REGULAR TWO-POINT BOUNDARY VALUE PROBLEMS
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1 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS PHIIP WWAKER Supposethtndrerelnumerswith<,echofp,q,ndwiscontinuous rel vlued function with domin[,], the functionphs continuous first derivtive, ech of p(x) nd w(x) is positive for ll x in[,], ech of M ij nd N ij is complex numer fori=1,2 ndj=1,2, nd the qudruples(m 11,M 12,N 11,N 12 ) nd(m 21,M 22,N 21,N 22 ) re linerly independent Suppose tht the opertor τ is given y τϕ= (pϕ ) qϕ whenever ϕ is twice differentile complex vlued function with domin[, ] We shll e concerned with finding the complex numers λ nd complex vlued functions ϕ such tht (1) τϕ=λwϕ on[,], (2) M 11 ϕ()+m 12 ϕ ()+N 11 ϕ()+n 12 ϕ ()=, nd (3) M 21 ϕ()+m 22 ϕ ()+N 21 ϕ()+n 22 ϕ ()= Remrk 1 The eqution (1) is equivlent to pϕ +p ϕ +(q+λw)ϕ= Thus (1) is regulr second order liner homogeneous differentil eqution Remrk 2 The zero function on[,] (ie the function ϕ such tht ϕ(x)= for ll x in[,]) is lwys solution to (1), (2), nd (3) Definition 1 The sttement tht λ is n eigenvlue for the prolem (1), (2), nd (3) mens tht there is complex vlued functionϕother thn the zero function on [,] tht stisfies (1), (2), nd (3) when λ = λ When this is the cse, the sttement tht ϕ is n eigenfunction corresponding to the eigenvlue λ mens tht ϕ is not the zero function on [,] nd ϕ stisfies (1), (2), nd (3) when λ=λ When λ is n eigenvlue, theeigenspce corresponding to λ consists of ll eigenfunctions corresponding to λ together with the zero function on[,] Remrk 3 Suppose tht λ is n eigenvlue for (1), (2), nd (3) Since the eigenspce corresponding to λ is suspce of the set of ll solutions to (1), the eigenspce is either one-dimensionl or two-dimensionl 1
2 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 2 Exmple 1 Consider the prolem ( is positive numer ) (4) ϕ =λϕ on[,], (5) ϕ()=, nd (6) ϕ()= In order to find the eigenvlues nd eigenfunctions we will consider three cses Cse 1: Suppose tht λ< Thenϕstisfies (4) only in cse ϕ(x)=c 1 cosh λx+c 2 sinh λx for some pir of numers c 1 nd c 2 nd ll x in [,] Since cosh() = 1 nd sinh()=,it follows tht (5) will lso hold only in cse Thus (4) nd (5) hold only in cse c 1 = (7) ϕ(x)=c 2 sinh λx for some numer c 2 Sincesinhz= only in cse z= nd λ>, it follows tht (4), (5), nd (6) hold only in cse (7) holds with c 2 = or Thus there re no negtive eigenvlues ϕ(x)= for llxin[,] Cse 2: Suppose tht λ= Thenϕstisfies (4) only in cse (8) ϕ(x)=c 1 +c 2 x for some pir of numersc 1 ndc 2 nd llxin[,] Thus it follows tht (4), (5), nd (6) hold only in cse (8) holds with c 1 =c 2 = or Thus zero is not n eigenvlue ϕ(x)= for llxin[,] Cse 3: Suppose tht λ> Thenϕstisfies (4) only in cse ϕ(x)=c 1 cos λx+c 2 sin λx for some pir of numers c 1 nd c 2 nd ll x in [,] Since cos() = 1 nd sin()=,it follows tht (5) will lso hold only in cse Thus (4) nd (5) hold only in cse c 1 = (9) ϕ(x)=c 2 sin λx for some numer c 2 Sincesinz= only in cse z is n integrl multiple of π nd λ>, it follows tht (4), (5), nd (6) hold withϕdifferent from the zero function only in cse (9) holds with c 2 = nd λ=kπ or λ=kπ / or 2 kπ (1) λ= for some positive integer k
