The implicit finite-difference method in matrix terms. Vrije Universiteit Amsterdam

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1 - in matrix terms André Ran Vrije Universiteit Amsterdam 1

2 methods Last week we did the explicit finite, based on forward difference in time. This week we consider so-called implicit finite s, based on a backward difference in time. 2

3 methods The discretization of the heat equation becomes u m+1 n u m n δτ = um+1 n+1 2um+1 n + u m+1 n 1 (δx) 2 for m = 0, 1,, with the initial conditions u 0 n = u(nδx, 0), and with boundary conditions as before at ( N δx, mδτ) and at (N + δx, mδτ). Put α = δτ (δx) 2. Rewrite the equation above as αu m+1 n+1 + (1 + 2α)um+1 n αu m+1 n 1 = um n. Note that we now need to compute the u m n in an implicit manner, we cannot work our way forward in time as we did before. Thus it is now necessary to use matrix-vector methods. 3

4 methods Let us put the equation αu m+1 n+1 + (1 + 2α)um+1 n αu m+1 n 1 = um n. with its initial and boundary conditions in matrix-vector form. Put u m+1 = u m+1 N +1. u m+1 N + 1 This vector contains the u-values on the grid that are at time level (m + 1) δτ, but not on the boundary. Put k = N + + N 1. Then u m+1 R k. 4

5 methods To compute we need to use the boundary points, that is, we need the vector b m = u m N u m N + This should be understood correctly: we want b m and u m to be vectors of the same size k! 5

6 methods With these notations the equation αu m+1 n+1 can be rewritten as + (1 + 2α)um+1 n αu m+1 n 1 = um n. (I + αt)u m+1 = u m + αb m where α > 0 and T is the symmetric tridiagonal Toeplitz matrix T =

7 methods We can now solve the equation: (I + αt)u m+1 = u m + αb m u m+1 = (I + αt) 1 (u m + αb m ). Since we know u 0, which is the given initial condition, we can compute u m+1 u 1 = (I + αt) 1 ( u 0 + αb 0), u 2 = (I + αt) 1 ( u 1 + αb 1) =. (I + αt) 2 u 0 + α(i + αt) 2 b 0 + α(i + αt) 1 b 1, = (I + αt) m 1 u 0 + α m (I + αt) j 1 b m j. j=0 7

8 Error propagation methods Assuming we make no errors in b m, errors we make at time step m, which are denoted by e m propagate as e m+1 = (I + αt) 1 e m. Iterating this: errors we make at time 0 have at time m + 1 the effect e m+1 = (I + αt) m 1 e 0. These will decay as m increases when the eigenvalues of (I + αt) 1 are in the interval ( 1, 1). 8

9 Eigenvalues methods To see this, argue as follows. Since T = T, we have that T = V DV 1, where V is an orthogonal matrix, D is diagonal with real numbers on the diagonal. Then (I + αt) m 1 = V (I + αd) m 1 V 1 Now since D is a diagonal matrix, so is I + αd. So, the right hand side will go to zero as m if and only if all eigenvalues of (I + αt) 1 are in the interval ( 1, 1). We show something stronger. 9

10 methods Recall α = δτ δx 2 > 0. Theorem For all α > 0 the eigenvalues of (I + αt) 1 are in (0, 1). 10

11 methods Recall α = δτ δx 2 > 0. Theorem For all α > 0 the eigenvalues of (I + αt) 1 are in (0, 1). Proof. Since T is positive definite (i.e., Tx, x > 0 for all non-zero x) and α > 0 all eigenvalues of M = I + αt are in (1, ). Indeed, Mx = λx (M I)x = (λ 1)x αtx = (λ 1)x Tx = λ 1 α x. Thus λ > 1. 10

12 methods Recall α = δτ δx 2 > 0. Theorem For all α > 0 the eigenvalues of (I + αt) 1 are in (0, 1). Proof. Since T is positive definite (i.e., Tx, x > 0 for all non-zero x) and α > 0 all eigenvalues of M = I + αt are in (1, ). Indeed, Mx = λx (M I)x = (λ 1)x αtx = (λ 1)x Tx = λ 1 α x. Thus λ > 1. Furthermore, Mx = λx M 1 x = 1 λ x. So all eigenvalues of (I + αt) 1 are in (0, 1). 10

