V. More General Eigenfunction Series. A. The Sturm-Liouville Family of Problems

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1 V. More Gener Eigenfunction Series c 214, Phiip D. Loewen A. The Sturm-Liouvie Fmiy of Probems Given n re interv, b], functions p(x), q(x), r(x), nd constnts c, c 1, d, d 1, consider (ODE) (p(x)y (x)) q(x)y(x) + λr(x)y(x) =, < x < b, BC() c y() + c 1 y () =, BC(b) d y(b) + d 1 y (b) =. (EVP) Assume wys tht p(x) > nd r(x) > for x, b], nd tht (c, c 1 ) (, ), (d, d 1 ) (, ). This is the prototype for Sturm-Liouvie eigenvue probem. Terminoogy. () A constnt λ is n eigenvue in (EVP) if using this prticur vue in (ODE) ows for some nonzero soution function y in (ODE)+(BC). (Any such function is ced n eigenfunction corresponding to λ.) (b) A nonzero function y defined on, b] is n eigenfunction in (EVP) if this prticur function stisfies both (BC) nd (ODE) for some choice of constnt λ. (Tht constnt is then ced the eigenvue corresponding to y.) Prcticities. () Testing some given number λ for the sttus of eigenvue in (EVP) is not esy. Typicy it requires soving the differenti eqution (ODE) nd ppying the boundry conditions (BC) to see if nontrivi function is comptibe. We hve seen exmpes, nd there wi be more. (b) Testing some given function y for the sttus of eigenfunction in (EVP) is very esy. Checking the boundry vues nd verifying (BC) is extremey simpe, nd substition into (ODE) reduces tht identity to n eqution in which the constnt λ is the ony unknown. Aterntivey, one coud rerrnge (ODE) ike this: λ = q(x)y(x) (p(x)y (x)), < x < b. ( ) r(x)y(x) When y(x) is given, everything on the right side is known. If tht rtio simpifies to produce constnt, then y is n eigenfunction nd the rtio is the eigenvue; if not, then y is not n eigenfunction. As consequence of () (b) here, it is often enough to remember ony the eigenfunctions in (EVP). The eigenvues re wys esiy ccessibe from quick ccution nogous to ( ). Exmpe. When the interv hs = nd the coefficient functions re the constnts p(x) = 1, q(x) =, r(x) = 1, the ODE becomes to get y + λy =. Mking the further choices (c, c 1 ) = (1, ) nd (d, d 1 ) = (1, ) turns the BC s into y() = = y(b). Thus the eigenvue probem tht produces the Fourier Sine Series fits into the Fie 214notes, version of 17 Juy 24, pge 1. Typeset t 22:2 Juy 17, 214.

2 2 PHILIP D. LOEWEN frmework bove. Ech function y n (x) = sin(nπx/b) is n eigenfunction, nd the corresponding eigenvue comes from λ n = y n(x) y(x) = sin(nπx/b) (nπ/b)2 sin(nπx/b) = n2 π 2 b 2. The eigenvue probems ssocited with the FCS, HPSS, nd HPCS hve Sturm- Liouvie form s we they just invove different choices of (c, c 1 ) nd (d, d 1 ). Derivtive Pcking. In (EVP), the derivtives re concentrted in the first term, (py ). Using the product rue eds to this equivent eqution: p(x)y (x) + p (x)y (x) q(x)y(x) + λr(x)y(x) =, < x < b. () The speci retionship between the coefficients of y nd y here is critic in deriving the foowing ist of fcts. (See beow for detis.) Notice how these correspond to known properties of the eigenfunction fmiies we hve redy worked with. Theorem (Sturm-Liouvie). Under the stted conditions governing (EVP), () A the eigenvues re re numbers. (b) The eigenvues form n infinite sequence λ 1 < λ 2 < λ 3 < such tht im λ n = +. n (c) For ech eigenvue λ n, soutions of (ODE)/(BC) with λ = λ n cn be expressed s constnt mutipes of singe representtive eigenfunction y n. Tht is, the eigenspce E(λ n ) is one-dimension.] (d) Eigenfunctions with distinct eigenvues re orthogon with weight r. Tht is, if y m, y n re eigenfunctions for eigenvues λ m, λ n, (y m, y n ) r def = y m (x)y n (x)r(x) dx = whenever λ m λ n. Note the presence of r(x) in this orthogonity retion. (e) Eigenfunction series give rbitrriy good men-squre pproximtions for ny Riemnn-integrbe f:, b] R. Sturm-Liouvie/Sef-Adjoint Form. When derivtives re condensed into the term (py ) s shown in (ODE), the differenti eqution is sid to be in Sturm- Liouvie Form or Sef-Adjoint Form. This rrngement is ide for integrtion by prts. Here s why. Suppose u(x) nd v(x) re ny differentibe functions defined on, b]. Integrtion by prts gives ( b u(x) p(x)v (x)) dx = u(x)(p(x)v (x)) (p(x)v (x))u (x) dx. (1) Swpping the etters u nd v eds to different-ooking, but equivent, sttement: ( b v(x) p(x)u (x)) dx = v(x)(p(x)u (x)) (p(x)u (x))v (x) dx. (2) x= x= Fie 214notes, version of 17 Juy 24, pge 2. Typeset t 22:2 Juy 17, 214.

