Ordinary Differential Equations (ode)

Transcription

1 Ordinary Differential Equations (ode) Numerical Methods for Solving Initial condition (ic) problems and Boundary value problems (bvp)

2 What is an ODE? =,,...,, yx, dx dx dx dx n n 1 n d y d y d y In general, f n n 1 n d y Example, y x y 4x dx = dx + 3 dx = 1 st order ODE: f ( x, y)

3 Initial condition (ic) problem ic refers to the type of condition the ODE has. An n th order ODE requires n conditions to have a solution. For an ic problem, the conditions must all be located at the initial independent variable. d y 3 = y t y+ 4 t ; 0< t < 4 dt dt y ( ) dt t= 0 = 0 0 = ( )

4 Numerical method for solving a 1 st order i.c. ode f t y y y dt = = (, ); ( 0) 0 Euler s (pronounced oiler s method) Derivation Consider Taylor s series, truncated after the nd term y t 0 t y t 0 t dt t= t ( + Δ ) y ( ) + ( Δ ) 0

5 Euler s method derivation (cont d) But, the derivative is given by the ode, f ( x, y ) ; y ( 0 ) y0 dx = = so, ( + Δ ) ( ) +, ( ) ( Δ ) y t0 t y t0 f t y t0 t At the t=0, the ic i.c. gives y 0 and f(0,y 0 ) which allows us to determine y ( t + Δ 0 t) And, subsequently, y( t + Δt) and so on 0

6 Example, dt IC problem, ( t ) y; y ( ) = 1 + ; 0 = 1 Analyticalsolution (forcomparison purposes) ( 1 t) dt; y( 0) 1 y = + = ( ) ( ) y = t+ t + C; y 0 = 1 C = 1 y t t 4 = + +

7 Solve using Euler s method ( +Δ ) ( ) +, ( ) ( Δ ) y t0 t y t0 f t y t0 t ( +Δ ) ( ) + ( + ) ( ) ( Δ ) y t0 t y 0 1 t y 0 t If Δ t = 0.1 ( t) ( ) ( ) yt0+δ = 1.1

8 Continue using MATLAB, >> y(1) = 1; >> t = 0; >> y(n+1) = y(n) + (1+*t)*sqrt(y(n))*0.1 y = >> n = n+1; >> t = t+0.1; >> y(n+1) = y(n) + (1+*t)*sqrt(y(n))*0.1 y = >> t = t+0.1; >> n = n+1; >> y(n+1) = y(n) + (1+*t)*sqrt(y(n))*0.1 1 y = >> t = t+0.1; >> n = n+1; >> y(n+1) = y(n) + (1+*t)*sqrt(y(n))*0.1 y = >> t = [0:.1:.4] t = >> yana = (t.^ + t + ).^/4 yana = >> plot(t,y,'+',t,yana,'o')

9 Solution of ic ode, (Euler s method, h = 01) 0.1) ( 1 t) y; y( 0) 1 dt = + =

10 Computer program for Euler s method function [t,y] = eulers(f, h, y0, t0, tf) i= 1; t(i) = t0; y(i) = y0; while t(i) < tf y(i+1) = y(i) + f(t(i),y(i))*h; t(i+1) = t(i)+h; i= i+1; end

11 From the command window >> f >> h = 0.01; >> t0 = 0; >> tf = ; >> y0 = 1; >> [t,y] = eulers(f, h, y0, t0, tf) >> yana = (t.^ + t + ).^/4; >> plot(t,y,'+',t,yana,'o')

12 Solution of ic ode, (Euler s method, h = 0.01) ( 1 t) y; y( 0) 1 dt = + =

13 Heun s method (Improvement on Euler s method.) We have seen that Euler s method can be made more by decreasing the size of h. Strategy Another way to improve accuracy is to use the slope, /dx, at the right side of the interval as well as at the left side.

