Notes on Sturm-Liouville Differential Equations

1 Notes on Sturm-Liouville Differential EquationsCharles Byrne of Mathematical SciencesUniversity of Massachusetts at LowellLowell, MA 01854, USAA pril 9, 20091 Recalling the Wave EquationThe one-dimensional wave equation is tt(x, t) =c2 xx(x, t),( )wherec >0 is the propagation speed. Separating variables, we seek a solution of theform (x, t) =f(t)y(x). Inserting this into Equation ( ), we getf (t)y(x) =c2f(t)y (x),orf (t)/f(t) =c2y (x)/y(x) = 2,where >0 is the separation constant. We then have the separated differentialequationsf (t) + 2f(t) = 0,( )andy (x) + 2c2y(x) = 0.( )The solutions to Equation ( ) arey(x) = sin( cx).For each arbitrary , the corresponding solution of Equation ( ) isf(t) = sin( t),1orf(t) = cos( t).In the vibrating string problem, the string is fixed at both ends,x= 0 andx=L, sothat (0, t) = (L, t) = 0,for allt. Therefore, we must havey(0) =y(L) = 0, so that the solutions must havethe formy(x) =Amsin( mcx)=Amsin( mLx),where m= cmL, for any positive integerm.

2 Therefore, the boundary conditions limitthe choices for the separation constant . In addition, if the string is not moving attimet= 0, thenf(t) = cos( mt).We want to focus on Equation ( ).Equation ( ) can be written asy (x) + y(x) = 0,( )which is aneigenvalue problem. What we have just seen is that the boundary condi-tionsy(0) =y(L) = 0 limit the possible values of for which there can be solutions:we must have = m=( mc)2=( mL)2,for some positive integerm. The corresponding solutionsym(x) = sin( mLx)are the vibrating string problem, we typically have the condition (x,0) =h(x),whereh(x) describes the initial position of the string. The problem that remains isto find a linear combination of the eigenfunctions that satisfies this additional initialcondition. Therefore, we need to find coefficientsAmso thath(x) = m=1 Amsin( mLx).( )Orthogonality will multiply the equationy m= mym2byynand the equationy n= nynbyymand subtract, to gety myn y nym= ( n m)(ymyn).

3 Usingy myn y nym= (yny m ymy n) ,and integrating, we get0 =yn(L)y m(L) ym(L)y n(L) yn(0)y m(0) ym(0)y n(0) = ( n m) L0ym(x)yn(x)dx,so that L0ym(x)yn(x)dx= 0,form6=n. Using this orthogonality of theym(x), we can easily find the OverviewIn what follows we shall study the Sturm-Liouville Equations , a class of second-orderordinary differential Equations that contains, as a special case, the eigenvalue problemin Equation ( ). As we shall see, the theory follows closely what we have justdiscovered about the one-dimensional wave equation. The general form for the Sturm-Liouville Problem isddx(p(x)y (x))+ w(x)y(x) = 0.( )As with the one-dimensional wave equation, boundary conditions, such asy(a) =y(b) = 0, wherea= andb= + are allowed, restrict the possible eigenvalues to an increasing sequence of positive numbers m. The corresponding eigenfunctionsym(x) will bew(x)-orthogonal, meaning that0 = baym(x)yn(x)w(x)dx,form6=n. As we shall see later, for various choices ofw(x) andp(x) and variouschoices ofaandb, we obtain several famous sets of orthogonal called the problemy (x) + y(x) = 03an eigenvalue problem, which suggests that a theory similar to that for matrices mightbe possible.

4 This leads to the notion ofself-adjointdifferential operators and helpsto motivate the particular form of Sturm-Liouville problems. As we shall see, thepleasant properties of the solutions of the boundary-value problem involving Equation( ) stem from the fact that the operatorLy=y is self-adjoint on functions thatare zero at the end points. Many of these properties hold, as well, for solutions toother self-adjoint problems, in particular, to solutions of Sturm-Liouville Self-Adjoint Linear Differential OperatorsSeparation of variables in partial differential Equations often leads to eigenvalue prob-lems associated with linear differential operators. Self-adjoint linear differential oper-ators, which generalize the notion of real symmetric matrices, are a convenient classof operators for which the theory of eigenvalue problems is particularly Self-Adjoint MatricesThe usual inner product for real (column) vectorsuandvis just the dot product,written variously as u, v =u v= any real square matrixAand any inner product, theadjointmatrixA is definedby the property Au, v = u, A v ,for alluandv.

