Seprtion of Vribles in Liner PDE Now we pply the theory of Hilbert spces to liner differentil equtions with prtil derivtives (PDE). We strt with prticulr exmple, the one-dimensionl (1D) wve eqution 2 u t 2 = c2 2 u x 2, (1) where physicl interprettions of the function u u(x, t) (of coordinte x nd time t), nd the velocity, c, depend on prticulr problem. To be specific, we cn think of u s perpendiculr displcement of string in xy plne, ttched by two ends; the x xis is long the string. Without loss of generlity, we set c = 1, since this just corresponds to properly re-scling t. The fct tht the two ends, x = nd x = b, of the string re fixed is expressed by the boundry conditions: u(, t) = u(b, t) =. (2) We lso need to specify prticulr initil conditions for the function nd its time derivtive: u(x, ) = g 1 (x), u(x, ) = g 2 (x). (3) Note n symmetry between the vribles x nd t. With respect to t we hve Cuchy-type problem with initil conditions, while with respect to x we re deling with boundry vlue problem. Tht is why we will tret x nd t differently. With respect to the vrible x, we will look t u(x, t) s if it is vector of the spce L 2 [, b] (tht depends on t s prmeter): In the vector nottion, Eq. (1) hs the form u(x, t) u(t). (4) d 2 u(t) = L u(t), (5) dt2 where L is liner opertor in our cse L = 2 / x 2 is the Lplce opertor. The following circumstnces re crucil: 1
(i) Within the Hilbert spce L 2 [, b] we cn introduce vector (but not Hilbert!) subspce ν of functions tht hve second derivtives nd stisfy the conditions (2). (ii) The subspce ν is dense in L 2 [, b], which mens tht ny vector in L 2 [, b] cn be pproximted with ny given ccurcy by vector from ν, nd, correspondingly, ny orthonorml bsis (ONB) in ν is utomticlly n ONB for the whole Hilbert spce L 2 [, b]. (iii) Opertor L is well-defined nd self-djoint in ν, nd thus there exists n ONB consisting of the eigenvectors of L. Comments. The fct tht ν is vector spce is esily seen. The sttement tht Lplce opertor = 2 / x 2 is self-djoint in ν is checked by doing integrls by prts with (2) tken into ccount (we use sub-script nottion for prtil derivtives): f g = b = f g x b f xg b + b b f g xx dx = f g x b f xx g dx = b f xg x dx = f xx g dx = f g. (6) Here f nd g re two functions belonging to ν, which by definition mens tht they obey the boundry conditions (2), nd thus the boundry terms in (6) re equl to zero. We hve lredy discussed tht two eigenvectors of self-djoint liner opertor re utomticlly orthogonl if they correspond to different eigenvlues. The eigenvectors of one nd the sme eigenvlue re not necessrily orthogonl, but they form vector subspce nd thus cn be orthonormlized by Grm-Schmidt procedure. It is serious problem, however, to prove the completeness of the system of eigenvectors of Hermitin opertor in n infinite-dimensionl spce. The proof goes beyond the present course, nd below we will be tking for grnted trusting our mthemticin collegues tht for ny self-djoint differentil opertor there exists n orthonormlized eigenvector set tht forms n ONB in L 2 [, b]. Once we know tht there exists n ONB of eigenvectors of, we cn find these vectors explicitly by just solving the second-order ordinry differentil eqution e (x) = λ e(x), (7) with proper boundry conditions, which in our cse re e() = e(b) =. (8) 2
The generl solution of (7) is sine/cosine function for negtive λ, e(x) = A sin(x λ + θ ) (λ < ), (9) n exponentil function for positive λ, e(x) = Be x λ + Ce x λ (λ > ) (1) nd liner function for λ =, e(x) = D + Ex (λ = ). (11) With n exponentil function, however, the only wy to stisfy the boundry conditions is to hve B = C =, which is trivil solution (zero vector). In the cse of λ = the solution is lso trivil: D = E =. Hence, ll the eigenvlues re negtive, nd we need to exmine the solution (9). Without loss of generlity, set = nd b = 1. This cn be lwys chieved by shifting nd re-scling x. Then e() = sin θ = θ =. (12) (The solution θ = π corresponds to just chnging the sign of A.) Now we need to stisfy the second boundry condition: e(1) = sin( λ) = λ = πm, (13) where m is n integer. Noting tht m m does not produce new liner independent vector, we hve to confine ourselves to m >. Hence, The eigenfunctions re λ m = π 2 m 2, m = 1, 2, 3,.... (14) e m (x) sin(πmx), m = 1, 2, 3,.... (15) They re utomticlly orthogonl, since there is only one function for ny eigenvlue. The normliztion condition is nd we finlly get e 2 m(x) dx = 1, (16) e m (x) = 2 sin(πmx), m = 1, 2, 3.... (17) 3