3 V. More Gener Eigenfunction Series 3 The integrs on the fr right in (1) nd (2) re equ, so these two equtions cn be combined to produce n identity tht shows the discrepncy introduced by moving the pckged-derivtive opertion from one fctor to the other: u(x) ( (p(x)v (x)) dx = ( p(x)u (x)) v(x) dx ( ) ] b p(x) u(x)v (x) u (x)v(x). (3) x= If the functions u nd v hppen to stisfy both BC() nd BC(b), the brcketed terms in (3) equ exercise!] nd the identity becomes u(x)(p(x)v (x)) dx = (p(x)u (x)) v(x) dx (ssuming (BC)). In prticur, if we choose eigenfunctions y m nd y n (with eigenvues λ m nd λ n ) for the functions u nd v in the ccution bove, we cn use (ODE) to repce the pckged derivtive terms: y m (x) ( p(x)y n (x) ) dx = ( ) y m (x) q(x)y n (x) λ n r(x)y n (x) dx = ( p(x)y m (x) ) y n (x) dx ( ) q(x)y m (x) λ m r(x)y m (x) y n (x) dx λ n y m (x)y n (x)r(x) dx = λ m y m (x)y n (x)r(x) dx. This is equivent to (λ m λ n ) y m (x)y n (x)r(x) dx =. (4) If we hve λ m λ n, this forces the integr to equ, nd tht proves sttement (d) in the Sturm-Liouvie theorem. Exercise Soution. To hnde the evuted terms shown in ine (3), rewrite the sttement tht both functions u nd v stisfy BC() s the pir of equtions In mtrix form, this is equivent to u() u () v() v () c u() + c 1 u () = c v() + c 1 v () =. ] c c 1 ] = ]. Since (c, c 1 ) (, ), this eqution cn ony hod if the mtrix on the eft is noninvertibe. Tht is, we must hve ( ]) u() u () = det v() v = u()v () u ()v(). () This shows tht the evuted term in (3) vnishes t the eft endpoint, x =. The sme ine of rgument ppies t the right endpoint, x = b. //// Fie 214notes, version of 17 Juy 24, pge 3. Typeset t 22:2 Juy 17, 214.

4 4 PHILIP D. LOEWEN Ccution (1) cn be interesting even when u nd v re equ. Suppose y n is some eigenfunction for (EVP). Using u = y n = v, nd expoiting the identity in (ODE), we hve y n (x) ( ) b y n (x) p(x)y n(x) dx = p(x)y n (x)y n(x) x= ( ) q(x)y n (x) λ n r(x)y n (x) dx = p(x)y n (x)y n(x) b x= Rerrnging this produces n eqution invoving the eigenvue λ n : λ n = p(x)y n(x) 2 dx p(x)y n(x) 2 dx ) ] (p(x)y n (x)2 + q(x)y n (x) 2 dx p(b)y n (b)y n (b) p()y n()y n (). (5) r(x)y n (x) 2 dx To see how usefu this cn be, consider the simpe constnt functions p(x) = 1, q(x) =, r(x) = 1 nd boundry conditions y() =, y(b) = tht we ssocite with Fourier Sine Series. Without knowing nything bout the eigenfunction y n, we cn see tht its corresponding eigenvue is λ n = y n(x) 2 dx. (6) y n (x) 2 dx The integrs on the top nd bottom of this frction must both be non-negtive, so λ n is gurnteed. In fct, λ n = woud require y n (x) = for x, nd tht (together with the BC s) woud force y n be the constnt function which is incomptibe with the hypothesis tht y n is nontrivi in the first pce. So we cn t hve λ n = fter. Therefore λ n > must hod for ech n. Of course this is no surprise for the FSS exmpe, but the possibiity of predicting eigenvue signs in probems were the form of the eigenfunctions is not known in dvnce cn be re sset. Prctice. In the Trench textbook, try exercise 13.2 #8. Arrnging Sturm-Liouvie Form. Differenti equtions encountered in the wid my not hve their derivtives pcked into the first term s shown in (ODE). To reorgnize given eqution into this form, use the expnded version () s trget: p(x)y + p (x)y q(x)y + λr(x)y =. Tht is, given ny second-order iner ODE, try to mtch it with the form bove by introducing new unknown p(x) nd then mtching the coefficients of y nd y. Exmpe. Find orthogonity weight for y + 4y + (4 + 9λ)y =. Fie 214notes, version of 17 Juy 24, pge 4. Typeset t 22:2 Juy 17, 214.

