unit14.mws

ORDINARY DIFFERENTIAL EQUATIONS POWERTOOL

Unit 14 -- Euler's Method

Prof. Douglas B. Meade

Industrial Mathematics Institute

Department of Mathematics

University of South Carolina

Columbia, SC 29208

URL: http://www.math.sc.edu/~meade/

E-mail: meade@math.sc.edu

Copyright © 2001 by Douglas B. MeadeAll rights reserved

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Outline of Unit 14

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Initialization

> restart;

> with( DEtools ):

> with( plots ):

> with( linalg ):

Warning, the name changecoords has been redefined

Warning, the name adjoint has been redefined

Warning, the protected names norm and trace have been redefined and unprotected

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14.A Explicit Implementation of Euler's Method

Euler's method for the solution of a first-order IVP

Diff(y,x) = F(x,y(x)) , y(x[0]) = y[0]

can be summarized by the formula

x[n+1] = x[n]+h

y[n+1] = y[h]+h*F(x[n],y[n])

where h is the stepsize.

A simple implementation of Euler's method that accepts the function F, initial time x[0] , initial position y[0] , stepsize h , and number of steps N as input would be

> Euler0 := proc( F, x0, y0, h, N )

> local i, L, X;

> X := evalf( [ x0, y0 ] );

> L := X;

> for i from 1 to N do

> X := X + [ h, h*F(op(X)) ];

> L := L, X;

> end do;

> return matrix( N+1, 2, [ L ] );

> end proc;

Euler0 := proc (F, x0, y0, h, N) local i, L, X; X :...
Euler0 := proc (F, x0, y0, h, N) local i, L, X; X :...
Euler0 := proc (F, x0, y0, h, N) local i, L, X; X :...
Euler0 := proc (F, x0, y0, h, N) local i, L, X; X :...

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As a test, compute the Euler's Method solution to

Diff(y,x) = -2*y , y(0) = 1

on the interval [ 0, 1 ] with h =0.1.

> Euler0( (x,y)->-2*y, 0, 1, 0.1, 10 );

matrix([[0., 1.], [.1, .8], [.2, .64], [.3, .512], ...

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A more sophisticated implementation of Euler's method would accept as input the ODE, the initial condition, the interval on which the solution should be computed, and the number of steps. In this case, the implementation could appear as

> Euler := proc( ode, ic, domain, N )

> local h, i, t, y, F, L, X;

> t := lhs(domain);

> y := op(0,lhs(ic));

> h := ( op(2,rhs(domain))-op(1,rhs(domain)) )/N;

> F := unapply( subs( y(t)=_y, solve( ode, diff(y(t),t) ) ), (t,_y) );

> X := evalf( [ op(lhs(ic)), rhs(ic) ] );

> L := X;

> for i from 1 to N do

> X := X + [ h, h*F(op(X)) ];

> L := L, X;

> end do;

> return matrix(N+2,2,[[t,y],L])

> end proc:

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14.A-1 Example 1

The approximate solution to the IVP

> ode1 := diff( y(x), x ) = -2*y(x);

> ic1 := y(0)=1;

ode1 := diff(y(x),x) = -2*y(x)

ic1 := y(0) = 1

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on the interval [0,1] by Euler's Method with 10 subdivisions would be obtained with the command

> Euler( ode1, ic1, x=0..1, 10 );

matrix([[x, y], [0., 1.], [.1000000000, .8000000000...

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14.A-2 Example 2

For a second example, use Euler's Method with N = 2, 4, and 8 subdivisions to find an approximate value for y(2) where

> ode2 := diff( y(t), t )*sin(t) + y(t) = 3;

> ic2 := y(1)=2;

ode2 := diff(y(t),t)*sin(t)+y(t) = 3

ic2 := y(1) = 2

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> a2 := 1;

> b2 := 2;

> N2 := [ 2, 4, 8 ];

a2 := 1

b2 := 2

N2 := [2, 4, 8]

> H2 := map( n->(b2-a2)/n, N2 );

H2 := [1/2, 1/4, 1/8]

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> sol2 := Euler( ode2, ic2, t=a2..b2, N2[1] ):

> Y2[1] := sol2[N2[1]+2,2];

Y2[1] := 2.797608323

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> sol2 := Euler( ode2, ic2, t=a2..b2, N2[2] ):

> Y2[2] := sol2[N2[2]+2,2];

Y2[2] := 2.710599580

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> sol2 := Euler( ode2, ic2, t=a2..b2, N2[3] ):

> Y2[3] := sol2[N2[3]+2,2];

Y2[3] := 2.677425038

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These results can be summarized in a table

> v0 := vector( [ 'h', 'N', 'Y(2)' ] ):

> v1 := vector( [ H2[1], N2[1], Y2[1] ] ):

> v2 := vector( [ H2[2], N2[2], Y2[2] ] ):

> v3 := vector( [ H2[3], N2[3], Y2[3] ] ):

> stackmatrix( v0, v1, v2, v3 );

matrix([[h, N, Y(2)], [1/2, 2, 2.797608323], [1/4, ...

