11.1: Eigenvalue Problems for y'' + λy = 0 (2024)

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    In Chapter 12 we’ll study partial differential equations that arise in problems of heat conduction, wave propagation, and potential theory. The purpose of this chapter is to develop tools required to solve these equations. In this section we consider the following problems, where \(\lambda\) is a real number and \(L>0\):

    • Problem 1: \(y''+\lambda y=0,\quad y(0)=0,\quad y(L)=0\)
    • Problem 2: \(y''+\lambda y=0,\quad y'(0)=0,\quad y'(L)=0\)
    • Problem 3: \(y''+\lambda y=0,\quad y(0)=0,\quad y'(L)=0\)
    • Problem 4: \(y''+\lambda y=0,\quad y'(0)=0,\quad y(L)=0\)
    • Problem 5: \(y''+\lambda y=0,\quad y(-L)=y(L), \quad y'(-L)=y'(L)\)

    In each problem the conditions following the differential equation are called boundary conditions. Note that the boundary conditions in Problem 5, unlike those in Problems 1-4, don’t require that \(y\) or \(y'\) be zero at the boundary points, but only that \(y\) have the same value at \(x=\pm L\), and that \(y'\) have the same value at \(x=\pm L\). We say that the boundary conditions in Problem 5 are periodic.

    Obviously, \(y\equiv0\) (the trivial solution) is a solution of Problems 1-5 for any value of \(\lambda\). For most values of \(\lambda\), there are no other solutions. The interesting question is this:

    For what values of \(\lambda\) does the problem have nontrivial solutions, and what are they?

    A value of \(\lambda\) for which the problem has a nontrivial solution is an eigenvalue of the problem, and the nontrivial solutions are \(\lambda\)-eigenfunctions, or eigenfunctions associated with \(\lambda\). Note that a nonzero constant multiple of a \(\lambda\)-eigenfunction is again a \(\lambda\)-eigenfunction.

    Problems 1-5 are called eigenvalue problems. Solving an eigenvalue problem means finding all its eigenvalues and associated eigenfunctions. We’ll take it as given here that all the eigenvalues of Problems 1-5 are real numbers. This is proved in a more general setting in Section 13.2.

    Theorem 11.1.1

    Problems \(1\)–\(5\) have no negative eigenvalues. Moreover\(,\) \(\lambda=0\) is an eigenvalue of Problems \(2\) and \(5,\) with associated eigenfunction \(y_0=1,\) but \(\lambda=0\) isn’t an eigenvalue of Problems \(1,\) \(3,\) or \(4\).

    Proof

    We consider Problems 1-4, and leave Problem 5 to you (Exercise 11.1.1). If \(y''+\lambda y=0\), then \(y(y''+\lambda y)=0\), so

    \[\int_0^L y(x)(y''(x)+\lambda y(x))\,dx=0;\nonumber \]

    therefore,

    \[\label{eq:11.1.1} \lambda\int_0^L y^2(x)\,dx=-\int_0^L y(x)y''(x)\, dx.\]

    Integration by parts yields

    \[\label{eq:11.1.2} \begin{array}{rcl} \int_0^L y(x)y''(x)\, dx &= \left.\ y(x)y'(x)\right| _{0}^{L} -\int_0^L (y'(x))^2\,dx \\ &=\ y(L)y'(L)-y(0)y'(0)-\int_0^L (y'(x))^2\,dx. \end{array}\]

    However, if \(y\) satisfies any of the boundary conditions of Problems 1-4, then

    \[y(L)y'(L)-y(0)y'(0)=0;\nonumber \]

    hence, Equation \ref{eq:11.1.1} and Equation \ref{eq:11.1.2} imply that

    \[\lambda\int_0^L y^2(x)\,dx=\int_0^L (y'(x))^2\, dx.\nonumber \]

    If \(y\not\equiv0\), then \(\int_0^L y^2(x)\,dx>0\). Therefore \(\lambda\ge0\) and, if \(\lambda=0\), then \(y'(x)=0\) for all \(x\) in \((0,L)\) (why?), and \(y\) is constant on \((0,L)\). Any constant function satisfies the boundary conditions of Problem 2, so \(\lambda=0\) is an eigenvalue of Problem 2 and any nonzero constant function is an associated eigenfunction. However, the only constant function that satisfies the boundary conditions of Problems \(1\), \(3\), or \(4\) is \(y\equiv0\). Therefore \(\lambda=0\) isn’t an eigenvalue of any of these problems.

