Lecture Notes 4: Fourier Series and PDE s

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1 Lecture Notes 4: Fourier Series and PDE s 1. Periodic Functions A function fx defined on R is caed a periodic function if there exists a number T > such that fx + T = fx, x R. 1.1 The smaest number T for which the reation 1.1 hods is caed the period of f or fundamenta period of f. Lemma 1.1. If f is T - periodic continuous function, then a+t Proof. Consider the function It is cear that a fxdx = F x = x+t x T fxdx. 1.2 fsds F x = fx + T fx =, x R f is T- periodic Thus F x is a constant function. Hence Let F a = a+t a fxdx = T 2. Functiona Series fxdx = F. f 1 x, f 2 x,..., f n x, be a sequence of functions defined on some interva I R. We say that the sequence 3.1 is convergent or pointwise convergent to a function fx on I if for each fixed point x I the number sequence {f n x} converges to the number fx. If at east for one point x the sequence fx is divergent, then we say that the sequence of functions {f n x} is divergent on I.. 1

2 2 A sequence of functions 3.1 is said to be uniformy convergent to a function fx on I if for each ε > there exists a number N ε depending on ε ony, such that f n x fx ε for a n N ε. For a given sequence of functions 3.1 the series f n x 2.2 is the foowing imit where im S Nx, N S N x = f 1 x + f 2 x f N x is caed the N-th partia sum of the series. If the sequence of partia sums {S N x} converges to some function sx on I, i.e. f n x = sx 2.3, then we say that the series 3.8 is convergent on I to sx. Otherwise the series 3.8 is caed divergent. If the sequence {S N x} is uniformy convergent to sx then we say that the series 3.8 is uniformy convergent. Theorem 2.1. If the functions f 1 x, f 2 x,..., f n x,... are continuous on an interva [a, b] and the series f n x is uniformy convergent on [a, b], then the sum of the series sx is a continuous function on [a, b]. Moreover the series obtained term by term integration of this series is aso convergent and b b f n xdx = sxdx. a a

3 3 Theorem 2.2. Weierstrass Theorem If the functions f 1 x, f 2 x,..., f n x,... are continuous on an interva [a, b], f n x a n, x [a, b], n = 1, 2,... and the series a n is convergent then the series f n x is uniformy convergent to some function that is continuous on [a, b]. 3. Euer s formuas and Fourier series A series of the form a 2 + a n cosnx + b n sinnx 3.1 is caed a trigonometric series. Question: Which functions have trigonometric series expansion. If fx has a trigonometric series expansion fx = a 2 + a n cosnx + b n sinnx 3.2 how to compute a, a 1,..., a n,..., b 1, b 2,...? Theorem 3.1. Suppose that f is 2π-periodic function and fx = a 2 + a n cosnx + b n sinnx, 3.3 where the series is converges uniformy on the rea axis. Then a n = 1 π fx cosnxdx, n =, 1, b n = 1 π fx sinnxdx, n = 1,

4 4 Proof. Reay since cosnxdx =, sinnxdx = and due to uniform convergence of the series we can integrate 3.1 over, π and get: fxdx = a π, Let us mutipy 3.1 by cosmx and integrate over, π. Taking into account we obtain cos 2 mxdx = π, a n = 1 π b n = 1 π sin 2 mxdx = π fx cosnxdx, n =, 1, fx sinnxdx, n = 1, Here we have used the fact that for each m the series cosmx a n cosnx+ b n sinnx and sinmx a n cosnx + b n sinnx are uniformy convergent. The series 3.1 where a n and b n are defined by 3.6 and 3.7 is caed the Fourier series of the function f, the numbers a n, b n are caed the Fourier coefficients of f. Piecewise continuous function. A function fx is caed piecewise continuous on [a, b], if im x b fx, im x a + fx exist f is continuous on a, b except at finitey many of points in a, b, where f has one-sided imits. Theorem 3.2. If 2π-periodic function fx and its derivative f x are piecewise continuous functions, then fx+ + fx 2 = a 2 + a n cosnx + b n sinnx 3.8 for each x R, where a n and b n are defined by 3.4 and 3.5.

