Chapter 6 Sturm-Liouville Problems - IIT Bombay

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].(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.

2 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. 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.

3 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. 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.

4 ThusAremains an eigenvalue with eigenfunction being any non-zero constant. Note that eigenvalue issimple. 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.

5 ( )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.( )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.

6 ( )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.( )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.

7 ( )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. 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.

8 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. 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.

9 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. 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.( )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 bothyandysatisfy the periodic boundaryconditions. Note that this is new argument that we use to replace the correspondignargument in part (1) of Remark 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 =.

10 Thatis, is real. Note that we have used self-adjointness of operatorLsomewhere!(2)The eigenfunctions of a periodic 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.( )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 ( ).Sivaji Ganesh SistaMA 417: Ordinary Differential Periodic SL-BVP(3)The eigenvalues of a regular SL-BVP are not simple, and Example illustrates thisfact. However, it will be interesting to know why the arguments given in point (3) ofRemark can not be modified.