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Chapter 11: Problem 1

determine the normalized eigenfunctions of the given problem. $$ y^{\prime \prime}+\lambda y=0, \quad y(0)=0, \quad y^{\prime}(1)=0 $$

Short Answer

Expert verified
Question: Determine the eigenvalues and normalized eigenfunctions for the given Sturm-Liouville problem: $$ y^{\prime \prime}+\lambda y=0, \quad y(0)=0, \quad y^{\prime}(1)=0 $$ Answer: The eigenvalues for the given problem are λ= $n^2\pi^2$ for n = 1, 2, 3, ... and the normalized eigenfunctions are $y_n(x) = \sqrt{2} \sin(n\pi x)$ for n = 1, 2, 3, ....

Step by step solution

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01

Setup the given eigenvalue problem

The given eigenvalue problem is: $$ y^{\prime \prime}+\lambda y=0, \quad y(0)=0, \quad y^{\prime}(1)=0 $$ We need to find the eigenvalues (λ) and the corresponding eigenfunctions (y) that satisfy this equation and the given boundary conditions.
02

Solve the differential equation for different cases of λ

Depending on the values of λ, we have three different cases to consider: 1. λ > 0 $$ y^{\prime \prime}+\lambda y=0 $$ The general solution is: $$ y(x) = A \sin(\sqrt{\lambda}x) + B \cos(\sqrt{\lambda}x) $$ 2. λ = 0 $$ y^{\prime \prime}=0 $$ The general solution is: $$ y(x) = Ax+B $$ 3. λ < 0 $$ y^{\prime \prime}-\left|\lambda\right| y=0 $$ The general solution is: $$ y(x) = A \sinh(\sqrt{\left|\lambda\right|}x) + B \cosh(\sqrt{\left|\lambda\right|}x) $$
03

Apply boundary conditions

Apply the boundary conditions \(y(0)=0\) and \(y^{\prime}(1)=0\) for each case: 1. λ > 0 - Applying \(y(0)=0\): $$ 0 = A \sin(0) + B \cos(0) \Rightarrow B=0 $$ - The solution becomes: $$ y(x) = A \sin(\sqrt{\lambda}x) $$ - Applying \(y^{\prime}(1)=0\): $$ 0 = A \sqrt{\lambda}\cos(\sqrt{\lambda}) $$ - This implies that \(\sqrt{\lambda} = n\pi\) for \(n=1,2,3,\ldots\) 2. λ = 0 - Applying \(y(0)=0\): $$ 0 = A(0)+B \Rightarrow B=0 $$ - The solution is trivial: $$ y(x) = 0 $$ 3. λ < 0 - No eigenfunctions exist for this case, since \(y^{\prime}(1)=0\) cannot be satisfied by the general solution.
04

Normalize the eigenfunctions

For λ > 0, we have the eigenfunctions \(y_n(x) = A\sin(n\pi x)\), for \(n=1,2,3,\ldots\). To normalize these eigenfunctions, we need to find \(A\) such that: $$ \int_{0}^{1} |y_n(x)|^2 dx = 1 $$ For \(y_n(x) = A\sin(n\pi x)\), the integral becomes: $$ \int_{0}^{1} A^2\sin^2(n\pi x) dx = 1 $$ Integrating and solving for \(A\), we get $$ A_n = \sqrt{2} $$ So, the normalized eigenfunctions are: $$ y_n(x) = \sqrt{2} \sin(n\pi x) \quad \text{for } n = 1,2,3,\ldots $$

