ORDINARY DIFFERENTIAL EQUATIONSStudent NotesENGR 351 Numerical Methods for EngineersSouthern Illinois University CarbondaleCollege of EngineeringDr. L.R. ChevalierDr. B.A. DeVantier

Photo Credit: Mr. Jeffrey Burdick

Ordinary Differential Equations…where to use them

The dissolution (solubilization) of a contaminant into groundwater is governed by the equation:

where kl is a lumped mass transfer coefficient and Cs is the maximum solubility of the contaminant into the water (a constant). Given C(0)=2 mg/L, Cs = 500 mg/L and kl = 0.1 day-1, estimate C(0.5) and C(1.0) using a numerical method for ODE’s.

CCkdtdC

sl

A mass balance for a chemical in a completely mixed reactor can be written as:

where V is the volume (10 m3), c is concentration (g/m3), F is the feed rate (200 g/min), Q is the flow rate (1 m3/min), and k is reaction rate (0.1 m3/g/min). If c(0)=0, solve the ODE for c(0.5) and c(1.0)

2kVcQcFdtdcV

Ordinary Differential Equations…where to use them

Before coming to an exam Friday afternoon, Mr. Jones forgot to place 24 cans of a refreshing beverage in the refrigerator. His guests are arriving in 5 minutes. So, of course he puts the beverage in the refrigerator immediately. The cans are initially at 75, and the refrigerator is at a constant temperature of 40.

Ordinary Differential Equations…where to use them

The rate of cooling is proportional to the difference in the temperature between the beverage and the surrounding air, as expressed by the following equation with k = 0.1/min.

Use a numerical method to determine the temperature of the beverage after 5 minutes and 10 minutes.

airTTkdtdT

Ordinary Differential Equations…where to use them

Ordinary Differential Equations• A differential equation defines a

relationship between an unknown function and one or more of its derivatives

• Physical problems using differential equations• electrical circuits• heat transfer• motion• contaminant transport

Ordinary Differential Equations• The derivatives are of the dependent

variable with respect to the independent variable

• First order differential equation with y as the dependent variable and x as the independent variable would be:• dy/dx = f(x,y)

Ordinary Differential Equations• A second order differential

equation would have the form:d ydx

f x y dydx

2

2

, ,}

does not necessarily have to includeall of these variables

Ordinary Differential Equations• An ordinary differential equation is

one with a single independent variable.

• Thus, the previous two equations are ordinary differential equations

• The following is not: dy

dxf x x y

11 2 , ,

Ordinary Differential Equations• The analytical solution of ordinary

differential equation as well as partial differential equations is called the “closed form solution”

• This solution requires that the constants of integration be evaluated using prescribed values of the independent variable(s).

Ordinary Differential Equations• An ordinary differential equation of

order n requires that n conditions be specified.

• Boundary conditions• Initial conditions

consider this beam where the deflection is zero at the boundariesx= 0 and x = LThese are boundary conditions

a

yo

P

In some cases, the specific behavior of a system(s)is known at a particular time. Consider how the deflection of a beam at x = a is shown at time t =0 to be equal to yo.Being interested in the response for t > 0, this is called the initial condition.

Ordinary Differential Equations• At best, only a few differential

equations can be solved analytically in a closed form.

• Solutions of most practical engineering problems involving differential equations require the use of numerical methods.

Scope of Lectures on ODE• One Step Methods

• Euler’s Method• Heun’s Method• Improved Polygon• Runge Kutta• Systems of ODE

• Boundary Value Problems

Specific Study Objectives• Understand the visual representation of

Euler’s, Heun’s and the improved polygon methods

• Understand the difference between local and global truncation errors

• Know the general form of the Runge-Kutta methods

• Understand the derivation of the second-order RK method and how it relates to the Taylor series expansion

Specific Study Objectives• Realize that there are an infinite number

of possible versions for second- and higher-order RK methods

• Know how to apply any of the RK methods to systems of equations

• Understand the difference between initial value and boundary value problems

Review of Analytical Solutiondydx

x

dy x dx

y x C

4

4

43

2

2

3

At this point lets consider initial conditions.

y(0)=1andy(0)=2

y x C

for y

C

then Cfor y

C

and C

430 1

14 0

31

0 2

24 0

32

3

3

3

What we see are differentvalues of C for the twodifferent initial conditions.

