ORDINARY DIFFERENTIAL EQUATIONS
ENGR 351
Numerical Methods for Engineers
Southern Illinois University Carbondale
College of Engineering
Dr. L.R. Chevalier
Dr. B.A. DeVantier
Ordinary Differential
Equations…
where to use them
The dissolution (solubilization) of a contaminant into
groundwater is governed by the equation:
dC
 kl C s  C 
dt
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.
Ordinary Differential
Equations…
where to use them
A mass balance for a chemical in a completely mixed
reactor can be written as:
dc
V
 F  Qc  kVc 2
dt
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)
Ordinary Differential
Equations…
where to use them
Before coming to an exam Friday afternoon, Mr.
Bringer forgot to place 24 cans of a refreshing
beverage in the refrigerator. His guest 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.
dT
  k T  Tair 
dt
Use a numerical method to determine the temperature
of the beverage after 5 minutes and 10 minutes.
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
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:
d2y
dy 

 f  x, y, 
2

dx 
dx
}
does not necessarily have to include
all 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
 f  x1 , x2 , y
dx1
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
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 boundaries
x= 0 and x = L
These are boundary conditions
consider this beam where the
deflection is zero at the boundaries
x= 0 and x = L
These are boundary conditions
P
a
yo
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
• Adaptive step size control
Scope of Lectures on ODE
• Boundary value problems
• Case studies
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 secondorder 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 higherorder 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 Solution
dy
 4x2
dx
2
dy

4
x
  dx
3
4x
y
C
3
At this point lets consider
initial conditions.
y(0)=1
and
y(0)=2
4x3
y
C
3
for y0  1
40
1
C
3
then C  1
3
for y0  2
40
2
C
3
and C  2
3
What we see are different
values of C for the two
different initial conditions.
The resulting equations
are:
4x3
y
1
3
4x3
y
2
3
16
y (0 )=1
y (0 )=2
12
y (0 )=3
y
y (0 )=4
8
4
0
0
0 .5
1
x
1 .5
2
2 .5
One Step Methods
• Focus is on solving ODE in the form
dy
 f  x , y
dx
yi 1  yi  h
h
y
yi+1
yi
slope = 
x
This is the same as saying:
new value = old value + slope x step size
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
For the initial condition y(1)=1, determine y
for h = 0.1 analytically and using Euler’s
method given:
dy
2
 4x
dx
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
Ex a m ple
12
A n a ly t ica l
Solu t ion
10
y
8
Nu m er ica l
Solu t ion
6
4
2
0
0
0 .5
1
1 .5
2
2 .5
x
....end of example
Higher Order Taylor Series
Methods
yi  1
f '  xi , yi  2
 yi  f  xi , yi  h 
h
2
• 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
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
• These modification actually belong to a
larger class of solution techniques called
Runge-Kutta which we will explore
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
Use this “average” slope
to predict yi+1
xi
xi+1
{
yi 1  yi 
f  xi , yi   f xi 1 , yi 1 
h
2
y
f  xi , yi   f xi 1 , yi 1 
yi 1  yi 
h
2
y
xi
xi+1
xi
xi+1
x
f  xi , yi   f xi 1 , yi 1 
yi 1  yi 
h
2
y
yi 1  yi  h
xi
xi+1
x
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
Improved Polygon Method
• 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
Runge-Kutta Methods
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
and/or programming assignments.
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:
{
yi 1  yi    xi , yi , hh
 is called the incremental function
, Incremental Function
can be interpreted as a representative
slope over the interval
  a1k1  a2 k 2  an k n
where the a ' s are constant and the k ' s are:
k1  f  xi , yi 
k 2  f  xi  p1h , yi  q11k1h
k 3  f  xi  p2 h , yi  q21k1h  q22 k 2 h

k n  f  xi  pn h , yi  qn 1,1k1h  qn 1, 2 k 2 h  qn 1, n 1k n 1h
  a1k1  a2 k 2  an k n
where the a ' s are constant and the k ' s are:
k1  f  xi , yi 
k 2  f  xi  p1h , yi  q11k1h
k 3  f  xi  p2 h , yi  q21k1h  q22 k 2 h

k n  f  xi  pn h , yi  qn 1,1k1h  qn 1, 2 k 2 h  qn 1, n 1k n 1h
NOTE:
k’s are recurrence relationships,
that is k1 appears in the equation for k2
which appears in the equation for k3
This recurrence makes RK methods efficient for
computer calculations
Second Order RK Methods
yi  1  yi   a1k1  a2 k 2 h
where
k1  f  xi , yi 
k 2  f  xi  p1h, yi  q11k1h
  a1k1  a2 k 2  an k n
where the a ' s are constant and the k ' s are:
k1  f  xi , yi 
k 2  f  xi  p1h , yi  q11k1h
k 3  f  xi  p2 h , yi  q21k1h  q22 k 2 h

