Conditions

Boundary conditions are constraints necessary to solve differential equations. They specify the value of the solution at given points, ensuring a unique solution to the problem.
In this problem, two boundary conditions are provided:

• \( y(0) = 0 \)
• \( y(2) = 0 \)

These conditions mean that the value of the function \( y \) must be zero both at the starting point \( x = 0 \) and at \( x = 2 \).
In physics, such boundary conditions can be visualized as fixed endpoints of a string or membrane, where no displacement occurs at those points. Applying these conditions helps in determining specific solutions known as eigenfunctions that fulfill both the differential equation and boundary conditions.

Eigenvalues

Eigenvalues in this context are the specific values of \( \lambda \) that lead to non-trivial solutions of the differential equation. These values
are crucial because they determine the behavior of the system or structure you are analyzing.
For the given differential equation, the eigenvalues are derived by setting up conditions for the solution to remain non-zero under the imposed boundary conditions. In mathematical terms, this requires solving for \( \lambda \) such that the solution to:

• \( y'' + \lambda y = 0 \)

satisfies \( y(0) = 0 \) and \( y(2) = 0 \).
In the example, the eigenvalues \( -\lambda_n \) are shown to be linked to squared terms involving \( n\pi \,/ 2 \), reflecting periodic conditions as seen in wave functions.

Characteristic Equation

The characteristic equation is derived from the differential equation and is key to finding the eigenvalues. For a linear equation like:

• \( y'' + \lambda y = 0 \)

We look for solutions of a form that satisfies this setup, usually involving exponential functions due to their derivatives behaving predictably.
By substituting a trial solution form such as \( y = e^{rx} \), a characteristic equation can be formed:

• \( r^2 + \lambda = 0 \)

This quadratic equation provides the roots \( r \) - which are generally complex due to the nature of \( \lambda \). Solving this equation gives the values of \( r \), from which particular solutions can be constructed.
These solutions can then be adapted to satisfy the boundary conditions, leading to equations that determine the exact eigenvalues and define the behavior of the function \( y \) over the domain.