Fourier Series in Compex nottion sin(x) = eix e ix i = i eix e ix cos(x) = eix + e ix So So '(x) = A 1 0 + nx nx A n cos + B n sin = A 1 0 + e inx= + e inx= A n = A 0 + 1 = C n = C n 1 n= 1 A n ib n C n e inx C n = e inx= + A n + ib n 8 A n >< >: < x < i B n e inx= e inx= e inx= ib n n > 0 A n+ib n n < 0 A 0 n = 0 Z e inx e imx = C n = 1 ( 0 m 6= n Z m = n '(x)e inx 1
Sturm-Liouvie (*) + q(x)u(x) = m(x)u(x) < x < b p(x) > 0 q(x) 0 m(x) > 0 n quntities re re. De nition 1 is the eigenvue n u is the eigenfunction De nition A homogeneous bounry conition is symmetric if p [fg 0 f 0 g] b = 0 Exmpes: Dirichet Neumnn perioic Robin (?) then u() + b u () = 0 r u(b) + b r u (b) = 0 p [fg 0 f 0 g] b = f() b g() b f()g() = 0 De nition 3 Inner prouct (u; v) = m(x)u(x)v(x) m > 0
Exmpe p = 1 q = 0 m = 1 Then with Dirichet conitions we hve u = u(x) u(0) = u() = 0 0 < x < If > 0 then Using u(0) = 0 we get n using u() = 0 u(x) = A cos( p x) + B sin( p x) u(x) = B sin( p x) n = u(x) = B sin( n x) So we hve n in nite number of eigenvues/eigenfunctions. Green s Ientities First Ientity: v(x) + v = b v Secon Ientity: v(x)+ p(x) v u(x) = p(x) u v + v b u In muti-imensions this generizes to (p = 1) ZZZ ZZ (ru rv + vu) V = D ZZZ ZZ (vu uv) V = D @D @D v @u @n S v @u @n u @v S @n 3
Proof. From the ivergence theorem ZZZ ZZ iv(f )V = F ns ZZZ So iv(v gr u) = ru rv + vu ZZZ (ru rv + vu) V = iv(v gr u)v ZZ ZZ = v gr u ns = v @u @n S Interchnge u n v n subtrct to get the secon ientity. 4
From (*) (1) () + q(x)u(x) = 1 m(x)u(x) p(x) v + q(x)v(x) = m(x)v(x) Mutipy (1) by v n () by u n subtrct n integrte v(x) + u(x) p(x) v = ( 1 ) m(x)u(x)v(x) Integrte by prts (Green s theorem in mutiimensions) v u p(x) v u + p v p If the bounry conitions re symmetric then u v b v u b v u = ( 1 ) = ( 1 ) m(x)u(x)v(x) m(x)u(x)v(x) ( 1 ) m(x)u(x)v(x) = 0 Hence, if 1 6= Theorem 4 For symmetric bounry conitions, if 1 6= then (u; v) = 0 If 1 = then we hve subspce n we cn choose n orthogon bsis. This is one by Grm Schmit 5
Grm-Schmit If f k (x)g is inery inepenent bsis then we cn construct n orthonorm bsis tht spns the sme spce. ' 1 (x) = 1 (x) ' (x) = (x) ( ; 1 ) (' 1 ;' 1 ) ' 1 Then (' ; ' 1 ) = ( ; ' 1 ) ( ;' 1 ) (' 1 ;' 1 ) (' 1; ' 1 ) = 0 ' k (x) = k (x) k 1 ( k ;' j ) (' j ;' j ) ' j j=1 Hence, we cn consier the soutions of (*) to be orthogon to ech other. So consier sequence of orthogon soutions of (*) f' k (x)g :Then if f(x) = n (f; ' m ) = n n ' n (x) n (' n ; ' m ) = m (' m ; ' m ) So m = (f; ' R b m) (' m ; ' m ) = m(x)f(x)' m(x) R b m(x)' m(x) Theorem 5 If p,q,m re re n the bounry conitions re symmetric then there re no compex eigenvues Proof. (1) () + q(x)u(x) = m(x)u(x) + q(x)u(x) = m(x)u(x) As before mutipy rst eqution by u, the secon by u, subtrct n integrte. Then u(x) + u(x) = m(x)u(x)u(x) 6
Agin integrte by prts n use the symmetry of the bounry conitions. m(x)ju(x)j = 0 So = i.e. is re If u is compex then its re n imginry components re soutions. (*) Negtive Eigenvues + q(x)u(x) = m(x)u(x) < x < b By Green s rst ientity we hve for v v(x) = v + b v Choose v = u n ssume symmetric bounry conitions. Then Using the ODE we get u(x) = p(x) u 0 So u(x) [q(x)u(x) m(x)u(x)] = p(x) u m(x)u (x) = = u q(x)u (x) + p(x) R b q(x)u (x) + R b u p(x) R b 0 m(x)u (x) We cn hve equity ony if q(x) = 0 n u = 0 Note: A these proofs work equy we in mutiimensions using Green s theorem inste of integrtion by prts. 7
Competeness Theorem 6 There re n in nite number of eigenvues for (*) n n! 1. Furthermore f(x) = c n ' n (x) c n = (f; ' n) (' n ; ' n ) Convergence: 1 ' n (x)!? f(x) De nition 7 Pointwise Convergence: N im N!1 f(x) ' n (x) = 0 De nition 8 Uniform Convergence: N im mx N!1 xb f(x) ' n (x) = 0 for every x for every x De nition 9 L (root men squre) im N!1 f(x) N ' n (x) = 0 Uniform convergence impies pointwise convergence. Uniform convergence impies root men squre convergence. Exmpes f(x) = x n 0 x 1 Then x n! ( 0 0 < x < 1 1 x = 1 8
