Copyleft 2005 by Media Lab 1 11 1 Ordinary Differential Equations Boundary Value Problems.

Copyleft Copyleft 2005 by Media Lab 2005 by Media Lab 11

Ordinary Differential EquationsOrdinary Differential EquationsBoundary Value ProblemsBoundary Value Problems


14.1 Shooting Method for Solving Linear BVPs14.1 Shooting Method for Solving Linear BVPs

• We investigate the second-order, two-point boundary-value problem of the form

with Dirichlet boundary conditions

Or Neuman boundary conditions

Or mixed boundary condition

),,( yyxfy bxa ,

,)( ay )(by

,)( ay )(by

,)()( 1 aycay )()( 2 bycby



• Simple Boundary Conditions

The approach is to solve the two IVPs

If the solution of the original two-point BVP is given by y(x) = u(x) + Av(x) is found from the requirement that y(b) = u(b) + Av(b) = yb

,),()()( bxaxryxqyxpy

,)()(

),()()(

uxqvxpv

xruxquxpu

,0)(

,)(

av

yau a

.1)(

,0)(

av

au

).()(

)()()( xv

bv

buyxuxy

b

,0)( bv

ba ybyyay )(,)(

)(

)(

bv

buyA

b



• EX14.1

• convert to the pair of initial-value problems

,2y

xy

,10)1( y ,0)2( y

vx

v

ux

u

2

,2

,0)1(

,10)1(

v

u

,0)1(

,0)1(

v

u

,21 ww ,2

2wx

xw

,43 ww ,2

44 wx

w

).()(

)(0)( 3

3

11 xw

bw

bwxwxy



• General Boundary Condition at x = b

The condition at x=b involves a linear combination of y(b) and y’(b)

The approach is to solve the two IVPs

If there is a unique solution, given by

)()()( xryxqyxpy

,)()(

),()()(

uxqvxpv

xruxquxpu

,0)(

,)(

av

yau a

.1)(

,0)(

av

au

).()()(

)()()()( xv

bcvbv

bcubuyxuxy

b

0)()( bcvbv

ba ybcybyyay )()(,)(



• General Boundary Condition at Both Ends of the Interval

the approach is to solve two IVPs

If , there is a unique solution, given by

)()()( xryxqyxpy

,)()( 1 ayaycay .)()( 2 bybycby

,)()(

),()()(

uxqvxpv

xruxquxpu

,1)(

,0)(

av

au

.)(

,)(

1cav

yau a

).()()(

)()()()(

2

2

xvbvcbv

bucbuyxuxy

b

0)()( 2 bvcbv



• Ex 14.4

• Ex 14.5

,11

2

1

2 222

xyx

yx

xy .0)1()1(,1)0( yyy

6/)63( 24 xxxy

,11

2

1

2 222

xyx

yx

xy .3)1()1(,0)0()0( yyyy

12/36/ 24 xxxy


14.2 Shooting Method for Solving NonLinear BVPs14.2 Shooting Method for Solving NonLinear BVPs

• Nonlinear Shooting Based on the Secant Method

use an iterative process based on the secant method presented in Chapter 2

the initial slope ‘t’, begin with u’(a) = t(1) = 0, error is m(1) Unless the absolute value of m(1) is less than the tolerance, we continue by solving eq.

),,,( yyxfy ,)( ayay ,0))(),(( bybyh

).1()2()1(

)2()1()1()(

imimim

itititit



• EX 14.6

y = 1/(x+1)

,2 yyy ,1)0( y .025.0)1()1( yy



• Nonlinear Shooting Using Newton’s Method

begin by solving the initial-value problem

Check for convergence:

if |m| < tol, stop;

Otherwise, update t:

),,,( yyxfy ,)( ayay .)( byby

),,,(),,(''

),,,(

uuxfvuuxvfv

uuxfu

uu

,0)(

,)(

av

yau a

.1)(

,)(

av

tau k

;),( bk ytbum

);,(/1 kkk tbvmtt



• EX 14.8

,][ 2

y

yy

,2)1( y,1)0( y

,][ 12 uuu ,1)0( u .)0( ktu

.][2),,(

,)1(][),,(

1

22

uuuuxf

uuuuxf

u

u

,][2][ 122

uuvuuvv ,0)0( v .1)0( v


14.3 Finite Difference Method for Solving Linear BVPs14.3 Finite Difference Method for Solving Linear BVPs

• Replace the derivatives in the differential equation by finite-difference approximations (discussed in Chapter 11).

• We now consider the general linear two-point boundary-value problem

• with boundary conditions

• To solve this problem using finite-differences, we divide the interval [a, b] into n subintervals, so that h=(b-a)/n. To approximate the function y(x) at the points we use the central difference formulas from Chapter 11:

,),()()( bxaxryxqyxpy

,)(,)( byay

,)1(, 11 hnaxhax n

h

yyxy

h

yyyxy ii

iiii

i 2)(,

2)( 11

211


Finite Difference Method for Solving Linear BVPsFinite Difference Method for Solving Linear BVPs

• Substituting these expressions into the BVP and writing as ____ as and as gives

• Further algebraic simplification leads to a tridiagonal system for the unknowns viz.

• where and .