The resulting equations are:

y x

y x

43

1

43

2

3

3

0

4

8

12

16

0 0.5 1 1.5 2 2.5x

yy (0)=1y (0)=2y (0)=3y (0)=4

One Step Methods

• Focus is on solving ODE in the form

dydx

f x y ,y

x

xi, yi

One Step Methods

• Focus is on solving ODE in the form

dydx

f x y

y y hi i

,

1

y

x

xi, yi

One Step Methods

• Focus is on solving ODE in the form

dydx

f x y

y y hi i

,

1

y

x

h

This is the same as saying:new value = old value + slope x step size

xi, yi

One Step Methods• Focus is on solving ODE in the form

dydx

f x y

y y hi i

,

1

y

x

slope = dy/dxThis is the same as saying:new value = old value + (dy/dx)(h)where h is the step size

yi

h

One Step Methods• Focus is on solving ODE in the form

dydx

f x y

y y hi i

,

1

y

x

slope = dy/dxyi

h

This is the same as saying:new value = old value + (dy/dx)(h)where h is the step size

yi+1

Euler’s Method

• The first derivative provides a direct estimate of the slope at xi

• The equation is applied iteratively, or one step at a time, over small distance in order to reduce the error

• Hence this is often referred to as Euler’s One-Step Method

Example

24xdxdy

For the initial condition y(1)=1, determine y for h = 0.1 analytically and using Euler’s method given:

STRATEGY

Strategy• Determine the analytical solution based on the

initial conditions y(1) = 1• i.e. x=1, y=1

• Determine xi+1 = xi + h• Recognize that f(x,y) = = 4x2

• yi+1 = yi + 4(xi)2h• Recognize the meaning of “one-step”

• yi+1 = yi + 4(xi)2h• yi+2 = yi+1 + 4(xi+1)2h• yi+3 = yi+2 + 4(xi+2)2h…..and so on

Error Analysis of Euler’s Method• Truncation error - caused by the nature

of the techniques employed to approximate values of y• local truncation error (from Taylor Series)• propagated truncation error• sum of the two = global truncation error

• Round off error - caused by the limited number of significant digits that can be retained by a computer or calculator

Higher Order Taylor Series Methods

• This is simple enough to implement with polynomials

• Not so trivial with more complicated ODE

• In particular, ODE that are functions of both dependent and independent variables require chain-rule differentiation

• Alternative one-step methods are needed

y y f x y h

f x yhi i i i

i i 1

2

2,

' ,

Modification of Euler’s Methods• A fundamental error in Euler’s method is

that the derivative at the beginning of the interval is assumed to apply across the entire interval

• Two simple modifications will be demonstrated graphically in order to give insight on the different strategies that can be employed

• These modification actually belong to a larger class of solution techniques called Runge-Kutta which we will explore and apply later

Heun’s Method• Determine the derivative for the

interval • the initial point• end point

• Use the average to obtain an improved estimate of the slope for the entire interval

y

xi xi+1

Take the slope at xiProject to get f(xi+1 )based on the step size h

h

x

y

Use this “average” slopeto predict yi+1

hyxfyxfyy iiiiii 2

,, 111

{xxi xi+1

y

Use this “average” slopeto predict yi+1

hyxfyxfyy iiiiii 2

,, 111

{xxi xi+1

y

y

xi xi+1

hyxfyxfyy iiiiii 2

,, 111

xxi xi+1

Euler’s

Heun’s

x

y

xxi xi+1

hyxfyxfyy iiiiii 2

,, 111

hyy ii 1

Improved Polygon Method

• Another modification of Euler’s Method

• Uses Euler’s to predict a value of y at the midpoint of the interval

• This predicted value is used to estimate the slope at the midpoint

• We then assume that this slope represents a valid approximation of the average slope for the entire interval