Now we hve n ONB nd cn look for solution of the vector eqution (5) in the form of the expnsion u(t) = m=1 u m (t) e m, (18) where u m (t) re time-dependent Fourier coefficients stisfying the reltion u m (t) = e m u(t). (19) We plug (18) into (5) nd (3) nd form n inner product with n eigenvector e m. Tking dvntge of the fct tht we re deling with the eigenvectors of the opertor L, we get ü m (t) = λ m u m (t). (2) Remrkbly enough, for ech u m (t) we hve n independent eqution (2) n ordinry second-order differentil eqution. The initil conditions for this eqution redily follow from (19): u m () = e m u() e m g 1 = u m () = e m u() e m g 2 = e m (x) g 1 (x) dx, (21) e m (x) g 2 (x) dx. (22) Since ll λ s re negtive, ech of Eqs. (2) is just hrmonic oscilltor with the frequency ω m = λ m = πm. (23) We hve lredy considered this problem in our course, nd thus just write down the nswer stisfying given initil conditions: u m (t) = u m () cos(ω m t) + ( u m ()/ω m ) sin(ω m t), (24) where for ech m the vlues of u m () nd u m () re obtined by doing integrls (21) nd (22), respectively. The solution to the whole problem thus reds: u(x, t) = [A m cos(πmt) + B m sin(πmt)] sin(πmx), (25) m=1 A m = 2 g 1 (x) sin(πmx)dx, (26) 4
B m = (2/πm) g 2 (x) sin(πmx)dx. (27) The solution (25) hs the form of sum of products of functions of t nd x. In this connection, the method is often referred to s the method of seprtion of vribles x nd t in our cse. Note, however, tht x nd t were treted bsolutely differently. Het eqution. Now consider the het eqution u t = κ 2 u x 2 + f, (28) where u u(x, t) is the temperture s function of coordinte x nd time t; the prmeter κ > is the therml diffusivity; f f(x, t) is the het trnsferred to (if f > )/removed from (if f < ) the system per unit time nd per unit length. [Below we set κ = 1, which cn lwys be done by rescling t nd f.] We ssume tht our system is finite: x [, b]. To complete the sttement of the problem, we need the initil condition u(x, ) = g(x), (29) nd the boundry conditions t x = nd x = b. To be specific, we ssume tht our system is thermlly isolted t both ends. This mens tht there is no het flow t x = nd x = b, nd since the het flow is proportionl to the temperture grdient we hve u x (, t) = u x (b, t) =. (3) Here we use convenient nottion u x u/ x. Mthemticlly, the procedure is quite similr to wht we did with the wve eqution. We identify the function u(x, t) nd f(x, t) with (timedependent) vectors: u(x, t) u(t), f(x, t) f(t). (31) In the vector nottion, Eq. (28) hs the form d u(t) = L u(t) + f(t), (32) dt where L = 2 / x 2 is the Lplce opertor. 5
Once gin we identify the crucil circumstnces: (i) Within the spce L 2 [, b] one cn introduce subspce ν of functions tht hve second derivtives nd stisfy the conditions (3). (ii) The subspce ν is dense in L 2 [, b], which mens tht ny vector in L 2 [, b] cn be pproximted with ny given ccurcy by vector from ν, nd, correspondingly, ny orthonorml bsis (ONB) in ν is utomticlly n ONB for the whole Hilbert spce L 2 [, b]. (iii) Opertor L is well-defined nd self-djoint in ν, nd thus there exists n ONB consisting of the eigenvectors of L. The fct tht ν is vector spce is redily seen. The sttement tht Lplce opertor = 2 / x 2 is self-djoint in ν is checked by doing integrls by prts with (3) tken into ccount (we use shorthnd nottion for prtil derivtives): f g = b = f g x b f xg b + b b f g xx dx = f g x b f xx g dx = b f xg x dx = f xx g dx = f g. (33) Here f nd g re two functions from ν. They obey the boundry conditions (3), nd thus the boundry terms in (33) re equl to zero. Now we construct n ONB of eigenvectors of by solving the secondorder ordinry differentil eqution e (x) = λ e(x), (34) with proper boundry conditions, which in our cse re e () = e (b) =. (35) The generl solution of (34) is sine/cosine function for negtive λ, e(x) = A cos(x λ + θ ) (λ < ), (36) n exponentil function for positive λ, e(x) = Be x λ + Ce x λ (λ > ) (37) nd liner function for λ =, e(x) = D + Ex (λ = ). (38) 6