13 methods This is good news. Remember, for the explicit method we had to keep α below 1 2. Now we have stability of the method for all α > 0. So, if we want to improve on the calculations by choosing ten times as many data points in the x-direction (which in financial terms is the variable S), we can do so without having to change δτ as well! However, we pay a price: we have to compute an inverse: (I + αt) 1 or we have to solve the equations (I + αt)u m+1 = u m + αb m. 11

14 Methods methods Ideas: Direct inversion of I + αt is possible in a fast way (based on the fact that it is a tri-diagonal positive definite Toeplitz matrix; via the Gohberg-Semencul formulas or the Levinson algorithm). This will not be discussed. 12

15 Methods methods Ideas: Direct inversion of I + αt is possible in a fast way (based on the fact that it is a tri-diagonal positive definite Toeplitz matrix; via the Gohberg-Semencul formulas or the Levinson algorithm). This will not be discussed. We can use the LU-decomposition of I + αt to solve the matrix-vector equations. This will be discussed today. 12

16 Methods methods Ideas: Direct inversion of I + αt is possible in a fast way (based on the fact that it is a tri-diagonal positive definite Toeplitz matrix; via the Gohberg-Semencul formulas or the Levinson algorithm). This will not be discussed. We can use the LU-decomposition of I + αt to solve the matrix-vector equations. This will be discussed today. We can use iteration methods to solve the equations. This will be discussed next time. 12

17 methods Idea is to write I + αt = L U where L is lower triangular and U is upper triangular. Because of the fact that T is tri-diagonal, L will be bi-diagonal and U as well. Write y 1 u l y 2 u L = , U = l k y k 1 u k l k y k 13

18 methods gives I + αt = L U y 1 = 1 + 2α, l 1 = α/y 1, u 1 = α, l 1 u 1 + y 2 = 1 + 2α y 2 = 1 + 2α α2 y 1, l 2 = α/y 2, u 2 = α. Continue this way: y 1 = 1 + 2α and for all j = 1, 2, y j+1 = 1 + 2α α2 y j, l j = α y j, u j = α. 14

19 methods Now (I + αt)u m+1 = u m + αb m translates to L Uu m+1 = u m + αb m. First solve Lq m = u m + αb m, then solve Uu m+1 = q m. Denote c m = u m + αb m = c 1.. c k, q m = q 1. q k. with k = N + N

20 methods Then Lq m = c m can be solved top-to-bottom: q m 1 = c m 1, q m j = c m j l j q m j 1 = c m j + α y j 1 q m j 1. And from that Uu m+1 = q m can be solved from the bottom up: u m+1 N + 1 = qm k, u m+1 = qm j + αu m+1 j N +1. y j N k y j This is easy to implement. 16

21 methods In the examples we compute the European put, as before with T = 0.5, σ = 0.4, r = 0.02, E = 20. Again, the light blue line is the exact value. The dark blue line is the approximation using the implicit difference. In all figures the left hand graph is the graph of the put, the right hand graph is the graph of its derivative, the. In all cases we compute using an algorithm that solves the heat equation between x = 2 and x = 1.5. In the graphs we take δτ = 10 4, and vary δx. The values for δx are, in this order δx = 0.05, and δx = The corresponding values for α are, respectively, α = 0.04, and α = 1. We see that the value of the put is nicely approximated (actually indistinguishable from the exact value), but the approximation of is poor for large δx. It improves greatly with decreasing δx, and we do not now need to worry about keeping δx below 1 2. In fact, with δx = 0.01 the computed value is indistinguishable from the theoretical value. 17

22 prijs Europese put Approximation of Delta methods prijs Europese put 0 Approximation of Delta Figure 1: δx = 0.05 (above), δx = 0.01 (below), δτ =