• Use this slope to extrapolate linearly from xi to xi+1 using Euler’s algorithm

Improved Polygon Method

y

xxi

f(xi)

y

xxi xi+1/2 xi+1

h/2

h

y

xxi xi+1/2

h/2

y

xxi xi+1/2

f(xi+1/2)

y

xxi xi+1/2

f’(xi+1/2)

y

xxi xi+1/2 xi+1

h

Extend your slopenow to get f(x i+1)

y

xxi xi+1/2 xi+1

f(xi+1)

Both Heun’s and the Improved Polygon Method have been introduced graphically. However, the algorithms used are not as straight forward as they can be.

Let’s review the Runge-Kutta Methods. Choices in values of variable will give us these methods and more. It is recommend that you use this algorithm on your homework.

Runge-Kutta Methods

Runge-Kutta Methods• RK methods achieve the accuracy of a

Taylor series approach without requiring the calculation of a higher derivative

• Many variations exist but all can be cast in the generalized form:

y y x y h hi i i i 1 , ,{

is called the incremental function

, Incremental Functioncan be interpreted as a representative slope over the interval

a k a k a kwhere the a s are constant and the k s arek f x yk f x p h y q k hk f x p h y q k h q k h

k f x p h y q k h q k h q k h

n n

i i

i i

i i

n i n i n n n n n

1 1 2 2

1

2 1 11 1

3 2 21 1 22 2

1 1 1 1 2 2 1 1 1

' ' :,

,,

, , , ,

a k a k a kwhere the a s are constant and the k s arek f x yk f x p h y q k hk f x p h y q k h q k h

k f x p h y q k h q k h q k h

n n

i i

i i

i i

n i n i n n n n n

1 1 2 2

1

2 1 11 1

3 2 21 1 22 2

1 1 1 1 2 2 1 1 1

' ' :,

,,

, , , ,

NOTE:k’s are recurrence relationships,that is k1 appears in the equation for k2which appears in the equation for k3This recurrence makes RK methods efficient for computer calculations

Second Order RK Methods

y y a k a k hwherek f x yk f x p h y q k h

i i

i i

i i

1 1 1 2 2

1

2 1 11 1

,,

a k a k a kwhere the a s are constant and the k s arek f x yk f x p h y q k hk f x p h y q k h q k h

k f x p h y q k h q k h q k h

n n

i i

i i

i i

n i n i n n n n n

1 1 2 2

1

2 1 11 1

3 2 21 1 22 2

1 1 1 1 2 2 1 1 1

' ' :,

,,

, , , ,

Second Order RK Methods

• We have to determine values for the constants a1, a2, p1 and q11

• To do this consider the Taylor series in terms of yi+1 and f(xi,yi)

y y a k a k h

y y f x y h f x yh

i i

i i i i i i

1 1 1 2 2

1

2

2, ' ,

f x yfx

fy

dydx

substituting

y y f x y hfx

fy

dydx

h

i i

i i i i

' ,

,

1

2

2

Now, f’(xi , yi ) must be determined by thechain rule for differentiation

The basic strategy underlying Runge-Kutta methodsis to use algebraic manipulations to solve for valuesof a1, a2, p1 and q11

y y a k a k h

y y f x y hfx

fy

dydx

hi i

i i i i

1 1 1 2 2

1

2

2,

By setting these two equations equal to each other andrecalling:

k f x y

k f x p h y q k hi i

i i

1

2 1 11 1

,,

we derive three equations to evaluate the four unknown constants

a a

a p

a q

1 2

2 1

2 11

11212

Because we have three equations with four unknowns,we must assume a value of one of the unknowns.

