MAGIC058 & MATH64062: Partial Differential Equations 1

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1 MAGIC58 & MATH646: Prti Differenti Equtions 1 Section 4 Fourier series 4.1 Preiminry definitions Definition: Periodic function A function f( is sid to be periodic, with period p if, for, f( + p = f( where p is positive constnt. For empe, ( is periodic with period π. We cn construct periodic functions such s {, f( =, f( + = f(., < This defines periodic function with period y Figure 4.1 An empe of periodic function. Definition: Piecewise continuous function A function f( is sid to be piecewise continuous on the interv [, b] if it hs t most finite number of finite discontinuities on [, b]. y b Figure 4. An empe of piecewise continuous function. Definition: Dirichet conditions Suppose function f( stisfies (i f( hs finite number of etrem over the interv [, ],

2 MAGIC58 & MATH646: Prti Differenti Equtions (ii f( nd f ( re piecewise continuous on [, ], (iii f( is periodic with period. Then f( is sid to stisfy the Dirichet conditions. Definition: Orthogonity of {(nπ/} nd {(nπ/} The sets of functions {(nπ/} nd {(nπ/} re both orthogon on the interv [, ]. This is defined s ( mπ {, m = n d = (4.1, m n nd Proof of (4.1 For m n, use the retion ( mπ {, m = n ( d =, m n. ( mπ = 1 { ((m n π ((m + n π }. So, the integr becomes 1 { ((m n π ((m + n π } d = [ 1 However, for m = n, the integr is ( (m nπ (m n π ( (m + nπ (m + n π ] =. d = 1 { ( } nπ 1 d = 1 [ ( ] nπ nπ = The proof of (4. is nogous. for m n, we use ( mπ = 1 { ((m n π + ((m + n π } etc. Note tht the nd functions re so orthogon to ech other (4. ( mπ d =, for m, n. (4.3

3 MAGIC58 & MATH646: Prti Differenti Equtions 3 4. The fu rnge Fourier series Suppose function f(, defined between, stisfies the Dirichet conditions. Then, the series 1 { } + n + b n, where the Fourier coefficients n nd b n re given by n = 1 f( d, n =, 1,,..., (4.4 b n = 1 f( d, n = 1,, 3,..., (4.5 converges to f( if is point of continuity, (4.6 f(++f( if is point of discontinuity. Thus, t point of continuity, we cn write f( = 1 { } + n + b n. (4.7 This is the fu rnge Fourier series representtion of f(. Heuristic proof Remember, the set of functions {(nπ/} is orthogon on the interv [, ]. Mutipying (4.7 by (mπ/ nd integrting (with respect to over the rnge to we hve f( + ( mπ d = 1 { n ( mπ d ( mπ ( mπ d + b n d where we hve ssumed tht the orders of summtion nd integrtion cn be interchnged. Note tht the m = cse is different from the other integrs in (4., which is why the fctor 1/ is required in (4.7. Then, ug (4. nd (4.3, we deduce tht ( mπ f( d = m, m =, 1,,.... Simiry, mutipying (4.7 by (mπ/ nd integrting over the rnge to, we hve ( mπ f( d = 1 ( mπ d ( mπ ( mπ } + { n d + b n d. },

4 MAGIC58 & MATH646: Prti Differenti Equtions 4 Ug (4. nd (4.3, we hve ( mπ f( d = b m, m =, 1,,.... Note tht, throughout the bove work, we hve ssumed tht it is OK to interchnge the order of summtion nd integrtion. In fct, this is ony vid if the sum on the right hnd side of (4.7 converges uniformy to f(. Empe Find the Fourier series representtion of the function defined by ( mπ f( d = b m, m =, 1,,.... f( = The Fourier coefficients re given by {, <, <, f( + = f(. n = 1 b n = 1 Thus, t point of continuity = 1 d =, d = 1 [ ] nπ 1 nπ = (nπ {(nπ 1} = n = d = 1 [ ] nπ + 1 nπ = nπ (nπ d (nπ {( 1n 1}, d = b n = ( 1n nπ. f( = 4 + π { 1 ( πn {( 1n 1} ( 1n nπ } n, but {( 1 n 1} = so we cn rerrnge the series into the form f( = 4 π { for n = 1, 3, 5,..., for n =, 4, 6,..., { ( (n 1π ( π(n 1 + ( 1n nπ } n.

