Tut7_Sol

MATH3240 Numerical Methods for Differential Equations Tutorial 7 (Mar. 23, 2023) 1 Recall: Most ordinary differential equations of higher order can be converted to a system of first-order ODEs. Consider the following system of first-order ODEs          x 0 1 ( t ) = f 1 ( t, x 1 , x 2 , . . . , x n ) , x 0 2 ( t ) = f 2 ( t, x 1 , x 2 , . . . , x n ) , . . . x 0 n ( t ) = f n ( t, x 1 , x 2 , . . . , x n ) , (1) with initial conditions x 1 ( t 0 ) = x 10 , x 2 ( t 0 ) = x 20 , . . . , x n ( t 0 ) = x n 0 . 1. Taylor-series methods. Choose a step size h , consider the following discrete points: t 0 < t 1 < t 2 < · · · < t m < . . . , t k = t 0 + kh, k = 1 , 2 , 3 . . . , Suppose x i ( t k ) is known. By Taylor expansion, we have x i ( t k +1 ) = x i ( t k + h ) = x i ( t k ) + hx 0 i ( t k ) + · · · + h m m ! x ( m ) i ( t k ) + O ( h m +1 ) , so computing x i ( t k + h ) needs x i ( t k ) , x 0 i ( t k ) , x 00 i ( t k ) , . . . which can be obtained from (1). 2. Autonomous ODEs. If we introduce a new variable x 0 ( t ) = t , (1) becomes X 0 ( t ) = F ( X ) with X =      x 0 ( t ) x 1 ( t ) . . . x n ( t )      , F ( X ) =      f 0 ( X ) f 1 ( X ) . . . f n ( X )      . (2) This system of ODEs is called an autonomous system of ODEs, where t does not appear explicitly in the right- hand side function F ( X ) . 3. Analytical solutions of systems of first-order ODEs. We first discuss some properties about the solutions to the following homogeneous case of ODEs X 0 ( t ) = AX ( t ) . Theorem 1. If λ is an eigenvalue of the matrix A and V is a corresponding eigenvector, then X ( t ) = e λt V is a solution to the ODE X 0 ( t ) = AX ( t ) . Theorem 2. If the n × n matrix A has n linearly independent eigenvectors V 1 , V 2 , . . . , V n corresponding to the n eigenvalues λ 1 , λ 2 , . . . , λ n then the space the space span { X 1 ( t ) , X 2 ( t ) , . . . , X n ( t ) } with X i ( t ) = e λ i t V i , contains all the solutions of the ODE system X 0 ( t ) = AX ( t ) . 1

2 Exercises: Please do star questions during the tutorial and finish the remaining after class. 1. The following implicit one-step method x n +1 = x n + h 2 ( f n +1 + f n ) is used to approximate the solution of x 0 = f ( t, x ) with x ( t 0 ) = x 0 . Assume that the third derivative of the solution is uniformly bounded, that is, | x (3) | ≤ M , for some constant M . (a) Show that the local truncation error τ satisfies | τ | ≤ 5 Mh 2 12 . (b) Further assume x (3) ( t ) is Riemann integrable, show that the local truncation error τ satisfies | τ | ≤ Mh 2 12 . Hint: Use Taylor Theorem with integral remainder. (c) Further assume that | ∂f ∂x | ≤ L . Show that | x ( t 1 ) - x 1 | ≤ Mh 3 6 provided hL ≤ 1 . Hint: Repeat the proof of Theorem 2.1 in the lecture notes. Solution. We have the local truncation error in the following form τ n = h - 1 x ( t n +1 ) - x ( t n ) - h 2 ( x 0 ( t n +1 ) + x 0 ( t n ) ) . (a) By Taylor expansion, x ( t n +1 ) = x ( t n ) + hx 0 ( t n ) + h 2 2 x 00 ( t n ) + h 3 6 x (3) ( ξ 1 ) , x ( t n ) = x ( t n ) , x 0 ( t n +1 ) = x 0 ( t n ) + hx 00 ( t n ) + h 2 2 x (3) ( ξ 2 ) , x 0 ( t n ) = x 0 ( t n ) . Consequently, with the help of | x (3) ( t ) | ≤ M , | τ n | = h 2 6 x (3) ( ξ 1 ) - h 2 4 x (3) ( ξ 2 ) ≤ h 2 6 M + h 2 4 M = 5 h 2 12 M. (b) By Taylor Theorem with integral reminder, we have x ( t n +1 ) = x ( t n ) + hx 0 ( t n ) + h 2 2 x 00 ( t n ) + 1 2 t n +1 t n ( t n +1 - t ) 2 x (3) ( t ) dt, x ( t n ) = x ( t n ) , x 0 ( t n +1 ) = x 0 ( t n ) + hx 00 ( t n ) + t n +1 t n ( t n +1 - t ) x (3) ( t ) dt, x 0 ( t n ) = x 0 ( t n ) . Hence with the help of | x (3) ( t ) | ≤ M , | τ n | = 1 2 h t n +1 t n ( ( t n +1 - t ) 2 - h ( t n +1 - t ) ) x (3) ( t ) dt ≤ M 2 h t n +1 t n ( t - t n )( t n +1 - t ) dt = h 2 12 M. 2

