Question

Solve the eigenvalue problem. $$ y^{\prime \prime}+\lambda y=0, \quad y(0)=0, \quad y^{\prime}(1)=0 $$

Short answer

Based on the given eigenvalue problem and boundary conditions, the eigenvalues are represented as: $$ \lambda_n = -n^2 \pi^2, \quad n \in \mathbb{Z} $$ And their corresponding eigenfunctions are: $$ y_n(x) = C_1 (e^{n\pi x} - e^{-n\pi x}) $$

Step-by-step solution

Write the given eigenvalue problem and boundary conditions

We have the following eigenvalue problem: $$ y^{\prime \prime}+\lambda y=0, \quad y(0)=0, \quad y^{\prime}(1)=0 $$ Our task is to find the eigenvalues (λ) and eigenfunctions that satisfy this problem.

Solve the given differential equation for two cases

First, let's find the general solution of the given differential equation. We need to consider two cases: 1. Case λ = 0: $$ y^{\prime \prime} = 0 \\ y(x) = Ax + B $$ 2. Case λ ≠ 0: Let's assume a solution of the form \(y(x) = e^{rx}\) and substitute it into the differential equation: $$ r^2 e^{rx} + \lambda e^{rx} = 0 \\ e^{rx}(r^2 + \lambda) = 0 $$ Since \(e^{rx}\) is never zero, we can solve for \(r^2 + \lambda = 0\): $$ r = \pm \sqrt{-\lambda} $$ Now, we can write the general solution as a linear combination of the two linearly independent solutions given by the values of \(r\): $$ y(x) = C_1 e^{\sqrt{-\lambda}x} + C_2 e^{-\sqrt{-\lambda}x} $$

Apply the boundary conditions and find the eigenvalues and eigenfunctions

Now we will apply the boundary conditions to both cases and find the eigenvalues and corresponding eigenfunctions. 1. Case λ = 0: Apply the boundary condition \(y(0)=0\): $$ y(0) = A(0) + B = 0 \implies B=0 \\ y(x) = Ax $$ Apply the second boundary condition \(y'(1)=0\), and its derivative is \(y'(x)=A\). We have: $$ y^{\prime}(1)=A = 0 $$ The trivial solution indicates that λ = 0 is not an eigenvalue. 2. Case λ ≠ 0: Apply the boundary condition \(y(0)=0\): $$ y(0) = C_1 e^{0} + C_2 e^{0} \implies C_1 + C_2 = 0 \\ C_2 = -C_1 $$ Now, substitute this into the general solution: $$ y(x) = C_1(e^{\sqrt{-\lambda}x} - e^{-\sqrt{-\lambda}x}) $$ Apply the second boundary condition \(y'(1)=0\). Calculate the derivative, \(y'(x) = \sqrt{-\lambda} C_1(e^{\sqrt{-\lambda}x} + e^{-\sqrt{-\lambda}x})\), and apply the boundary condition: $$ y^{\prime}(1) = \sqrt{-\lambda} C_1(e^{\sqrt{-\lambda}} + e^{-\sqrt{-\lambda}}) = 0 $$ For non-trivial solutions, \(C_1\) must be non-zero. Thus, the only way for the above equation to be zero is when the term in the parenthesis is zero, that is: $$ \sinh(\sqrt{-\lambda}) = 0 \\ \sqrt{-\lambda} = n\pi, \quad n \in \mathbb{Z} $$ Our eigenvalues are: $$ \lambda_n = -n^2 \pi^2, \quad n \in \mathbb{Z} $$ The corresponding eigenfunctions are: $$ y_n(x) = C_1 (e^{n\pi x} - e^{-n\pi x}) $$

Final Answer

The eigenvalues for the given eigenvalue problem are: $$ \lambda_n = -n^2 \pi^2, \quad n \in \mathbb{Z} $$ And the corresponding eigenfunctions are: $$ y_n(x) = C_1 (e^{n\pi x} - e^{-n\pi x}) $$

Key concepts

Boundary Conditions

When solving differential equations, boundary conditions play a crucial role. They help in determining the specific solutions that satisfy the problem. In our eigenvalue problem, the boundary conditions are given as \(y(0)=0\) and \(y'(1)=0\).

• **\(y(0)=0\):** This means that when \(x=0\), the function \(y\) must be zero. This boundary condition helps to eliminate any constant terms that might arise in our solution.
• **\(y'(1)=0\):** At \(x=1\), the derivative of the function \(y\), which represents the slope, must be zero. This condition often relates to physical scenarios like a maximum or minimum point, or forces that keep a string in a certain position.

Boundary conditions are essential in finding the eigenvalues and eigenfunctions as they ensure the uniqueness of the solution. Each condition provides a specific requirement that your solution must meet, leading to the determination of the values of constants involved in the solution.

Differential Equations

Differential equations involve functions and their derivatives. They describe various phenomena, including physical, chemical, or biological processes. Our given differential equation is \(y'' + \lambda y = 0\). This is a second-order linear differential equation. The "order" of the differential equation indicates the highest derivative present, which, in this case, is the second derivative \(y''\).
To solve differential equations like this, we follow a process:

• **Assuming Solutions:** For non-zero \(\lambda\), assume solutions in exponential form, like \(y(x) = e^{rx}\).
• **Characteristic Equation:** Substitute these assumed solutions back into the differential equation to find a characteristic equation. In our case, this becomes \(r^2 + \lambda = 0\).
• **Solving the Characteristic Equation:** Solve for \(r\) to find the general solution to the equation. Here, \(r = \pm \sqrt{-\lambda}\), leading to solutions that involve trigonometric functions due to the imaginary roots.

The process of finding the solution involves determining the constants with the help of boundary conditions. This is critical as not all solutions are applicable; only those that meet given conditions are valid.

Eigenvalues and Eigenfunctions

Eigenvalues and eigenfunctions are central to many mathematical problems involving linear transformations. In our case, they help solve the differential equation and its boundary conditions. An *eigenvalue* (\(\lambda\)) is a scalar that, when used in a differential equation, results in non-zero solutions that satisfy given boundary conditions. The associated *eigenfunction* is that specific solution or function that corresponds to an eigenvalue.
Here's how we extract them:

• **Trial and Error:** For \(\lambda = 0\), the solutions fail to satisfy boundary conditions in a non-trivial way, indicating that \(\lambda = 0\) is not an eigenvalue.
• **Solving for Non-zero \(\lambda\):** Assume \(\lambda eq 0\) and solve for \(y(x) = C_1(e^{\sqrt{-\lambda}x} - e^{-\sqrt{-\lambda}x})\). This involves re-expressing in hyperbolic or trigonometric functions.
• **Apply Boundary Conditions:** Use the boundary conditions to solve for \(\lambda\), leading to \(\lambda_n = -n^2\pi^2\).
• **Resulting Eigenfunctions:** The resulting eigenfunctions come in unique forms, satisfying the conditions, such as \(y_n(x) = C_1 (e^{n\pi x} - e^{-n\pi x})\).

Eigenvalues and eigenfunctions provide a complete set of solutions for various values of \(n\), helping in the analysis of systems described by differential equations. It also relates closely to vibrational modes and frequencies in physical systems.