Question

Find the eigenvalues and eigenfunctions of the given boundary value problem. Assume that all eigenvalues are real. $y^{\prime \prime }+\lambda y=0,y(0)=0,y^{\prime }(\pi )=0$

Short answer

Answer: The eigenvalues for the given boundary value problem are ${\lambda }_n=((2n+1)/2{)}^2$ and the corresponding eigenfunctions are $y_n(x)=\sin (((2n+1)/2)x)$, where $n$ is an integer.

Step-by-step solution

Solve the differential equation

To solve the differential equation $y^{\prime \prime }+\lambda y=0$, we'll first consider two cases: $\lambda >0$ and $\lambda <0$. For $\lambda >0$, let $\lambda =k^2$. The differential equation becomes: $y^{\prime \prime }+k^{2}y=0$. The general solution is given by: $y(x)=A\sin (kx)+B\cos (kx)$. For $\lambda <0$, let $\lambda =-k^2$. The differential equation becomes: $y^{\prime \prime }-k^{2}y=0$. The general solution is given by: $y(x)=A\sinh (kx)+B\cosh (kx)$.

Apply the boundary conditions

Using the boundary condition $y(0)=0$, we get: For $\lambda >0$: $y(0)=A\sin (0)+B\cos (0)=0$. Since $\cos (0)=1$, we have $B=0$. The equation becomes $y(x)=A\sin (kx)$. For $\lambda <0$: $y(0)=A\sinh (0)+B\cosh (0)=0$. Since $\cosh (0)=1$, we have $B=0$. The equation becomes $y(x)=A\sinh (kx)$. Now, we'll apply the second boundary condition: $y^{\prime }(\pi )=0$. For $\lambda >0$: The derivative of $y(x)$ is $y^{\prime }(x)=Ak\cos (kx)$. Using the boundary condition, $Ak\cos (k\pi )=0$. Since $A$ and $k$ cannot both be zero, we must have $\cos (k\pi )=0$. This condition is satisfied for $k=(2n+1)/2$, where $n$ is an integer. Hence, the eigenvalues are ${\lambda }_n=((2n+1)/2{)}^2$. For $\lambda <0$: The derivative of $y(x)$ is $y^{\prime }(x)=Ak\cosh (kx)$. Using the boundary condition, $Ak\cosh (k\pi )=0$. However, $\cosh (kx)$ is always positive and non-zero, which means no negative eigenvalues can satisfy this condition.

Determine the eigenfunctions

For the eigenvalues ${\lambda }_n=((2n+1)/2{)}^2$, the corresponding eigenfunctions are given by $y_n(x)=A_n\sin (((2n+1)/2)x)$, where $A_n$ is an arbitrary constant. Eigenfunctions are determined up to a constant, so we can set $A_n=1$. Therefore, the eigenfunctions are $y_n(x)=\sin (((2n+1)/2)x)$. In conclusion, the eigenvalues for the given boundary value problem are ${\lambda }_n=((2n+1)/2{)}^2$ and the corresponding eigenfunctions are $y_n(x)=\sin (((2n+1)/2)x)$.

Key concepts

Differential Equation

A differential equation is an equation that relates a function with its derivatives. It expresses a relationship between a dependent variable and one or more of its derivatives with respect to independent variables. In this exercise, we have the second-order differential equation:

• $y^{″}+\lambda y=0$

Here, $y^{″}$ is the second derivative of $y$ with respect to $x$, and $\lambda$ is a constant parameter. The goal is to find solutions $y(x)$, which are known as the general solutions. These solutions provide valuable information about the behavior of the system described by the differential equation. The solutions vary depending on whether the parameter $\lambda$ is positive or negative, leading to different functional forms such as sine and cosine functions for $\lambda >0$ or hyperbolic functions for $\lambda <0$.

Boundary Value Problem

A boundary value problem is a type of differential equation problem where you need to find a solution that satisfies certain conditions at specified values of the independent variable, known as boundary conditions. In this situation, the boundary conditions provided are:

• $y(0)=0$
• $y^{\prime }(\pi )=0$

These conditions are used to determine specific solutions, known as eigenfunctions, from the general solution of the differential equation. The given boundary conditions ensure that the solution is unique to the setup. These types of problems are common in mathematical physics, especially when dealing with scenarios such as vibrations, heat conduction, or quantum mechanics.

Real Eigenvalues

Eigenvalues play a crucial role in solving differential equations associated with linear systems. They are special numbers associated with a system that can have profound implications for its behavior. In this exercise, only real eigenvalues are considered — this means they take values in the set of real numbers, avoiding any complex numbers. For the boundary value problem given, the solution shows that all eigenvalues are real and positive:

• ${\lambda }_n=((2n+1)/2{)}^2$

These eigenvalues arise from satisfying the boundary conditions where $n$ is an integer. Real eigenvalues ensure that the eigenfunctions are composed of standard trigonometric functions, which are real-valued and applicable in physical situations.

Trigonometric Functions

Trigonometric functions like sine and cosine are often inherent in solutions to differential equations, especially those involving periodic or wave-like phenomena. In this problem, the solutions to the differential equation when $\lambda >0$ are constructed using the sine function:

• $y_n(x)=\sin (((2n+1)/2)x)$

These functions naturally satisfy the boundary and eigenvalue conditions. Trigonometric functions are immensely useful due to their properties:

• Periodicity, which is beneficial in modeling cyclical processes.
• Well-defined derivatives, making them ideal for forming differential equations.
• Ability to represent a wide range of physical phenomena, from sound waves to oscillations in mechanical systems.

Through the use of trigonometric functions, solutions can easily model real-world systems that exhibit regular, repeating behaviors.