Fourier Series in Complex notation. cos(x) = eix + e ix 2. A n cos + B n sin l. i 2 B n. e inx=l + A n + ib n 2 8. ( 0 m 6= n. C n = 1 2l.

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1 Fourier Series in Compex nottion sin(x) = eix e ix i = i eix e ix cos(x) = eix + e ix So So '(x) = A nx nx A n cos + B n sin = A e inx= + e inx= A n = A = 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

2 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

3 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 @n 3

4 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 S Interchnge u n v n subtrct to get the secon ientity. 4

5 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

6 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

7 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

8 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

9 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

10 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 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

11 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

12 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 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

13 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