5 V. More Gener Eigenfunction Series 5 Soution. Mutipy the given eqution by p(x). Tht mkes the y coefficient excty wht we wnt: p(x)y + 4p(x)y + (4 + 9λ)p(x)y =. To mtch the coefficient on y, we wnt p (x) = 4p(x). This is simpe differenti eqution for p(x). The gener soution is p(x) = Ae 4x, A R. Any constnt A wi work; we tke A = 1 becuse it s simpe. (Admittedy, A = woud be simper, but tht choice woud reduce our origin differenti eqution to = profound oss of informtion bout y!) Then the given eqution is equivent to = e 4x y + 4e 4x y + (4 + 9λ)e 4x y = ( e 4x y ) + 4e 4x y + λ ( 9e 4x) y =. It foows tht eigenfunctions on the interv < x < b, with distinct eigenvues, re orthogon with weight r(x) = 9e 4x : y m (x)y n (x)9e 4x dx = whenever m n. (You cn drop the fctor 9 here nd the sttement remins true.) //// Exmpe. Find the orthogonity weight for Lguerre s eqution, xy + (1 x)y + λy =. Soution. Know the terminoogy: In the nottion of (EVP) bove, orthogonity weight mens function r(x). Mutipy through Lguerr s eqution by p(x)/x to mtch the y term with the form in (): ( ) 1 x p(x)y + p(x)y + λ x ( p(x) To mtch the y -term, we need p (x) = 1 x p(x), i.e., x x ) y =. dp p = 1 x dx n(p) = nx x + C p(x) = Axe x. x Choose A = 1 for simpicity, to get = xe x y + (1 x)e x y + λe x y = ( xe x y ) + λe x y. Hence the orthogonity weight is r(x) = e x. This function is not cery visibe in the given eqution, but it s very importnt. In ny version of (EVP) bsed on this ODE with some interv, b], eigenfunctions y m nd y n wi hve the orthogonity property y m (x)y n (x)e x dx = whenever m n. (Think bout this: the sttement just mde ony works if is not point in the interv, b]. Why?) //// Fie 214notes, version of 17 Juy 24, pge 5. Typeset t 22:2 Juy 17, 214.

6 6 PHILIP D. LOEWEN Prctice. In the Trench textbook, try exercises 13.2 #1 7. Exmpe (Euer Series from Lpce Eqn). For the foowing eigenvue probem bsed on n Euer-type eqution, x 2 y (x) + xy (x) + λy(x) =, 1 < x < b, y(1) = = y(b), ( nπ ) the eigenfunctions turn out to be (prctice!) y n (x) = sin nb nx, with corresponding eigenvues ( ) 2 nπ λ n =, n = 1, 2, 3,.... nb From FSS theory pus chnge of vribes, or from gener SL theory, we hve the orthogonity retion = 1 y m (x)y n (x) dx x, m n. Orthogonity nd Coefficient Extrction. Suppose we know fu ist of representtive eigenfunctions y 1, y 2,... in probem (EVP). Then, et f(x) be given on, b]. Wht coefficients c n wi mke correct eigenfunction series identity beow? f(x) = c n y n (x), < x < b. ( ) Orthogonity provides the nswer. Pick ny prticur subscript k, mutipy ( ) by y k (x)r(x) nd integrte: f(x)y k (x)r(x) dx = c n y n (x)y k (x)r(x) dx = c k y k (x) 2 r(x) dx. Thnks to orthogonity, terms in the sum except the one where n = k equ zero. This expins the second step bove, nd eds to c k = f(x)y k(x)r(x) dx y k(x) 2 r(x) dx. Normiztion. When choosing the eigenfunction y k for λ k, if you set the constnts right you cn rrnge y k (x) 2 r(x) dx = 1. This mkes the formu for c k bove itte cener. Eigenfunctions chosen to stisfy ( ) re ced normized. In our course, normizing eigenfunctions is wste of time: it mkes no difference to the series soutions we ccute. Don t bother! ( ) Fie 214notes, version of 17 Juy 24, pge 6. Typeset t 22:2 Juy 17, 214.

7 V. More Gener Eigenfunction Series 7 Consider this eigenvue probem: B. Extended Exmpe 1 (ODE) x 2 y + xy + λy =, 1 < x < e, (BC) y(1) = = y(e). () Express the eqution in stndrd form nd find the orthogonity retion. Get r(x) = 1/x, deduce tht if y m nd y n re eigenfunctions with different eigenvues, then = e 1 y m (x)y n (x) dx x. (b) Find eigenvues nd eigenfunctions. Grinding cse-by-cse nysis eds to λ n = n 2 π 2, y n (x) = sin(nπ nx), n = 1, 2,.... (c) Tsk 1: Chnge of Bsis/Coefficient Extrction] Given f(x), find formus for the coefficients in f(x) = c n y n (x) = c n sin(nπ nx), 1 < x < e. Evute the c n s expicity for f(x) = Genery, c n = (f, y n) r = (y n, y n ) r = 2 e 1 { nx, 1 < x < e,, otherwise. e 1 f(x) sin(nπ nx) x 1 dx e 1 sin2 (nπ nx) x 1 dx f(x) sin(nπ nx) x 1 dx. For the specific f in question, spitting gives e c n = 2 (nx) sin(nπ nx) dx /2 1 x = 2 u sin(nπu) du = 2 ( ) nπ (nπ) sin 1 ( ) nπ 2 2 nπ cos. 2 n c n 2/π 2 1/2π 2/9π 2 1/4π 2/25π 2 1/6π 2/49π 2 (d) Tsk 2: Eqution-Soving] Use suitbe eigenfunction series to sove x 2 y + xy + 7y = f(x), 1 < x < e; y(1) = = y(e), Fie 214notes, version of 17 Juy 24, pge 7. Typeset t 22:2 Juy 17, 214.