Then we have the following iterative formula:      x 0 ,k +1 = x 0 ,k + h, x 1 ,k +1 = x 1 ,k + hx 2 ,k + h 2 2 ( x 0 ,k sin( x 2 ,k ) - cos( x 1 ,k ) + 25) , x 2 ,k +1 = x 2 ,k + h ( x 0 ,k sin( x 2 ,k ) - cos( x 1 ,k ) + 25) + h 2 2 (sin( x 2 ,k ) + x 0 ,k cos( x 2 ,k ) + sin( x 1 ,k ) x 2 ,k ) with initial value      x 0 , 0 = 0 , x 1 , 0 = 5 , x 2 , 0 = 3 . (c) One iteration: x 0 , 1 = h x 1 , 1 = x 1 , 0 + hx 2 , 0 + h 2 2 ( x 0 , 0 sin( x 2 , 0 ) - cos( x 1 , 0 ) + 25) , = 5 + 3 h + h 2 2 (25 - cos(5)) . x 2 , 1 = x 2 , 0 + h ( x 0 , 0 sin( x 2 , 0 ) - cos( x 1 , 0 ) + 25) + h 2 2 (sin( x 2 , 0 ) + x 0 , 0 cos( x 2 , 0 ) + sin( x 1 , 0 ) x 2 , 0 ) = 3 + h (25 - cos(5)) + h 2 2 (sin(3) + 3 sin(5)) . 3. Consider the following two ODE systems X 0 ( t ) = A i X ( t ) + ω i ( t ) i = 1 , 2 with A 1 =   - 1 0 - 6 0 1 0 0 0 2   ω 1 =   sin t e - t 2   X 1 (0) =   1 4 7   ; A 2 =   - 1 0 0 0 2 0 3 1 2   ω 2 =   t 3 t 2 t   X 2 (0) =   2 5 8   (a) Check whether the diagonalization approach works or not. (b) Solve the systems that the diagonalization approach works. Solution. (a) For the first ODE system A 1 =   - 1 0 - 6 0 1 0 0 0 2   D 1 =   - 1 0 0 0 1 0 0 0 2   P 1 =   1 0 2 0 1 0 0 0 - 1   We get eigenvalues λ 1 = - 1 , λ 2 = λ 3 = 2 and three eigenvectors V 1 =   1 0 0   V 2 =   0 1 0   V 3 =   2 0 - 1   Now we can transform the ODE system into a diagonal system by using the tranformation X ( t ) = P 1 Y ( t ) . The original equation becomes a diagonal Y 0 ( t ) = D 1 Y ( t ) = G ( t ) with initial value:   y 1 (0) y 2 (0) y 3 (0)   = Y (0) = P - 1 1 X (0) =   15 4 - 7   4

where G ( t ) = P - 1 1 ω 1 ( t ) =   4 + sin( t ) e - t - 2   Solving the diagonal system, we obtain   y 1 ( t ) y 2 ( t ) y 3 ( t )   =   23 2 e - t + 4 + sin t - cos t 2 9 2 e t - 1 2 e - t - 8 e 2 t + 1   Finally, the original solution is   x 1 ( t ) x 2 ( t ) x 3 ( t )   = X ( t ) = P 1 Y ( t ) =   23 2 e - t - 16 e 2 t + 6 + sin t - cos t 2 9 2 e t - 1 2 e - t 8 e 2 t - 1   (b) For the second ODE system A 2 =   - 1 0 0 0 2 0 3 1 2   D 2 =   - 1 0 0 0 2 1 0 0 2   P 2 =   1 0 0 0 0 1 - 1 1 0   We get eigenvalues λ 1 = - 1 , λ 2 = λ 3 = 2 and three eigenvectors W 1 =   1 - 3 0   W 2 =   0 0 1   W 3 =   0 0 0   Only two linearly independent eigenvectors exists, the diagonalization approach fails with this ODE system. 5