With n exponentil function, however, the only wy to stisfy the boundry conditions is to hve B = C =, which is trivil solution (zero vector). In the cse of λ = we do hve non-trivil eigenvector: E =, D. Hence, ll the eigenvlues except for λ = re negtive, nd we need to exmine the solution (36) to find them with corresponding eigenvectors. For definiteness, set = nd b = 1. Then e () = sin θ = θ =. (39) (The solution θ = π corresponds to just chnging the sign of A.) Now we need to stisfy the second boundry condition: e (1) = sin( λ) = λ = πm, (4) where m is n integer. Noting tht m m does not produce new liner independent vector, we hve to confine ourselves to m. Hence, The eigenfunctions re λ m = π 2 m 2, m =, 1, 2,.... (41) e m (x) cos(πmx), m =, 1, 2,.... (42) They re utomticlly orthogonl, since there is only one function for ny eigenvlue. The normliztion condition is e 2 m(x) dx = 1, (43) nd we finlly get e (x) = 1, (44) e m (x) = 2 cos(πmx), m = 1, 2, 3.... (45) Now we hve n ONB nd cn look for solution of the vector eqution (32) in the form of the expnsion u(t) = m= u m (t) e m, (46) where u m (t) re time-dependent Fourier coefficients. We plug this into (32) nd (29) nd form n inner product with n eigenvector e m. Tking dvntge of the fct tht we re deling with the eigenvectors of the opertor L, we get u m (t) = λ m u m (t) + f m (t), (47) 7
where nd f m (t) = e m f = u m () = e m g = e m (x) f(x, t) dx, (48) e m (x) g(x) dx. (49) For ech u m (t) we hve n independent eqution (47) n ordinry firstorder differentil eqution with the initil condition (49). For simplicity, consider the cse f =. The solution for u m is nd we ultimtely hve u(x, t) = g + u m (t) = u m () e λmt, (5) m=1 g = g m e π2 m 2t cos(πmx), (51) g(x) dx, (52) g m = 2 g(x) cos(πmx) dx, m = 1, 2, 3,... (53) Consider now the cse of time-independent f f(x), nd find the symptotic solution t t. For ny m > the symptotic solution of Eq. (47) is time-independent: = λ m u m + f m u m = f m /λ m (m > ). (54) The cse of m = is specil one, since λ =. Here we hve u (t) = f u (t) = g + f t. (55) If f f(x)dx =, then u is time-independent nd is equl to g it just remembers its initil condition. But if f, then, symptoticlly, there is liner increse/decrese of u. We cn summrize the bove-discussed tretment of liner prtil differentil equtions with Hermitin differentil opertor in the following lgorithm. 8
Step 1. Not necessry, but strongly recommended. Shift nd/or rescle the vrible x in such wy tht [, b] [, 1], or [, b] [ 1, 1]. Rescle the vrible t so tht to remove (reduce the number of) dimensionl constnts (like κ, c 2, etc.). Step 2. Check whether the boundry conditions re cnonicl or not. Cnonicl boundry conditions imply tht (i) the set of functions obeying them is vector spce, (ii) sptil differentil opertor is Hermitin opertor in this vector spce. If boundry conditions re not cnonicl, render them cnonicl by substituting u(x, t) = ũ(x, t) + u 1 (x, t) with n pproprite u 1 (x, t). Step 3. Construct eigenvector ONB nd find corresponding eigenvlues of the sptil differentil opertor. Mke sure tht there is no double counting of one nd the sme eigenvector. If there re more thn one vectors corresponding to one nd the sme eigenvlue, nd these re not orthogonl, orthonormlize them by Grm-Schmidt procedure. Step 4. Look for the solution of the given PDE in the form of the Fourier series with respect to obtined ONB. Plug the expnsion into PDE, tke into ccount tht the bsis vectors re the eigenvectors of the sptil differentil opertor, nd form the inner products of the l.h.s. nd r.h.s. of PDE with the bsis vectors. This will yield n independent ordinry differentil eqution for ech Fourier coefficient s function of time. If there is no time-dependence, the equtions re just lgebric ones. Solve them nd go directly to the step 7. Step 5. Obtin the initil conditions for these differentil equtions by forming inner products of the bsis vectors with the initil condition(s). Step 6. Solve the differentil equtions with the initil conditions. Step 7. Write down the nswer. Restore originl units of t nd x, if necessry. Problem 19. Consider the het eqution u t = u xx, (56) u = u(x, t), x [, 1], with the boundry conditions (note the derivtive in the first one) u x (, t) =, u(1, t) = (57) 9
