Chapter 6 Sturm-Liouville Problems

1 Chapter 6 Sturm-Liouville ProblemsDefinition ( Sturm-Liouville Boundary Value problem (SL-BVP))With the notationL[y] ddx[p(x)dydx]+q(x)y,( )consider the Sturm-Liouville equationL[y] + r(x)y= 0,( )wherep >0,r 0, andp, q, rare continuous functions on interval[a, b]; along with the boundaryconditionsa1y(a) +a2p(a)y (a) = 0, b1y(b) +b2p(b)y (b) = 0,( )wherea21+a226= 0andb21+b226= problem of finding a complex number if any, such that the BVP( )-( )with = , hasa non-trivial solution is called a Sturm-Liouville Eigen Value problem (SL-EVP). Such a value is called an eigenvalue and the corresponding non-trivial solutionsy(.; )are called ,(i)An SL-EVP is called aregular SL-EVPifp >0andr >0on[a, b].(ii)An SL-EVP is called asingular SL-EVPif (i)p >0on(a, b)andp(a) = 0 =p(b), and (ii)r 0on[a, b].

2 (iii)Ifp(a) =p(b),p >0andr >0on[a, b],p, q, rare continuous functions on[a, b], then solvingSturm- liouville equation( )coupled with boundary conditionsy(a) =y(b), y (a) =y (b),( )is called aperiodic are not going to discuss singular SL-BVPs. Before we discuss further, let us completely studytwo examples that are representatives of their class of Two examplesExample R, solvey + y= 0, y(0) = 0, y ( ) = 0.( )For reasons that will be clear later on, it is enough to consider Two examplesCase <0. Then = 2, where is real and non-zero. The general solution of ODE in( )is given byy(x) =Ae x+Be x( )Thisysatisfies boundary conditions in( )if and only ifA=B= 0. That is,y 0. Therefore,there are no negative = 0.

3 In this case, it easily follows that trivial solution is the only solution ofy = 0, y(0) = 0, y ( ) = 0.( )Thus,0is not an >0. Then = 2, where is real and non-zero. The general solution of ODE in( )is given byy(x) =Acos( x) +Bsin( x)( )Thisysatisfies boundary conditions in( )if and only ifA= 0andBcos( ) = 0. ButBcos( ) = 0if and only if, eitherB= 0orcos( ) = conditionA= 0andB= 0meansy 0. This does not yield any eigenvalue. Ify6 0, thenb6= 0. Thuscos( ) = 0should hold. This last equation has solutions given by =2n 12, forn= 0, 1, 2, .. Thus eigenvalues are given by n=2n 12, n= 0, 1, 2, ..( )and the corresponding eigenfunctions are given by n(x) =Bsin(2n 12x), n= 0, 1, 2, ..( )Note: All the eigenvalues are positive.

4 The eigenfunctions corresponding to each eigenvalue forma one dimensional vector space and so the eigenfunctions are unique upto a constant R, solvey + y= 0, y(0) y( ) = 0, y (0) y ( ) = 0.( )This is not a SL-BVP. It is a mixed boundary condition unlike the separated BC above. Theseboundary conditions are called periodic boundary <0. Then = 2, where is real and non-zero. In this case, it can be easilyverified that trivial solution is the only solution of the BVP( ).Case = 0. In this case, general solution of ODE in( )is given byy(x) =A+Bx( )Thisysatisfies the BCs in( )if and only ifB= 0. ThusAremains an eigenvalue with eigenfunction being any non-zero constant. Note that eigenvalue issimple.

5 An eigenvalue is called simple eigenvalue if the corresponding eigenspace is of dimensionone, otherwise eigenvalue is called multiple >0. Then = 2, where is real and non-zero. The general solution of ODE in( )is given byy(x) =Acos( x) +Bsin( x)( )Thisysatisfies boundary conditions in( )if and only ifAsin( ) +B(1 cos( ))= 0,A(1 cos( )) Bsin( ) = 417: Ordinary Differential EquationsSivaji Ganesh SistaChapter 6 : Sturm-Liouville Problems55 This has non-trivial solution for the pair(A, B)if and only if sin( ) 1 cos( )1 cos( ) sin( ) = 0.( )That is,cos( ) = 1. This further implies that = 2nwithn N, and hence = 4n2withn positive eigenvalues are given by n= 4n2, n N.( )and the eigenfunctions corresponding to nare given by n(x) = cos (2nx), n(x) = sin (2nx), n N.