8 8 PHILIP D. LOEWEN with f s given in prt (c) bove. Notice tht y n (x) = sin(nπ nx) gives y n (x) = nπ x cos(nπ nx), y n (x) = nπ x 2 cos(nπ nx) n2 π 2 x 2 sin(nπ nx), so x 2 y n + xy n = n2 π 2 sin(nπ nx) = n 2 π 2 y n. This recpitutes the eigenvue/eigenfunction retionship nd is usefu when guessing y = n y n for the desired soution. Substitution gives x 2 y + xy + 7y = = n x 2 y n + xy n + 7y ] n n n 2 π ] y n. Therefore n 7 n 2 π 2] must mtch the coefficient c n computed bove. Answer ( ) c n n = 7 n 2 π 2 = 1 2 nπ 7 n 2 π 2 (nπ) 2 sin 2 1 nπ cos ( nπ 2 )]. (e) Tsk 3: Dynmics] Sove for u = u(x, t) in this het conduction probem: (PDE) u t = x 2 u xx + xu x, 1 < x < e, t >, with f s bove. (BC) u(1, t) = = u(e, t), t >, (IC) u(x, ) = f(x), 1 < x < e, Answer: Seprtion of vribes eds to the eigenvue probem stted bove, hence to the postute u(x, t) = T n (t) sin(nπ nx). To compete the soution, it suffices to find the functions T n (t). First initiize T n () = c n (see prt (c) bove), then propgte T n = n2 π 2 T n, so with c n s before. u(x, t) = c n e n2 π 2t sin(nπ nx), Fie 214notes, version of 17 Juy 24, pge 8. Typeset t 22:2 Juy 17, 214.

9 V. More Gener Eigenfunction Series 9 Consider this eigenvue probem: C. Extended Exmpe 2 (ODE) y + λy =, < x < 1, (BC) y () = = y(1) y (1). () Express the eqution in sef-djoint form nd find the orthogonity retion. The eqution redy is in sef-djoint (Sturm-Liouvie) form, with p(x) = 1, q(x) =, nd r(x) = 1. In prticur, ny two eigenfunctions y m nd y n hving different eigenvues λ m nd λ n wi obey y m (x)y n (x) dx =. (b) Find eigenvues nd eigenfunctions. Guessing y = e sx for constnt s C eds to soution whenever s 2 + λ =. Three types of s-vues emerge, depending on the sign of λ. Cse λ < : Define α = λ > to get s 2 = α 2, so s = ±α: y = Ae αx + Be αx, A, B R. The eft BC gives = y () = (A B)α. Since α >, B = A nd The right BC requires y = A e αx + e αx], y = αa e αx e αx]. ( ) = y(1) y (1) = A e α + e α α(e α e α ) ]. For nontriviity (see ( )) we must hve A, so this requires e α + e α = α(e α e α 1 ), i.e., α = eα e α e α + e. ( ) α Express this st condition s L(α) = R(α) for the functions L nd R defined by 3 L(α) = 1/α, R(α) = eα e α e α + e α α Grphic soution for α Fie 214notes, version of 17 Juy 24, pge 9. Typeset t 22:2 Juy 17, 214.

10 1 PHILIP D. LOEWEN Potting the curves z = L(α) nd z = R(α) for α > nd ooking for n intersection point gives one soution: c this α , with corresponding eigenvue λ = α A representtive eigenfunction is obtined by tking A = 1 nd α = α in ( ): y (x) = e α x + e α x, α given by ( ). Cse λ = : When λ = the gener soution of the given ODE is y(x) = A + Bx. The eft BC gives = y () = B, so y = A. The right BC gives = y(1) y (1) = A, so y =. Ony the trivi soution obeys both BC s, so λ = is not n eigenvue. Cse λ > : Write ω = λ >, so tht the chrcteristic eqution is s 2 = ω 2, with soutions s = ±iω. Here y = A cos(ωx) + B sin(ωx), A, B R, nd the eft BC requires = y () = Bω. Since ω >, this gives B = nd Now the right BC requires y = A cos(ωx), y = ωa sin(ωx), A R. ( ) = y(1) y (1) = A cos ω + ω sin ω]. For nontriviity (see ( )) we must hve A, so this requires cos ω = ω sin ω i.e., 1 ω = sin ω cos ω. ( ) Express this st condition s L(ω) = R(ω) for the functions L nd R defined by L(ω) = 1/ω, R(ω) = tnω ω 1 ω 2 ω 3 ω Grphic soution for ω 1, ω 2,... Fie 214notes, version of 17 Juy 24, pge 1. Typeset t 22:2 Juy 17, 214.

