Chapter 5 Sturm-Liouville Theory - Texas Tech …

1 Chapter 5 Sturm-Liouville Oscillation and Separation TheoryConsider the di erential equationa2(x)y00+a1(x)y0+a0(x)y= 0( )wherea2(x) is not zero for allx2[a;b],ai(x)2C[a;b]. Rewrite ( ) in the formy00+a1a2y0+a0a2y=y00+p(x)y0+q(x)y=0D e nek(x)=eRp(s)ds;Thenddx(k(x)y0)+k(x)q(x) y=0or(ky0)0+g(x)y= 0( )De ne the di erential operatorsL(y)=(ky0)0+g(x)yM(y)=a2y00+a1y 0+a0y( )The adjoint of M is de ned byM(y)=(a2y)00 (a1y)0+a0y12 Chapter 5. Sturm-Liouville THEORYM(y)=a2y00+(2a02 a1)y0+(a002 a01+a0)y:After some manipulation it is easy to show thatvM(u) uM(v)=[(a1 a02)vu+a2(vu0 uv0)]0:This result is calledLaGrange's identityand we rewrite it asvM(u) uM(v)=ddxP(u;v):By an integration we obtainGreen's formula,Zba vM(u) uM(v) dx=P(u;v)jbaIfM(u)=M(u), the equationM(u) = 0 is said to be self-adjoint.

2 HenceM(u)=0isself-adjoint ifa02=a1:In this case Lagrange's identity becomesvM(u) uM(v)=[a2(vu0 uv0)]0=[a2(x)N(v;u)]0andM(u)=(a2u0)0+a0u Clearly the operatorLde ned by ( ) is self-adjoint and the discussion preceding ( )shows every general linear equation can be put into self adjoint Separation theoremsTheorem [ sturm Separation Theorem]1. A nontrivial solution ofM(y)=0can have at most a nite number of zeros on[a;b]:2. All zeros of a solution are Ifu1(x),u2(x)are linearly independent solutions ofM(y)=0then between any twozeros ofu1(x)there is precisely one zero ofu2(x). Suppose there exists in nitely many zeros,fzng, select a subsequencefxngsuchthatxn!bx. Then0 = limn!1y(xn)=y(bx):Also,y0(bx) = limn!

3 1y(xn) y(bx)xn bx=0and soy(x) = 0 by OSCILLATION AND SEPARATION THEORY32. The proof of this was a much earlier Supposex0,x1are consecutive zeros ofu1(x), and assume thatx0<x1. Thenu2(x0)6=0 andu2(x1)6= 0, or elseW(u1;u2)(xi)=0;i=0;1:So without loss of generalityassumeu1(x)>0;x2(x0;x1);u2(x0)>0:NowW(u1;u2)(x0)= u01(x0)u2(x0)W(u1;u2)(x1)= u01(x1)u2(x1)Sinceu01(x0)>0;W(x0)<0:Beca use the Wronskian ofu1;u2cannot change sign,W(x1)<0:Butu01(x1)<0 so this would require thatu2(x1)<0:Henceu2(x) mustvanish in (x0;x1):If we apply the argument with the roles ofu1andu2interchanged, we see that betweentwo consecutive zeros ofu2there must be a zero ofu1(x). Hence the zeros ofu1andu2must speaking, the sturm Separation theorem states that linearly independent solu-tions have the same number of zeros.

4 If we consider two di erent equations, for exampley00+y=0;y00+4y=0then solutions of the second equation oscillate more rapidly than those of the rst. Moregenerally, sturm Comparison theorems address the rate of oscillation of solutions of di Oscillation TheoryHere we shall consider equations of the formL(y)=(ky0)0+gy=0;a<x<bwherek2C1(a;b);g2C0[a;b];andk>0on[a;b].Theorem (y)=(ky0)0+giy=0whereg2>g1andg2is not identically equaltog1on any subinterval of(a;b).IfL1(u1)=0andL2(u2)=0;then between any twoconsecutive zeros ofu1(x)there is a zero ofu2(x):4 Chapter 5. Sturm-Liouville (x1)=u1(x2) = 0 andu2(x)6= 0 on (x1;x2). Without loss of generalitytakeu1;u2>0on(x1;x2) Lagrange's identity givesu2L1(u1) u1L1(u2)=ddx(k(x)(u2u01 u1u02):We also haveL1(u2) L2(u2)=(g1 g2)u2:Henceu2L1(u1) u1[L2(u2)+(g1 g2)u2]=(k(x)(u2u01 u1u02))0ork(x)(u2u01 u1u02)jx2x1=Zx2x1(g2 g1)u1u2dx >0:However, the left hand side reduces tok(x2)u2(x2)u01(x2) k(x1)u2(x1)u01(x1):(*)Sinceu2(x2) 0;u01(x2)<0 andu2(x1) 0;u01(x1)>0 the above expression is nonpositiveand hence we obtain a that we assumedu2(x1)=0;then in the above proof expression ( ) becomesk(x2)u01(x2)u2(x2):Ifu2(x2) 0, then again ( ) 0 and a contradiction would be obtained.)