3 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 3 Aswewillseeelow, lleigenvluesof(4),(5),nd(6)musterelnumers Thus λisneigenvlueonlyincse(1)holds,ndwhen(1)holds,ϕiscorresponding eigenfunction only in cse ϕ(x)=csin kπx for some numer c= nd ll x in[,] From this, it follows tht ech eigenspce is one dimensionl Remrk 4 Here is procedure for finding the eigenvlues nd eigenfunctions of the prolem consisting of (1), (2), nd (3) For ech complex numerλ, let(u λ,v λ ) e linerly independent pir of solution to (1) Then ϕ stisfies (1) only in cse ϕ(x)=c 1 u λ (x)+c 2 v λ (x) for ll x in[,] for some pir of complex numers (c 1,c 2 ) Moreover, ϕ is different from the zero function only in cse t lest one of c 1 nd c 2 is different from zero When ϕ= c 1 u λ +c 2 v λ then so ϕ (x)=c 1 u λ(x)+c 2 v λ(x) for ll x in[,] ϕ(x) ϕ (x) c1 =Φ λ (x) c 2 for ll x in[,] uλ (x) v Φ λ (x)= λ (x) u λ (x) v λ (x) Conditions (2) nd (3) together re equivlent to ϕ() ϕ() M ϕ +N () ϕ = () (11) M = nd (12) N = M11 M 12 M 21 M 22 N11 N 12 N 21 N 22 Thus ϕ will stisfy (1), (2), (3) only in cse ϕ=c 1 u λ +c 2 v λ nd c1 c MΦ λ () +NΦ c λ () = 2 c 2 or c D(λ) = c 2 MΦ λ ()+NΦ λ () Moreover, there will e solution t lest one of c 1 nd c 2 is not zero only in cse (λ)= (λ)=detd(λ)
4 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 4 We hve estlished the following Theorem 1 The eigenvlues of the prolem (1), (2), nd (3) re the zeros of the function, nd if (λ )= then ϕ is n eigenfunction corresponding to the eigenvlue λ only in cse ϕ=c 1 u λ +c 2 v λ nd t lest one of the numers c 1 nd c 2 is different from zero nd c D(λ ) = c 2 From this it follows tht if λ is n eigenvlue, the corresponding eigenspce is two-dimensionl when D(λ )= nd is one-dimensionl when D(λ )= Definition 2 The sttement tht the prolem (1), (2), nd (3) is self-djoint mens tht (τf)g= whenever ech of f nd g is twice continuously differentile complex-vlued function with domin[,] nd ech of f nd g stisfies the conditions (2) nd (3) Exmple2 Considertheprolem(4),(5),nd(6)giveninthelstexmple Here τϕ= ϕ So if ech of f nd g hs continuous second derivtive nd ech stisfies (5) nd (6), (τf)g = fτg f g= [f g] + f g =+ = [fg ] + f( g )= Thus the prolem (4), (5), nd (6) is self-djoint f( g )= The following theorem gives strightforwrd test for self-djointness f g fτg Theorem 2 The prolem (1), (2), nd (3) is self-djoint only in cse 1 p()m M 1 =p()n N M nd N re given y (11) nd (12) Rememer tht w is continuous positive vlued function defined on[, ] Definition 3 Whenever ech of f nd g is piecewise continuous complex-vlued function defined on [,], the inner product of f nd g is denoted y < f,g > nd is defined y <f,g>= f(x)g(x)w(x)dx
5 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 5 Definition 4 Suppose tht ech of f nd g is piecewise continuous complexvlued function defined on [,] The sttement tht f nd g re orthogonl mens tht <f,g>= Suppose tht ϕ k,ϕ k+1,ϕ k+2, is sequence of piecewise continuous complexvlued functions defined on [] The sttement tht ϕ k,ϕ,ϕ k+1 k +2, is orthogonl mens tht <ϕ i,ϕ j >= whenever i=j Theorem 3 If the prolem (1), (2), nd (3) is self-djoint, then ll eigenvlues re rel nd eigenfunctions corresponding to different eigenvlues re orthogonl Proof Suppose tht λ is n eigenvlue nd ϕ is corresponding eigenfunction Then λ<ϕ,ϕ> = <λϕ,ϕ>= = ϕλwϕ=λ λwϕϕ= (τϕ)ϕ= ϕϕw=λ<ϕ,ϕ> ϕτϕ Since ϕ is continuous nd not the zero function it follows tht <ϕ,ϕ>= Thus λ=λ showing tht λ is rel Suppose now tht λ nd µ re eigenvlues, λ=µ, ϕ is n eigenfunction corresponding to λ, nd ψ is n eeigenfunction