5 Since, for the dot product, we have Au, v = (Au)Tv=uT(ATv) = u, ATv ,it follows thatA =ATfor this inner product. Therefore, the matrices that areself-adjoint for the usual inner product are just the symmetric nand mare distinct eigenvalues of a real symmetric matrixAthen theircorresponding eigenvectors,unandum, are orthogonal: we have(Aun)Tum=uTnATum=uTn(Aum) = muTnum,and(Aun)Tum= n6= m, it follows thatuTnum= Self-Adjoint OperatorsWe want to extend this idea of being self-adjoint to linear differential operators andinner products of any inner product on functions, written y, z , and any linear operator onthese functions,Ly, the adjoint ofLis defined by the identity Ly, z = y, L z ,for all functionsyandz. The operatorLisself-adjointon a certain class of functionsifL =Lfor those The OperatorDy=y For example, consider the linear differential operatorDy=dydx. We take for the innerproduct of two functionsy(x) andz(x) the integral y, z = 10y(x)z(x) functionsyandzthat are zero at the end points, we have, using integration byparts, Dy, z = 10y (x)z(x)dx= 10y(x)z (x)dx= y, D z ,from which we conclude thatD z= The OperatorLy=y Now consider the linear differential operatorLy=d2ydx2.

6 Using the same inner product,restricting to functionsyandzthat are zero at the end points, and again usingintegration by parts, we find thatL z=d2zdx2=Lz; therefore, we say that thisoperator isself-adjoint. Self-adjoint operators generalize real symmetric General Second-Order Linear ODE sWe are concerned, in these Notes , with second-order linear differential Equations with(possibly) non-constant coefficients, that is, differential Equations of the forma2(x)y (x) +a1(x)y (x) +a0(x)y(x) = 0.( )Now we ask when the linear differential operatorLy= [a2(x)y +a1(x)y ]5is self-adjoint. Once again, we consider functions that are zero at end pointsx=aandx=band define the inner product ofyandzto be y, z = bay(x)z(x) integration by parts several times, we find thatL z=a2(x)z + (2a 2(x) a1(x))z + (a 2(x) a 1(x)) , if it is the case thata 2(x) =a1(x), thenL =LandLis self-adjoint. Inthis case, we can write Equation ( ) as(a2(x)y (x)) +a0(x)y(x) = 0,which has the form of the Sturm-Liouville problem,ddx(p(x)y (x))+w(x)y(x) = similar calculation shows that, for any weight functionw(x)>0, the linear differ-ential operatorT y=1w(x)(p(x)y (x)) is self-adjoint with respect to the inner product defined by y, z = bay(x)z(x)w(x) we can write Equation ( ) as1w(x)(p(x)y (x)) + y(x) = 0,this tells us that we are dealing with an eigenvalue problem associated with a self-adjoint linear differential Qualitative Analysis of ODEWe are interested in second-order linear differential Equations with possibly varyingcoefficients, as given in equation ( ), which we can also write asy +P(x)y +Q(x)y= 0.

7 ( )Although we can find explicit solutions of Equation ( ) in special cases, such asy +y= 0,( )generally, we will not be able to do this. Instead, we can try to answer certainquestions about the behavior of the solution, without actually finding the solution;such an approach is calledqualitative analysis. The discussion here is based on thatin Simmons [1]. A Simple ExampleWe know that the solution to Equation ( ) satisfyingy(0) = 0, andy (0) = 1 isy(x) = sinx; withy(0) = 1 andy (0) = 0, the solution isy(x) = cosx. But, supposethat we did not know these solutions; what could we find out without solving forthem?Suppose thaty(x) =s(x) satisfies Equation ( ), withs(0) = 0,s( ) = 0, ands (0) = 1. As the graph ofs(x) leaves the point (0,0) withxincreasing, the slope isinitiallys (0) = 1, so the graph climbs above thex-axis. But sincey (x) = y(x), thesecond derivative is negative fory(x)>0, and becomes increasingly so asy(x) climbshigher; therefore, the derivative is decreasing froms (0) = 1, eventually equaling zero,at sayx=m, and continuing to become negative.