5 5 Exampe 3.3. Consider the function Φx given on [, π] by { π + x, x [, ] φx = π x, x [, π]. φx is piecewise smooth. a = 1 π a n = 2 π = 2 π φxdx = 2 π π xdx = 2 π π x cosnxdx == 2 = 2 nπ 1 x ] π [πx x2 2 = 2π π = π cosnxdx 2 π n sinnx dx = 2 nπ [x sinnx]π + 2 nπ [ 1 ] π n cosnx = 2 [ 1 nπ n 1 ] n cosnπ x cosnxdx = 2 nπ sinnxdx 1 n 1n 2 n 2 Thus a n = if n is and even number, and a n = 2 n, if n is and odd number. 2 φx = π cos2k 1x π 2k 1 2 k=1 Exampe 3.4. Using Fourier series show that π 2 8 = Functions of any period p = 2. a n = 1 fx = a 2 + a n cos fx cos x dx, b n = 1 x + b n sin x 3.9 fx sin x dx Even and Odd Functions. If a function fx is an even function, then b n = 2 fx sin x dx =

6 6 and its Fourier series has the form fx = a 2 + a n cos x, 3.11 where a n = 2 fx cos x dx, n =, 1, 2, If a function fx is an odd function, then a n = 2 fx cos x dx = and its Fourier series has the form fx = b n sin x x, 3.13 where b n = 2 fx sin x dx, n = 1, 2, Let fx be defined on [, ]. We define the even periodic extension f e of f as foows f e x = f x, if x [, ], and f e x = f e x + 2, x R. An odd periodic extension f of f is defined as foows f x = f x, if x [, ], and f x = f x + 2, x R. Exampe 3.5. Find Fourier series expansion for fx = 1 x 2, x [ 1, 1]. a n = a = x 2 dx = 4 3, 1 x 2 cosnπxdx = 2 1 x 2 nπ sinnπx dx = 2 nπ dx = x 1 nπ nπ cosnπx 1 1 cosnπxdx+ 2x sinnπxdx = nπ 2x cosnπx 1 = 4 n 2 π 2 1n.

7 7 Thus we have fx = π 2 1 n+1 n 2 cosnπx. 4. Riemann -Lebesgue Lemma Proposition 4.1. If fx is a continuous function, then im I n = im fx cosnxdx =, n n Proof of 4.1. Since cos α = cosα + π we have I n := im J n = im fx sinnxdx =. 4.1 n n fx cosnxdx = fx cos [x + π ] n n dx. Making change of variabes x + π n = y and using the Lemma 1.1 we obtain Hence we have + π 2 fx cosnxdx = fy π + π n cosnydy = fy π n cosnydy 2 I n + I n = 2 fx cosnxdx = [fx fx π ] cosnxdx n = 2 fx cosnxdx fx fx π n dx. The function f is continuous on [, π] thus it is uniformy continuous on [, π]. Therefore the integra in the right hand side of R tends to zero as n. So I n as n. Simiary we can show that J n as n. Probem. Let fx be 2π -periodic and f x is continuous function. Show that 1 1 a n = o, b n = o. n n

8 8 5. Besse inequaity and mean vaue approximation. Theorem 5.1. Besse inequaity If fx is a piecewise continuous function on, π, then the foowing inequaity caeed the Besse inequaity hods true a a 2 n + b 2 n 1 f 2 xdx, 5.1 π where a, a n and b n, n = 1, 2,... are Fourier coefficients of f. Proof. It is cear that E N = f 2 xdx 2 By using Euer formuas we get S N xfxdx = a 2 fxdx + N S N xfxdx + S 2 Nxdx. 5.2 fx [a k cosnx + b k sinnx] dx [ 1 = π 2 a2 + a a 2 N + b b 2 N [ a [S N x] 2 2 dx = π 2 + a a 2 N + b b 2 N Substituting 5.3 and 5.4 into 5.2 we obtain: [ π E N = f 2 a 2 N ] xdx π 2 + a 2 n + b 2 n. or a 2 N 2 + a 2 n + b 2 n 1 π We cam pass to the imit as N and get 5.1. ], 5.3 ]. 5.4 f 2 xdx. 5.5 By using the Besse inequaity we can prove the foowing theorem Theorem 5.2. Suppose that f is continuous 2π-periodic function and f is piecewise continuous function. Then the Fourier series of f converges absoutey and uniformy to the function f.