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boundary Value Problem
In the realm of differential equations, a Boundary Value Problem (BVP) is where a differential equation is accompanied by a set of constraints, known as boundary conditions. The importance of BVPs lies in their ability to model real-world scenarios such as the vibrations of a drumhead or temperature distribution along a rod. These problems typically involve finding a function that satisfies the differential equation within a certain domain, and also meets the specific boundary conditions set at the extremes of this domain.
In the exercise provided, the boundary conditions are set as follows: \( y(0) = 0 \) and \( y'(1) = 0 \). These conditions must be satisfied by the solution to the differential equation \( y'' + \lambda y = 0 \). In essence, these constraints define the acceptable solutions, which is why they are crucial in solving the problem effectively.
Eigenvalue Problem
An eigenvalue problem in differential equations involves determining both the eigenvalues and eigenfunctions of a given system. The eigenvalues represent specific powers or intensities of the system, while the eigenfunctions describe how these values are manifested in the system.
In the provided differential equation \( y'' + \lambda y = 0 \), finding the eigenvalues (\( \lambda \)) and the corresponding eigenfunctions (\( y(x) \)) involves analyzing different cases for \( \lambda \): positive, zero, and negative. Each case provides a unique solution set governed by the boundary conditions:
  • For \( \lambda > 0 \), the solution is trigonometric, with eigenfunctions taking the form \( y(x) = A \sin(\sqrt{\lambda}x) \).
  • For \( \lambda = 0 \), the solution is linear, but leads to a trivial result (y = 0).
  • For \( \lambda < 0 \), no viable eigenfunctions exist due to boundary condition limitations.
Determining the eigenvalues requires applying the boundary condition \( y'(1) = 0 \). This condition transforms into the equation \( \sqrt{\lambda} = n\pi \), providing a discrete set of eigenvalues \( \lambda = (n\pi)^2 \), where \( n \) is a positive integer.
Normalization of Eigenfunctions
Normalization is a process of adjusting the magnitude of eigenfunctions to ensure they conform to specific criteria, notably that their total area under the curve within a given domain equals one. This adjustment makes them easier to work with and interprets in physical contexts where amplitudes or probabilities are critical.
In the problem at hand, the normalization requirement is set as \( \int_{0}^{1} |y_n(x)|^2 dx = 1 \). For the eigenfunctions determined, \( y_n(x) = A \sin(n\pi x) \), we integrate \( A^2 \sin^2(n\pi x) \) from 0 to 1. Solving this integral gives us the value of \( A^2 = 2 \), thus \( A = \sqrt{2} \).
This results in the normalized eigenfunctions \( y_n(x) = \sqrt{2} \sin(n\pi x) \). Normalizing in this manner ensures that each eigenfunction can be used appropriately in complex calculations or simulations without altering the system's intrinsic properties.

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Most popular questions from this chapter

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determine the normalized eigenfunctions of the given problem. $$ y^{\prime \prime}+\lambda y=0, \quad y^{\prime}(0)=0, \quad y^{\prime}(1)=0 $$

Consider the Sturm-Liouville problem $$ -\left[p(x) y^{\prime}\right]^{\prime}+q(x) y=\lambda r(x) y $$ $$ a_{1} y(0)+a_{2} y^{\prime}(0)=0, \quad b_{1} y(1)+b_{2} y^{\prime}(1)=0 $$ where \(p, q,\) and \(r\) satisfy the conditions stated in the text. (a) Show that if \(\lambda\) is an eigenvalue and \(\phi\) a corresponding eigenfunction, then $$ \lambda \int_{0}^{1} r \phi^{2} d x=\int_{0}^{1}\left(p \phi^{2}+q \phi^{2}\right) d x+\frac{b_{1}}{b_{2}} p(1) \phi^{2}(1)-\frac{a_{1}}{a_{2}} p(0) \phi^{2}(0) $$ provided that \(a_{2} \neq 0\) and \(b_{2} \neq 0 .\) How must this result be modified if \(a_{2}=0\) or \(b_{2}=0\) ? (b) Show that if \(q(x) \geq 0\) and if \(b_{1} / b_{2}\) and \(-a_{1} / a_{2}\) are nonnegative, then the eigenvalue \(\lambda\) is nonnegative. (c) Under the conditions of part (b) show that the eigenvalue \(\lambda\) is strictly positive unless \(q(x)=0\) for each \(x\) in \(0 \leq x \leq 1\) and also \(a_{1}=b_{1}=0\)
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