f n (x) = (1 x)x n 1 = x n 1 x n N N f n (x) = (x n 1 x n ) = 1 x N! 1 s N! 1 so we hve pointwise convergence. However mx 1 1 x N = mx x N = 1 6= 0 0x1 So we on t hve uniform convergence. For L we hve Z 1 So we hve L convergence. 0 0x1 x N 1 = N + 1! 0 9
Theorem 10 If f; f 0 ; f 00 exist n re continuous in x b i.e. fc [; b] f stis es the bounry conitions then f(x) = 1 n ' n (x) Theorem 11 If R b f (x) < 1 P then f(x) = 1 n ' n (x) converges in L Theorem 1 For sine n cosine series ony. If f is continuous on x b f 0 is piecewise continuous converges uniformy Then the series converges pointwise. If f n f 0 re piecewise continuous then f(x+) + f(x ) n ' n (x)! Theorem 13 Integrtion: If formy f(x) $ A 1 0 + nx nx A n cos +B n sin Then xr f(y)y = A 0 (x+)+ 1 An nt n sin not necessriy convergent x Bn nt n cos is convergent We note tht for i erentition it is the opposite i.e. the erivtive of convergent series my not converge. Exmpe: expning x in sine series we hve x = Di erentiting we get 1 = 1 n ( 1)n+1 sin nx 1 nx ( 1) n+1 cos 0 x 0 x This is certiny NOT the cosine series of 1 which is just 1. In fct this series oes not converge! We begin with the proof of convergence in est squres. Restting the theorem we hve 10
Theorem 14 If ' n re the eigenfunctions of Sturm-Liouvie probem with symmetric bounry conitions n jjf jj < 1. Then jjf N n ' n jj! 0 n = (f; ' n) (' n ; ' n ) Theorem 15 Let ' n be n orthogon P set n jjfjj < 1, Then the choice of constnts c n tht minimizes jjf c n ' n jj is c n = n Proof. Assume for simpicity tht quntities re re E N = jjf c n ' n jj = R jf(x) c n ' n (x)j = R jf(x)j R c n f(x)'n (x) + P n Z P c n c m m ' n (x)' m (x) = jjfjj c n (f; ' n ) + c njj' n (x)jj = jjfjj + jj' n (x)jj (f; ' c n ) n jj' n (x)jj (f; ' n ) jj' n (x)jj To minimize we cn ony "py" with c n. Since the mie term is positive we minimize E N if c n = (f; ' n) jj' n (x)jj = n Then E N = jjfjj (f; ' n ) jj' n (x)jj = jjfjj A njj' n (x)jj 0 So we hve Besse s inequity. If jjfjj < 1 then (f; ' n ) jj' n (x)jj jjfjj Prsev s Equity Theorem 16 The Fourier Series converges to f(x) in L if n ony if (f; ' n ) jj' n (x)jj = njj' n (x)jj = jjfjj De nition 17 A sequence f' n (x)g is compete if Prsev s equity hos whenever jjfjj < 1 11
Riemnn-Lebesque Theorem Theorem 18 If () f C 1 or (b) jjfjj L < 1 Then im n!1 R f(x) ( sin( nx ) cos( nx ) = 0 Proof. 1. integrtion by prts. In Fourier series B n = R f(x) sin( nx ) by Besse s inequity B n! 0 Exmpe Consier f(x) = 1 on (0; ). We n tht So by Prsev s equity 1 = P n o R 1 = P 0 P n o 4 n sin(nx) n o 1 n = 8 4 n Pointwise Convergence If f(x) = A 1 0 + A n cos (nx) + B n sin (nx) A n = 1 R f(y) cos(ny)y n = 0; 1; ; 3::: B n = 1 R f(y) sin(ny)y n = 1; ; 3::: Dirichet kerne 1
Consier the prti sum where S N = A N 0 + A n cos (nx) + B n sin (nx) " # = 1 R N 1 + (cos(nx) cos(ny) + sin(nx) sin(ny)) f(y)y " # = 1 R N 1 + cos(nx ny) f(y)y = 1 R K N (x y)f(y)y K N () = 1 + N cos(n) = sin(n + 1 ) sin( ) Proof. Use cos() = ein +e n get geometric series. in Now et = y x. Then S N = 1 R K N ()f(x + ) S N (x) f(x) = 1 R K N () [f(x + ) f(x)] = 1 R g() sin(n + 1 f(x + ) f(x) ) where g() = sin( ) Let n () = sin(n + 1 ). By Besse s inequity we hve 1 j(g; n )j jj n jj = 1 1 R j(g; n )j jjgjj [f(x + ) f(x)] = sin ( ) By L hopit s rue the integrn is nite t = 0. Hence it is boune everywhere n the integr exists. Since the sum converges ech term much pproch zero n so R j(g; n )j = g() sin(n + 1 )! 0 Gibbs Phenomen If inste we re intereste in uniform convergence we nee to nyze im mx js N (x) f(x)j N!1 x One cn show tht if the function f(x) hs iscontinuity t x = x 0 then in fct this imit is nonzero n is bout 9% of the size of the jump on either sie. 13