)( ixp)(, ii xqp ,iq )( ixr ir

iiiii

iiii ryq

h

yyp

h

yyy

2

2 112

11

,,, 11 nyy

,1,,1,2

1)2(2

1 21

21

nirhy

hpyqhy

hp iiiiiii

)(0 ayy )(byyn


Finite Difference Method for Solving Linear BVPsFinite Difference Method for Solving Linear BVPs

• Expanding this expression into the full system gives

,2

1)2( 2

1

,2

1)2(2

1

, 2

1)2(2

1

, 2

1)2(2

1

,2

1 2

1)2(

112

112

21

22

12222

321

21

21

22

32222

11

112

21112

hprhyqhy

hp

rhyh

pyqhyh

p

rhyh

pyqhyh

p

rhyh

pyqhyh

p

hprhy

hpyqh

nnnnnn

nnnnnnn

iiiiiii


Example 14.9 A Finite-Difference ProblemExample 14.9 A Finite-Difference Problem

• Use the finite-difference method to solve the problem

• with y(0)=y(4)=0 and n=4 subintervals.• Using the central difference formula for the second derivative, we find that the differential equation becomes the system.

• For this example, h = 1 ,i =1, y = 0, and i = 3, y = 0. Substituting this values, we obtain

4,x0 ),4( xxyy

1,2,3.i ),4(2

)(2

11

iii

iiii xxy

h

yyyxy

).43(320

),42(22

),41(102

323

2123

112

yyy

yyyy

yyy



• Combining like terms and simplifying gives

• Solving, we find that y_1=13/7, y_2 = 18/7, and y_3 = 13/7.• We note for comparison that the exact solution of this problem is

,33

,4 3

,3 3

32

321

21

yy

yyy

yy

.24)1(2)1(2 2

44

4

44

4

xxeee

ee

ee

ey xx




Example 14.10 A Matlab Script for a Linear FDP.Example 14.10 A Matlab Script for a Linear FDP.• The Matlab script that follows solves the BVP.

cos(3).1.0)3( ,1.0)0( ,22 3eyyyyy function S_linear_FDaa = 0; bb = 3; n = 300;p = 2*ones(1, n-1); q = -2*ones(1, n-1); r = zeros(1, n-1);ya = 0.1; yb = 0.1*exp(3)*cos(3);h = (bb-aa)/n; h2 = h/2; hh= h*h;x = linspace(aa+h, bb, n);a = zeros(1, n-1); b = a;a(1:n-2) = 1 - p(1, 1:n-2)*h2; d = -(2 + hh*q);b(2:n-1) = 1 + p(1, 2:n-1)*h2; c(1) = hh*r(1) - (1+p(1)*h2)*ya;c(2:n-2) = hh*r(2:n-2);c(n-1) = hh*r(n-1) - (1 - p(n-1)*h2)*yb;y = Thomas(a, d, b, c)xx = [aa x]; yy = [ya y yb];out = [xx' yy']; disp(out)plot(xx, yy), grid on, hold onplot(xx, 0.1*exp(xx).*cos(xx))hold off


Example 14.10 A Matlab Script for a Linear FDP.Example 14.10 A Matlab Script for a Linear FDP.


14.4 FDM for Solving Nonlinear BVPs14.4 FDM for Solving Nonlinear BVPs

• We consider the nonlinear ODE-BVP of the form

• Assume that there are constants and such that

• Use a finite-difference grid with spacing and let denote the result of evaluating at using for . • The ODE then becomes the system

• An explicit iteration scheme, analogous to the SOR method

• where and The process will converge for

cos(3).1.0)3( ,1.0)0( ,22 3eyyyyy *

* Q,Q *P*

** |),,(| ),,(0 PyyxfandQyyxfQ yy

,/2 *Ph iff ix )2/()( 11 hyy ii iy

.02

211

i

iii fh

yyy

,2)1(2

1 211 iiiii fhyyyy

0y .ny .2/*2Qhw


Example 14.12 Solving a Nonlinear BVP by Using FDMExample 14.12 Solving a Nonlinear BVP by Using FDM

• Consider again the nonlinear BVP

• We illustrate the use of the iterative procedure just outlined by taking a grid with h = ¼. The general form of the difference equation is

• where

.2)1( ,1)0( ,

2

yyy

yy

,2)1(2

1 211 iiiii fhyyyy

.

)1(4

2)2/()(2

2111

21

211

i

iiii

i

iii yh

yyyy

y

hyyf



• Substituting the rightmost expression for f_i into the equation for y_i, we obtain

• The computed solution after 10 iterations agrees very closely with the exact solution.

.4

22

)1(2

1 2111

21

11

i

iiiiiiii y

yyyyyyyy

function S_nonlinear_FDya = 1; yb = 2; a = 0; b = 1;max_it = 10; n = 4; w = 0.1; ww = 1/(2*(1+w)); h = (b-a)/ny(1:n-1) = 1for k = 1:max_it y(1) = ww*(ya+2*w*y(1)+y(2)+(ya^2-2*ya*y(2)+y(2)^2)/(4*y(1))); y(2) = ww*(y(1)+2*w*y(2)+y(3)+(y(1)^2-2*y(1)*y(3)+y(3)^2)/(4*y(2))); y(3) = ww*(y(2)+2*w*y(3)+yb+(y(2)^2-2*y(2)*yb+yb^2)/(4*y(3)));endx = [ a a+h a+2*h a+3*h b ]; z = [ ya y yb ];plot(x, z), hold on, zz = sqrt(3*x+1); plot(x, zz), hold off



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Copyleft 2005 by Media Lab 1 11 1 Ordinary Differential Equations Boundary Value Problems.

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