14 y Euler s dx + x i 1 y Heun s dx x i + dx x dx i xi + 1 X i X i+1

15 Heun s method takes the average of the slopes on the left and right sides of the interval using two step method known as a predictor corrector. Predictor y predict ( t0 + Δt) y( t0) + f t, y( t0) ( Δt) Corrector ( ) ( ) { f ( ) ( ) } t, y t0 + f t+δ t, ypredict t0 +Δt ( ) ycorrect t0 +Δt y t0 + Δt

16 Example dt IC problem, ( t ) y; y ( ) Analytical, = 1+ ; 0 = 1 1 y t t 4 = + + ( + Δ ) ( ) +, ( ) ( Δ ) y predict t0 t y t0 f t y t0 t ( + Δ ) ( ) + ( + ) ( ) ( Δ ) y t0 t y 0 1 t y 0 t predict

17 As before, t y( ) Δ = 0.1, 0 = 1 ( ) ( ) ( ) y predict t0 + Δt = 1.1 ( ) ( ) { f ( ) ( ) } t, y t0 + f t+ Δ t, ypredict t0 + Δt ( ) ycorrect t0 +Δt y t0 + Δt ( ) ( ) {( 1+ t ) ( ) ( ) } 0 y t0 +Δt y predict Δt ycorrect ( t0 +Δt) y( t0) + ( Δt ) ( ) ( ) {( 1+ 0) ( ) 1.1} ( ) ycorrect t0 +Δt = 1.113

18 MATLAB to solve some more steps >> n = 1; >> t = 0; >> y(1) = 1; >> yp(n+1) = y(n)+(1+*t)*sqrt(y(n))*0 ( yp = >> y(n+1) = y(n) + ((1+*t)*sqrt(y(n))+(1+*(t+0.1))*sqrt(yp(n+1)))/*0.1 y = >> n = n+1; >> t = t+0.1; >> yp(n+1) = y(n) + (1+*t)*sqrt(y(n))*0.1 yp = ( 1) ( ) ((1 *t)* t( ( )) (1 *(t 0 1))* t( ( 1)))/*0 1 >> y(n+1) = y(n) + ((1+*t)*sqrt(y(n))+(1+*(t+0.1))*sqrt(yp(n+1)))/*0.1 y = Etc >> y(n+1) = y(n) + ((1+*t)*sqrt(y(n))+(1+*(t+0.1))*sqrt(yp(n+1)))/*0.1 y =

19 MATLAB (cont d) >> t = [0:.1:.4] t = >> yana = (t.^ + t + ).^/4 yana = >> plot(t,y,'+',t,yana,'o')

20 Solution of ic ode, (Heun s method, h = 01) 0.1) ( 1 t) y; y( 0) 1 dt = + =

21 Runge Kutta methods Heun s method could be represented as where 1 1 yi+ = yi + k + k k = ( i, i) (, ) f t y k = f t + h y + kh i i 1 This is known as a nd order Runge Kutta method.

22 Extending this concept, the classical l l4 th order Runge Kutta method can be devised. where 1 ( ) y = i 1 y + + i k1 k k3 k4 h ( i i) ( i, i ) ( i, i ) (, ) k1 = f t, y k = f t + h y + kh 1 k = f t + h y + kh 3 k = f t + h y + k h 4 i i 3

23 Example problem 1 ; 0 1 dt = + = Solve ( t) y y( ) using the 4 th order Runge Kutta method. Use a step size of h = 0.4. Define k i for I = 1,,3,4 = ( + ) = ( + ) = k1 1 t y k = 1+ t+ h yi + kh 1 = = k3 = 1+ t+ h yi + kh = = k4 = 1 + t + h yi + k3 h = ( 1 + ( ) ) =

24 Plug k values, y i, and h into 1 y = i 1 y + + i ( k1 k k3 k4) h ( ) y i + 1 = = Heun method: k k 1 = 1.0 ( ) = ( = y i + 1 = = Analytical: y i+1 =

25 Using MATLAB function, ODE45 MATLAB has a number of ic ode solvers. One of the more popular solvers is known as ode45. To learn more about this particular solver as well as others, use the MATLAB help facility.

26 In a nutshell As a simple example, let s use ode45 to solve, Steps: ( 1 t) y; y( 0) 1 dt = + = 1. Create m file that conveys the ode and ic(s) to MATLAB. Call the m file from the command window.