corresponding to µ Thus λ<ϕ,ψ> = <λϕ,ψ>= = ϕτψ= λwϕψ= ϕµwψ= (λ µ)<ϕ,ψ>=, nd sinceλ=µ, it follows tht <ϕ,ψ>= (τϕ)ψ ϕµwψ=µ<ϕ,ψ> Theorem 4 Suppose tht the prolem (1), (2), nd (3) is self-djoint There will e infinitely mny eigenvlues nd they cn e rrnged in nondecresing sequence λ k,λ k+1,λ k+2, with lim k λ k= When necessry, the Grm-Schmidt process cn e used to convert linerly independent sequence into n orthogonl one The following theorem gives the process for linerly independent pir Theorem 5 Suppose tht λ is n eigenvlue nd the corresponding eigenspce is two-dimensionl An orthogonl sis for this eigenspce is(α, β) nd α=u λ β=v λ <v λ,α> <α,α> α
6 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 6 Definition 5 Suppose tht the prolem (1), (2), nd (3) is self-djoint A proper listing of eigenvlues nd eigenfunctions for the prolem consists of nondecresing sequence of eigenvluesλ k,λ k+1,λ k+2, in which ech eigenvlue is listed exctly the numer of times tht is the dimension of the corresponding eigenspce nd n orthogonl sequence of eigenfunctions ϕ k,ϕ k+1,ϕ k+2, in which ϕ j is n eigenfunction corresponding toλ j for j=k,k +1,k +2, Theorem 6 If the prolem (1), (2), nd (3) is self-djoint, then there is proper listing of eigenvlues nd eigenfunctions for the prolem Definition 6 The sttement tht the prolem (1), (2), nd (3) is Sturm- iouville prolem mens tht the conditions (2) nd (3) re equivlent to ones of the form M 11 ϕ()+m 12 ϕ () = nd N 21 ϕ()+n 22 ϕ () = ech of M 11,M 12, N 21, ndn 22 is rel, t lest ofm 11 ndm 12 is not zero, nd t lest one on N 21 nd N 22 is not zero Theorem 7 All Sturm-iouville prolems re self-djoint nd hve eigenspces tht re ll one-dimensionl Definition7 Suppose tht the prolem (1), (2), nd (3) is self-djoint nd{λ k } k=k nd {ϕ k } k=k is proper listing of eigenvlues nd eigenfunctions When f is function tht is piecewise continuous on [,], the expnsion of f in terms of {ϕ k } k=k is the sequence of functions{s n } n=k given y S n (x)= n <f,ϕ k > <ϕ k=k k,ϕ k > ϕ k(x) fol ll x in[,] nd n=k,k +1,k +2, Theorem8 Suppose tht the prolem (1), (2), nd (3) is self-djoint nd{λ k } k=k nd{ϕ k } k=k is proper listing of eigenfunctions nd eigenvlues, nd suppose tht f is function tht is piecewise continuous on[] (i) It follows tht{c k } k=k is sequence of numers nd f = c k ϕ k k=k with convergence in the men (ie only in cse lim <f n c k ϕ n k,f k=k n k=k c k ϕ k >= ) c k = <f,ϕ k > <ϕ k,ϕ k > for k=k,k +1,k +2, (ii) If f is piecewise smooth nd then c k = <f,ϕ k > <ϕ k,ϕ k > for k=k,k +1,k +2, lim n n <f,ϕ k > <ϕ k=k k,ϕ k > ϕ k(x)= 1 2 [f(x+)+f(x )]
7 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 7 for ech x with < x < (iii) If f hs continuous second derivtive, stisfies the oundry conditions (2) nd (3), nd n <f,ϕ S n (x)= k > <ϕ k=k k,ϕ k > ϕ k(x) fol ll x in[,] nd n=k,k +1,k +2,, then{s n }converges uniformly to f on[,] Theorem 9 (The Ryleigh Quotient) Suppose tht the prolem (1), (2), nd (3) is self-djoint, tht λ is n eigenvlue, nd tht ϕ is corresponding eigenfunction It follows tht λ= [p()ϕ ()ϕ() p()ϕ ()ϕ()]+ (p ϕ 2 q ϕ 2 ) ( ϕ 2 w) Thus if q(x) for ll x in [,] nd the oundry conditions (2) nd (3) imply [p()ϕ ()ϕ() p()ϕ ()ϕ()], it follows tht λ If it is lso true tht the non-zero constnt functions fil to stisfy either (2) or (3) then nd it follows tht λ> (p ϕ 2 )> Theorem 1 Suppose tht the prolem (1), (2), nd (3) is