ODE

jiangyushan

Pendulum

L

m

mg

As a example of a system that is nonlinear, consider the swinging pendulum shown above. When the mass of the pendulum is small in comparison with the mass m at the end of the pendulum, the equation of motion of the pendulum is as follows.

22

2sin 0 (1.1)

d dmL mLg

dt dt

Here is the angular position of the pendulum measured relative to vertical, m is mass at the end of the pendulum, L is the length of the pendulum, is the coefficient of viscous friction, and is the acceleration due to gravity. Again, this is a second-order equation in the form of (1.1)

g

Pendulum

If we define the state vector to be , then from above (1.1) we get the following first-order system.

[ , / ]Tx d dt

12

21 22

(1.2)

sin ( )

dxx

dtdx g

x xdt L mL

This is an autonomous system because there is no explicit dependence on t. Furthermore, it is nonlinear due to the presence of the term.1sin x

22

2sin 0 (1.1)

d dmL mLg

dt dt

Predator-prey Ecological System

Dynamic systems occur in many fields of study. Consider, for example, the problem of modeling the population levels of a predator-prey pair of species. Let denote the population level of the prey, and let denote the population level of the predator. Suppose and are expressed in units of ,say, thousands. The following simplified model of population growth is referred to as the Lotka-Volterra system.

1x2x

1x 2x

11 1 2 1

22 2 1 2

( )(1.3)

( )

dxb c x x

dtdx

b c x xdt

Here the parameter denotes the normalized growth rate of the prey when the predator is not present . Similarly, denotes the rate at which the predator population decrease in the absence of prey .

The term represents the decrease in the prey population as a result of the actions by the predator,

and the term represents the increase in the predator population as a result of the availability of prey.

Predator-prey Ecological System

1b2( 0)x

2b2( 0)x

1 1 2c x x

2 1 2c x x

11 1 2 1

22 2 1 2

( )(1.3)

( )

dxb c x x

dtdx

b c x xdt

This is a nonlinear system due to the presence of the product terms. As we shall see, it has a periodic solution in which the population levels of the predator and prey go through ecological cycles. It can be shown that amplitude of the cycle depends on the initial conditions, while the period of the cycle is

Predator-prey Ecological System

1 2x x

1 2

2(1.4)T

bb

INITIAL AND BOUNDARY PROBLEM

By itself, a differential equation does not uniquely determine a solution; additional side conditions must be imposed on the solution to make it unique. These side conditions prescribe values that the solution or its derivatives must have at some specified point or points. If all of the side conditions are specified at the same point, then we have an initial value problem, which we call it an Initial Value Problem. If the side conditions are specified at more than one point, then we have a Boundary Value Problem.

Euler’s Method

The example of dynamic systems introduced earlier represent special cases of the following general first-order nonlinear system where is an n×1 vector. ( )x t

( , ), ( ) (1.5)dx

f t x x adt

we restrict our consideration to systems for which the right-hand side function is sufficiently smooth that Equation (1.5) has a unique solution satisfying the initial condition,

Sufficient conditions on to ensure the existence of a unique solution over can be found in (Vid78).

( )x a ( , )f t x

[ , ]

Euler’s Method

We are interested in estimating for where the are equality spaced over the interval .That is, where the step size is

( )kx t 0 k m

kt [ , ] kt kh h

(1.6)hm

Suppose the value of is known. This is certainly true

for because To find in terms

of we multiply both sides of Equation (1.5) by

and then integrate from . This yields the

following reformulation of (1.5) as an integral equation.

( )kx t

0k 0( ) ( )x t x a 1( )kx t

( )kx t

dt1k kt to t

Euler’s Method

11( ) ( ) ( , ( )) (1.7)

k

k

tk k

tx t x t f x d

The problem with applying (1.7) directly is that we do not k

now the value of for ,and without it we c

an not evaluate the integral. However, if the stepsize , is

sufficiently small, we can approximate the integrand over t

he interval ,by it value at the start of

the interval.

( )x 1k kt t

h

[ , ]k kt t h [ ( )]k kf t x t

Euler’s Method

In this case, the integral in (1.7) simplifies to .If

denotes the approximate solution obtained in this

manner, this yields the following solution formula, which

is called Euler’s method.

Euler’s method has a local truncation error of order

and the global truncation error is of order

[ ( )]k khf t x t

( )k kx x t

1 ( , ), 0 (1.8)k k k kx x hf t x k m

2( )O h

( )O h

An Example for Euler’s method

To illustrate the use of Euler’s method, consider the

following simple one-dimensional first-order system.

Here the constant . This is a one-dimensional

linear system whose exact solution is .

Applying Euler’s method in (1.8) ,we have

, (0)dx

cx x adt

0c

( ) exp( )x t a ct

1 ( , ) (1 )k k k k k k kx x hf t x x hcx hc x

An Example for Euler’s method

This difference equation is simple enough that we can write

a closed-form expression for the solution. If then

Recall that the exact solution is a decaying exponential

that approaches zero in the steady state. The Euler

estimate of the solution will go to zero as

approaches infinity only if or

0x a

(1 ) , 0k kx hc a k

k

|1 | 1hc 0 2 /h c

RUNGE-KUTTA METHODS1

2 1

3 2

4 3

1 1 2 3 4

( , )

( / 2, / 2)

( / 2, / 2)

( , )

( 2 2 )6

k k

k k

k k

k k

k k

q f t x

q f t h x hq

q f t h x hq

q f t h x hq

hx x q q q q

The coefficients of the fourth-order Runge-Kutta method a

re chosen to ensure that its local truncation error is of ord

er , and its global truncation error is of order .

5( )O h 4( )O h

Consider the predator-prey equations discussion in (1.3).

For convenience, suppose the parameters of the system

are . Using the fourth-order Runge-Kutta

method to solve this system from to using an

initial condition of

An Example for Runge-Kutta method

11 1 2 1

22 2 1 2

( )(1.3)

( )

dxb c x x

dtdx

b c x xdt

[1,1] , [1,1]T Tb c

0 10

(0) [0.5,0.5]Tx

Objectives

• Know how to convert a higher-order differential equation into an equivalent system of first-order equations.

• Understand the difference between initial and boundary conditions.

• Understand the relationship between local and global truncation error.

• Be able to apply the Runge-Kutta single-step solution methods.

• Know how to adjust the step size to control the local truncation error.

• Understand how ordinary differential equation techniques can be used to solve practical engineering problems.

• Understand the relative strengths and weaknesses of each computational method and know which are most applicable for a given problem.

Exercise: Chemical Reactor

uq

V

qx