    Example 11.1.1

    Solve the eigenvalue problem

    \[\label{eq:11.1.3} y''+\lambda y=0,\quad y(0)=0,\quad y(L)=0.\]

    Solution

    From Theorem 11.1.1, any eigenvalues of Equation \ref{eq:11.1.3} must be positive. If \(y\) satisfies Equation \ref{eq:11.1.3} with \(\lambda>0\), then

    \[y=c_1\cos\sqrt\lambda\, x+c_2\sin\sqrt\lambda\, x,\nonumber \]

    where \(c_1\) and \(c_2\) are constants. The boundary condition \(y(0)=0\) implies that \(c_1=0\). Therefore \(y=c_2\sin\sqrt\lambda\, x\). Now the boundary condition \(y(L)=0\) implies that \(c_2\sin\sqrt\lambda\, L=0\). To make \(c_2\sin\sqrt\lambda\, L=0\) with \(c_2\ne0\), we must choose \(\sqrt\lambda=n\pi/L\), where \(n\) is a positive integer. Therefore \(\lambda_n=n^2\pi^2/L^2\) is an eigenvalue and

    \[y_n=\sin{n\pi x\over L}\nonumber \]

    is an associated eigenfunction.

    For future reference, we state the result of Example 11.1.1 as a theorem.

    Theorem 11.1.2

    The eigenvalue problem

    \[y''+\lambda y=0,\quad y(0)=0,\quad y(L)=0\nonumber \]

    has infinitely many positive eigenvalues \(\lambda_n=n^2\pi^2/L^2\), with associated eigenfunctions

    \[y_n=\sin {n\pi x\over L},\quad n=1,2,3,\dots.\nonumber\]

    There are no other eigenvalues.

    We leave it to you to prove the next theorem about Problem 2 by an argument like that of Example 11.1.1 (Exercise 11.1.17).

    Theorem 11.1.3

    The eigenvalue problem

    \[y''+\lambda y=0,\quad y'(0)=0,\quad y'(L)=0\nonumber \]

    has the eigenvalue \(\lambda_0=0\), with associated eigenfunction \(y_0=1\), and infinitely many positive eigenvalues \(\lambda_n=n^2\pi^2/L^2\), with associated eigenfunctions

    \[y_n=\cos {n\pi x\over L}, n=1,2,3\dots.\nonumber\]

    There are no other eigenvalues.

    Example 11.1.2

    Solve the eigenvalue problem

    \[\label{eq:11.1.4} y''+\lambda y=0,\quad y(0)=0,\quad y'(L)=0.\]

    Solution:

    From Theorem 11.1.1, any eigenvalues of Equation \ref{eq:11.1.4} must be positive. If \(y\) satisfies Equation \ref{eq:11.1.4} with \(\lambda>0\), then

    \[y=c_1\cos\sqrt\lambda\, x+c_2\sin\sqrt\lambda\, x,\nonumber \]

    where \(c_1\) and \(c_2\) are constants. The boundary condition \(y(0)=0\) implies that \(c_1=0\). Therefore \(y=c_2\sin\sqrt\lambda\, x\). Hence, \(y'=c_2\sqrt\lambda\cos\sqrt\lambda\,x\) and the boundary condition \(y'(L)=0\) implies that \(c_2\cos\sqrt\lambda\,L=0\). To make \(c_2\cos\sqrt\lambda\,L=0\) with \(c_2\ne0\) we must choose

    \[\sqrt\lambda={(2n-1)\pi\over2L},\nonumber \]

    where \(n\) is a positive integer. Then \(\lambda_n=(2n-1)^2\pi^2/4L^2\) is an eigenvalue and

    \[y_n=\sin{(2n-1)\pi x\over2L}\nonumber \]

    is an associated eigenfunction.

    For future reference, we state the result of Example 11.1.2 as a theorem.

    Theorem 11.1.4

    The eigenvalue problem

    \[y''+\lambda y=0,\quad y(0)=0,\quad y'(L)=0\nonumber \]

    has infinitely many positive eigenvalues \(\lambda_n=(2n-1)^2\pi^2/4L^2,\) with associated eigenfunctions

    \[y_n=\sin{(2n-1)\pi x\over2L},\quad n=1,2,3,\dots.\nonumber \]

    There are no other eigenvalues.

    We leave it to you to prove the next theorem about Problem 4 by an argument like that of Example 11.1.2 (Exercise 11.1.18).

