Question

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

Short answer

Question: Determine the eigenvalues (λ) and corresponding eigenfunctions (y) that satisfy the second-order ordinary differential equation (ODE) with boundary conditions: $$ y^{\prime\prime}(x) = -\lambda y(x) $$ with boundary conditions $$ y(0) = 0 \quad \text{and} \quad y'(\pi) = 0 $$ Answer: The eigenvalues of the given eigenvalue problem are: $$ \lambda_k = -k^2 $$ where \(k\) is an integer. The corresponding eigenfunctions are: $$ y_k(x) = c_1 (e^{ikx} - e^{-ikx}) $$ where \(c_1\) is a constant.

Step-by-step solution

Solve the ODE

To solve the given ODE, we first rewrite the equation as follows: $$ y^{\prime\prime}(x) = -\lambda y(x) $$ Since this is a linear homogeneous ODE, we can guess a solution in the form of \(y = e^{rx}\). Then, substitute this into the ODE to get: $$ r^2 e^{rx} = -\lambda e^{rx} $$ Dividing both sides by \(e^{rx}\), we get the auxiliary equation: $$ r^2 = -\lambda $$ We now find the roots of the auxiliary equation.

Find the roots of the auxiliary equation

The roots of the auxiliary equation are: $$ r_1 = \sqrt{-\lambda} \quad \text{and} \quad r_2 = -\sqrt{-\lambda} $$ Using these roots, we can write the general solution of the ODE as: $$ y(x) = c_1 e^{\sqrt{-\lambda} x} + c_2 e^{-\sqrt{-\lambda} x} $$ where \(c_1\) and \(c_2\) are constants.

Apply the first boundary condition

The first boundary condition is \(y(0) = 0\). Substitute this into the general solution to find: $$ 0 = c_1 e^{\sqrt{-\lambda} (0)} + c_2 e^{-\sqrt{-\lambda} (0)} $$ Since both exponential terms are equal to 1, we find that \(c_1 + c_2 = 0\), which implies \(c_1 = -c_2\).

Apply the second boundary condition

The second boundary condition is \(y'(\pi) = 0\). First, find the derivative of the general solution: $$ y'(x) = \sqrt{-\lambda} c_1 e^{\sqrt{-\lambda} x} - \sqrt{-\lambda} c_2 e^{-\sqrt{-\lambda} x} $$ Substitute the condition \(y'(\pi) = 0\) into the derivative: $$ 0 = \sqrt{-\lambda} c_1 e^{\sqrt{-\lambda} \pi} - \sqrt{-\lambda} c_2 e^{-\sqrt{-\lambda} \pi} $$ Using the result \(c_1 = -c_2\), we can rewrite the equation above as: $$ 0 = \sqrt{-\lambda} c_1 (e^{\sqrt{-\lambda} \pi} - e^{-\sqrt{-\lambda} \pi}) $$ Since \(c_1\) cannot be zero, otherwise the eigenfunction would be a trivial solution, we must have: $$ e^{\sqrt{-\lambda} \pi} - e^{-\sqrt{-\lambda} \pi} = 0 $$

Find the eigenvalues

From the previous step, we have the equation for the eigenvalues: $$ e^{\sqrt{-\lambda} \pi} - e^{-\sqrt{-\lambda} \pi} = 0 $$ This equation is true when \(\sqrt{-\lambda} = ik\), where \(k\) is an integer and \(i\) is the imaginary unit. Therefore, the eigenvalues are given by: $$ \lambda_k = -k^2 $$

Find the eigenfunctions

Now that we have the eigenvalues, we can find the corresponding eigenfunctions by substituting the eigenvalues back into the general solution. For each \(\lambda_k = -k^2\), the eigenfunctions are given by: $$ y_k(x) = c_1 e^{ikx} + c_2 e^{-ikx} $$ Using the result \(c_1 = -c_2\), we can rewrite the eigenfunctions as: $$ y_k(x) = c_1 (e^{ikx} - e^{-ikx}) $$ These eigenfunctions solve the given eigenvalue problem.