11 V. More Gener Eigenfunction Series 11 Potting the curves z = L(ω) nd z = R(ω) for ω > nd ooking for intersection points gives n infinite fmiy of soutions: c these ω 1, ω 2,..., with corresponding eigenvues λ n = ωn 2. Representtive eigenfunctions re obtined by tking A = 1 nd ω = ω n in ( ): y n (x) = cos(ω n x), ω n given by ( ), n = 1, 2,.... Since L(ω) s ω, the points where L(ω) = R(ω) get coser nd coser to zeros of R(o) = tnω s ω increses. These zeros re known: they re the integer mutipes of π. Thinking crefuy bout the subscripting scheme introduced bove (see the sketch) eds to the pproximtion ω n nπ, so λ n n 2 π 2, n 1. (c) Tsk 1: Chnge of Bsis/Coefficient Extrction] Outine the construction for gener eigenfunction expnsion f(x) = c n y n (x) nd then evute the coefficients for f(x) = n { 1, x < 1/2,, 1/2 x 1. The gener summtion formu must invove the eigenfunctions for the whoe probem. Our numbering scheme strts with n = : λ = α 2 < whie for n 1, λ n = ωn 2 >. So in our cse the eigenfunction expnsion ooks ike f(x) = c n y n (x) = c y (x) + c 1 y 1 (x) + c 2 y 2 (x) + n= = c e α x + e α x ] + c 1 cos(ω 1 x) + c 2 cos(ω 2 x) +. For ny fixed k, mutipying cross the given identity by y k (x)r(x) = y k (x) nd integrting the resut ets us hrness the power of the orthogonity retion in prt (): f(x)y k (x)r(x) dx = = c k = c n y n (x)y k (x)r(x) dx = c k y k (x) 2 r(x) dx n= f(x)y k (x)r(x) dx, k =, 1, 2,... y k (x) 2 r(x) dx The boxed formu works in ny Sturm-Liouvie probem; here, we hve r(x) = 1. To mtch the subscripts in the origin expnsion bove, copy the coefficient formu with k = n: c n = f(x)y n (x) dx, n =, 1, 2,.... y n (x) 2 dx Fie 214notes, version of 17 Juy 24, pge 11. Typeset t 22:2 Juy 17, 214.

12 12 PHILIP D. LOEWEN Since ech y n is known function, the denomintors here cn be evuted with no reference to whtever function f(x) we hppen to be considering: n 1 = y (x) 2 dx = y n (x) 2 dx = e 2α x e 2α x ] dx = e2α e 2α, 2α 2α cos 2 (ω n x) dx = sin(2ω n) 4ω n. (In contrst to the fmous Big Four eigenvue expnsions, the vue of the denomintor integr now depends on the counter, n.) For the specific f in question, the numertors cn be evuted by spitting: f(x)y (x) dx = f(x)y n (x) dx = /2 /2 e α x + e α x ] dx = eα /2 1 α e α/2 1 α, cos(ω n x) dx = sin(ω n/2) ω n, n = 1, 2,.... Combining numertors nd denomintors gives the foowing resuts. They re not pretty, but they re competey expicit rec tht α = , ω 1 = , ω 2 = , etc., re known constnts. c = α 1 e α /2 e ] α / = 2 e α /2 e ] α /2 2 α 1 e 2α e 2α ] 4α + e 2α e 2α, ωn 1 c n = sin(ω n/2) (1/2) + (1/4)ωn 1 sin(2ω n ) = 4 sin(ω n/2), n = 1, 2,.... 2ω n + sin(2ω n ) (d) Tsk 2: Eqution-Soving] Use suitbe eigenfunction series to sove with f s bove. y 17y = f(x), < x < 1; y () = = y(1) y (1), Here it s critic tht the BC s nd differenti opertor for y hve the sme form s the BC s nd differenti opertor in the origin eigenvue probem. This mkes it sfe to postute series soution of the form y(x) = K n y n (x), ( ) n= using the eigenfunctions found bove, becuse ny choice of constnts K n wi respect the BC s. Finding the constnts K n tht so respects the given ODE wi sove the probem. Notice tht y n(x) = λ n y n (x) for ech n, so using ( ) in the given eqution gives f(x) = y 17y = = n= λ n 17]K n y n (x), n= K n y n (x) 17K ny n (x)] Fie 214notes, version of 17 Juy 24, pge 12. Typeset t 22:2 Juy 17, 214.

13 V. More Gener Eigenfunction Series 13 This is n eigenfunction expnsion for f. We know such expnsions must be unique, nd we redy hve one in hnd from prt (c) bove. This forces K (λ + 17) = c, i.e., K = c λ + 17 = c 17 α 2, K n (λ + 17) = c n, i.e., K n = c n λ n + 17 = c 17 + ωn 2, n = 1, 2, 3,..., with c, c 1,... s found in prt (c) bove. (e) Tsk 3: Dynmics Wve Type] Sove this wve-motion probem, using f(x) s given bove: (PDE) u tt = c 2 u xx, < x < 1, t >, (BC) u x (, t) =, u(1, t) u x (1, t) =, t >, (IC) u(x, ) = f(x), < x < 1, (IC) u t (x, ) =, < x < 1. Eigen-nysis: Since PDE/BC re both homogeneous, we try seprtion of vribes. Putting u(x, t) = X(x)T(t) into PDE/BC produces n eigenvue probem for X(x) of excty the form shown bove. Postute: So we postute series soution of the form u(x, t) = T n (t)y n (x), n= using the eigenfunctions redy identified. Here the functions T (t), T 1 (t),... re not yet known: finding them wi compete the soution. Initiize: The initi conditions require f(x) = u(x, ), so T n () = c n s ccuted in prt (c). Aso, = u t (x, ) gives T n () = for ech n. Propgte: Pugging the postuted series into the PDE produces differenti eqution for ech coefficient function. T n (t) + c 2 λ n T n (t) =, n =, 1, 2,.... For n =, this wi be unstbe becuse λ = α 2 < : using the initi informtion found bove gives T (t) c 2 α 2 T (t) = = T (t) = c e α ct + e α ct ] /2. For n 1, this wi be hrmonic oscitor eqution becuse λ n = ωn 2 initi dt ed to T n (t) + c 2 ω 2 n T n(t) = = T n (t) = c n cos(ω n ct). >. Our Report: The fu soution wi be u(x, t) = c 2 e α ct + e α ct ] e α x + e α x ] + c n cos(ω n ct) cos(ω n x), with the constnts c, c 1,... pproprite for f coming from prt (c) bove. Nonhomogeneous het exmpe not shown.] Fie 214notes, version of 17 Juy 24, pge 13. Typeset t 22:2 Juy 17, 214.