Suppose we specify a value for a2.

What would the equations be?

a a

p qa

1 2

1 112

11

2

Because we can choose an infinite number of valuesfor a2 there are an infinite number of second orderRK methods.

Every solution would yield exactly the same resultif the solution to the ODE were quadratic, linear or a constant.

Lets review three of the most commonly used andpreferred versions.

y y a k a k hwherek f x yk f x p h y q k ha a

a p

a q

i i

i i

i i

1 1 1 2 2

1

2 1 11 1

1 2

2 1

2 11

11212

,,

Consider the following:

Case 1: a2 = 1/2

Case 2: a2 = 1

These two methodshave been previouslystudied.

What are they?

a a

a p

a q

p qa

y y k k h

wherek f x yk f x h y k h

i i

i i

i i

1 2

2 1

2 11

1 112

1 1 2

1

2 1

1 1 1 2 1 21212

12

1

12

12

/ /

,,

Case 1: a2 = 1/2

This is Heun’s Method

Note that k1 is the slope atthe beginning of the interval and k2 is the slope at the end of the interval.

y y a k a k hwherek f x yk f x p h y q k h

i i

i i

i i

1 1 1 2 2

1

2 1 11 1

,,

a a

a p

a q

p qa

y y k hwherek f x y

k f x h y k h

i i

i i

i i

1 2

2 1

2 11

1 112

1 2

1

2 1

1 1 1 01212

12

12

12

12

,

,

y y a k a k hwherek f x yk f x p h y q k h

i i

i i

i i

1 1 1 2 2

1

2 1 11 1

,,

Case 2: a2 = 1

This is the Improved Polygon Method (also called Mid-Point Technique).

Ralston’s Method

Ralston (1962) and Ralston and Rabinowitiz (1978)determined that choosing a2 = 2/3 provides a minimum bound on the truncation error for the second order RKalgorithms.

This results in a1 = 1/3 and p1 = q11 = 3/4

y y k k h

wherek f x y

k f x h y k h

i i

i i

i i

1 1 2

1

2 1

13

23

34

34

,

,

Example

1.0

11..11..

4 2

hsizestepyeixatyCI

yxdxdy Evaluate the following

ODE using Heun’s Methods (e.g. a2 = ½)

STRATEGY

Strategy• Calculate k1=f(x,y) = =

4x2y• Use initial values x=1, y=1

• Calculate the x and y values for k2• x= xi + h• y= yi + 4(xi)2(yi)h

• Calculate k2 = 4(xi+1 )2(yi+1)• Calculate

• yi+1 = yi +0.5(k1 + k2)h• Start process again using

this value of y and x+h as the new initial values

a a

a p

a q

p qa

y y k k h

wherek f x y

k f x h y k h

i i

i i

i i

1 2

2 1

2 11

1 112

1 1 2

1

2 1

1 1 1 2 1 21212

12

1

12

12

/ /

,

,

Third Order Runge-Kutta Methods• Derivation is similar to the one for the second-

order • Results in six equations and eight unknowns.• One common version results in the following

y y k k k h

wherek f x y

k f x h y k h

k f x h y hk hk

i i

i i

i i

i i

1 1 2 3

1

2 1

3 1 2

16

4

12

12

2

,

,

,

Note the third term

NOTE: if the derivative is a function of x only, this reduces to Simpson’s 1/3 Rule

Fourth Order Runge Kutta• The most popular• The following is sometimes called the

classical fourth-order RK method

y y k k k k h

wherek f x y

k f x h y k h

k f x h y hk

k f x h y hk

i i

i i

i i

i i

i i

1 1 2 3 4

1

2 1

3 2

4 3

16

2 2

12

12

12

12

,

,

,

,

• Note that for ODE that are a function of x alone that this is also the equivalent of Simpson’s 1/3 Rule