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Implementations of Euler's Method for a first-order system are not significantly different or more difficult, but will not be considered at this time.

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14.B dsolve and Euler's Method

While it is not difficult to implement Euler's Method in Maple, there is no real reason to do so. The dsolve command can be used to obtain approximate solutions to IVPs for first-order ODEs, including systems.

To illustrate, revisit the two examples considered in the previous section.

14.B-1 Example 1 (revisited)

When an explicit table of values is needed, it is necessary to provide a list of values of the independent variable at which the approximate solution should be reported.

> x0 := 0;

> h1 := 0.1;

> N1 := 10;

x0 := 0

h1 := .1

N1 := 10

> x_list := vector( N1+1, i -> x0 + (i-1)*h1 );

x_list := vector([0., .1, .2, .3, .4, .5, .6, .7, ....

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Then, the table of approximate solution values computed using Euler's Method is

> dsolve( { ode1, ic1 }, y(x), type=numeric,

> method=classical, stepsize=h1, value=x_list );

matrix([[vector([x, y(x)])], [matrix([[0., 1.], [.1...

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If the results are to be plotted, then the dsolve and odeplot commands can be used as follows

> sol := dsolve( { ode1, ic1 }, y(x), type=numeric,

> method=classical, stepsize=h1 ):

> odeplot( sol, [x,y(x)], 0..1 );

[Maple Plot]

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14.B-2 Example 2 (revisited)

> sol2 := dsolve( { ode2, ic2 }, y(t), type=numeric, method=classical, stepsize=H2[1] ):

> Y2[1] := eval( y(t), sol2(2) );

Y2[1] := 2.79760832314891060

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> sol2 := dsolve( { ode2, ic2 }, y(t), type=numeric, method=classical, stepsize=H2[2] ):

> Y2[2] := eval( y(t), sol2(2) );

Y2[2] := 2.71059958025902015

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> sol2 := dsolve( { ode2, ic2 }, y(t), type=numeric, method=classical, stepsize=H2[3] ):

> Y2[3] := eval( y(t), sol2(2) );

Y2[3] := 2.67742503870978908

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These results can be summarized in a table

> v0 := vector( [ 'h', 'N', 'Y(2)' ] ):

> v1 := vector( [ H2[1], N2[1], Y2[1] ] ):

> v2 := vector( [ H2[2], N2[2], Y2[2] ] ):

> v3 := vector( [ H2[3], N2[3], Y2[3] ] ):

> stackmatrix( v0, v1, v2, v3 );

matrix([[h, N, Y(2)], [1/2, 2, 2.797608323], [1/4, ...

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14.C Example 3

The final example illustrates the use of the full range of Maple tools to obtain, visualize, and analyze an approximate solution to an IVP obtained by Euler's Method.

Consider the problem of obtaining a solution to the IVP

> ode3 := diff( y(t) , t ) = sin( y(t) ) ;

> ic3 := y(0) = 1;

ode3 := diff(y(t),t) = sin(y(t))

ic3 := y(0) = 1

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on the interval [0,8].

Euler's Method with N = 4 subdivisions yields

> sol3e := dsolve( { ode3, ic3 }, y(t), type=numeric,

> method=classical, stepsize=2 ):

> euler_plot := odeplot( sol3e, [t,y(t)], 0..8, style=line,

> color=BLUE, numpoints=4 ):

> display( euler_plot );

[Maple Plot]

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To determine if this approximation is reasonable, superimpose this solution on the slope field.

> slope_field := DEplot( ode3, y(t), t=0..8, y=0..4, arrows=SMALL ):

> display( [ slope_field, euler_plot ] );

[Maple Plot]

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On the interval [0,2] the Euler solution does not look too bad. However, on [2,4] the Euler solution does not follow the slope field and is a much poorer approximation to the true solution. Note also that the Euler solution repeatedly crosses the equilibrium solution y(t) = Pi . To obtain a reasonable approximation on the entire interval using Euler's Method, a smaller stepsize is required.

> sol3e2 := dsolve( { ode3, ic3 }, y(t), type=numeric,

> method=classical, stepsize=0.5 ):

> euler_plot2 := odeplot( sol3e2, [t,y(t)], 0..8, style=line,

> color=CYAN, numpoints=16 ):

> display( [ slope_field, euler_plot2 ] );

[Maple Plot]

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To complete the evaluation of this approximate solution, the DEplot command is used to include the (Maple-generated approximate) solution curve for this initial condition.

> sol_plot := DEplot( ode3, y(t), t=0..8, [ [ic3] ],

> arrows=NONE, linecolor=GREEN ):

> display( [ slope_field, sol_plot, euler_plot, euler_plot2 ] );

[Maple Plot]

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