Math 257: Finite difference methods
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1 Math 257: Fiite differece methods 1 Fiite Differeces Remember the defiitio of a derivative f f(x + ) f(x) (x) = lim 0 Also recall Taylor s formula: (1) f(x + ) = f(x) + f (x) + 2 f (x) + 3 f (3) (x) +... (2) 2! 3! or, with istead of +: Forward differece f(x ) = f(x) f (x) + 2 2! f (x) 3 f (3) (x) +... (3) 3! O a computer, derivatives are approximated by fiite differece expressios; rearragig (2) gives the forward differece approximatio f(x + ) f(x) = f (x) + O(), (4) where O() meas terms of order, ie. terms which have size similar to or smaller tha whe is small. 1 So the expressio o the left approximates the derivative of f at x, ad has a error of size ; the approximatio is said to be first order accurate. 1 The strict mathematical defiitio of O(δ) is to say that it represets terms y such that ( y ) lim is fiite. δ 0 δ So δ 2 ad δ are O(δ), but δ 1/2 or 1, for istace, are ot. Aother commoly used otatio is o(δ), which represets terms y for which ( y ) lim = 0. δ 0 δ I other words, o(δ) represets terms that are strictly smaller tha δ as δ 0. So δ 2 is o(δ), but δ is ot. 1
2 Backward differece Rearragig (3) similarly gives the backward differece approximatio f(x) f(x ) which is also first order accurate, sice the error is of order. Cetered differece = f (x) + O(), (5) Combiig (2) ad (3) gives the cetered differece approximatio f(x + ) f(x ) 2 = f (x) + O( 2 ), (6) which is secod order accurate, because the error this time is of order 2. Secod derivative, cetered differece Addig (2) ad (3) gives f(x + ) + f(x ) = 2f(x) + 2 f (x) f (4) (x) +... (7) Rearragig this therefore gives the cetered differece approximatio to the secod derivative: f(x + ) 2f(x) + f(x ) 2 = f (x) + O( 2 ), (8) which is secod order accurate. 2 Heat Equatio Dirichlet boudary coditios To fid a umerical solutio to the heat equatio t = 2 u α2, 0 < x < L, (9) x2 u(0, t) = A, u(l, t) = B, u(x, 0) = f(x), (10) approximate the time derivative usig forward differeces, ad the spatial derivative usig cetered differeces; u(x, t + t) u(x, t) t 2 u(x +, t) 2u(x, t) + u(x, t) = α + O( t, 2 ). (11) 2 2
3 t t u k 1 u k u k +1 u 2 1 u 2 0 u 2 N u 2 N+1 u 1 1 u 1 0 u 1 N u 1 N+1 u 0 0 u 0 1 u 0 2 u 0 N x 0 x 1 x 2 x N x = 0 x = L Figure 1: Mesh poits ad fiite differece stecil for the heat equatio. Blue poits are prescribed the iitial coditio, red poits are prescribed by the boudary coditios. For Neuma boudary coditios, fictioal poits at x = ad x = L + ca be used to facilitate the method. This approximatio is secod order accurate i space ad first order accurate i time. The use of the forward differece meas the method is explicit, because it gives a explicit formula for u(x, t + t) depedig oly o the values of u at time t. Divide the iterval 0 < x < L ito N + 1 evely spaced poits, with spacig ; ie. x =, for = 0, 1,..., N, ad divide time ito discrete levels t k = k t for k = 0, 1,.... The seek the solutio by fidig the discrete values From (11), these satisfy the equatios u k t u k = u(x, t k ). (12) = α 2 uk +1 2u k + u k 1 2, (13) or, rearragig, = u k + α2 t ( ) u k u k + u k 1. (14) If the values of u at timestep k are kow, this formula gives all the values at timestep k + 1, ad it ca the be iterated agai ad agai to march forward i time. The iitial coditio gives u 0 = f(x ), (15) 3
4 for all, ad the boudary coditios require u k 0 = A, u k N = B, (16) for all values of k > 0 (equatio (14) oly has to be solved for 1 N 1). Stability Note, this method will oly be stable, provided the coditio is satisfied; otherwise it will ot work. Neuma boudary coditios α 2 t 2 1 2, (17) To apply (0, t) = C, (18) x istead of (10), otice that the cetered differece versio of this boudary coditio would be u(0 +, t) u(0, t) = C, (19) 2 ad therefore u(, t) = u(, t) 2C. (20) x = is outside the domai of iterest, but kowig the value of u(, t) there allows the discretised equatio (14) to be used also for x = 0 ( = 0), ad it becomes 0 = u k 0 + α2 t 2 ( 2u k 1 2u k 0 2C ). (21) So i this case we must solve (14) for 1 N 1, ad (21) for = 0 at each timestep. If there is a derivative coditio at x = L the same procedure is followed ad a equatio similar to (21) must be solved for = N too. A hady way to implemet this type of boudary coditio, which eables the same formula (14) to be used for all poits, is to itroduce fictioal mesh poits for = 1 ad = N +1, ad to prescribe the value u k 1 give by (20) (or the equivalet for u k N+1 ) at those poits. 4
