Numerical Analysis II. Problem Sheet 12

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1 P. Grohs S. Hosseini Ž. Kereta Spring Term 2015 Numerical Analysis II ETH Zürich D-MATH Problem Sheet 12 Problem 12.1 The Exponential Runge-Kutta Method Consider the exponential Runge Kutta single-step method k i := φ(cha) ( i 1 ) f(u i ) + ha c ij k j, i = 1,..., s, j=1 i 1 u i := y 0 + h a ij k j i = 1,..., s, (12.1.1) Ψ h y 0 := y 0 + h j=1 s b i k i. i=1 for the autonomous differential equation ẏ = f(y), where A := Df(y 0 ) and (12.1a) φ(z) = exp(z) 1. z Show that the two-step exponential Runge Kutta method, with the parameter values c = 1 2, a 21 = α, c 21 = 3 4 α2 α, b 1 = 1 1 3α, b 2 2 = 1 3α, (12.1.2) 2 and α free to be chosen, solves the linear inhomogeneous differential equation ẏ = Ay + g, A R d d, g R d exactly. Solution: We must show that the method applied to the linear inhomogeneous differential equation ẏ = Ay + g, A R d d, g R d gives us the exact solution y(t) = exp(at)y 0 + t 0 exp(a(t τ))g dτ = exp(at)y 0 + (e ta 1)A 1 g. The procedure of the 2 step Runge Kutta method is (for the linear case, Df(y 0 ) = A) u 1 = y 0 k 1 = φ(cha)f(y 0 ) = φ(cha)(ay 0 + g) u 2 = y 0 + ha 21 k 1 k 2 = φ(cha)(f(y 0 + ha 21 k 1 ) + hac 21 k 1 ) = φ(cha)(ay 0 + h (a 21 + c 21 ) }{{} = 3 4 α2 Ak 1 + g), Problem Sheet 12 Page 1 Problem 12.1

2 and 2 Ψ h y 0 = y 0 + h b i k i. i=1 With the parameter values we get c = 1 2, a 21 = α, c 21 = 3 4 α2 α, b 1 = 1 1 3α 2, b 2 = 1 3α 2, ( Ψ h y 0 = y 0 + hφ(cha) (1 1 3α )(Ay g)+ ) 1 3α (Ay hα2 Ak 1 + g) ( = y 0 + hφ(cha) Ay 0 + g + h ) 4 Ak 1 ( = y 0 + hφ(cha) Ay 0 + g + h ) 4 Aφ(chA)(Ay 0 + g) = y 0 + 2(exp (ha/2) I)A 1 (Ay 0 + g + ) h 2 AA 1 (exp (ha/2) I)(Ay 0 + g) = y 0 + 2(exp (ha/2) I)y 0 + 2(exp (ha/2) I)A 1 g+ (exp (ha/2) I)A 1 A(exp ha/2 I)(y 0 + A 1 g) = exp(ah)y 0 + (e ha 1)A 1 g. (12.1b) Implement a MATLAB function function [t,y] = exponentialrk2(f,df,y0,t,h,alpha) which solves the autonomous differential equation ẏ = f(y), y(0) = y 0 with the 2-step Runge Kutta method (12.1.1) and parameter values (12.1.2). HINT: Even though the method (12.1.1) solves autonomous differential equations, write your implementation exponentialrk2 for the general system ẏ = f(t, y) so it is compatible with MATLAB ODE-solvers. Solution: See Listing 12.1 Listing 12.1: Implementation of exponentialrk2.m from subproblem (12.1b) 1 f u n c t i o n [t,y] = ExponentialRK2(f,df,y0,T,h,alpha) 2 % this code solves the differential equation y =F(y) and Jacobi DF using a 3 % 2-stage Exponential Runge-Kutta method. Problem Sheet 12 Page 2 Problem 12.1