    Theorem 11.1.5

    The eigenvalue problem

    \[y''+\lambda y=0,\quad y'(0)=0,\quad y(L)=0\nonumber \]

    has infinitely many positive eigenvalues \(\lambda_n=(2n-1)^2\pi^2/4L^2,\) with associated eigenfunctions

    \[y_n=\cos{(2n-1)\pi x\over2L},\quad n=1,2,3,\dots.\nonumber\]

    There are no other eigenvalues.

    Example 11.1.3

    Solve the eigenvalue problem

    \[\label{eq:11.1.5} y''+\lambda y=0,\quad y(-L)=y(L),\quad y'(-L)=y'(L).\]

    Solution

    From Theorem 11.1.1, \(\lambda=0\) is an eigenvalue of Equation \ref{eq:11.1.5} with associated eigenfunction \(y_0=1\), and any other eigenvalues must be positive. If \(y\) satisfies Equation \ref{eq:11.1.5} with \(\lambda>0\), then

    \[\label{eq:11.1.6} y=c_1\cos\sqrt\lambda\, x+c_2\sin\sqrt\lambda\, x,\]

    where \(c_1\) and \(c_2\) are constants. The boundary condition \(y(-L)=y(L)\) implies that

    \[\label{eq:11.1.7} c_1\cos(-\sqrt\lambda\,L)+c_2\sin(-\sqrt\lambda\,L)=c_1\cos \sqrt\lambda\,L+c_2\sin \sqrt\lambda\,L.\]

    Since

    \[\label{eq:11.1.8} \cos(-\sqrt\lambda\,L)=\cos \sqrt\lambda\,L\quad \text{and} \quad \sin(-\sqrt\lambda\,L)=-\sin \sqrt\lambda\,L,\]

    Equation \ref{eq:11.1.7} implies that

    \[\label{eq:11.1.9} c_2\sin \sqrt\lambda\,L=0.\]

    Differentiating Equation \ref{eq:11.1.6} yields

    \[y'=\sqrt\lambda\left(-c_1\sin\sqrt\lambda x+c_2\cos\sqrt\lambda x\right).\nonumber\]

    The boundary condition \(y'(-L)=y'(L)\) implies that

    \[-c_1\sin(-\sqrt\lambda\,L)+c_2\cos(-\sqrt\lambda\,L)=-c_1\sin \sqrt\lambda\,L+c_2\cos \sqrt\lambda\,L,\nonumber\]

    and Equation \ref{eq:11.1.8} implies that

    \[\label{eq:11.1.10} c_1\sin \sqrt\lambda\,L=0.\]

    Eqns. Equation \ref{eq:11.1.9} and Equation \ref{eq:11.1.10} imply that \(c_1=c_2=0\) unless \(\sqrt\lambda =n\pi /L\), where \(n\) is a positive integer. In this case Equation \ref{eq:11.1.9} and Equation \ref{eq:11.1.10} both hold for arbitrary \(c_1\) and \(c_2\). The eigenvalue determined in this way is \(\lambda_n=n^2\pi^2/L^2\), and each such eigenvalue has the linearly independent associated eigenfunctions

    \[\cos {n\pi x\over L} \quad \text{and} \quad \sin{ n\pi x\over L}. \nonumber\]

    For future reference we state the result of Example 11.1.3 as a theorem.

    Theorem 11.1.6

    The eigenvalue problem

    \[y''+\lambda y=0,\quad y(-L)=y(L),\quad y'(-L)=y'(L),\nonumber\]

    has the eigenvalue \(\lambda_0=0\), with associated eigenfunction \(y_0=1\) and infinitely many positive eigenvalues \(\lambda_n=n^2\pi^2/L^2,\) with associated eigenfunctions

    \[y_{1n}=\cos {n\pi x\over L} \quad \text{and} \quad y_{2n}=\sin {n\pi x\over L},\quad n=1,2,3,\dots.\nonumber\]

    There are no other eigenvalues.

    Orthogonality

    We say that two integrable functions \(f\) and \(g\) are orthogonal on an interval \([a,b]\) if

    \[\int_a^bf(x)g(x)\,dx=0.\nonumber\]

    More generally, we say that the functions \(\phi_1\), \(\phi_2\), …, \(\phi_n\), …(finitely or infinitely many) are orthogonal on \([a,b]\) if

    \[\int_a^b\phi_i(x)\phi_j(x)\,dx=0\quad \text{whenever} \quad i\ne j.\nonumber\]

    The importance of orthogonality will become clear when we study Fourier series in the next two sections.

