Sturm-Liouville Theory

1 LECTURE 12 Sturm-Liouville TheoryIn the two preceding lectures I demonstrated the utility of Fourier series in solving PDE/BVPs. As we llnow see, Fourier series are just the tip of the iceberg of the Theory and utility of special functions. Beforepreceding with the general Theory , let me state clearly the basic properties of Fourier series we intend togeneralize:(i) The Fourier sine functions{sin(n Lx)|n= 1,2,3,..}and cosine functions{cos(n L)x|n= 0,1,2,..}are solutions of a second order linear homogeneous differential equation(1)y = ysatisfying certain linear homogenous boundary conditions: viz, the Fourier-sine functions satisfy(2a)y(0) = 0 =y(L)and the Fourier-cosine functions satisfy(2b)y (0) = 0 =y (L).(ii) The Fourier sine and cosine functions obey certain orthogonality relations2L L0sin(n Lx)sin(m Lx)dx={1 ifn=m0 ifn6=m2L L0cos(n Lx)cos(m Lx)dx={1 ifn=m0 ifn6=m(iii) Any continuous, differentiable function on [0,L] can be expressed in terms of a Fourier-sine orFourier-cosine expansionf(x) =a02+ n=1ancos(n Lx)= n=1bnsin(n Lx)Moreover, the orthogonality relations (ii) allow us to determine the coefficientsanandbnan=2L L0f(x) cos(n Lx)dxbn=2L L0f(x) sin(n Lx)dx1.}}

2 Sturm-Liouville problemis a second order homogeneous linear differential equationof the form(3)ddx[p(x)dydx] q(x)y+ r(x)y= 0, x [a,b]551. Sturm-Liouville PROBLEMS56together with boundary conditions of the form 1y(a) + 2y (b) = 0(4a) 1y(a) + 2y (b) = 0(4b)In what follow, is to be regarded as a constant parameter and we shall always assume thatp(x)andr(x)are positive functions on the interval[a,b]p(x)>0 x [a,b](5)r(x)>0 x [a,b] (x) = 1r(x) = 1q(x) = 0a= 0b=L 1= 1= 1 2= 2= 0, we see that the corresponding Sturm-Liouville problemy = yy(0) = 0y(L) = 0has as its solutionsy(x) = sin(n Lx)and =(n L)2n= 1,2,3,..Note, in particular, that the solution of the Sturm-Liouville problem only exists for certan values of ; thesevalues are eigenvalues of the Sturm-Liouville (Linear differential operator notation).In what follows we shall denote byLthe lineardifferential operator (6)L=ddx[p(x)ddx] q(x)which acts on a functiony(x) byL[y] =ddx[p(x)dydx] q(x)yIn terms ofL, the differential equation of a Sturm-Liouville problem can be expressedL[y] = r(x) (Lagrange s identity).

3 Suppose and are two functions satisfying boundary conditions ofthe form 1y(a) + 2y (a) = 0(7a) 1y(b) + 2y (b) = 0(7b)then ba(L[ ] L[ ])dx= 01. Sturm-Liouville have baL[ ] dx= ba(ddx[pd dx] q )dx= baddx[pd dx] dx baq dx=pd dx ba ba(pd dx)d dxdx baq dx=pd dx ba bad dx(pd dx)dx baq dx=pd dx ba ( pd dx) ba+ ba ddx(pd dx)dx baq dx=p( )|ba+ ba L[ ]dv(*)where we have twice utilized the integration by parts formula badfdxgdx=fg|ba bafdgdxdxNow consider the first term in the last linep( )|baSince and satisfy (7a) and (7b) we have assuming 1, 2, 1, 26= 0 (a) = 2 1 (a), (a) = 2 1 (a) (b) = 2 1 (b), (b) = 2 1 (b)and (a) (a) (a) (a) = 2 1 (a) (a) (a)( 2 1 (a))= 0 (b) (b) (b) (b) = 2 1 (a) (b) (a)( 2 1 (b))= 0 Hencep(u v uv )|10=p(1) (u (1)v(1) u(1)v (1)) p(0) (u (0)v(0) u(0)v (0))=p(1) 0 p(0) 0= 0We can hence conclude from (*) that baL[ ] dx= 0 + ba L[ ]dxand the theorem follows.

4 Recall that, for example, the Fourier-sine functions sin ( x) not only satisfy certain a second order lineardifferential equationf = 2fbut they also satisfy certain linear homogeneous boundary conditionsf(0) = 0 =f(L)1. Sturm-Liouville PROBLEMS58exactly when the parameter is tuned to the boundary conditionsf = 2ff(0) = 0f(L) = 0 = { =n Lf(x) = sin(n Lx)What may seem a little surprising at first is that the fact that the Fourier-sine functions are solutions to aSturm- liouville problem is also responsible for their orthogonality properties2L L0sin(n Lx)sin(m Lx)dx={1 ifn=m0 ifn6= and are two solutions of a Sturm-Liouville problem corresponding to different valuesof the parameter then 10 (x) (x)r(x)dx= SupposeL[ ] = 1r(x) L[ ] = 2r(x) with 16= 2. Then since , satisfy the Strum- liouville boundary conditions, we have by Theorem 10L[ ] dx= 10 (L[ ])dxand because they satisfy Sturm-Liouville -type differential equations 10( 1r(x) (x)) (x)dx= 10 (x) ( 2r(x) (x))dxor( 1 2) 10 (x) (x)r(x)dx= 0By hypothesis, 16= 2and so we must conclude 10 (x) (x)r(x)dx= 0 As remarked above, it is common to write (in shorthand) the differential equation of a sturm -Liouvilleproblem as(8)L[y] = r(x)yand refer to the parameter (corresponding to a particular solutiony(x)) as an eigenvalue of the differentialoperatorL.}}