14 14 PHILIP D. LOEWEN D. Extended Exmpe 3 Boyce/DiPrim (7/e) theme exmpe: 11.1 #8; 11.2 #4, 11; 11.3 #4. Discuss: (ODE) y + λy =, < x < 1, (BC()) y () =, (BC(1)) y(1) + y (1) =. Soution. A eigenvues re positive. To see this, first rue out λ = by soving y (x) = nd ppying the BC s to get the unique soution y(x) =. Then imgine tht some eigenvue-eigenfunction pir λ, y stisfies the conditions bove. Mutipy through (ODE) by y, rerrnge, nd integrte by prts to get 1 1 ] 1 λ y(x) 2 dx = y(x)y (x) dx = y(x)y (x) (y (x)) 2 dx. Pug in y () = nd y (1) = y(1) to rrive t λ y(x) 2 dx = y(1) 2 + x= (y (x)) 2 dx. Cery this requires λ, nd we know λ redy. So et ω = λ, etc., nd find tht y n (x) = cos(ω n x) is n eigenfunction for λ n if nd ony if ω n is the nth positive soution of 1 ω = tnω. Sketching the functions of ω in this eqution heps us predict tht ω 1 π/4, ω 2 (π, 3π/2), nd ω n (n 1)π for n 1. 5 z=1/ω nd z=tn(ω), ω> 4 3 z ω Eigenfunction Expnsion. Probem: Find constnts c n such tht x = c n cos(ω n x), < x < 1, for the constnts ω n identified bove. Fie 214notes, version of 17 Juy 214, pge 14. Typeset t 22:2 Juy 17, 214.

15 V. More Gener Eigenfunction Series 15 Soution. The orthogonity of the eigenfunctions y n = cos(ω n x) is critic here. The usu derivtion, bsed on eds to = y m (x)y n (x) dx = c m = cos(ω m x) cos(ω n x) dx whenever m n, Stndrd integrtion by prts wi give top = bot = Thus we hve m=1 f(x)y m (x) dx. y m (x) 2 dx x cos(ω m x) dx = ω k sin ω k + cos ω k 1, cos 2 (ω m x) dx = c m = ω 2 k 1 + cos(2ω m x) 2 ( ) 2 cos ω + ω sin ω 1 ω ω + sin ω cos ω dx = ω m + sin(ω m ) cos(ω m ) 2ω m. ω=ωm The textbook nswer (B/DiP, 11.3 #11) sys ( ) cosωm 1 x = m 1 + sin 2 cos(ω m x), where m = ω m 1 + sin 2. ω m This suggests n expnsion simir to ours, but with coefficients ] 2 2 (2 cos ω 1) m 1 + sin 2 = ω m ω ( sin 2 ω ) ω=ω m. These re the sme vues we ccuted, even though they ook quite different. To expin this, remember tht 1/ω m = tnω m, so ω m sin ω m = cos ω m nd hence ω m + sin ω m cos ω m = ω m (1 + sin 2 ω m ). PDE BVP. It s no troube to invent prti differenti eqution probem for which the ingredients bove turn out to be key. (PDE) u t = α 2 u xx, < x < 1, t >, (BC) u x (, t) =, u(1, t) + u x (1, t) =, t >, (IC) u(x, ) = x, < x < 1. Seprtion of vribes wi show tht u(x, t) = X(x)T(t) stisfies (PDE)+(BC) if nd ony if X(x) hppens to be one of the eigenfunctions found bove. So we postute soution of the form u(x, t) = T n (t) cos(ω n x), ω 2 m Fie 214notes, version of 17 Juy 214, pge 15. Typeset t 22:2 Juy 17, 214.

16 16 PHILIP D. LOEWEN initiize T n () = c n s detied bove, nd then propgte by pugging the series form into the (PDE). The resut: u(x, t) = c n e ω2 n α2t cos(ω n x), c n = ( ) 2 cos ω + ω sin ω 1 ω ω + sinω cos ω. ω=ωn Fie 214notes, version of 17 Juy 214, pge 16. Typeset t 22:2 Juy 17, 214.