    Example 11.1.4

    Show that the eigenfunctions

    \[\label{eq:11.1.11} 1,\, \cos{\pi x\over L},\, \sin{\pi x\over L}, \, \cos{2\pi x\over L}, \, \sin{2\pi x\over L},\dots, \cos{n\pi x\over L}, \, \sin{n\pi x\over L},\dots\]

    of Problem 5 are orthogonal on \([-L,L]\).

    Solution

    We must show that

    \[\label{eq:11.1.12} \int_{-L}^L f(x)g(x)\,dx=0\]

    whenever \(f\) and \(g\) are distinct functions from Equation \ref{eq:11.1.11}. If \(r\) is any nonzero integer, then

    \[\label{eq:11.1.13} \int_{-L}^L\cos{r\pi x\over L}\,dx ={L\over r\pi}\sin{r\pi x\over L}\bigg|_{-L}^L=0.\]

    and

    \[\int_{-L}^L\sin{r\pi x\over L}\,dx =-{L\over r\pi}\cos{r\pi x\over L}\bigg|_{-L}^L=0.\nonumber\]

    Therefore Equation \ref{eq:11.1.12} holds if \(f\equiv1\) and \(g\) is any other function in Equation \ref{eq:11.1.11}.

    If \(f(x)=\cos m\pi x/L\) and \(g(x)=\cos n\pi x/L\) where \(m\) and \(n\) are distinct positive integers, then

    \[\label{eq:11.1.14} \int_{-L}^L f(x)g(x)\,dx=\int_{-L}^L\cos{m\pi x\over L} \cos{n\pi x\over L}\,dx.\]

    To evaluate this integral, we use the identity

    \[\cos A\cos B={1\over2}[\cos(A-B)+\cos(A+B)]\nonumber\]

    with \(A=m\pi x/L\) and \(B=n\pi x/L\). Then Equation \ref{eq:11.1.14} becomes

    \[\int_{-L}^L f(x)g(x)\,dx={1\over2}\left[\int_{-L}^L\cos{(m-n)\pi x\over L}\,dx +\int_{-L}^L\cos{(m+n)\pi x\over L}\,dx\right].\nonumber\]

    Since \(m-n\) and \(m+n\) are both nonzero integers, Equation \ref{eq:11.1.13} implies that the integrals on the right are both zero. Therefore Equation \ref{eq:11.1.12} is true in this case.

    If \(f(x)=\sin m\pi x/L\) and \(g(x)=\sin n\pi x/L\) where \(m\) and \(n\) are distinct positive integers, then

    \[\label{eq:11.1.15} \int_{-L}^L f(x)g(x)\,dx=\int_{-L}^L\sin{m\pi x\over L} \sin{n\pi x\over L}\,dx.\]

    To evaluate this integral, we use the identity

    \[\sin A\sin B={1\over2}[\cos(A-B)-\cos(A+B)]\nonumber\]

    with \(A=m\pi x/L\) and \(B=n\pi x/L\). Then Equation \ref{eq:11.1.15} becomes

    \[\int_{-L}^L f(x)g(x)\,dx={1\over2}\left[\int_{-L}^L\cos{(m-n)\pi x\over L}\,dx -\int_{-L}^L\cos{(m+n)\pi x\over L}\,dx\right]=0.\nonumber\]

    If \(f(x)=\sin m\pi x/L\) and \(g(x)=\cos n\pi x/L\) where \(m\) and \(n\) are positive integers (not necessarily distinct), then

    \[\int_{-L}^L f(x)g(x)\,dx=\int_{-L}^L\sin{m\pi x\over L} \cos{n\pi x\over L}\,dx=0\nonumber\]

    because the integrand is an odd function and the limits are symmetric about \(x=0\).

    Exercises 11.1.19-11.1.22 ask you to verify that the eigenfunctions of Problems 1-4 are orthogonal on \([0,L]\). However, this also follows from a general theorem that we'll prove in Chapter 13.

    11.1: Eigenvalue Problems for y'' + λy = 0 (2024)

    FAQs

    What is the eigenvalue problem function? ›

    As a rule, an eigenvalue problem is represented by a homogeneous equation with a parameter. The values of the parameter such that the equation has nontrivial solutions are called eigenvalues, and the corresponding solutions are called eigenfunctions. The simplest eigenvalue problems were considered by Euler.