3 4 % 5 6 i f nargin < 6; 7 alpha = 1; 8 end 9 10 % Initialize variables 11 N = c e i l ((T(2)-T(1))/h); % number of steps 12 h = (T(2)-T(1))/N; % adjust step size to fit interval length 13 d = s i z e (y0,1); % dimension of solution 14 t = l i n s p a c e (T(1),T(2),N+1); % time grid 15 y = nan(d,n+1); % solution 16 y(:,1) = y0; % set initial value % Compute solution 19 f o r j=1:n 20 y(:,j+1) = ExponentialRK2Step(f,df,y(:,j),t,h,alpha); 21 end end f u n c t i o n y = ExponentialRK2Step(f,df,y0,t,h,alpha) c = 0.5; 28 a21 = alpha; 29 c21 = 0.75*alphaˆ2 - alpha; b = nan( s i z e (y0)); 32 b(2) = 1/(3*alphaˆ2); 33 b(1) = 1 - b(2); DF = df(t,y0); phi = ( expm(c*h*df)-eye( s i z e (DF)) )/(c*h*df); u = nan( s i z e (y0)); 40 k = nan( s i z e (y0)); u(:,1) = y0; 43 k(:,1) = phi*f(t,u(:,1)); u(:,2) = y0 + h*a21*k(:,1); 46 k(:,2) = phi*( f(t,u(:,2)) + h*df*c21*k(:,1) ); y = y0 + h*(b(1)*k(:,1) + b(2)*k(:,2)); end Problem Sheet 12 Page 3 Problem 12.1

4 (12.1c) Write a MATLAB function which determines the convergence rate of the method numerically. For your numerical experiment, solve Zeeman s heartbeat model l = l 3 + α heart l p ṗ = βl (12.1.3) with the parameter values α heart = 3, β = 0.01, initial values (p(0), l(0)) = (0, 1) up to the end point T = 100. Use the step sizes h = 1, k = 2,..., 6, and as a reference solution, use the 2 k results given by ode15s with very high accuracy, f.e. AbsTol=1e-12 and RelTol=1e-12. Solution: See Listing 12.2 for the implementation. See Figure 12.1 and Figure 12.2 for the solution graph and the convergence loglog plot respectively. We get an estimated convergence rate of f u n c t i o n ExponentialRK2Conv Listing 12.2: Implementation of subproblem (12.1c) 2 % this code calculates the order of exponentialrk2 using a reference 3 % solution for Zeeman s Heartbeat model obtained using ode23s. 4 5 a l p h a = 0.75; % free-parameter in ExponentialRK beta = 0.01; 8 heartalpha = 3; 9 10 % Zeeman s Heartbeat model: Example 11, Section f [ -y(1)ˆ3 + heartalpha*y(1) - y(2); 12 beta*y(1)]; df [ -3*y(1)ˆ2 + heartalpha, -1; 15 beta, 0 ]; y0 = [1; 0]; 18 T = [0 100]; 19 h = 2.ˆ-[2:6]; options = odeset( RelTol,1e-12, AbsTol,1e-12, Jacobian,df); 22 [tgood,ygood] = ode23s(f,t,y0,options); disp( ode23s done... ); ygood = ygood. ; % matlab compatibility err = nan( s i z e (h)); f o r j=1: l e n g t h (h) Problem Sheet 12 Page 4 Problem 12.1