5 MAGIC58 & MATH646: Prti Differenti Equtions 5 y Figure 4.3 An empe of periodic function. This is the Fourier series representtion for f( s sketched bove. Note, t point of discontinuity (i.e. t = ±, ±3,... the sum converges to /. (Proof eft to reder. Points to remember (i The Dirichet conditions re sufficient but not necessry for the Fourier series to converge to f( t point of continuity. There re functions, which we my come cross, which re not Dirichet but do hve Fourier series representtions. (ii In the bove empe, the given function is periodic. If the given function is defined on the interv [, ] ony, then the Fourier series representtion is equ to f( on this interv nd defines periodic etension of f( for. (iii In the bove empe, the given function is defined on the fu rnge [, ] nd is neither even or odd on this interv. Thus, we need the fu rnge Fourier series to represent this function. However, if function is either even of odd then (4.7 reduces to hf-rnge Fourier series. 4.3 Hf-rnge Fourier series Suppose we re given n even function, f(, defined on [, ]. then f( (nπ/ is so even but f( (nπ/ is odd. It foows tht (4.4 nd (4.5 reduce to n = f( d, b n =. (4.8 Thus, the Fourier series representtion for n even function contins no e terms. Aterntivey, if f( is odd, then f( (nπ/ is odd nd f( (nπ/ is even. Hence, (4.4 nd (4.5 reduce to n =, b n = f( d. (4.9 The Fourier series representtion of n odd function contins no ine terms. So, if we re given function, f(, which is defined ony on the hf-rnge interv [, ], then we cn use the bove informtion to construct both n even nd odd periodic etension to f( for. This is wht we did in section 3 for the pucked string.

6 MAGIC58 & MATH646: Prti Differenti Equtions 6 Empe Obtin the hf-rnge ( ine series nd (b e series for the function Sketch both hf-rnge series. ( Hf-rnge ine series f( = <. n = d = [ ] nπ [ = nπ nπ ] nπ d = [(nπ 1], (nπ nd therefore For the cse n =, nd so (4.7 becomes n = (nπ (( 1n 1 for n = 1,, 3,.... = f( = + π d = [ ( 1 n 1 n ] =,, which cn be rerrnged to obtin f( = 4 π The even etension of f( is sketched beow. ( 1 (n 1π (n 1. (4.1 f( y Figure 4.4 An empe of n even periodic etension of function.

7 MAGIC58 & MATH646: Prti Differenti Equtions 7 (b Hf-rnge e series nd therefore which gives b n = d = nπ (nπ, b n = nπ ( 1n for n = 1,, 3,..., f( = π ( 1 n The odd etension of f( is sketched beow f( y n. ( Figure 4.5 An empe of n odd periodic etension of function. Both (4.1 nd (4.11 represent the function f( = on the interv [, ]. On the etension < <, (4.1 nd (4.11 re different s shown. Importnt note: the series (4.1 converges much more rpidy (s 1/n thn (4.11 (s ( 1 n /n. Why is this? Finy, we cn note tht in (4.1, if we put = then we get or = 4 π so this is wy of determining infinite sums. In (4.11, et = /, which gives = π 1 (n 1, 1 (n 1 = π 8, ( 1 n n

8 MAGIC58 & MATH646: Prti Differenti Equtions 8 so = ( 1 n+1 n 1 = π 4. There re mny infinite series which my be determined by this mens. 4.4 Prsev s identity Suppose the Fourier series corresponding to f( converges uniformy to f( on the interv [, ]. Then, This is Prsev s identity. Proof On the interv [, ], we hve 1 {f(} d = + ( n + b n. (4.1 f( = + ( n + b n. Mutipy by f( nd integrte over the rnge to to obtin {f(} d = = + f(d + ( n f( d + b n ( n + b n, ce the integrs inside the sum equte to n nd b n. Prsev s identity is simper if f( is even: f( d {f(} d = + n, (4.13 or if f( is odd: {f(} d = b n. (4.14 Empe Tke the Fourier series empe we used previousy, i.e. f( =, < <.

9 MAGIC58 & MATH646: Prti Differenti Equtions 9 (i Tking the Fourier ine series, we hve then (4.13 gives n = (nπ (( 1n 1, = d = + 4 (nπ 4 (( 1n 1, which, ce (( 1 n 1 = ( 1 n ( 1 n + 1 = (1 ( 1 n, gives, on dividing by, or 3 = π 4 1 (n (n 1 4 = π4 96. A fin note on the theory of Fourier series is tht the Fourier series representtion (4.7 cn be differentited to obtin Fourier series representtion for new function f ( s ong s the sum converges for (see empe sheet. 4.5 Boundry vue probems We now return to ook t some boundry vue probems. First, consider the twodimension diffusion eqution ( u t = ν u + u, (4.15 y which governs the temperture distribution (u(, y, t in two-dimension sb (or rectngur pte. There re two stndrd simpifictions one cn mke. (i Suppose the ter surfces of the sb re insuted, nd the vertic width is sm: y u y = i.e. insuted Figure 4.6 u = y A sb tht is insuted on the top nd bottom surfces. Then, no het fows in the y-direction, nd throughout the br / y =. Thus, (4.15 becomes u t = ν u, (4.16