5 This language, of course, is derived from the nomenclature of linear algebra where ifAis ann nmatrix,vis ann 1 column vector and is a number such thatAv= vthenvis called an eigenvector ofAand is the corresponding eigenvalue ofA. In fact, it is commonto use such linear-algebraic-like nomenclature throughout Sturm-Liouville Theory ; and henceforth we shallrefer to the functionsy(x) that satisfy (8) aseigenfunctionsofLand the numbers as the the eigenvalues of a Sturm-Liouville problem are Sturm-Liouville the following inner product on the space of complex-valued solutions of a sturm -Liouvilleproblem (8) (for various ) u,v = 10u(x)v(x)dxSupposeu(x) =v(x) =R(x) +iI(x)that is, suppose we setuequal tovand split it up into its real and imaginary parts. Then u,u = 10(R(x) +iI(x)) (R(x) iI(x))dx= 10[(R(x))2+ (I(x))2]dxNote now that the integrand is always non-negative.

6 By a well-known result from Calculus, the integral of acontinuous non-negative function over a finite interval is always non-negative, and zero only if the functionis identically equal to zero. We conclude that for any continuous, non-zero, complex-valued functionu(x) 10u(x)u(x)dx >0 Now consider equation obtained from the conclusion of Theorem by settingv=u,(9) 10L[u]vdx= 10uL[u]dxand supposeuis a Sturm-Liouville eigenfunction:(10)L[u] = rufor some CTaking the complex conjugate of this equation, we have, becausep(x),q(x), andr(x) are all assumed tobe real-valued functions,L[u] = rv= L[u] = ru= ddx[p(x)dudx] q(x)u= ru= ddx[p(x)dudx] q(x)u= ruor(11)L[u] = ruAnd so plugging the right hand sides of (10) and (11) into (9) we can conclude that 10 (ru)u dx= 10 u(ru)dxor(12)( ) 10ruudx= 0We have argued above thatu(x)u(x) is a non-negative function, and by assumptionr(x) is a positivefunction on [0,1].

7 Thus, the integrand is a non-negative function and0< 10ruudxso long asu(x) is not identically zero. We can therefore conclude from (12) that = 0= R. 1. Sturm-Liouville PROBLEMS60In Linear Algebra, we say that an eigenvalue of a matrixAhas multiplicitymif the dimension of thecorresponding eigenspace ism; that is to say,m= dimNullSp(A I)In Sturm-Liouville Theory , we say that the multiplicity of an eigenvalue of a Sturm-Liouville problemL[ ] = r(x) (x)a1 (0) +a2 (0) = 0b1 (1) +b2 (1) = 0if there are exactlymlinearly independent solutions for that value of . eigenvalues of a Sturm-Liouville problem are all of multiplicity one. Moreover, theeigenvalues form an infinite sequence and can be ordered according to increasing magnitude:{eigenvalues}={ 1, 2, 3,..}, 1< 2< 3< The ordering of the eigenvalues , and the fact that they multiplicity free, establishes a certain canonicalordering of the corresponding Sturm-Liouville eigenfunctions.

8 We denote by nthe solution of the Sturm-Liouville problem with = n. Actually, this only determines n(x) up to a constant multiple (because if (x) satisfies (3), (4a), (4b) then so does any constant multiple of (x)). It is common practise to removethis ambiguity by demanding in addition 10 n(x) n(x)r(x)dx= 1which fixes (x) up to a scalar factor of the formei .Moreover, the facts that the eigenvalues are all real and multiplicity free, also implies that we can choosethe n(x) to be real-valued functions ofx(because the complex span of the functions n(x) must coincidewith that of n(x)). Thus, we can choose the eigenfunctions of a Sturm-Liouville problem to be real-valuedfunctions 10 n(x) n(x)r(x)dx= 1, 2, 3,..be the normalized eigenfunctions of a Sturm-Liouville problem and supposefis a piecewise continuous function on[0,1]. Then ifcn:= 10f(x) n(x)r(x)dxthe series n=1cn n(x)converges tof(x+) +f(x )2at each point on(0,1).

9 Heref(x ) := lim 0+f(x )Note that at any pointxwherefis continuousf(x+) +f(x )2=f(x).1. Sturm-Liouville PROBLEMS61In summary, any time you have a differential equation of the formddx[p(x)dydx] q(x)y= r(x)ywith homogeneous boundary conditions of the forma1y(0) +a2y (0) = 0 =b1y(1) +b2y (1)Then: Solutions will exist for only a discrete (but otherwise infinite) set of values for the parameter ,all of which are real numbers. There is a unique lowest eigenvalue 0and the other eigenvaluescan be totally ordered 0< 1< 2< 3< For each n, there is a solution nofddx[p(x)dydx] q(x)y= nr(x)ythat is unique up to a scalar factor. The solutions n,n= 0,1,2,..can be normalized such that 10 n(x) m(x)r(x)dx={1 ifn=m0 otherwise Any continuous functionf(x) on the interval [0,1] can be expanded in terms of the sturm -Liouvilleeigenfunctions n,n= 0,1,2,,..f(x) = n=0an n(x)with the coefficientsanbeing determined byan= 10f(x) n(x)r(x)dxIff(x) is merely piece-wise continuous on [0,1], one has insteadf(x+) +f(x )2= n=0an n(x)with the coefficientsandetermined by the same formula.}