17 V. More Gener Eigenfunction Series 17 E. Fu Fourier Series (FFS) Periodic Functions. A function f: R R is periodic if there is some number k comptibe with n identity of the form f(x + k) = f(x), x R. This k is ced the period of f, nd the phrse f is k-periodic succincty describes the sitution. Lemm. Let c, c 1, c 2 be re constnts; et k >. For ny soution y of the differenti eqution c 2 y + c 1 y + c y =, x R, the foowing sttements re equivent: (i) The function y is k-periodic. (ii) For ech nd every open interv (, b) with b = k, one hs y() = y(b) nd y () = y (b). ( ) (iii) There exists some open interv (, b) such tht ( ) hods. Proof. (i ii) Assume (i). This mens tht y(x+k) = y(x) for re x. Differentite both sides: y (x+k) = y (x) for re x. To prove (ii), pick ny open interv (, b) with b = k. Then pug x = into the identities bove, noting tht + k = b: This proves (ii). y(b) = y(), y (b) = y (). (ii iii) Obvious. Assume (ii). One of the intervs comptibe with the setup in (ii) is (, k), so we know y() = y(k) nd y () = y (k). So the interv (, k) shows tht sttement (iii) hods.] (iii i) Assume (iii). Tht is, ssume tht some prticur vues of nd b = + k mke ine ( ) vid. Define u(x) = y(x + k). This identity impies u (x) = y (x + k) nd u (x) = y (x + k). Since the given differenti eqution bout y is true t every point, we hve c 2 u (x) + c 1 u (x) + c u(x) = c 2 y (x + k) + c 1 y (x + k) + c y(x + k) =. Aso, from ( ), u() = y( + k) = y(b) = y(), u () = y ( + k) = y (b) = y (). This shows tht the function u stisfies the sme ODE nd IC s s the function y. Thnks to the uniqueness theorem for ODE s, the functions u nd y must be identic. Tht is, y(x + k) = u(x) = y(x), x R. This proves (i). //// Fie 214notes, version of 17 Juy 214, pge 17. Typeset t 22:2 Juy 17, 214.

18 18 PHILIP D. LOEWEN An Eigenvue Probem. Given constnt k >, we seek nontrivi k-periodic soutions of y + λy =, x R. Which constnts λ work here, nd wht re the corresponding soutions? Thnks to the Lemm, the property of being k-periodic is equivent to the property of stisfying pir of repeting BC s ike y(k) = y() nd y (k) = y (). Any x-vues seprted by k cn be used to set this up, nd symmetric pproch turns out to be convenient ter. So define = k/2 nd set up the probem (ODE) y + λy =, x R, (BC) y() = y( ), y () = y ( ). This probem does not hve Sturm-Liouvie form, becuse its boundry conditions rete behviour t two different endpoints, wheres the Sturm-Liouvie theory requires seprte BC s t ech end: { c y( ) + c 1 y ( ) =, d y() + d 1 y () =. Since (BC) bove is not comptibe with the Sturm-Liouvie setup, we shoud not be stonished tht some of the min concusions of Sturm-Liouvie theory re not vibe in this cse. Some S-L Properties Persist. For the FFS eigenprobem (ODE)+(BC) bove, 1. A the eigenvues re re numbers. 2. There re no negtive eigenvues (review ine (6) in Section A, bove). 3. Eigenfunctions for distinct eigenvues re orthogon on < x <. Home Prctice: Adpt erier proofs to justify these sttements.] Some S-L Properties re Lost. In prticur, the eigenspces re not onedimension. Let s find the eigenfunctions in deti. Cse λ =. When λ =, soutions of (ODE)/(BC) re mutipes of y (x) = 1. Abstrct sttement: Eigenspce E() is one-dimension, with bsis {y }. Cse λ >. Define ω = λ >. Then the gener soution of (ODE) is y(x) = A cos(ωx) + B sin(ωx); note y (x) = Bω cos(ωx) Aω sin(ωx). The BC s require = y() y( ) = 2B sin(ω), = y () y ( ) = 2Aω sin(ω). Fie 214notes, version of 17 Juy 214, pge 18. Typeset t 22:2 Juy 17, 214.

19 V. More Gener Eigenfunction Series 19 If sin(ω), we get A = B = nd y(x) = is the trivi soution. However, if sin(ω) =, i.e., ω = nπ/ for n N, then both BC equtions bove reduce to = no mtter wht constnts A nd B we choose. Therefore the positive eigenvues re precisey λ n = ω 2 n = n2 π 2, n = 1, 2,..., 2 nd for ech integer n 1, ech nd every nonzero function of the form ( ) ( ) nπx nπx y(x) = A cos + B sin, A, B R is n eigenfunction for λ = λ n. In other words, the set of eigenfunctions for λ n consists of (nonzero) iner combintions of the two functions ( ) ( ) y n (x) def nπx = cos, ψ n (x) def nπx = sin. Notice tht (y n, ψ n ) = cos sin dx = (odd symmetry). In bstrct terminoogy, we hve for ech integer n 1 tht the eigenspce E(λ n ) is two-dimension, with orthogon bsis {y n, ψ n }. Eigenfunction Series. Given some function f(x) defined for < x <, we might hope to for n identity of the form ( ) ( )] nπx nπx f(x) = 1 2 y (x) + n cos + b n sin. ( ) For such n identity to hod, orthogonity requires f(x) sin dx b n = ( ) = 1 f(x) sin dx, n = 1, 2,..., nπx sin 2 dx n = 1 f(x) cos dx, n =, 1, 2,.... Convergence. Is ( ) vid identity between functions? Mosty. Given f(x), we cn ccute the coefficients n nd b n from the integrs bove, nd use the resuts to define f(x) = The resuting function f wi obey n cos ( ) ( nπx nπx + b n sin )]. Fie 214notes, version of 17 Juy 214, pge 19. Typeset t 22:2 Juy 17, 214.