9 9 Proof. Let us cacuate the Fourier coefficients of f : α = 1 π f xdx = 1 fπ f =, π α n = 1 π f x cosnxdx = 1 π cosnxfx x=π +n x= fx sinnxdx = nb n, β n = 1 π f x sinnxdx = 1 π sinnxfx x=π So we have Empoying the inequaity x= n fx cosnxdx = na n. α n = nb n, n =, 1, 2,..., β n = na n, n = 1, 2,..., 5.6 ab 1 4 a2 + b 2 we obtain from 5.6 the foowing inequity a n + b n = 1 n β n + 1 n α n 1 2n 2 + α2 n + β 2 n. Due to the Besse inequaity the series αn 2 + βn 2 is convergent. Hence the series a n + b n is aso convergent. Therefore the series a 2 + a n cos nπ x + b n sin nπ x is uniformy convergent to a continuous function f. where a n and b n are Fourier coefficients of the function f

10 1 Exampe 5.3. Assume that the Fourier series of fx on [, π] converges to fx and can be integrated term by term. Mutipy a 2 + a n cos nπ x + b n sin nπ x by fx and integrate the obtained reation from to π to derive the identity 1 π f 2 xdx = a2 π 2 + a 2 n + b 2 n. 5.7 This identity is caed the Parseva identity. 6. Heat Equation. Method of Separation of Variabes We consider the probem u t = a 2 u xx, x,, t >, 6.1 ux, = fx, x [, ], 6.2 u, t = u, t =, t, 6.3 We assume that the soution of the probem has the form ux, t = XxT t, where Xx and T t are nonzero functions. Substituting into 6.1 we get XxT t = a 2 X xt t. Dividing both sides of the ast equaity by a 2 XxT t we obtain T t a 2 T t = X x Xx 6.4 Since the eft hand side of 6.4 depends ony on t and the right hand side depend ony on x each side of this equaity can ony be equa to some constant. Thus T t a 2 T t = X x Xx = λ, λ = constant or T t = λa 2 T t 6.5 X x + λxx =. 6.6

11 It foows from 3 that X = X =. 6.7 So we have to find the vaues of λ for which the equation 6.6 has nonzero soution which satisfy 6.7. The vaues of for which 6.6 has nonzero soution satisfying 6.7 are caed eigenvaues of the probem 6.6,6.7 When = 6.7, the corresponding soutions - eigenfunctions of 6.6,6.7. When = the equation has a genera soution Xx = Ax + B This function satisfies 6.7 just for A = B =. Thus λ = is not an eigenvaue. If λ < then genera soution of 6.6 has the form Xx = C 1 e λ x + C 1 e λ x. It is easy to see that this function satisfies 6.7 just when C 1 = C 2 =. So 6.6,6.7 has not negative eigenvaues. If > then the genera soution of 6.6 has the form Substituting into 6.7 we obtain Xx = C 1 cos λx + C 2 sin λx. X = C 1 =, X = C 2 sin λ = The second equaity hods for = nπ, n = ±1, ±2,... Thus the numbers λ n = n2 π 2, n = 1, 2,... 2 are eigenvaues of the probem 6.6,6.7, and the functions X n x = sin x, n = 1, 2,... are the corresponding eigenfunctions. It is easy to see that the genera soution of 6.5 for = n has the form T n t = D n e a2 λ n t, n = 1, 2,... Hence for each n = 1, 2,... the function e a2 λ n t sin x satisfies 6.1,6.3. Since the equation 6.6 is a inear equation for each N u N x, t = N 11 D n e a2 λ n nπ t sin x, 6.8