5 3 Wave equatio For the wave equatio, 2 u t = 2 u 2 c2, x2 0 < x < L, (22) u(0, t) = 0, u(l, t) = 0, (23) u(x, 0) = f(x), (x, 0) = g(x), (24) t discretise x ito N + 1 evely spaced mesh poits x =, discretise time ito evely spaced time levels t k = k t, ad seek the solutio at these mesh poits; u k = u(x, t k ). Usig cetered differece approximatios to the two secod derivatives, the discrete equatio is 2u k + u k 1 t 2 This ca be rearraged to give = c 2 uk +1 2u k + k O( 2, t 2 ). (25) = 2u k u k 1 + c2 t 2 ( ) u k u k + k 1, (26) which gives a explicit method to calculate i terms of the values at the previous two timesteps k ad k 1. This method is secod order accurate i space ad time - it is sometimes referred to as the leap-frog method. Boudary coditios To apply Dirichlet boudary coditios (23), the values of 0 ad N are simply prescribed to be 0; there is o eed to solve a equatio for these ed poits. If the boudary coditios are Neuma, o the other had: (0, t) = 0, x (L, t) = 0, (27) x the (26) ca still be solved for = 0 ad = N if we use the cetered differece approximatios to the boudary coditios i order to determie the fictioal values u k 1 ad u k N+1 respectively. At x = 0, for istace, the discretised coditio (27) is u k 1 u k 1 2 = 0, (28) ad therefore u k 1 = u k 1. Similarly u k N+1 = uk N 1. 5
6 t t u k 1 u k u k +1 u k 1 u 2 1 u 2 0 u 2 N u 2 N+1 u 1 1 u 1 0 u 1 N u 1 N+1 u 0 0 u 0 1 u 0 2 u 0 N x 0 x 1 x 2 x N x = 0 x = L Figure 2: Mesh poits ad fiite differece stecil for the wave equatio. Iitial coditios To apply the iitial coditios, ote that to use (26) for the first timestep k = 1, we eed to kow the values of u 0 ad also u 1. The values of u 0 follow from the iitial coditio o u(x, 0); u 0 = f(x ). (29) The values of u 1 come from cosiderig the cetered differece approximatio to the derivative coditio (24); from which u(x, 0 + t) u(x, 0 t) 2 t = g(x), (30) u 1 = u 1 2 tg(x ). (31) Substitutig this ito the discrete equatio (26), ad rearragig, gives the formula u 1 = u c 2 t 2 ( ) u u tg(x ), (32) for the first timestep. Oce the solutio has bee iitialised i this way, all subsequet timesteps ca be made usig (26). 6
7 Stability This method is stable provided c t 1, (33) which is ofte called the Courat-Friedrichs-Levy (or CFL) coditio. 4 Laplace s equatio Laplace s equatio o a square domai is with boudary coditios, for istace, 2 u x + 2 u = 0, 0 < x < L, 0 < y < L, (34) 2 y2 u(0, y) = 0, u(l, y) = 0, u(x, 0) = f(x), u(x, L) = 0. (35) Choose a mesh with spacig i the x directio, y i the y directio (ofte it is sesible to choose = y), so grid poits are x = for = 0, 1,... N, ad y m = m y for m = 0, 1,... M. The seek solutio values u m = u(x, y m ). Usig secod order accurate cetered differeces for the two derivatives, the discrete equatio is u +1 m 2u m + u 1 m 2 ad if = y this simplifies to + u m+1 2u m + u m 1 y 2 = O( 2, y 2 ), (36) u m = 1 4 (u +1 m + u 1 m + u m+1 + u m 1 ) (37) Note this equatio shows that the solutio has the property that the value at each poit (x, y m ) is the average of the values at its four eighbourig poits. This is a importat property of Laplace s equatio. Boudary coditios The boudary coditios (35) determie the values u m o each of the four boudaries, so (37) does ot have to be solved at these mesh poits: u 0 m = 0, u N m = 0, u 0 = f(x ), u M = 0. (38) Neuma boudary coditios ca be icorporated by calculatig values for u 1 m (for istace), as for the heat equatio. 7
8 y = L y M u 0 M u N M u m+1 y u 1 m u m u +1 m u m 1 y 1 u 0 1 y = 0 y 0 u 0 0 u 1 0 u 2 0 u N 0 x 0 x 1 x 2 x N x = 0 x = L Figure 3: Mesh poits ad fiite differece stecil for Laplace s equatio. Jacobi Iteratio Note (37) is ot a explicit formula, sice the solutio at each poit depeds o the ukow values at other poits, ad it is therefore harder to solve tha the previous explicit methods. Oe way to do it, however, is to take a guess at the solutio (eg. u m = 0 everywhere), ad the go through each of the mesh poits updatig the solutio accordig to (37) by takig the values of the previous guess o the right had side. Iteratig this process may times, the successive approximatios will hopefully chage by less ad less, ad the values they coverge to provide the solutio. Give a iitial guess u (0) m, the successive iteratios u (1) m,..., u (k) m, u (k+1) m,..., are give by u (k+1) m = 1 ( ) u (k) +1 m + u (k) 1 m + u (k) m+1 + u (k) m 1. (39) 4 Here the superscript (k) deotes the values for the kth iteratio. The process is cotiued util the chage betwee successive iteratios is less tha some desired tolerace. 8
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