27 Example 1 st order ode (higher order odes can also be solved) ( 1 t) y; y( 0) 1 dt = + = M file (lifted from MATLAB example) function = example1(t,y) = zeros(1,1); 1) % a column vector (1) = (1+*t)*sqrt(y(1));

28 Call m file from MATLAB command window [TY] [T,Y] = ode45(@example1,[0 0.4],[1]); 04][1]); where [T,Y] example1 [0 04] 0.4] is the output from ode45 is the name of the m file tspan is the length of time over which the ode is solved [1] is the value of the i. c., y(0)

29 MATLAB ode45 solution >> [T,Y] = ode45(@example1,[0 0.4],[1]); >> size(y) ans = 41 1 >> Y(41,1) ans = Whi h i th th lti l t l tt Which is the same as the analytical, at least to four places.

30 Higher order ODE ic problems MATLAB ode45 can also solve higher h order ode s as well as systems of ode s with multiple dependant variables. Consider a second order ode with two ic s. d y ( 0) + + y = e y( ) = = dt dt dx 1.5t , 0 1, 0 This equation is non homogeneous and linear, but ode45 can solve non linear ode s as well.

31 Converting an n th order ode to n 1 st order odes. Generic n th order ode, =,,...,,, y, t n n 1 n d y d y d y d y f n n 1 n dt dt dt dt dt Define variables y 1, y,, y n 1, y n y y y 1... y = y 1 = dt d y1 3 = = n dt dt n d y = = n dt dt n 1

32 Or, dt dt... 1 dt n dt = y = y n 1 = = 3 y n f( y, y,..., y, y, t) n n 1 1 The nth order n n 1 n ode, f d y d y d y d y =,,...,,,, n y t n 1 n dt dt dt dt dt has been converted into n 1 st order ode s.

33 Example, Consider the previous example ode with ic s, d y ( ) 1.5t y = 10 e, y ( 0 ) = 1, = 0 dt dt dx Define, y y ( 1) ( ) y ( 1 ) dx

34 Rearrange ode, d y ( 0 ) t 0 = 3.5y 10 e, y ( 0) = 1, = 0 dt dt dx Convert to two 1 st order ode s ( 1 ) dx ( ) dt ( ) ( ) = y ;@ t = 0, y 1= t ( ) ( ) t ( ) = 3y.55 y 1 10 e t = 0, y = 0

35 Create m file for system of 1 st order odes function = example(t,y) = zeros(,1); % a column vector (1) = y(); () = 3*y().5*y(1) 10*exp( 1.5*t); Comparing this m file to that from the 1 st example, we see the in zeros(,1) indicates that there are 1 st order odes to solve. The variable (1) means the derivative of variable y(1) with respect to t.

36 In the command window, enter the statement, [T,Y] = ode45(@example,[0 10],[1,0]); Also check out the size of the output, Y. >> [TY] [T,Y] = ode45(@example,[0 10],[1,0]); 0]) >> size(y) ans = 105 Y(105 ) has the solution at 105 time steps Y(105,) has the solution at 105 time steps for both y(1) and y(), y and y.

37 Analytical solution of the ode. Solve d y ( 0) + + y = e y( ) = = dt dt dx 1.5t , 0 1, 0 ( ) ( ) y t = + t t e 1.5t

38 Compare numerical solution to analytical >> plot(t,y(:,1),'o') >> grid on >> hold on >> t = linspace(0,10); >> y = ( *t 5*t.^).*exp( 1.5*t); >> plot(t,y) solution of ode, y" + 3y' +.5y = -10exp(-1.5t); y(0) = 1, y'(0) = 0 numerical analytical y(t) t

39 Two Point Boundary Value Problems (pbvp) With a higher order ode ( or greater), boundary conditions can be set at the lower end of the interval and also at the higher end of the interval. For example, d y y = 100sinh 5 dx dx 1 y( 0) = 0; y = 5 ( x )

40 For pbvp s, the boundary condition has to be satisfied at both y(0) and y(x end ). Two methods for solving the pbvp. 1. Shooting method.. Finite difference method.

41 Shooting method: Problem: ode has boundary conditions, y(0) and y(x f ), but Runge Kutta method requires initial conditions, y(0) & y (0) (0). Strategy: vary y (0) until the correct boundary condition for y(x f ) = y f is obtained.