34

23

12

1

43211

,21,

21

21,

21

,

2261

hkyhxfk

hkyhxfk

hkyhxfk

yxfkwhere

hkkkkyy

ii

ii

ii

ii

ii

Example

Use 4th Order RK to solve the following differential equation:

dydx

xyx

I C y

1

1 12 . .

using an interval of h = 0.1

STRATEGY

Strategy• Calculate k1 = f(x,y)• Determine the x and y value for k2, then

calculate k2

• Determine the x and y value for k3, then calculate k3

• Determine the x and y value for k4, then calculate k4

• Estimate • yi+1 = yi +1/6(k1 + 2k2+2k3+k4)h

Higher Order RK Methods

• When more accurate results are required, Bucher’s (1964) fifth order RK method is recommended

• There is a similarity to Boole’s Rule• The gain in accuracy is offset by

added computational effort and complexity

Systems of Equations• Many practical problems in engineering

and science require the solution of a system of simultaneous differential equations

dydx

f x y y y

dydx

f x y y y

dydx

f x y y y

n

n

nn n

11 1 2

22 1 2

1 2

, , , ,

, , , ,

, , , ,

• Solution requires n initial conditions• All the methods for single equations

can be used• The procedure involves applying the

one-step technique for every equation at each step before proceeding to the next step

dydx

f x y y y

dydx

f x y y y

dydx

f x y y y

n

n

nn n

11 1 2

22 1 2

1 2

, , , ,

, , , ,

, , , ,

Boundary Value Problems

• Recall that the solution to an nth order ODE requires n conditions

• If all the conditions are specified at the same value of the independent variable, then we are dealing with an initial value problem

• Problems so far have been devoted to this type of problem

Boundary Value Problems

• In contrast, we may also have conditions a different value of the independent variable.

• These are often specified at the extreme point or boundaries of as system and customarily referred to as boundary value problems

• To approaches to the solution• shooting method• finite difference approach

General Methods for Boundary Value ProblemsThe conservation of heat can be used to develop a heat balance for a long, thin rod. If the rod is not insulated along its length and the system is at steady state. The equation that results is: d Tdx

h T Ta

2

2 0 '

T1 T2

Ta

Ta

T1 T2

Ta

Ta

d Tdx

h T Ta

2

2 0 '

Clearly this second orderODE needs 2 conditions.This can be satisfied byknowing the temperatureat the boundaries,i.e. T1 and T2

T(0) = T1

T(L) = T2

d Tdx

h T Ta

2

2 0 '

T(0) = T1

T(L) = T2

Use these conditions to solvethe equation analytically.

For a 10 m rod with Ta = 20T(0) = 40T(10) = 200h’ = 0.01

T e ex x 7345 5345 200 1 0 1. .. .

Now that we have an analytical solution, lets evaluate ourtwo proposed numerical methods.

Shooting Method

d Tdx

h T T

dTdx

z

dzdx

h T T

a

a

2

2 0

'

'

Given:

We need an initial valueof z.

For the shooting method, guessan initial value.

Guessing z(0) = 10

dzdx

h T Ta '

Using a fourth-order RK method with a step sizeof 2, T(10) = 168.38

This differs from the BC T(10) = 200

Making another guess, z(0) = 20

T(10) = 285.90

Because the original ODE is linear, the estimates of z(0) are linearly related.

Guessing z(0) = 10

Using a linear interpolation formula between the valuesof z(0), determine a new value of z(0)

Recall:

first estimate z(0) = 10 T(20) = 168.38 second estimate z(0)=20 T(20) = 285.90

What is z(0) that would give us T(20)=200?

150

200

250

300

0 5 10 15 20 25

z(0)

T(20)

z 0 10 20 1028590 168 38

200 168 38 12 69

. .

. .

150

200

250

300

0 5 10 15 20 25

z(0)

T(20)

z 0 10 20 1028590 168 38

200 168 38 12 69

. .