self-djoint If ech M ij nd ech N ij is rel numer, then ech eigenvlue hs corresponding rel vlued eigenfunction Remrk 5 In the specil cse τ is given y τϕ= ϕ, w(x)=1, the prolem (1), (2), nd (3) is self-djoint, λ is n eigenvlue, nd ϕ is corresponding rel vlued eigenfunction, the Ryleigh Quotient ecomes λ= [ϕ ()ϕ() ϕ ()ϕ()]+ (ϕ ) 2 (ϕ)2 Remrk 6 In the specil cse τ is given y τϕ= ϕ nd w(x)=1, eqution (1) is equivlent to ϕ +λϕ=, nd we will let the linerly independent pir of solutions(u λ,v λ ) e given y cosh λx when λ< u λ (x)= 1 when λ= cos λx when λ> nd v λ (x)= sinh λx when λ< x when λ= sin λx when λ>
8 λ=kπ or λ= kπ REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 8 With this definition of(u λ,v λ ), note tht cosh λx sinh λx Φ λ (x)= when λ<, λsinh λx λcosh λx 1 x Φ λ (x)= when λ=,nd cos λx sin λx Φ λ (x)= λsin λx λcos when λ> λx Exmple 3 Consider the prolem ϕ = λϕ on[,], ϕ() =, nd ϕ() = This is Sturm-iouville prolem, so it is self-djoint If λ is n eigenvlue nd ϕ is corresponding rel vlued eigenfunction then (Ryleigh Quotient) λ = [ϕ () ϕ () )]+ (ϕ ) 2, so (ϕ)2 λ = (ϕ ) 2 (ϕ)2 Thus ll eigenvlues re nonnegtive The nonzero constnt functions do not stisfy the oundry conditions, so ll eigenvlues re positive The oundry conditions re equivlent to ϕ() ϕ() M ϕ +N () ϕ = () When λ>, So nd M = λ + nd N = cos λ sin λ (λ)=detsin λ cos λ sin λ λsin λ λcos λ From this we see tht λ is n eigenvlue only in cse 2 for some positive integer k Note tht 2 kπ D( )= ( 1) k
9 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 9 so only in cse 2 kπ c D( ) = c 2 c 1 = From this it follows tht ϕ is n eigenfunction corresponding to the eigenvlue kπ 2 only in cse ϕ(x)=c 2 sin kπx for ll x in[,] nd some numer c 2 = Bsed on these oservtions, it follows tht proper listing of eigenvlues nd eigenfunctions for this prolem is{λ k } k=1 nd {ϕ k} k=1 2 kπ λ k = for k=1,2,3, nd ϕ k (x)=sin kπx for k=1,2,3, nd x Computtion shows tht <ϕ k,ϕ k >= Exmple 4 Consider the prolem (ϕ k ) 2 = 2 ϕ = λϕ on[,], ϕ () =, nd ϕ () = for k=1,2,3, This is Sturm-iouville prolem, so it is self-djoint If λ is n eigenvlue nd ϕ is corresponding rel vlued eigenfunction then (Ryleigh Quotient) λ = [ ϕ() ϕ()]+ (ϕ ) 2, so (ϕ)2 λ = (ϕ ) 2 (ϕ)2 Thus ll eigenvlues re nonnegtive The nonzero constnt functions do stisfy the oundry conditions, so ll tht we cn conclude t this point is tht ll eigenvlues re nonnegtive The oundry conditions re equivlent to ϕ() ϕ() M ϕ +N () ϕ = () When λ>, M = λ + nd N = cos λ sin λ λsin λ λcos λ
10 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 1 So λ λsin λ λcos λ nd (λ)=detλsin λ From this we see tht λ is positive eigenvlue only in cse 2 for some positive integer k λ=kπ or λ= kπ Note tht when k is positive integer, 2 kπ D( )= kπ kπ, ( 1)k so 2 kπ c D( ) = c 2 only in cse c 2 = From this it follows tht ϕ is n eigenfunction corresponding to the eigenvlue kπ 2 when k is positive integer only in cse ϕ(x)=ccos kπx for some numer c= nd llxin[,] When λ=, D()= + nd (λ)= ()= Thus zero is n eigenvlue Note tht c D() = c 2 1 = only in cse c 2 =,so ϕ is n eigenfunction corresponding to the eigenvlue zero only in cse ϕ(x)=c for some numer c= nd llxin[,] Bsed on these oservtions, it follows tht proper listing of eigenvlues nd eigenfunctions for this prolem is{λ k } k= nd {ϕ k} k= λ =, ϕ (x)=1 for x, 2 kπ λ k = for k=1,2,3,, nd ϕ k (x)=cos kπx for k=1,2,3, nd x Computtion shows tht <ϕ,ϕ >= (ϕ ) 2 =, nd