5 Thus could be restated: Ifk2C1(a;b);g2C0[a;b];k >0on[a;b] andu1(a)=u2(a)=0andu1(x1)=0;a<x1<b;then there existsz;a<z<x1such thatu2(z)=0:Thusu2(x)has at least as many zeros asu1(x)on[a;b]:A more general version of this theorem isTheorem ;q2C0[a;b]andz(x)is a non trivial solution ofz00+q(x)z=0wherez(a)=z(b)=0:IfZba(p q)z2dx OSCILLATION AND SEPARATION THEORY5then a nontrivial solution ofy00+p(x)y=0y(a)=0has a zero in the interval(a;b] (x)6= 0 in (a;b]:Thenz(z00+qz)=0z2y(y00+py)=0orzz00 z2y00y=z2(p q)orzy(yz0 zy0)0=z2(p q):Now note thatlimx!az(x)y(x)= limz!az0(a)y0(a)and sincez0(a)6=0;y0(a)6=0;z0;y02C[a;b];this limit exists and is nite. Hence it makessense to writeZbazy(yz0 zy0)0dx=Zbaz2(p q)dx 0:Now integrate by parts and sincez(b)=0;y(b)6=0;z(a)=y(a)=0;zy(yz0 zy0)jba Zba(yz0 zy0)(zy)0=Zbaz2(p q)dx 0;( )or Zba(yz0 zy0)2y2dx 0or0 Zba(yz0 zy0)2y2dx:The right hand side is identically zero ify(x)=cz(x) in which case the result is triviallytrue.))

6 So ify(x)6=cz(x);the right hand side is positive and we get a contradiction. Hencey(x) must vanish in (a;b]:6 Chapter 5. Sturm-Liouville THEORYThe proof shows that ifp(x)6=q(x) thenZbaz2(p q)dx >0:In this casey(x) must have a zero in (a;b). If not, then just as before we could derive (*) bydividing byy(x) and the boundary term in (*) would vanish sincey(b) = 0, and we wouldobtainZba(yz0 zy0)2y2dx <0;which is a conclude with a generalization of these [a;b];gi2C[a;b]withki>0:Ifzis a nontrivial solution of(k1z)0+g1z=0z(a)=z(b)=0;andyis a non trivial solution of(k2y0)0+g2y=0y(a)=0andZba(k1 k2)(z0)2+(g2 g1)z2dx 0;theny(x)has a zero in(a;b]. If the inequality is strict, the zero is in the open interval(a;b) the rst equation byzand subtract the second multiplied by (z2=y)toobtain the Picone formula,Zba(k1 k2)(z0)2+(g2 g1)z2dx+Zbak2(yz0 zy0y)2dx=zy(k1yz0 k2y0z)jbaand proceed as an immediate consequence of this theorem we obtainTheorem [ sturm -Picone Theorem]Suppose(k1z0)0+g1z=0(k2y0)0+g2y= 0whereg2 g1andk1 k2>0andg26 g1,k26 k1on[a;b]andz(a)=z(b)=0:theny(x)has a zero in(a;b) BOUNDARY VALUE Boundary Value ProblemsWe consider the problem of solvingM(y)=a2y00+a1y0+a0y=f(x); a<x<b( )subject to the boundary conditionsB1(y)= 11y(a)+ 12y0(a)+ 11y(b)+ 12y0(b)= 1( )B2(y)= 21y(a)+ 22y0(a)+ 21y(b)+ 22y0(b)= 2:Herea2;a1.))

7 A02C[a;b];a2(x)6=0; ij; ij; iare constants. Equations ( ) and ( )constitute what is called a boundary value problem(BVP). If 11= 12= 0 and 21= 22= 0, the boundary conditions are separated. If 12= 12= 0 and 21= 21= 0, and 1= 2= 0 the boundary conditions are periodic. If 12= 11= 12= 21= 21= 22=0we have the initial conditionsy(a)= 1 11;y0(a)= 2 21:Note that the boundary operatorsBiare linear. We refer toff(x); 1; 2gas the data of the BVP ( )-( ) with dataf0; 1; 2ghas a unique solution i theBVP with dataf0;0;0ghas only the trivial rst that we know that solutions to the BVP ( ), ( ) are unique andsupposey1;y2are linearly independent solutions of ( ). Ifuis any solution ofM(y)=0,with dataf0; 1; 2g, then, since every solution must be a linear combination ofy1andy2,there exist unique 1; 2so thatu= 1y1+ 2y2:Then 1B1(y1)+ 2B1(y2)= 1 1B2(y1)+ 2B2(y2)= 2;or B1(y1)B1(y2)B2(y1)B2(y2) 1 2 = 1 2 ( )Since 1; 2are unique,det B1(y1)B1(y2)B2(y1)B2(y2) 6=0:( )8 Chapter 5.