12 12 where D n, n = 1,..., N are arbitrary constants aso satisfies 6.1,6.3. Next we try to satisfy the initia condition 6.2: u N x, = N D n sin x = fx We see that the soution of the probem has this form just when the initia function is inear combination of functions Let us consider the series Let us note that if sin λ 1 x,..., sin λ n x, λ k = k2 π 2 2. ux, t = D n e a2 λ n nπ t sin x. 6.9 D n <, then this series is uniformy convergent on [, ] [, T ], T >. The function ux, t defined by 6.9 satisfies the boundary conditions 6.3 since each term of the series satisfies these conditions. It foows from 6.9 that ux, t satisfies the initia condition 6.2 fx = ux, = D n sin x 6.1 iff D n = f n = 2 fx sin x dx. Theorem 6.1. If fx is continuous on [, L], f x is piecewise continuous on [, L], f = f = then the function ux, t = f n e a2 λ n nπ t sin x 6.11 satisfies To prove this theorem we need the foowing proposition Proposition 6.2. Assume that the functions v n x, t, n = 1, 2,... are continuous Q T = [a, b] [t, T ] and v n x, t a n, x, t Q T, n = 1, 2,...

13 where the sequence of positive numbers {a n } is so that the series a n is convergent. Then the series v n x, t is absoutey and uniformy convergent on Q T. Moreover the function is continuous on Q T. If v n x, t t b n, and then the series vx, t = v n x, t 2 v n x, t x2 d n, x, t Q T, n = 1, 2,... b n <, v n x, t and t d n <, 2 v n x, t x2 uniformy converge to v t x, t and v xx x, t in Q T. Moreover these functions are continuous in Q T. Proof of Theorem 6.1. Since f is piecewise smooth the series f n is convergent. Thus the Proposition 6.2 impies that the function ux, t is continuous on [, ] [,. Let us show that the series u n 2 u n a x, t and b x, t 6.12 t x2 are convergent uniformy for t t, x [, ], where t is an arbitrary positive number. Continuity of f on [, ] impies boundedness of the sequence {f n }. So there exists M > so that f n M, n = 1, 2,... Thus for each t t and x [, ] we have u n x, t t = f n a 2 n2 π 2 2 e a 2 n 2 π 2 2 t nπ sin x M 1 n 2 e a 2 n 2 π 2 2 t, 6.13 where M 1 = Ma 2 π u n x, t x2 = n 2 π 2 f n 2 It is easy to see that the series e 13 a 2 n 2 π 2 2 t nπ sin x n 2 M 1 a 2 n 2 π 2 a 2 e 2 t, 6.14 n 2 M 1 a 2 n 2 π 2 a 2 e 2 t is convergent. Therefore due to the Proposition 6.2 the function ux, t defined by 7.1 is a soution of the probem

14 14 Probem 6.3. Let the series a 2 n and b 2 n be convergent. Show that a n b n 6.1. Nonhomogeneous Equation. a 2 n 1/2 b 2 n 1/2 u t = a 2 u xx + hx, t, x,, t >, 6.15 ux, = fx, x [, ], 6.16 u, t = u, t =, t, 6.17 We ook for soution of the probem in the form ux, t = u n t sin x We expand hx, t hx, t = h n t sin x By using 6.18 we obtain from 6.15 [ u n t + λ n a 2 u n t h n t ] sin λ n x =. This equaity hods iff 6.18 u nt + λ n a 2 u n t = h n t, n = 1, 2, Taking into accoun the initia condition 6.16 we obtain ux, = h n sin x = fx = f n sin It foows then x. u n = f n, n = 1, 2, The initia vaue probem 6.19,6.21 has the soution u n t = e λ na 2t f n + t e λ na 2 t s h n sds

15 Inserting the expression for u n t into 6.18 we get [ t ] ux, t = f n e λ na 2t + e λ na 2 t s h n sds sin Let us reca that λ n = n2 π x Probem 6.4. Show that the probem has a unique soution. Hint. Assume that vx, t is another soution of this probem, i.e. Then show that the function v t = a 2 v xx, x,, t >, vx, = fx, which is a soution of the probem satisfies 1 d 2 dt x [, ], v, t = v, t =, t, wx, t = ux, t vx, t w t = a 2 w xx, x,, t >, wx, =, x [, ], w, t = w, t =, t [wx, t] 2 dx + a 2 [w x x, t] 2 dx = Nonhomogeneous boundary conditions. Let us consider the probem u t = a 2 u xx, x,, t >, 6.22 ux, = fx, x [, ], 6.23 u, t = A, u, t = B, t, 6.24 where A, B are given constants. The soution of the probem ux, t is a summ of two functions vx, t and W x, where W is a soution of the probem { W x =, x,, 6.25 W = A, W = B