. .

150

200

250

300

0 5 10 15 20 25

z(0)

T(20)

d Tdx

h T T

dTdx

z

dzdx

h T T

a

a

2

2 0

'

'

We can now use this to solve the first order ODE

For nonlinear boundary value problems, linear interpolation will not necessarily result in an accurate estimation. One alternative is to apply three applications of the shooting method and use quadratic interpolation..

0

50

100

150

200

250

0 5 10

distance (m)

TAnaly ticalSolution

ShootingMethod

Finite Difference MethodsThe finite divided difference approximation forthe 2nd derivative can be substituted into the governing equation.

d Tdx

T T Tx

d Tdx

h T T

T T Tx

h T T

i i i

a

i i ia i

2

21 1

2

2

2

1 12

2

0

2 0

'

'

T T Tx

h T T

T h x T T h x T

i i ia i

i i i a

1 12

12

12

2 0

2

'

' '

Collect terms

We can now apply this equation to each interior nodeon the rod.

Divide the rod into a grid, and consider a “node” to beat each division. i.e.. x = 2m

T1 T2

L = 10 m

x = 2 m

aiii TxhTTxhT 21

21 ''2

T(0) T(10)L = 10 m

x = 2 m

Consider the previous problem:L = 10 m Ta = 20T(0) = 40T(10) = 200h’ = 0.01

We need to solve for thetemperature at the interiornodes (4 unknowns). Apply the governingequation at these nodes (4equations).What is the matrix?

aiii TxhTTxhT 21

21 ''2

T(0) T(10)x=0 2 4 6 8 10

i=0 1 2 3 4 5

Notice the labeling for numbering x and i

aiii TxhTTxhT 21

21 ''2

T(0) T(10)x=0 2 4 6 8 10

i=0 1 2 3 4 5

40 200

Note also that the dependent values are known at the boundaries (hence the term boundary value problem)

aiii TxhTTxhT 21

21 ''2

T(0) T(10)x=0 2 4 6 8 10

i=0 1 2 3 4 5

40 200Apply the governing equation at node 1

8.4004.28.004.240

''2

21

21

221

20

TTTT

TxhTTxhT a

aiii TxhTTxhT 21

21 ''2

T(0) T(10)x=0 2 4 6 8 10

i=0 1 2 3 4 5

40 200Apply the equation at node 2

8.004.2

''2

321

232

21

TTTTxhTTxhT a

aiii TxhTTxhT 21

21 ''2

T(0) T(10)x=0 2 4 6 8 10

i=0 1 2 3 4 5

40 200

We get a similar equation at node 3

8.004.2

''2

432

243

22

TTTTxhTTxhT a

aiii TxhTTxhT 21

21 ''2

T(0) T(10)x=0 2 4 6 8 10

i=0 1 2 3 4 5

40 200

8.20004.28.020004.2

''2

43

43

254

23

TTTT

TxhTTxhT a

At node 4, we consider theboundary at the right.

For the four interior nodes, we get the following4 x 4 matrix

48.15954.12478.9397.65

8.2008.08.08.40

04.2100104.210

0104.2100104.2

4

3

2

1

TT

TTTT

0

50

100

150

200

250

0 5 10

distance (m)

TAnaly ticalSolutionShootingMethodFiniteDifference

Example

Consider the previous example, but with x=1. What is the matrix?

Specific Study Objectives• Understand the visual representation of

Euler’s, Heun’s and the improved polygon methods.

• Understand the difference between local and global truncation errors

• Know the general form of the Runge-Kutta methods.

• Understand the derivation of the second-order RK method and how it relates to the Taylor series expansion.

Specific Study Objectives• Realize that there are an infinite number

of possible versions for second- and higher-order RK methods

• Know how to apply any of the RK methods to systems of equations

• Understand the difference between initial value and boundary value problems