8 Sturm-Liouville THEORYNow let us suppose thatwsolves the boundary value problem with dataf0;0;0gand wewritew= 1y1+ 2y2( )then as above we get B1(y1)B1(y2)B2(y1)B2(y2) 1 2 = 00 and since the coe cient matrix is nonsingular we must have 1= 2= 0 by ( ).Conversely, if the BVP with dataf0;0;0ghas only the trivial solution then from ( )we conclude that ( ) holds and hence ( ) has a unique BVPM(y)=f; x2(a;b)B1(y)= 1;B2(y)= 2has a unique solution if the BVP with dataf0,0,0ghas only the trivial ;u2be linearly independent solutions ofM(y) = 0 and letypbe a particularsolution ofM(y)=f:Seek a solution of the BVP in the formy= 1u1+ 2u2+yp:Solve for 1; 2such thatB1(y)= 1B1(u1)+ 2B1(u2)+B1(yp)= 1B2(y)= 1B2(u1)+ 2B2(u2)+B2(yp)= 2or B1(u1)B1(u2)B2(u1)B2(u2) 1 2 = 1 B1(yp) 2 B2(yp) :Since the homogeneous problem has only the trivial solution the matrix is invertible and so 1 2 = B1(u1)B1(u2)B2(u1)B2(u2) 1 1 B1(yp) 2 B2(yp) Sturm-Liouville BOUNDARY VALUE Sturm-Liouville Boundary Value ProblemsIn paractice one often encounters a second order di erential equation in so-called self-adjointform and generally one nds that the most common boundary conditions are either separatedor periodic.

9 A second order operatorLis in self-adjoint form ifL(y)=(ky0)0+g(x)y:We are particularly interested in BVP's of the formL(y)+ p(x)y=0; a<x<b;( )B1(y)=0;B2(y)=0;( )wherek;k0;g;pare real and continuous on [a;b], andk;p >0on[a;b]. The correpondingseparated boundary conditions are given byB1(y)= 1y(a)+ 2y0(a) = 0( )B2(y)= 1y(b)+ 2y0(b)=0:( )The BVP ( )-( ) is called a Regular Sturm-Liouville Eigenvalue Problem. The valuesof for which the BVP has a nontrivial solution are called a generalLy=a0y00+a1y0+a2y, the BVPL(y)+ p(x)y=0B1(y)=0;B2(y)=0is said to be self-adjoint providedZba[uL(v) vL(u)]dx=0for allu;vthat satisfy the above boundary conditions ( )-( ).Theorem BVP corresponding to the regular SLBVPL(u)+ pu=0 1y(a)+ 2y0(a)=0 1y(b)+ 2y0(b)=0is 5.

10 Sturm-Liouville by parts yields the so-called Green's formulaZba[vL(u) uL(v)]dx=k(x)(u0v v0u)jba P(u;v)jba( )Ifuandvsatisfy the boundary conditions atx=a, then v(a)v0(a)u(a)u0(a) 1 2 = 00 :Since 21+ 226=0,wehaveu0(a)v(a) v0(a)u(a)=0:In the same way we see that ifuandvsatisfy the conditions atx=b, thenu0(b)v(b) v0(b)u(b)=0:From this we see thatP(u;v)jba= 0 and the result that the above proof shows, in general, that ifu;vsatisfy separated 's, thenP(u;v)jba= 0. The next theorem states that \eigenfunctions corresponding to di erenteigenvalues are orthogonal with respect to the weightp(x)."Theorem ( ;u);( ;v)denote an eigenpair of the RSLBVP,L(y)+ p(x)y=0 1y(a)+ 2y0(a)=0 1y(b)+ 2y0(b)=0:Then( )Zbau(x)v(x)p(x)dx=0; ,uandvare orthogonal with respect to the Green's formula ( ), we know that( )Zbauvp(x)dx=Zba[u(L(v) v(L(u)]dx==k(x)(u0v v0u)jba=0:Theorem eigenvalues of the RSLBVP are Sturm-Liouville BOUNDARY VALUE (u)+ pu=0;thenL(u)+ pu=0;orL(u)+ pu=0and clearlyB1(y)=0=B2(y):Thus ( ; ) is an eigenpair and so( )Zbauupdx=0;or( )Zbajuj2pdx=0:Sincep>0, andjuj26 0 (sinceuis an eigenfunction), we must have = :An eigenvalue is said to be simple if the dimension of the null space ofL is one, ,the diemnsion off':L '=0gis one.))