Key concepts

Boundary Conditions

Boundary conditions are critical in solving differential equations, as they provide the additional information needed to find specific solutions out of the possible general solutions. In the given eigenvalue problem, we have two boundary conditions:

• The first boundary condition is given as: \( y(0) = 0 \). This condition implies that the function value at \( x = 0 \) must be zero. It helps us determine the relationship between the constants in the general solution.
• The second boundary condition is: \( y'(\pi) = 0 \). This specifies that the derivative of the function at \( x = \pi \) must be zero. This helps further refine the solution and pinpoint the appropriate eigenvalues that satisfy the equation.

Boundary conditions serve to specify requirements that the solution must satisfy, making it possible to determine unique solutions for differential equations that have infinite possible solutions otherwise.

Linear Homogeneous Differential Equations

A linear homogeneous differential equation is an equation in which the dependent variable and its derivatives appear linearly, and all terms are of the form that includes the unknown function and its derivatives, with no standalone constants. In this problem, the given differential equation: \[ y'' + \lambda y = 0 \] falls into this category.Key characteristics of linear homogeneous differential equations include:

• They usually allow solutions found through superposition - that means a linear combination of basic solutions fulfills the original equation.
• They are characterized by having a zero on one side of the equation, simplifying the task of combining solutions linearly.
• Such equations often arise in the context of physical and engineering problems where natural frequencies (eigenvalues) need to be determined.

Recognizing that an equation is linear and homogeneous helps in identifying suitable methods for obtaining solutions and solving various types of boundary value problems.

General Solution of Differential Equations

The general solution of a differential equation is a solution that incorporates arbitrary constants, representing the complete family of potential solutions to the differential equation, prior to applying boundary conditions or initial conditions.For the differential equation \( y'' + \lambda y = 0 \), assuming \( y = e^{rx} \), we derive an auxiliary equation \( r^2 = -\lambda \). Solving this helps us find roots, either real or complex, and form the general solution:\[ y(x) = c_1 e^{\sqrt{-\lambda} x} + c_2 e^{-\sqrt{-\lambda} x} \] Here:

• \( c_1 \) and \( c_2 \) are arbitrary constants determined through boundary conditions.
• The form includes general exponential terms due to the complex roots derived from the auxiliary equation.

Applying boundary conditions lets us find specific values for these constants and establish a distinct solution for the problem at hand.

Eigenfunctions

Eigenfunctions are specific non-zero solutions to a differential equation involving an operator that corresponds to the eigenvalues, typically representing modes of a system like vibration modes in physical applications. In the provided problem, after solving for eigenvalues, we obtain eigenfunctions linked to each possible eigenvalue.Given:

• The eigenfunctions corresponding to the eigenvalues \( \lambda_k = -k^2 \) are expressed as:\[ y_k(x) = c_1 (e^{ikx} - e^{-ikx}) \]
• The form involves sinusoidal solutions, a result of the complex nature of \( \sqrt{-\lambda} = ik \).
• These functions define wave-like solutions common in periodic systems.

Eigenfunctions are essential in determining the mode shapes of differential equations and linked to the unique solutions of boundary value problems.

Auxiliary Equation

In differential equations, an auxiliary equation is derived to simplify finding the general solution of the differential equation. It is typically formed by substituting an exponential trial solution and extracting an algebraic equation that guides finding possible solutions.For the differential equation \( y'' + \lambda y = 0 \), by assuming a solution form \( y = e^{rx} \), it leads us to the auxiliary equation:\[ r^2 = -\lambda \]Key points about auxiliary equations include:

• They help in determining the characteristic roots, which can be real or complex-based on \( \lambda \).
• The solutions of the auxiliary equation allow constructing the general solution which incorporates exponential or sinusoidal components, based on the value of \( \lambda \).

Thus, the auxiliary equation plays a fundamental role in identifying the basic structure of solutions to differential equations and makes handling complex parts easier for further analysis.