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ORDINARY DIFFERENTIAL EQUATIONS Student Notes

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ORDINARY DIFFERENTIAL EQUATIONS Student Notes

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ORDINARY DIFFERENTIAL EQUATIONSStudent Notes

ENGR 351

Numerical Methods for Engineers

Southern Illinois University Carbondale

College of Engineering

Dr. L.R. Chevalier

Dr. B.A. DeVantier

Photo Credit: Mr. Jeffrey Burdick

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.

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)

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.

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.

- 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

- 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)

- A second order differential equation would have the form:

}

does not necessarily have to include

all of these variables

- An ordinary differential equation is one with a single independent variable.
- Thus, the previous two equations are ordinary differential equations
- The following is not:

- 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).

- 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

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.

- 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.

- One Step Methods
- Euler’s Method
- Heun’s Method
- Improved Polygon
- Runge Kutta
- Systems of ODE

- Boundary Value Problems

- 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

- 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

At this point lets consider

initial conditions.

y(0)=1

and

y(0)=2

What we see are different

values of C for the two

different initial conditions.

The resulting equations

are:

- Focus is on solving ODE in the form

y

xi, yi

x

- Focus is on solving ODE in the form

y

xi, yi

x

- Focus is on solving ODE in the form

h

y

xi, yi

x

This is the same as saying:

new value = old value + slope x step size

- Focus is on solving ODE in the form

y

yi

slope = f = dy/dx

This is the same as saying:

new value = old value + (dy/dx)(h)

where h is the step size

x

h

- Focus is on solving ODE in the form

y

yi+1

yi

slope = f = dy/dx

This is the same as saying:

new value = old value + (dy/dx)(h)

where h is the step size

x

h

- 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

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

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) = f = 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

- 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

- 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

- 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

- 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

Take the slope at xi

Project to get f(xi+1 )

based on the step size h

h

x

xixi+1

y

Use this “average” slope

to predict yi+1

x

xixi+1

{

y

Use this “average” slope

to predict yi+1

x

xixi+1

{

y

y

x

xixi+1

Euler’s

x

xi xi+1

Heun’s

y

x

xi xi+1

- 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

y

f(xi)

x

xi

y

h/2

h

x

xixi+1/2 xi+1

y

h/2

x

xixi+1/2

y

f(xi+1/2)

x

xixi+1/2

y

f’(xi+1/2)

x

xixi+1/2

y

Extend your slope

now to get f(x i+1)

h

x

xixi+1/2 xi+1

y

f(xi+1)

x

xixi+1/2xi+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.

- 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:

{

f is called the incremental function

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

- 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)

Now, f’(xi , yi ) must be determined by the

chain rule for differentiation

The basic strategy underlying Runge-Kutta methods

is to use algebraic manipulations to solve for values

of a1, a2, p1 and q11

By setting these two equations equal to each other and

recalling:

we derive three equations to evaluate the four unknown

constants

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?

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.

Consider the following:

Case 1: a2 = 1/2

Case 2: a2 = 1

These two methods

have been previously

studied.

What are they?

Case 1: a2 = 1/2

This is Heun’s Method

Note that k1 is the slope at

the beginning of the interval and k2 is the slope at the

end of the interval.

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 RK

algorithms.

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

Evaluate the following

ODE using Heun’s

Methods (e.g. a2 = ½)

STRATEGY

- Calculate k1=f(x,y) = f = 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

- Derivation is similar to the one for the second-order
- Results in six equations and eight unknowns.
- One common version results in the following

Note the third term

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

- The most popular
- The following is sometimes called the classical fourth-order RK method

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

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

using an interval of h = 0.1

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

- 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

- Many practical problems in engineering and science require the solution of a system of simultaneous differential equations

- 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

- 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

- 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

Ta

T1

T2

Ta

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:

Ta

T1

T2

Ta

Clearly this second order

ODE needs 2 conditions.

This can be satisfied by

knowing the temperature

at the boundaries,

i.e. T1 and T2

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(0) = T1

T(L) = T2

Now that we have an analytical solution, lets evaluate our

two proposed numerical methods.

Given:

We need an initial value

of z.

For the shooting method, guess

an initial value.

Guessing z(0) = 10

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?

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..

The finite divided difference approximation for

the 2nd derivative can be substituted into the

governing equation.

D x = 2 m

T1

T2

L = 10 m

Collect terms

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.. D x = 2m

D x = 2 m

T(0)

T(10)

L = 10 m

Consider the previous problem:

L = 10 m

Ta = 20

T(0) = 40

T(10) = 200

h’ = 0.01

We need to solve for the

temperature at the interior

nodes (4 unknowns).

Apply the governing

equation at these nodes (4

equations).

What is the matrix?

x=0 2 4 6 8 10

T(0)

T(10)

i=0 1 2 3 4 5

Notice the labeling for numbering Dx and i

x=0 2 4 6 8 10

T(0)

T(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)

x=0 2 4 6 8 10

T(0)

T(10)

i=0 1 2 3 4 5

40

200

Apply the governing equation at node 1

x=0 2 4 6 8 10

T(0)

T(10)

i=0 1 2 3 4 5

40

200

Apply the equation at node 2

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

x=0 2 4 6 8 10

T(0)

T(10)

i=0 1 2 3 4 5

At node 4, we consider the

boundary at the right.

40

200

For the four interior nodes, we get the following

4 x 4 matrix

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

- 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.

- 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