16 16 and v is a soution of the probem v t = a 2 v xx, x,, t >, vx, = fx W x, x [, ], u, t = u, t =, t, 6.26 It is cear that W x = A + 1 B Ax is a soution of the probem 6.25 Hence the soution of the probe is the function where ux, t = q n = 2 Next we consider the foowing probem q n e n2 a 2t nπ sin x + A + 1 B Ax, fx A 1 B Ax sin x. u t = a 2 u xx, x,, t >, 6.27 ux, = fx, x [, ], 6.28 u, t = φt, u, t = ψt, t, 6.29 we ook the soution of the probem in the form For u t we have ux, t = u n t sin λn x 6.3 where u t x, t = u xx x, t = g n t = 2 u nt sin λn x 6.31 g n t sin λn x, 6.32 u xx x, t sin λn x dx.

17 17 Integrating by parts we obtain g n t = 2 = [ ] u x sin λn x [ 2 λn ux, t cos λn 2 λn x ] 2 λ n u x x, t cos n x dx = 2 λn [u, t u, t cosnπ] λ n u n t. Empoying the boundary conditions 6.29 we obtain Thus 6.32 impies u xx x, t = ux, t sin λn x dx g n t = 2 λn [φt ψt cosnπ] λ n u n t [ 2 λn φt 2 ] λ n 1 n ψt n u n t sin λn x By using the ast reation and 6.31 in6.27 we obtain [ 2a u nt a 2 2 λ n φt 2a2 λ n 1 n ψt aλ n u n t Therefore the Fourier coefficients u n t satisfy u nt = a 2 [ 2 λn φt 2 λ n 1 n ψt λ n u n t The function u wi satisfy the initia condition 6.28 iff We sove the initia vaue probem 6.33,6.34 and get u n t = f n e a2 λ nt 2a2 λ n ] sin λn x =. ], n = 1, 2, u n = f n, n = 1, 2, t e a2 λ nt s [ 1 n ψs φs] ds, n = 1, 2, So the soution of the probem has the form 6.3, where u n t, n = 1, 2,... are defined by Wave Equation In this section we study the wave equation. The first probem is the initia boundary vaue probem: u tt = c 2 u xx, x,, t >, 7.1 ux, = fx, u t x, = gx, x [, ], 7.2 u, t = u, t =, t, 7.3 We assume that the soution of the probem has the form ux, t = XxT t,

18 18 where Xx and T t are nonzero functions. Substituting into 7.1 we get XxT t = a 2 X xt t. Dividing both sides of the ast equaity by c 2 XxT t we obtain T t c 2 T t = X x Xx 7.4 Since the eft hand side of 7.4 depends ony on t and the right hand side depend ony on x each side of this equaity can ony be equa to some constant. Thus T t c 2 T t = X x Xx = λ, λ = constant or T t = λc 2 T t 7.5 It foows from 7.3 that X x + λxx =. 7.6 X = X =. 7.7 So we have to sove the eigenvaue probem 7.6,7.7. We have seen that the numbers λ n = n2 π 2, n = 1, 2,... 2 are eigenvaues of the probem 7.6,7.7, and the functions X n x = sin x, n = 1, 2,... are the corresponding eigenfunctions. It is easy to see that the genera soution of 7.5 for = λ n has the form T n t = A n cosc λ n t + B n sinc λ n t, n = 1, 2,... It is easy to see that for each N the function N [ u N x, t = A n cosc λ n t + B n sinc ] λ n t sin x, 7.8 where A n, B n, n = 1,..., N are arbitrary constants satisfies 7.1,7.3. But this function may satisfy the initia condition 7.2 ony when f is a inear