8 The functions(x) will be zeroagain atx= , and, by symmetry, we havem= lety(x) =c(x) solve Equation ( ), but withc(0) = 1, andc (0) = (x) =s(x) satisfies Equation ( ), so doesy(x) =s (x), withs (0) = 1 ands (0) = 0. Therefore,c(x) =s (x). Since the derivative of the functions(x)2+c(x)2is zero, this function must be equal to one for allx. In the section that follows, weshall investigate the zeros of sturm Separation TheoremTheorem (x)andy2(x)be linearly independent solutions of Equation( ). Then their zeros are distinct and occur :Since, for eachx,a=b= 0 is the only solution of the systemay1(x) +by2(x) = 0,ay 1(x) +by 2(x) = 0,the Wronskian,W(y1, y2) =y1(x)y 2(x) y2(x)y 1(x),which is the determinant of this two-by-two linear system of Equations , can neverbe zero, so must have constant sign, asxvaries. Therefore, the two functionsy1(x)andy2(x) have no common zero. Suppose thaty2(x1) =y2(x2) = 0, withx1< x2successive zeros ofy2(x).

9 Suppose, in addition, thaty2(x)>0 in the interval (x1, x2).Therefore, we havey 2(x1)>0 andy 2(x2)<0. It follows thaty1(x1) andy1(x2) haveopposite signs, and there must be a zero From Standard to Normal FormEquation ( ) is called thestandard formof the differential equation. To put theequation intonormal form, by which we mean an equation of the formu (x) +q(x)u(x) = 0,( )we writey(x) =u(x)v(x). Inserting this product into Equation ( ), we obtainvu + (2v +P v)u+ (v +P v +Qv)u= exp( 12 P dx),the coefficient ofu becomes zero. Now we setq(x) =Q(x) 14P(x)2 12P (x),to getu (x) +q(x)u(x) = can be shown that, ifq(x)<0 andu(x) satisfies Equation ( ), then eitheru(x) = 0, for allx, oru(x) has at most one zero. Since we are interested in oscillatorysolutions, we restrictq(x) to be (eventually) positive. Withq(x)>0 and 1q(x)dx= ,the solutionu(x) will have infinitely many zeros, but only finitely many on anybounded sturm Comparison TheoremSolutions toy + 4y= 0oscillate faster than solutions of Equation ( ).

10 This leads to the sturm ComparisonTheorem:Theorem +q(x)y= 0andz +r(x)z= 0, with0< r(x)< q(x), for allx. Then between any two zeros ofz(x)is a zero ofy(x). Bessel s EquationBessel s Equation isx2y +xy + (x2 p2)y= 0.( )In normal form, it becomesu +(1 +1 4p24x2)u= 0.( )Information about the zeros of solutions of Bessel s Equation can be obtained byusing sturm s Comparison Theorem and comparing with solutions of Equation ( ).5 Sturm-Liouville EquationsThe Sturm-Liouville Equations have the formddx(p(x)dydx)+ w(x)y= 0.( )Here we assume thatp(x)>0 andw(x)>0 are continuous, andp (x) is Special CasesThe problem of the vibrations of a hanging chain leads to the equation x(gx y x)= 2y t2,( )and, after separating variables, toddx(gxdudx)+ u= 0.( )The problem of the vibrations of a non-homogeneous string leads to 2y x2=m(x)T 2y t2,( )and, after separating the variables, tou + m(x)u= 0,( )withu(0) =u( ) = Normal FormWe can put an equation in the Sturm-Liouville form into normal form by first writingit in standard form.