10 MAGIC58 & MATH646: Prti Differenti Equtions 1 i.e. the one-dimension diffusion eqution. (ii Suppose we re given het conduction probem, e.g. sb insuted on three sides nd heted on the fourth, nd we eve it to sette down to stedy stte. Therefore, eventuy u = u(, y ony so / t =. Then the diffusion eqution reduces to Lpce s eqution u = u + u y = (4.17 We sh sove some boundry vue probems invoving both (4.16 nd (4.17. We usuy sove by seprtion of vribes, s we did with the pucked string. We note tht the possibe boundry conditions on the horizont fces re (i u = f( = given temperture profie on fce, (ii u y = = insuted fce. Simiry, on the vertic fces, either the temperture is specified (s function of y, or it is insuted ( u/ =. Empe A br of ength hs initi temperture distribution u(, = f(. The fces t = nd = re mintined t zero temperture. Find the temperture distribution in the br for t >. The boundry vue probem is written s with u t = ν u, We try seprbe soution which turns the PDE into the form u(, t = u(, t =, u(, = f(. where λ is n rbitrry constnt. Hence, u = X(T (t, T T = λ = ν X X, which gives X + λ ν X =, T + λ T =, T = Ae λt, ( λ ν X = C ( λ ν + D.

11 MAGIC58 & MATH646: Prti Differenti Equtions 11 The boundry condition u(, t = gives X( = C =, nd the boundry condition u(, t = gives ( λ ν X( = D = which impies tht λ ν = nπ = λ = nπ ν, n = 1,, 3,.... Thus, n eigensoution hs the form nd the fu soution is u = u n (, t = A n e n π νt/ u n (, t =, ( A n e n π νt/ nπ. To determine the A n, we need the initi conditions. Hence u(, = nd so, from Fourier series theory A n = Suppose f( = u constnt, then A n = A n = f(, f( d. u d = u nπ (( 1n 1 nd so the soution of the boundry vue probem is u(, t = 4u π ( e {(n 1 π νt/ } 1 (n 1π n 1. Note tht the soution decys to zero s t.

12 MAGIC58 & MATH646: Prti Differenti Equtions 1 Empe A rectngur pte, ength nd bredth b, hs insuted fces. Three edges re hed t zero temperture whist the fourth (on u = hs temperture u(, = f(. Determine the stedy stte temperture distribution in the pte. y u = b u = u = u = u = f( Figure 4.7 A sb tht in insuted on fces with n initi temperture distribution on the bottom fce. The boundry vue probem is written s ong with u + u y = (i u(, y =, (ii u(, y =, (iii u(, b =, (iv u(, = f(. We ook for seprbe soution of the form Substituting this into the PDE we obtin u(, y = X(Y (y. X X = λ = Y Y. Note tht we choose the prmeter to be λ (s before for resons tht wi be cer soon. In this cse X( is periodic but Y (y is not. We hve the foowing two ODEs, X + λ X = = X( = A (λ + B (λ, Y λ Y = = Y (y = C h(λy + D h(λy.

13 MAGIC58 & MATH646: Prti Differenti Equtions 13 We now ppy boundry condition (i to get u(, y = = X( = = A =, nd boundry condition (ii gives u(, y = = X( = = B (λ =, so tht, for non-trivi soution, λ = nπ. Hence, we see why the sign of the constnt ( λ ws chosen to give oscitory behviour in ; otherwise the second condition coud not be stisfied. This gives X( = B. Ug boundry condition (iii gives u(, b = = Y (b = = C h y giving + D h y =, D = C h b h b. Hence, { Y (y = C h y h b ( nπ } h ( h nπb y, or C Y (y = h ( h nπb (b y. Therefore, the eigensoutions re of the form nd the gener soution is u n (, y = A ( nπ n h ( h nπb (b y, u(, y = A ( nπ n h ( h nπb (b y. We finy use the st boundry condition to determine the coefficients A n. Thus, condition (iv gives u(, = A n = f(

14 MAGIC58 & MATH646: Prti Differenti Equtions 14 which, we know from previousy, gives A n = f( d. Tke specific cse, sy f( = u constnt. Then, A n = u d = u nπ (1 ( 1n. So, the soution of the boundry vue probem is u(, y = u 1 π n (1 ( 1n ( nπ h ( h nπb (b y, which cn be rewritten s u(, y = 4u π (n 1 h ( (n 1π ( (n 1π ( (n 1π h (b y. b I. D. Abrhms