19 combination of finitey many eigenfunctions. To satisfy the initia condition 7.2 we consider the series [ ux, t = A n cosc λ n t + B n sinc ] λ n t sin x. 7.9 Let us note that if A n < and B n <, then this series is uniformy convergent on [, ] [, T ], T >. The function ux, t defined by 7.9 satisfies the boundary conditions 7.3 since each term of the series satisfies these conditions. It foows from 7.9 that ux, t satisfies the initia conditions 7.2 fx = ux, = gx = u t x, = A n sin x cnπ B n sin x iff A n = f n = 2 fx sin x dx, and B n = cnπ g n = 2 gx sin cnπ x dx, Simiar to the corresponding theorem for the heat equation we can prove the foowing Theorem 7.1. If fx, gx, g x, f x, f x are continuous g x, f x are piecewise continuous on [, L], then the function ux, t = f = f =, f = f =, g = g = is a soution of [ f n cos c λ n t + g n c sin c λ n t ] sin λ n 19 x 7.1

20 Nonhomogeneous Equation. u tt = c 2 u xx + hx, t, x,, t >, 7.11 ux, = fx, u t x, = gx, x [, ], 7.12 We ook for soution of the probem in the form ux, t = u n t sin x We expand hx, t u, t = u, t =, t, 7.13 hx, t = h n t sin x By using 7.14 we obtain from 7.11 [ u n t + λ n a 2 u n t h n t ] sin λ n x =. This equaity hod iff 7.14 u nt + λ n c 2 u n t = h n t, n = 1, 2, Taking into account the initia condition 7.12 we obtain ux, = A n sin x = fx = f n sin x. It foows then u t x, = cnπ B n sin x = gx = g n sin x. A n = f n, B n = cnπ g n, n = 1, 2, The initia vaue probem 7.15,7.16 has the soution u n t = f n cos c λ n t + g n c sin c λ n t n Inserting the expression for u n t into 7.14 we get ux, t = Remember that λ n = n2 π t [ f n cos c λ n t + g n c sin c λ n t ] sin x λ n t + [ cnπ ] sin t s h n sds [ sin c ] λ n t s h n sds sin x 7.17 Probem 7.2. Show that the probem has a unique soution.

21 Hint. Assume that v is another soution of the probem. Then observe that wx, t satisfies w tt = c 2 w xx, x,, t >, 7.18 wx, =, w t x, =, x [, ], 7.19 w, t = w, t =, t. 7.2 Mutipy the equation by w t, integrate the obtained reation with respect to x over the interva, and obtain the equaity d dt [wt x, t] 2 + [w x x, t] 2 dx = The Cauchy probem for the wave equation. Now we consider the initia vaue probem for the wave equation, i.e. we woud ike to find soution of the equation under the initia conditions u tt = c 2 u xx, t >, x,, 7.21 ux, = fx; u t x, = gx, x,, 7.22 where f and g are given numbers and c > is a given number. To sove the probem we make the foowing change of variabes ξ = x ct, By using the chain rue we find η = x + ct. u t = u ξ ξ t + u η η t = au ξ + cu η = cu η u ξ, u tt = cu ηξ ξ t + u ηη η t u ξξ ξ t u ξη η t = c 2 u ξξ 2u ξη + u ηη, 7.23 u x = u ξ ξ x + u η η x = u ξ + u η, u xx = u ξξ ξ x + u ξη η x + u ηξ ξ x + u ηη η x = u ξξ + 2u ξη + u ηη 7.24 Bu using 7.23 and 7.24 in 7.21 we find u ξη = It is cear that for each differentiabe functions u 1, u 2 the function uξ, η = u 1 ξ + u 2 η, isa soution of Hence the function ux, t = u 1 x ct + u 2 x + ct

22 22 is a soution of Bu using the initia conditions 7.22 we get ux, = u 1 x + u 2 x = fx, 7.27 u t x, = cu 1x + cu 2x = gx. Integrating the ast equaity we obtain u 2 x u 1 x = 1 gsds + C c x Soving the system of equations 7.27, 7.28 we obtain u 1 x = 1 2 fx 1 2c x x x gsds C 2, u 2 x = 1 2 fx + 1 gsds + C 2c x 2. Iserting the vaues of u 1, u 2 from the ast two equaities into 7.26 we get or ux, t = 1 1 ϕx ct 2 2a ux, t = x ct x fx ct + fx + ct 2 x gsds fx + ct + 1 2c The ast equaity is caed the D Aembert formua. x+ct x gsds + 1 x+ct gsds, a x ct