    What is the eigenvalue problem in a differential equation? ›

    The eigenvalue problem for such an A (with boundary conditions) is to find all the possible eigenvalues of A. In other words, we have to find all of the numbers λ such that there is a solution of the equation AX = λX for some function X (X = 0) that satisfies the boundary conditions at 0 and at l.

    What is the Sturm Liouville problem? ›

    The goals of a given Sturm–Liouville problem are: To find the λ for which there exists a non-trivial solution to the problem. Such values λ are called the eigenvalues of the problem. For each eigenvalue λ, to find the corresponding solution.

    How to solve an eigenvalue problem? ›

    For the general eigenvalue problem, we are given an n × n matrix, A, and we introduce a sequence of transformations that transform the eigenproblem for A onto equivalent eigenproblems for matrices Ar, where Ar → U (U upper triangular) as r → ∞. This is an Iterative method.

    Is lambda 0 an eigenvalue? ›

    P is singular, so λ = 0 is an eigenvalue.

    What is an example of an eigenvalue equation? ›

    Example: Find the eigenvalues and associated eigenvectors of the matrix A =   7 0 −3 −9 −2 3 18 0 −8  . = −(2 + λ)[(7 − λ)(−8 − λ) + 54] = −(λ + 2)(λ2 + λ − 2) = −(λ + 2)2(λ − 1). Thus A has two distinct eigenvalues, λ1 = −2 and λ3 = 1. (Note that we might say λ2 = −2, since, as a root, −2 has multiplicity two.

    How to find the eigenvalues of a function? ›

    The eigenvalue problem is to determine the solution to the equation Av = λv, where A is an n -by- n matrix, v is a column vector of length n , and λ is a scalar. The values of λ that satisfy the equation are the eigenvalues.

    What is the Sturm-Liouville eigenvalue problem? ›

    The problem of finding a complex number µ if any, such that the BVP (6.2)-(6.3) with λ = µ, has a non-trivial solution is called a Sturm-Liouville Eigen Value Problem (SL-EVP). Such a value µ is called an eigenvalue and the corresponding non-trivial solutions y(.; µ) are called eigenfunctions.

    What is the strum theory? ›

    Sturm's theorem expresses the number of distinct real roots of p located in an interval in terms of the number of changes of signs of the values of the Sturm sequence at the bounds of the interval.

    What are the applications of Sturm-Liouville problem? ›

    The most straightforward applications of the Sturm-Liouville problem bring about the various Fourier series, and more sophisticated applications lead to the generalizations of Fourier series involving Bessel functions, Hermite polynomials, and other special functions.

    Can eigenvalue be 0? ›

    Eigenvalues and eigenvectors are only for square matrices. Eigenvectors are by definition nonzero. Eigenvalues may be equal to zero.

    What is an example of an eigenvalue and an eigenfunction? ›

    The positive eigenvalues are then, λn=(n2)2=n24n=1,2,3,… λ n = ( n 2 ) 2 = n 2 4 n = 1 , 2 , 3 , … and the eigenfunctions that correspond to these eigenvalues are, yn(x)=sin(nx2)n=1,2,3,… y n ( x ) = sin ⁡ ( n x 2 ) n = 1 , 2 , 3 , …

    What is the easiest way to find eigenvalues and eigenvectors? ›

    How to Calculate Eigenvalues and Eigenvectors? For any square matrix A, To find eigenvalues: Solve the characteristic equation |A - λI| = 0 for λ. To find the eigenvectors: Solve the equation (A - λI) v = O for v.

    What is the eigen problem? ›

    The problem is to find the numbers, called eigenvalues, and their matching vectors, called eigenvectors. This is extremely general—it is used in differential equations (because solutions to linear differential equations form linear spaces!) and described in detail in linear algebra.

    What is an eigenvalue equation? ›

    Eigen' is a German word that means 'proper' or 'characteristic'. Therefore, the term eigenvalue can be termed as characteristic value, characteristic root, proper values or latent roots as well. In simple words, the eigenvalue is a scalar that is used to transform the eigenvector. The basic equation is. Ax = λx.

    What is the meaning of eigen function? ›

    The formal definition of an eigenfunction for a linear operator L is a function f(x) that satisfies the equation: L[f(x)] = λf(x) where: L is a linear operator. f(x) is the eigenfunction.

    What is the purpose of eigen value? ›

    Eigenvectors and eigenvalues are used to reduce noise in data. They can help us improve efficiency in computationally intensive tasks. They also eliminate features that have a strong correlation between them and also help in reducing over-fitting.

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