11 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 11 <ϕ k,ϕ k >= Exmple 5 Consider the prolem (ϕ k ) 2 = 2 ϕ = λϕ on[,], ϕ( ) = ϕ(), nd ϕ ( ) = ϕ () for k=1,2,3, This is not Sturm-iouville prolem; however it is self-djoint Ifλis n eigenvlue nd ϕ is corresponding rel vlued eigenfunction then (Ryleigh Quotient) λ = [ϕ ( ) ϕ( ) ϕ () ϕ()]+ (ϕ ) 2, so (ϕ)2 λ = [ϕ () ϕ() ϕ () ϕ()]+ (ϕ ) 2,so (ϕ)2 λ = (ϕ ) 2 (ϕ)2 Thus ll eigenvlues re nonnegtive The nonzero constnt functions do stisfy the oundry conditions, so ll tht we cn conclude t this point is tht ll eigenvlues re nonnegtive The oundry conditions re equivlent to ϕ( ) ϕ() M ϕ +N ( ) ϕ = () When λ>, So nd M= nd N= cos λ sin λ λsin λ λcos λ sin λ 2 λsin λ (λ)=det4 λsin 2 λ From this we see tht λ is positive eigenvlue only in cse 2 for some positive integer k cos λ sin λ λsin λ λcos λ λ=kπ or λ= kπ When k is positive integer, 2 kπ D( )=
12 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 12 so the eigenspce corresponding to kπ 2 is two dimensionl nd corresponding linerly independent pir of eigenfunctions is(u, v) u(x)=cos kπx ndv(x)=sinkπx Computtion shows tht <u,v>=; this pir is lredy orthogonl When λ=, D()= nd (λ)= ()= Thus zero is n eigenvlue Note tht c D() = c 2 1 = 2 only in cse c 2 =, so ϕ is n eigenfunction corresponding to the eigenvlue zero only in cse ϕ(x)=c for some numer c= nd llxin[,] Thus proper listing of eigenvlues nd eigenfunctions for this prolem is{λ k } k= nd {ϕ k } k= λ =, ϕ (x)=1 for x, 2 kπ λ 2k 1 = λ 2k = for k=1,2,3,, ϕ 2k 1 (x) = cos kπx, nd ϕ 2k (x)=sinkπx for k=1,2,3, nd x Computtion shows tht <ϕ,ϕ >= <ϕ k,ϕ k >= Exmple 6 Consider the prolem (ϕ ) 2 =2, nd (ϕ k ) 2 = for k=1,2,3, ϕ = λϕ on[,1], ϕ() ϕ () =, nd ϕ(1) = This is Sturm-iouville prolem, so it is self-djoint nd ll eigenspces re onedimensionl If λ is n eigenvlue nd ϕ is corresponding rel vlued eigenfunction then (Ryleigh Quotient) λ = [(ϕ())2 ϕ (1) ]+ 1 (ϕ ) 2 1, so (ϕ)2 λ = (ϕ())2 + 1 (ϕ ) 2 1 (ϕ)2
13 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 13 Thus ll eigenvlues re nonnegtive The nonzero constnt functions do not stisfy the oundry conditions, so ll eigenvlues re positive The oundry conditions re equivlent to ϕ() ϕ(1) M ϕ +N () ϕ = (1) When λ>, 1 1 So nd M = 1 1 λ ndn = + 1 λ cos λ sin λ (λ)=detsin λ+ λcos λ From this we see tht λ is n eigenvlue only in cse ρ is positive numer such tht λ=ρ 2 sinρ+ρcosρ= cos λ sin λ λsin λ λcos λ or such tht tnρ= ρ (For ech positive integer k there is exctly one solution to this eqution etween kπ π 2 ndkπ+π See pge 21 of the text Newton s method (ook it up) cn 2 e used to pproximte the zeros of f nf(ρ)=sinρ+ρcosρ) Note tht when λ is n eigenvlue, then 1 λ cos λ sin = λ so c D(λ) = c 2 1 λ cos λ λcos λ only in cse c 1 λc 2 = From this it follows tht ϕ is corresponding eigenfunction only in cse ϕ(x)=ccos λx+ c sin λx λ for ll x in[,1] nd some numer c= Bsed on these oservtions, it follows tht proper listing of eigenvlues nd eigenfunctions for this prolem is{λ k } k=1 nd {ϕ k} k=1 λ k = ρ 2 k, ρ k is the kth positive numer such tht sinρ+ρcosρ=,
14 REGUAR TWO-POINT BOUNDARY VAUE PROBEMS 14 nd ϕ k (x)=cos λ k x+ 1 λk sin λ k x for ll x in [,1] Numericl pproximtions for the first three eigenvlues re s follows k ρ k λ k D M, U H E-mil ddress: pwlker@mthuhedu
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