Fourier Series Solution of the Heat Equation

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1 Fourier Series Soluion of he Hea Equaion Physical Applicaion; he Hea Equaion In he early nineeenh cenury Joseph Fourier, a French scienis and mahemaician who had accompanied Napoleon on his Egypian campaign, inroduced he idea of expanding funcions in rigonomeric series as a device o solve he hea conducion equaion he had developed in his reaise Theorie Analyique de la Chaleur Analyic Theory of Hea. The subjec was grealy exended beyond Fourier s original conribuions by laer mahemaicians bu has coninued o bear Fourier s name. Our reamen of Fourier series would clearly be incomplee wihou reference o he subjec of hea conducion which led o Fourier s discovery. We consider a long hin bar of hea conducing maerial; he lengh coordinae may be aken o be x in he inerval [0,]. We will suppose ha he specific hea per uni lengh a scalar indicaing he capaciy of a uni lengh of he maerial o hold hea is σ and he hea conduciviy is κ. We le N be a posiive ineger, h = /N, and x k = kh, k = 0,1,2,,N. For he presen discussion we hink of he bar formed ino a ring. Thus x 0 = 0 is idenified wih x n = and we use he wraparound convenion wih respec o indices k lying ouside he range 0 hrough N. We suppose he emperaure in he subinerval I k = [x k 1,x k ] a a given ime can be adequaely approximaed by he scalar funcion T k. The hea conained in I k is hen h σ T k. The hea conduciviy coefficien expresses he relaionship beween he rae of flow of hea and he emperaure differenial per uni lengh, i.e., T x. Since our model is spaially discree so far, we approximae T x x k by he difference quoien T k T k 1 h. The rae of hea flow ino I k is hen h σ dt k d while he flow of hea ino I k from I k+1 and I k 1 is κ h T k+1 T k 1

2 and κ h T k 1 T k, respecively. Assuming hea is conserved we obain h σ dt k = κ d h T k+1 2T k + T k 1. Dividing by h we have σ dt k d = κ T k+1 2T k + T k 1 h 2. If we assume he acual hea disribuion is a funcion Tx, of boh ime,, and space, x, he fracion on he righ, a second difference divided by h 2, may be regarded as an approximaion o 2 T x x 2 k. eing N, h = /N ends o zero and, in he limi, we obain for Tx, he parial differenial equaion σ T x, = κ 2 T x,, x [0,], > 0. x2 If here are exernal hea sources, or losses which we can represen by a funcion fx, he equaion is augmened o he more general form σ T x, = κ 2 T x, + fx,. x2 In addiion o he parial differenial equaion i is necessary o give an iniial hea disribuion. We may suppose he iniial insan corresponds o 0 = 0 and hen sipulae he iniial daa Tx, 0 = T 0 x wih he iniial hea disribuion funcion T 0 x a leas piecewise coninuous as a funcion of x. Sophisicaed mahemaical echniques are required o show ha parial differenial equaions such as hese acually have soluions, ha he soluions are uniquely deermined by he given iniial daa, ec. These quesions are imporan bu we do no pursue hem here; our 2

3 ask is o describe a mehod for soluion of equaions of his ype wih he use of Fourier series echniques. The Mehod of Separaion of Variables e us divide he parial differenial equaion shown earlier by he posiive number σ, define κ/σ α and rename αfx, as fx, again. Then we have T x, = α 2 T x, + fx,. x2 We begin wih he homogeneous case fx, 0. To implemen he mehod of separaion of variables we wrie Tx, = z yx, hus expressing Tx, as he produc of a funcion of and a funcion of x. Using ż o denoe dz d and y, y o denoe dy dx, d 2 y, respecively, we dx 2 obain ż yx = α z y x. Assuming z, yx are non-zero, we hen have ż α z = y x yx. Since he lef hand side is a consan wih respec o x and he righ hand side is a consan wih respec o, boh sides mus, in fac, be consan. I urns ou ha consan should be aken o be non-posiive, so we indicae i as ω 2 ; hus ż α z = y x yx = ω2 and we hen have wo ordinary differenial equaions ż = α ω 2 z, y x = ω 2 yx. We firs deal wih he second equaion, wriing i as y x + ω 2 yx = 0. 3

4 The general soluion of his equaion akes he form yx = c cos ωx + d sin ωx. Since we wan yx o be periodic wih period he choices for ω are ω = 2πk, k = 0,1,2,. The choice k = 0 is only useful for he cosine; cos 0 = 1. Indexing he coefficiens c, d o correspond o he indicaed choices of ω, we have soluions for he y equaion in he forms c 0 a consan; c k cos 2πkx + d k sin 2πkx, k = 1,2,. Now, for each indicaed choice ω = 2πk he z equaion akes he form ż = α π2 k 2 z which has he general soluion 2 z = c exp α π2 k 2 Absorbing he consan c appearing here ino he earlier c k, d k we have soluions of he homogeneous parial differenial equaion in he form 2. Tx, = exp α π2 k 2 2 Tx, = c 0, c k cos 2πkx + d k sin 2πkx,k = 1,2,. Since we are working a his poin wih a linear homogeneous equaion, any linear combinaion of hese soluions will also be a soluion.

5 This means we can represen a whole family of soluions, involving an infinie number of parameers, in he form Tx, = c 0 + exp α π2 k 2 2 c k cos 2πkx + d k sin 2πkx. I should be noed ha his expression is a represenaion of Tx, in he form of a Fourier series wih coefficiens depending on he ime, : Tx, = c 0 + c k cos 2πkx + d k sin 2πkx, where c k = c k exp α π2 k 2, d k = d k exp α π2 k 2 2 The coefficiens c k, d k, k = 1,2,3, in he above represenaion of Tx, remain undeermined, of course, o precisely he exen ha he consans c k, d k remain undeermined. In order o obain definie values for hese coefficiens i is necessary o use he iniial emperaure disribuion T 0 x. This funcion has a Fourier series represenaion T 0 x = a 0 + a k cos 2πkx 2 + b k cos 2πkx, where a 0 = 1 T 0x dx, a k = 2 2πkx cos 0 0 T 0x dx, b k = 2 sin 2πkx 0 T 0x dx. To obain agreemen a = 0 beween our Fourier series represenaion of Tx, 0 and his Fourier series represenaion of T 0 x we require, since exp α π2 k 2 0 = 1, 2 c 0 = a 0, c k = a k, d k = b k, k = 1,2,3,. 5.

6 Thus we have, in fac Tx, = a 0 + exp α π2 k 2 2 a k cos 2πkx + b k sin 2πkx, where a 0, a k, b k, k = 1,2,3, are he Fourier coefficiens of he iniial emperaure disribuion T 0 x. Example 1 In he secion on real Fourier series we have compued he coefficiens of he Fourier series for he funcion fx = π x π on he inerval [0,2π]. Taking his funcion as he iniial emperaure disribuion T 0 x for he spaially periodic hea conducion process discussed above, we have T 0 x = π 2 k = 1 k odd cos kx. πk2 Accordingly, he soluion of he periodic hea equaion wih his iniial emperaure disribuion is Tx, = π 2 k = 1 k odd πk 2 e k2 cos kx. Since he non-negaive odd inegers can be represened as k = 2j + 1, j = 0,1,2,..., we can wrie his wihou he qualifier k odd in he form Tx, = π 2 j=0 π2j e 2j+12 cos2j + 1x. 6

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