6 ( )Note: All the eigenvalues are non-negative. There are two linearly independent eigenfunctions,namelycos (2nx)andsin (2nx)corresponding to each positive eigenvalue n= 4n2. Comparethese properties with that of previous Regular SL-BVPWe noted some properties of the SL-BVP Example These properties hold for general RegularSL-BVPs as record here some of the properties of regular SL-BVPs.(1)The eigenvalues, if any, of a regular SL-BVP are :Suppose Cis an eigenvalue of a regular SL-BVP and letybe correspondingeigenfunction. That is,L[y] + r(x)y= 0, a1y(a) +a2p(a)y (a) = 0, b1y(b) +b2p(b)y (b) = 0.( )Taking the complex conjugates, we getL[y] + r(x)y= 0, a1y(a) +a2p(a)y (a) = 0, b1y(b) +b2p(b)y (b) = 0.

7 ( )Multiply the ODE in ( ) wtihy, and multiply that of ( ) withy, and subtractingone from the other yields[p(y y y y)] + ( )ryy= 0( )Integrating the last equality yields[p(y y y y)] ba= ( ) bar(x)|y(x)|2dx.( )But LHS of the last equation is zero, since we have bothb1y(b) +b2p(b)y (b) = 0andb1y(b) +b2p(b)y (b) = 0, we also know thatb21+b226= 0, and hence a certaindeterminant associated is we have( ) bar|y|2dy= 0.( )Sincey, being an eigenfunction, is not identically equal to zero, and integral of non-negative function (sincer >0) is not zero, the only possibility is that = . Thatis, is real. Note that we have used self-adjointness of operatorLsomewhere!Sivaji Ganesh SistaMA 417: Ordinary Differential Regular SL-BVP(2)The eigenfunctions of a regular SL-BVP corresponding to distinct eigenvalues areothogonal weight functionron[a, b], that is, ifuandvare eigenfunctionscorresponding to distinct eigenvalues and respectively, then bar(x)u(x)v(x)dx= 0.

8 ( )Proof :As in the previous proof, writing down the equations satisfied byuandv, and multi-plying the equation foruwithvand vice versa, finally subtracting one from another,we get[p(u v v u)] + ( )ruv= 0( )Integrating the last equality yields[p(u v v u)] ba= ( ) bar(x)u(x)v(x)dx.( )Reasoning exactly as in the previous proof, LHS of the above equality is zero. Since 6= , we get the desired ( ).(3)The eigenvalues of a regular SL-BVP are simple. Thus an eigenfunction correspondingto an eigenvalue is unique up to a constant :Let 1and 2be two eigenfunctions corresponding to the same eigenvalue .We recall from the section on Green s functions (the identity ( ) ) here:By Lagrange s identity ( ), we getddx[p( 1 2 1 2)]= 0.

9 This impliesp( 1 2 1 2) c,a constant.( )Since 1and 2satisfy the boundary conditionU1[y] = 0, we get the following 1(a) 1(a) 2(a) 2(a) = 0.( )Since( 1 2 1 2) is the wronskian of two solutions of a second order ODE, it isidentically equal to zero. From here, it follows that 1and 2differ by a the above remark, we only analysed the properties of an eigenvalue or of two eigenfunctionscorresponding to distinct eigenvalues. We have not proved the existence of eigenvalues for a regularSL-BVP so far. We are not going to do this, since the result follows easily from a much generaltheory of a subject known asFunctional analysis, to be more specific, the topic is calledspectraltheory. Some references where a proof can be found are books on functional analysis , Yosidaand also books on ODE byCoddington & Levinson, Hartmanor evenbooks on PDE byWeinberger.

10 We will only state the self-adjoint regular SL-BVP has an infinite sequence of real eigenvalues( n)n N,that are simple satisfying 1< 2< < n< ..( )withlimn n .Exercise >0be a real number. Find eigenvalues and corresponding eigenvectors of theregular SL-BVP posed on the interval[0,1]y + y= 0, y(0) = 0, y(1) +hy (1) = 417: Ordinary Differential EquationsSivaji Ganesh SistaChapter 6 : Sturm-Liouville Periodic SL-BVPFor a periodic SL-BVP also, eigenvalues are real, eigenfunctions corresponding to distinct eigen-values are orthogonal weight functionr, but eigenvalues need not be simple. We recordthese in the following record here some of the properties of periodic SL-BVPs.(1)The eigenvalues, if any, of a regular SL-BVP are :Suppose Cis an eigenvalue of a regular SL-BVP and letybe correspondingeigenfunction.