k n  f  xi  pn h , yi  qn 1,1k1h  qn 1, 2 k 2 h  qn 1, n 1k n 1h
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)
yi  1  yi  a1k1  a2 k 2 h
h2
yi  1  yi  f  xi , yi h  f '  xi , yi 
2
Now, f’(xi , yi ) must be determined by the
chain rule for differentiation
f f dy
f '  xi , yi  

x y dx
substituting
 f f dy  h 2
yi  1  yi  f  xi , yi h   

 x y dx  2
The basic strategy underlying Runge-Kutta methods
is to use algebraic manipulations to solve for values
of a1, a2, p1 and q11
yi 1  yi  a1k1  a2 k2 h
 f f dy  h 2
yi 1  yi  f  xi , yi h   

 x y dx  2
By setting these two equations equal to each other and
recalling:
k1  f  xi , yi 
k2  f  xi  p1h, yi  q11k1h
we derive three equations to evaluate the four unknown
constants
a1  a2  1
1
2
1
a2 q11 
2
a2 p1 
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?
a1  1  a2
1
p1  q11 
2a2
Because we can choose an infinite number of values
for a2 there are an infinite number of second order
RK methods.
Every solution would yield exactly the same result
if the solution to the ODE were quadratic, linear or a
constant.
Lets review three of the most commonly used and
preferred versions.
y i 1  y i   a 1 k 1  a 2 k 2  h
Consider the following:
where
k 1  f  xi , yi 
k 2  f  xi  p1h, yi  q11 k1h
a1  a 2  1
1
a 2 p1 
2
1
a 2 q11 
2
Case 1: a2 = 1/2
Case 2: a2 = 1
These two methods
have been previously
studied.
What are they?
Case 1: a2 = 1/2
a1  1  a2  1  1 / 2  1 / 2
1
a2 p1 
2
1
a2 q11 
2
1
p1  q11 
1
2a2
1 
1
yi  1  yi   k1  k 2  h
2
2 
where
k1  f  xi , yi 
k 2  f  xi  h, yi  k1h
This is Heun’s Method with
a single corrector.
Note that k1 is the slope at
the beginning of the interval
and k2 is the slope at the
end of the interval.
yi  1  yi  a1k1  a2 k 2 h
where
k1  f  xi , yi 
k 2  f  xi  p1h, yi  q11k1h
a1  1  a2  1  1  0
Case 2: a2 = 1
1
2
1
a2 q11 
2
This is the Improved
Polygon Method.
a2 p1 
1
1
p1  q11 

2 a2 2
yi  1  yi  k 2 h
where
k1  f  xi , yi 
1
1 

k 2  f  xi  h, yi  k1h

2
2 
yi  1  yi  a1k1  a2 k 2 h
where
k1  f  xi , yi 
k 2  f  xi  p1h, yi  q11k1h
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 RK
algorithms.
This results in a1 = 1/3 and p1 = q11 = 3/4
1
2 

yi 1  yi   k1  k 2  h
3
3 
where
k1  f  xi , yi 
3
3 

k 2  f  xi  h, yi  k1h

4
4 
Example
dy
 4x 2 y
dx
I .C. y  1 at x  1 i.e. y 1  1
step size h  0.1
Evaluate the following
ODE using Heun’s
Methods
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
1

yi  1  yi    k 1  4 k 2  k 3   h
6

where
k 1  f  xi , yi 
Note the third term
1
1 

k 2  f  xi  h, yi  k1h


2
2
k 3  f  xi  h, yi  hk1  2hk 2 
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
1

yi  1  yi    k 1  2 k 2  2 k 3  k 4   h
6

where
k 1  f  xi , yi 
1
1


k 2  f  xi  h, yi  k1h


2
2
1
1


k 3  f  xi  h, yi  hk 2 


2
2
k 4  f  xi  h, yi  hk 3 
• Note that for ODE that are a function of x alone that
this is also the equivalent of Simpson’s 1/3 Rule
1

yi 1  yi   k1  2k 2  2k3  k 4  h
6

where
k1  f  xi , yi 
1
1
k 2  f  xi  h, yi  k1h 
2
2 

1
1


k3  f  xi  h, yi  hk2 
2
2


k 4  f  xi  h, yi  hk3 
Example
Use 4th Order RK to solve the following differential equation:
dy
xy

dx 1  x 2
I . C. y1  1
using an interval of h = 0.1
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
dy1
 f 1  x , y1 , y2 , , yn 
dx
dy2
 f 2  x , y1 , y2 , , yn 
dx

dyn
 f n  x , y1 , y2 , , yn 
dx
• 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
dy1
 f 1  x , y1 , y 2 , , y n 
dx
dy2
 f 2  x , y1 , y2 , , yn 
dx