5 31 [t,y] = exponentialrk2(f,df,y0,t,h(j), a l p h a); 32 err(j) = norm( y(1,end) - ygood(1,end) ); 33 end f i g u r e ; 36 p l o t (t,y(1,:), ro ); 37 hold on; 38 p l o t (tgood, ygood(1,:), g-, LineWidth,1.1); 39 t i t l e ( Solution to Zeeman s Heartbeat Model using Exponential RK2 ); 40 x l a b e l ( t ); 41 y l a b e l ( y(t) ); 42 legend([ Exponential RK2, h= num2str(h(end))], ode23s ); f i g u r e ; 45 l o g l o g (h,err, bx: ) 46 grid on; 47 t i t l e ( Log-Log Plot of Error against Step-Size for Exponential RK2 ); 48 x l a b e l ( h ); 49 y l a b e l ( error ); fit = p o l y f i t ( l o g (h), l o g (err),1); 52 f p r i n t f ( calculated numerical convergence orders: %f\n,fit(1)); Problem 12.2 Robustness of L-Stable Method In this exercise we investigate the robustness of L-stable 1-step methods for the scalar linear model problem [NUMODE, Eq. (3.1.1)]. Robustness denotes the special property of a numerical integrator, namely that the integration error can be reasonably estimated independently of a parameter in the differential equation. We apply the two step Radau-2-method ( [NUMODE, Eq. (3.6.4)]) to the model problem ẏ = λy, y(0) = 1. (12.2.1) Hence we construct an equidistant mesh with step size h = 1 N on the time interval [0, 1]. (12.2a) Determine an analytic expression for e N (λ) := y N y(1), (12.2.2) by expressing the numerical solution using the stability function ( [NUMODE, Thm ]) of the method. Solution: By [NUMODE, Eq. (3.1.16)] we have y N = (Ψ h ) N y(0) = S(hλ) N y(0) = S( λ N )N, Problem Sheet 12 Page 5 Problem 12.2

6 Solution to Zeeman s Heartbeat Model using Exponential RK2 Exponential RK2, h= ode23s y(t) t Figure 12.1: subproblem (12.1c), Solution graph 10 4 Log Log Plot of Error against Step Size for Exponential RK error h Figure 12.2: subproblem (12.1c), convergence rate plot Problem Sheet 12 Page 6 Problem 12.2

7 where S( ) denotes the stability function of the Radau-2-method. This gives e N (λ) = S( λ N )N e λ. For the stability function [NUMODE, Thm ] implies S(z) = z z z2. (12.2b) Write a MATLAB function which determines a lower bound for by sampling on the interval [0, N]. Solution: See Listing f u n c t i o n e = sup_e(n) e = sup e(n), sup e N (λ) 0 λ N Listing 12.3: subproblem (12.2b) 2 % determine lower bound for the sup over e_n(lambda) for 3 % 0 <= lambda <= N 4 % 5 % Input: 6 % N: number of steps, see above 7 % 8 % Output: 9 % e: lower bound for the sup 10 % 11 % call: 12 % e = sup_e(10); % stability function of the Radau-2- method 15 S (1 + z/3)./ (1 + (z/2-2).* z/3); % generate net for the sampling 18 lambda = l i n s p a c e (0, N, 5000); % find max 21 e = max( abs( S(lambda/N).ˆN - exp(lambda) ) ); (12.2c) Give a constant C, independent of h and λ, such that y N y(1) Ch. (12.2.3) Solution: No C with those properties exists because sup e N (λ) e N (N) = S(1) N e N for N = 1. h 0 λ N Problem Sheet 12 Page 7 Problem 12.2

8 (12.2d) State a 1-step method which when applied to (12.2.1) with uniform time step size h has the property (12.2.3) with C independent of h and λ. Solution: The exponential Euler method has the desired property since its stability function is given by S(z) = exp(z) and hence Problem 12.3 e N (λ) = S( λ N )N exp(λ) = exp( λ N )N exp(λ) = 0. Exact Quadrature on P 2s 2 Implies Positive Weights It is often important to know whether the coefficients b i in the Butcher-scheme of a RK single step method are positive. Let the quadrature formula 1 0 f(x) dx s b i f(c i ) i=1 c i c j be exact for polynomials f P 2s 2. Show that the weights b i are necessarily positive. HINT: Define to the s different supporting points suitable polynomials p i (x) P 2s 2 with p i (x) 0. Solution: We consider the polynomials p k (x) = s i=1,i k (x c i ) 2 0 x. Then we have 0 < 1 0 p k (x) dx = s b i p k (c i ) = b k p k (c k ). i=1 So that b k > 0, because p k (c k ) > 0. Published on 12 May To be submitted by 19 May References [NUMODE] Lecture Slides for the course Numerical Methods for Ordinary Differential Equations, SVN revision # Last modified on May 15, 2015 Problem Sheet 12 Page 8 Problem 12.3