nd the initil condition u(x, ) = 1. (58) () Mke sure tht the boundry conditions re cnonicl. (b) Construct the orthonorml bsis of the eigenfunctions of the Lplce opertor in the spce of functions obeying the boundry conditions (57). (c) Find the solution u(x, t) of the problem (56)-(58) in the form of the Fourier series in terms of the constructed ONB. Problem 2. Find the solution u(x), x [, 1] in the form of the Fourier series in terms of the eigenfunctions of the Lplce opertor of the sttionry het eqution u xx + f(x) =, (59) { 1, x [,.5), f(x) =, x [.5, 1], (6) u() = u(1) =. (61) Problem 21. The wvefunction ψ(x, t) of 1D quntum prticle living on ring of circumference L obeys Schrödinger eqution i hψ t = h2 ψ xx 2m, x [, L], (62) with periodic boundry conditions (becuse of the ring topology): At t = the wvefunction is ψ(, t) = ψ(l, t) ψ x (, t) = ψ x (L, t). (63) ψ(x, ) = { (2/L) 1/2, x [, L/2],, x (L/2, L). (64) Solve for ψ(x, t). Do not forget tht in contrst to het nd wve equtions, the function ψ is complex. Now we would like to generlize the bove-discussed tretment. There re three issues to be ddressed: (i) the form of the time-differentil term, (ii) the form of the differentil opertor L, nd (iii) the form of the boundry conditions under which we cn introduce vector subspce ν for the solutions with the opertor L being Hermitin for ll u ν. Form of the time-differentil term. The only requirement here is tht this 1
term is liner. Otherwise, the series expnsion mkes little sense. Hence, we cn work with ny eqution of the form D u(t) = L u(t) + f(t), (65) where D is ny liner time-differentil opertor, or just zero. Sturm-Liouville opertor. The generl form of second-order rel differentil opertor which proves Hermitin under pproprite boundry conditions is s follows. L = 1 [ w(x) x p(x) ] x q(x), (66) where w, p, nd q re rel, nd lso w >. It is the Sturm-Liouville opertor. The inner product now is defined with the weight function w(x), nd tht is why we need the requirement w >. Doing the integrls by prts in complete nlogy with Eq. (6) we get f L g = pf g x b pf xg b + Lf g. (67) We see tht the opertor L is Hermitin in the subspce ν if for ny f, g ν the boundry terms in the r.h.s. of (67) re zero. This cn be chieved if f ν the following condition is stisfied [ (f() = ) or (f x () = ) or (f x ()/f() = ξ ) or (p() = ) ] nd [ (f(b) = ) or (f x (b) = ) or (f x (b)/f(b) = ξ b ) or (p(b) = ) ]. (68) Here ξ nd ξ b re f-independent rel constnts. In the cse p() = [p(b) = ] the boundry condition lso implies tht both f() nd f x () [f(b) nd f x (b)] re finite t x [x b], nd this is relly condition, becuse t p() = [p(b) = ] divergent solutions pper. This boundry condition excludes them. The opertor L is Hermitin lso in the periodic cse: p() = p(b), f() = f(b), f x () = f x (b). (69) We will refer to the boundry conditions (68) nd (69) s cnonicl. If either (68) or (69) tkes plce, the eigenfunctions e j (x) nd eigenvlues λ j of the opertor L re found by solving the Sturm-Liouville eqution x p u qu = λwu. (7) x 11