dyn
 f n  x , y1 , y2 , , yn 
dx
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 Problems
The 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 2T
 h '  Ta  T   0
2
dx
Ta
T1
T2
Ta
Ta
T1
d 2T
 h '  Ta  T   0
2
dx
Clearly this second order
ODE needs 2 conditions.
This can be satisfied by
knowing the temperature
at the boundaries,
i.e. T1 and T2
T2
Ta
T(0) = T1
T(L) = T2
d 2T
 h '  Ta  T   0
2
dx
T(0) = T1
T(L) = T2
Use these conditions to solve
the equation analytically.
For a 10 m rod with
Ta = 20
T(0) = 40
T(10) = 200
h’ = 0.01
T  73.45e0.1x  53.45e0.1x  20
Now that we have an analytical solution, lets evaluate our
two proposed numerical methods.
Shooting Method
Given:
d 2T
 h '  Ta  T   0
2
dx
dT
z
dx
dz
 h '  Ta  T 
dx
We need an initial value
of z.
For the shooting method, guess
an initial value.
Guessing z(0) = 10
dz
 h '  Ta  T 
dx
Guessing z(0) = 10
Using a fourth-order RK method with a step size
of 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.
Using a linear interpolation formula between the values
of 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?
T(20)
300
250
200
150
0
5
10
15
z(0)
20
25
T(20)
300
250
200
150
0
5
10
15
20
25
z(0)
20  10
z 0  10 
 200  168.38  12.69
285.90  168.38
We can now use
this to solve the first
order ODE
d 2T
 h '  Ta  T   0
2
dx
dT
z
dx
dz
 h '  Ta  T 
dx
250
150
Sh oot in g
Met h od
T
2 00
A n a ly t ica l
Solu t ion
1 00
50
0
0
5
10
dist a n ce (m )
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..
Finite Difference Methods
The finite divided difference approximation for
the 2nd derivative can be substituted into the
governing equation.
d 2 T Ti 1  2 Ti  Ti 1

2
dx
x 2
d 2T
 h ' Ta  T   0
2
dx
Ti 1  2 Ti  Ti 1
 h ' Ta  Ti   0
2
x
Collect terms
Ti 1  2 Ti  Ti 1
 h ' Ta  Ti   0
2
x
 Ti 1   2  h ' x 2  Ti  Ti 1  h ' x 2 Ta
We can now apply this equation to each interior node
on the rod.
Divide the rod into a grid, and consider a “node” to be
at each division. i.e..  x = 2m
x=2m
T1
T2
L = 10 m
 Ti 1  2  h' x 2 Ti  Ti 1  h' x 2Ta
x=2m
T(0)
T(10)
L = 10 m
Consider the previous problem:
L = 10 m
We need to solve for the
Ta = 20
temperature at the interior
T(0) = 40
T(10) = 200
nodes (4 unknowns).
h’ = 0.01
Apply the governing
equation at these nodes (4
equations).
What is the matrix?
 Ti 1  2  h' x 2 Ti  Ti 1  h' x 2Ta
x=0
2
4
6
8
10
T(0)
T(10)
i=0
1
2
3
4
5
Notice the labeling for numbering x and i
 Ti 1  2  h' x 2 Ti  Ti 1  h' x 2Ta
x=0
2
4
6
8
10
T(0)
T(10)
i=0
40
1
2
3
4
5
200
Note also that the dependent values are
known at the boundaries (hence the term
boundary value problem)
 Ti 1  2  h' x 2 Ti  Ti 1  h' x 2Ta
x=0
2
4
6
8
10
T(0)
T(10)
i=0
40
1
2
3
4
5
200
Apply the governing equation at node 1
 T0  2  h' x 2 T1  T2  h' x 2Ta
 40  2.04T1  T2  0.8
2.04T1  T2  40.8
 Ti 1  2  h' x 2 Ti  Ti 1  h' x 2Ta
x=0
2
4
6
8
10
T(0)
T(10)
i=0
40
1
2
3
4
Apply the equation at node 2
 T1  2  h' x 2 T2  T3  h' x 2Ta
 T1  2.04T2  T3  0.8
5
200
 Ti 1  2  h' x 2 Ti  Ti 1  h' x 2Ta
x=0
2
4
6
8
10
T(0)
T(10)
i=0
1
2
3
4
5
40
200
We get a similar equation at node 3
 T2  2  h' x 2 T3  T4  h' x 2Ta
 T2  2.04T3  T4  0.8
 Ti 1  2  h' x 2 Ti  Ti 1  h' x 2Ta
x=0
2
4
6
8
10
T(0)
T(10)
i=0
40
1
2
3
4
At node 4, we consider the
boundary at the right.
 T3  2  h' x 2 T4  T5  h' x 2Ta
 T3  2.04T4  200  0.8
 T3  2.04T4  200.8
5
200
For the four interior nodes, we get the following
4 x 4 matrix
0
0  T1   40.8 
2.04  1
  1 2.04  1
0  T2   0.8 

   

T
0

1
2
.
04

1
0
.
8

 3  

 0
 T  200.8
0

1
2
.
04

 4  

T T  65.97 93.78 124.54 159.48
250
A n a ly t ica l
Solu t ion
2 00
Sh oot in g
Met h od
Fin it e
Differ en ce
T
150
1 00
50
0
0
5
dist a n ce (m )
10
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 secondorder 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 higherorder 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
… end of lecture on ODE