20 2 PHILIP D. LOEWEN (i) f(x) = f(x) t ech point x (, ) where f is continuous; (ii) f is 2-periodic; (iii) f is jump-verging t every point. If the origin function f hs properties (ii) (iii), then f wi recover f excty. More genery, properties (i) (iii) chrcterize the best pproximte reconstruction of f tht we cn resonby expect from series with the given form. Even Functions. If the given f: (, ) R hppens to be even, the FFS coefficients described bove cn be expressed s foows: n = 1 f(x) cos dx = 2 f(x) cos dx (even symmetry), b n = 1 f(x) sin dx = (odd symmetry). Since b n = for ech n, the FFS of f invoves ony cosine terms. Further, by symmetry, the FFS coefficient formu for n mtches excty the FCS formu bsed in the interv < x <. In short, the FFS nd the FCS mount to the sme thing for even functions. Odd Functions. If f: (, ) R hppens to be odd, the FFS coefficients described bove cn be expressed s foows: n = 1 f(x) cos dx = (odd symmetry), b n = 1 f(x) sin dx = 2 f(x) sin dx (even symmetry). Since n = for ech n, the FFS of f invoves ony sine terms. Further, by symmetry, the FFS coefficients b n re identic with the FSS coefficients ccuted using the interv < x <. In short, the FFS nd the FSS mount to the sme thing for odd functions. Simir sttements ppy to the HPSS nd HPCS. Thus four pointwise convergence theorems described bove cn be derived s ppictions of the singe resut for the FFS cse. Any resut for FFS hs immedite ppictions for other types. Convergence Anysis. Most textbook presenttion of the Fourier theory strt with the FFS nd then derive the Big Four s consequences. Here we py this deveopment in the reverse order. Suppose > is given nd f: (, ) R. For such n f, define two new functions: f even (x) = 1 f(x) + f( x)], 2 f odd(x) = f(x) f even (x) = 1 f(x) f( x)]. 2 These new functions re nmed fter their symmetry properties: f even ( x) = f even (x), f odd ( x) = f odd (x), so f even is n even function, so f odd is n odd function. Fie 214notes, version of 17 Juy 214, pge 2. Typeset t 22:2 Juy 17, 214.

21 V. More Gener Eigenfunction Series 21 Aso, it s esy to check the identity f(x) = f even (x) + f odd (x), < x <. Intuitivey the definitions bove spit ny given f into the sum of n even function nd n odd function. And we know ot bout Fourier expnsions of functions with symmetry properties. () Since f odd is odd on (, ), its Fourier Sine Series correcty reproduces it (t continuity points) not just in (, ) but indeed in of (, ). Aso, b n = 2 = 1 = 1 = 1 f odd (x) sin dx f odd (x) sin dx f(x) f even (x)]sin f(x) sin dx. dx (b) Since f even is even on (, ), its Fourier Cosine Series correcty reproduces it (t continuity points) not just in (, ) but indeed in of (, ). Aso, n = 2 = 1 = 1 = 1 f even (x) cos dx f even (x) cos dx f(x) f odd (x)]cos f(x) cos dx dx Using the coefficients ccuted bove, we hve the FCS nd FSS reconstructions f even (x) = 2 + n cos, f odd (x) = b n sin. Fie 214notes, version of 17 Juy 214, pge 21. Typeset t 22:2 Juy 17, 214.

22 22 PHILIP D. LOEWEN It foows tht for every continuity point x of f in the interv (, ), f(x) = f even (x) + f odd (x) = f even (x) + f odd (x) = f(x) = 2 + ( ( ) ( )) nπx nπx def n cos + b n sin = f(x), where n = 1 f(x) cos dx, n =, 1, 2,..., b n = 1 f(x) sin dx, n = 1, 2,.... More genery, f is piecewise continuous, 2π-periodic nd jump-verging. Usefu Misceny. 1. DC Offset. For integers n 1, cos dx = (by ccution, or becuse (y, y n ) = ) sin dx = (by odd symmetry, or becuse (y, ψ n ) = ) It foows tht f(x) dx = 2 + ( n cos ( ) ( nπx nπx + b n sin )) ] dx = 2 2. Tht is, 2 = 1 f(x) dx. 2 This shows tht the constnt term in the FS (or the FCS) reves the verge vue of the input function f over one period. 2. Coefficient Decy. For ny resonbe (i.e., piecewise smooth) function f, there is constnt M (independent of n) such tht both n M n nd b n M n for n 1. In prticur, n nd b n s n. Fie 214notes, version of 17 Juy 214, pge 22. Typeset t 22:2 Juy 17, 214.

23 V. More Gener Eigenfunction Series 23 Proof ( n ony; b n simir): Integrte by prts to get n = f(x) cos dx ( ) ( )] nπx = f(x) sin nπ = n = + ( ) nπx sin f (x) dx nπ 1 f (x) dx nπ = n 1 nπ 1 f (x) dx. ( ) ( ) nπx sin nπ f (x) dx This gives the sttement bove, with M = 1 π f (x) dx. Fie 214notes, version of 17 Juy 214, pge 23. Typeset t 22:2 Juy 17, 214.