Note (check) tht ny opertor L = r(x) 2 x 2 + s(x) + z(x) (71) x with rel r, s, z nd s, r > cn be written in the form (66), with [ x ] p(x) = exp dx s(x )/r(x ), w = p/r, q = zw. (72) Below we list some chrcteristic 1D equtions which cn be solved by the method described bove under the cnonicl boundry conditions. u t = κu xx + f(x, t) (Het/diffusion eqution), (73) iu t = u xx + q(x) u (Schrödinger eqution) (74) (we use the units h = 2m = 1), u tt = c 2 u xx + f(x, t) (Wve eqution), (75) where c is the wve velocity nd f is n externl force, nd u xx = f(x) (Poisson eqution). (76) Non-cnonicl boundry conditions. If the boundry conditions re noncnonicl, then generic prescription is to subtrct from the function u(x, t) some prticulr function u 1 (x, t), so tht the difference ũ(x, t) = u(x, t) u 1 (x, t) stisfies one of the cnonicl boundry conditions. Clerly, this subtrction will result only in chnging the form of the function f(x, t). We illustrte this ide by the following exmple. Let Now if we write where then for the function ũ(x, t) we get u(, t) = µ(t), u x (b, t) = η(t). (77) u(x, t) = ũ(x, t) + u 1 (x, t), (78) u 1 (x, t) = µ(t) + (x ) η(t), (79) ũ(, t) =, ũ x (b, t) =, (8) 12
which re the cnonicl boundry conditions. supposed to stisfy the het eqution In prticulr, if u(x, t) is then for the function ũ we will hve u t = u xx + f, (81) ũ t = ũ xx + f, (82) with Tht is f(x, t) = f(x, t) + 2 u 1 x 2 u 1 t. (83) f(x, t) = f(x, t) µ(t) (x ) η(t). (84) Issues of Convergence. Gibbs Phenomenon When constructing ONB of eigenfunctions of Sturm-Liouville opertor, we re deling with vector spce ν defined by the (cnonicl) boundry conditions of given eqution. Is this requirement relly crucil? Indeed, ny ONB in the Hilbert spce L 2 [, b] cn be used for expnding ny function L 2 [, b]. So why don t we use ONB corresponding to, sy, periodic boundry conditions for expnding function u stisfying the conditions u() =, u x (b) =, or vice vers? The point is tht the convergence in this cse will be, generlly speking, lmost everywhere, rther thn every where. And this is relly crucil, becuse the centrl step of the lgorithm of solving PDE is the interchnging the orders of summtion nd differentition when cting with the Sturm- Liouville opertor on the Fourier series. This is legitimte only if the series converges everywhere rther thn lmost everywhere. Gibbs phenomenon. Consider the solution (51)-(53) to the problem (28)- (3) in the cse when the initil condition reds g(x) = { 1, x [,.5],, x (.5, 1]. (85) A comment is in order here concerning the consistency of this discontinuous initil condition with PDE implying existence of the derivtives. Actully, 13
the initil condition (85) is understood s limit of smooth, but rbitrrily steep function. And the solution (51)-(53) is just wht we need for tking this limit, since it contins only the integrls of the function g(x), which re well defined even for discontinuous function. Even if the evolution of u(x) strts from stepwise initil condition (85), t ny finite t the solution is smooth. However, the convergence of the series towrds this smooth solution is rther peculir. Let us explore this convergence. Explicitly doing the integrls, we get g = 1/2, (86) /2 g m = 2 cos(πmx) dx = [here n is integer] nd rrive t the finl nswer { ( 1) n (2/πm), m = 2n + 1,, m = 2n (87) u(x, t) = 1/2 + (2/π) n= ( 1) n 2n + 1 e π2 (2n+1) 2t cos[(2n + 1)πx]. (88) In Figs. 1 nd 2 we plot some prtil sums of the series (88), with n mx being the mximl n in the sum. We see tht t t = the convergence is not homogeneous: No mtter how lrge is n mx, there is lwys finite mplitude overshoot in the vicinity of the point x =.5. Such n overshoot is generic for Fourier expnsions of stepwise functions. It is clled Gibbs phenomenon. In Fig. 2 we present truncted series t very smll, but finite t. We see qulittive difference. While t smll enough n mx the result is indistinguishble from tht of t =, t lrger n mx Gibbs phenomenon disppers. A similr sitution with Gibbs phenomenon rises when the initil condition does not stisfy one or both boundry conditions. See, for exmple, problem 2. The solution will stisfy the boundry conditions t ny t >, but not t t =. Correspondingly, t the boundry t which the boundry condition is not stisfied t t =, the Gibbs phenomenon in the Fourier series (but not in rel life!) will be seen in the limit of smll enough times. 14
Figure 1: The t = cse. 15
Figure 2: The t =.6 cse. At this time moment the cross-over from Gibbs-phenomenon behvior to smooth behvior tkes plce t n mx 25. 16