42 Analogy: Firing a mortar. The angle of the mortar relative to the vertical is varied until the projectile hits the target. Technique: Vary the y (0) until y(x f ) = y f. Rootfinding problem, find y (0) such that f(y (0)) = y(x f ) y f = 0. Use the secant method to speed up convergence. k+ 1 ( 0) y ( 0) y = k ( 1 y ( 0 ) y ( 0 ) ) f y ( 0 ) k k ( k ) ( k ( ) ) ( k 1 0 ( 0) ) f y f y

43 Example: Consider the ode, pbvp: d y y = 100sinh 5 dx dx y 1 ( 0) = 0; y = 5 ( x) The analytical solution is ( ) y = x e + e x e x x x

44 For the numerical solution, solve using ode45 as if the problem is an ic ode, but iterate as described above to determine the required y (0) that causes y(0.5) = 5. The m file that describes the ode: function = example(x,y) = zeros(,1); % a column vector (1) = y(); () = 10*y() 5*y(1) + 100*sinh(5*x);

45 Thecommand window, >> xp0 = 0; >> xp1 = 10; %this is and initial guess for y (0) >> Yright = 5; >> toll = ; >> [xp, Yrhs, iter] = shoot(xp0,xp1,yright,toll) xp = Yrhs = iter =

46 A program employing the secant method, function [xp, Yrhs, iter] = shoot(xp0,xp1,yright,toll) [X,Y] = ode45(@example,[0.5],[0 xp0]); Y0 = Y(size(Y,1),1) Yright; [X,Y] = ode45(@example,[0.5],[0 xp1]); Y1 = Y(size(Y,1),1) Yright; iter = 0; while abs(y1 Y0)>tollY0)>toll iter = iter + 1; xp = xp1 Y1*(xp1 xp0)/(y1 Y0); [X,Y] = ode45(@example,[0.5],[0 xp]); Y = Y(size(Y,1),1) Yright; xp0 = xp1; xp1 = xp; Y0 = Y1; Y1 = Y; end plot(x,y(:,1)) % Plot the numerical result. xp = xp; Yrhs = Y + Yright;

47 Plot the analytical, >> x = linspace(0,.5,10) >> y = ( *x).*exp( 5*x) + exp(5*x)/ 5*x.^.*exp( 5*x) >> hold on >> plot(x,y,'+') '')

48 6 solution of y'' + 10y' + 5y = 100sinh(5x); y(0) = 0, Y(0.5) = 5 5 numerical analytical 4 3 y x

49 Finite difference method (for linear pbvp ode s) Basic strategy: 1. Consider a grid consisting of n nodes. Substitute finite difference expressions for derivatives in ode. This yields a system of n linear algebraic equations. 3. Solve for the resulting matrix for y(x).

50 Example: Given: d y y = 100sinh 5 dx dx 1 y( 0) = 0; y = 5 ( x ) i=1 i=n At nodes 1 and n, b.c. s are given. At nodes n 1, apply finite it difference version of ode.

51 Finite it difference ode: d y y = 100sinh 5 dx dx 1 y( 0) = 0; y = 5 ( x ) yi 1 yi + yi + 1 yi+ 1 yi y 100sinh 5 i = Δx Δx y1 = 0; y n = 5 Multiply by Δx ( x ) ( ) ( ) y y + y + 5Δx y y + 5Δ x y = 100Δx sinh 5x i 1 i i+ 1 i+ 1 i 1 i i y1 = 0; y n = 5 i

52 Solve for y i in terms of everything else: ( ) ( ) ( ) ( ) y 5Δ x y = y + y + 5Δx y y 100Δx sinh 5x y = i i i i 1 i+ 1 i+ 1 i 1 i y + y + 5Δx y y 100Δx sinh 5x 5Δx i 1 i+ 1 i+ 1 i 1 i Solve for y i using excel: 1 st, set Δ x = 0.05 to create 10 intervals between x = 0 and x = 0.5. Note that n = 11.

53 nd, define y i in nodes n = to n = 10 in excel in terms of x i, y i 1, and y i+1 i1using finite difference equation. The formula for n = is =(B3+D3+5*(C B)*(D3 B3) 100*(C B)^*SINH(5*C))/( 5*(C B)^) B)*(D3 B3) B)^*SINH(5*C))/( 5*(C B)^) Copy this formula to cells You will alsoneed to turn on excel s facility to iterate: MS Office button, excel options, formulas, enable iterative calculations.

54 Result: Analytical: for cell C4: =( *C 0.5)*EXP( 5*C)+0.5*EXP(5*C) 5*C^*EXP( 5*C) i xi yi yana

55 Result: Plot: yi yana