Chapter 6
Numerical Methods for Ordinary
Differential Equations
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6.1 The Initial Value Problem: Background
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Example 6.1
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Example 6.2
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Example 6.3
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Example 6.4
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Methods

To solve differential equation problem:
–
If f is “smooth enough”, then
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
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Two ways of expressing “smooth enough”
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a solution will exist and be unique
we will be able to approximate it accurately with a wide
variety of method
Lipschitz continuity
Smooth and uniformly monotone decreasing
Definitions 6.1 and 6.2
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Example 6.5
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Example 6.5 (con.)
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Theorem 6.1

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Proof: See Goldstine’s book
6.2 Euler’s Method
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
We have treated Euler’s method in Chapter 2.
There are two main derivations about Euler’s
method
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–
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Geometric derivation
Analytic derivation
Geometric Derivation
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Analytic Derivation
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Error Estimation for Euler’s Method
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
By the analytic derivation, we have

The residual for Euler’s method:

The truncation error:
Example 6.6
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6.3 Analysis of Euler’s Method
O(h)
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Proof: pp. 323-324 (You can study it by yourselves.)
Theorem 6.4
Theorem 6.4 ( Error Estimate for Euler ' s Method , VersionII )
Let f be smooth and uniformly monotone decreasing in y , and
assume that the solution y  C 2 ([ t0 , T ]) for some T  t0 .
Then for h sufficient ly small,
max y (t k )  y k  C0 y (t0 )  y 0  Ch y"  ,[ t ,T ]
t k T
0
where C0  1, C0  0 as k  , and
C

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O(h)
1
2m
Proof: pp. 324-325 (You can study it by yourselves.)
Discussion

Both error theorems show that Euler’s method
is only first-order accurate (O(h)).
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–
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If f is only Lipschitz continuous, then the constants
multiplying the initial error and the mash parameter
can be quite large, rapidly growing.
If f is smooth and uniformly monotone decreasing in y,
then the constants in the error estimate are bounded
for all n.
Discussion

How is the initial error affected by f ?
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–
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If f is monotone decreasing in y, then the effect of the
initial error decreases rapidly as the computation
progresses.
If f is only Lipschitz continuous, then any initial error
that is made could be amplified to something
exponentially large.
6.4 Variants of Euler’s Method


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Euler’s method is not the only or even the best
scheme for approximating solutions to initial
value problems.
Several ideas can be considered based on
some simple extensions of one derivation of
Euler’s method.
Variants of Euler’s Method

We start with the differential equation
And replace the derivative with the simple
difference quotient derived in (2.1)
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What happens if we use other approximations
to the derivative?
Variants of Euler’s Method

If we use
O(h)
then we get the backward Euler method
Method 1

If we use
then we get the midpoint method
Method 2
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O(h2)
Variants of Euler’s Method

If we use the methods based on interpolation
(Section 4.5)
O(h2)
O(h2)
then we get two numerical methods
Method 3
Method 4
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Variants of Euler’s Method

By integrating the differential equation:
(6.23)
and apply the trapezoid rule to (6.23) to get
O(h3)
Thus
Method 5
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(6.25)
Variants of Euler’s Method

We can use a midpoint rule approximation to
integrating (6.23)
O(h3)
and get
Method 6
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Discussion


What about these method? Are any of them any good?
Observations
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–
–
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Methods 2, 3, and 4 are all based on derivative approximations
that are O(h2), thus they are more accurate than Euler method
and method 1 (O(h)).
Similarly, methods 5 and 6 are also more accurate.
Methods 2, 3, and 4 are not single-step methods, but multistep
methods. They depend on information from more than one
previous approximate value of the unknown function.
Discussion

Observations (con.)
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–
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Concerning methods 1, 4, and 5, all of these
formulas involve
we cannot explicitly
solve for the new approximate values
Thus
these methods are called implicit methods.
Methods 2 and 3 are called explicit methods.
6.4.1 The Residual and Truncation Error
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Definition 6.3
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Example 6.7
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Example 6.8
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Definition 6.4
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6.4.2 Implicit Methods and PredictorCorrector Schemes

How to get the value of yn+1? Using Newton’s method or
the secant method or a fixed point iteration.
Let y = yn+1
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F(y)
F’(y)
F(y)
h
F(y+h)-F(y)
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Predictor-corrector idea
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Can we use a much cruder (coarse) means of
estimating yn+1?
Example 6.11
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Example 6.12
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Discussion

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Generally speaking, unless the differential
equation is very sensitive to changes in the data,
a simple predictor-corrector method will be just
as good as the more time-consuming process of
solving for the exact values of yn+1 that satisfies
the implicit recursion.
Discussion


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If the differential equation is linear, we can entirely avoid
the problem of implicitness.
Write the general linear ODE as
6.4.3 Starting Values and Multistep
Method
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How to find the starting values?
Example 6.15 (con.)
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Example 6.15 (con.)
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6.4.4 The Midpoint Method and Weak
Stability
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Discussion

What is going on here?
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–
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–
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It is the weakly stability problem.
The problem is not caused by rounding error.
The problem is inherent in the midpoint method and
would occur in exact arithmetic.
Why? (pp. 340-342)
6.5 Single-step Method: Runge-Kutta
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The Runge-Kutta family of methods is one of the most
popular families of accurate solvers for initial value
problems.

Consider the more general method:
Residual
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
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Rewrite the formula of R, we get
Solution 1
Solution 2
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Solution 3
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Runge-Kutta Method
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Example 6.16
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Example 6.16 (con.)
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One major drawback of the
Runge-Kutta methods is that
they require more evaluations
of the function f than other
methods.
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6.6 Multistep Methods
6.6.1 The Adams families
 Adams families include the following most popular
ones:
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–
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Explicit methods: Adams-Bashforth families
Implicit methods: Adams-Moulton families
Adams-Bashforth Methods
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Discussion

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If we assume a uniform grid with mash spacing h, then
the formulas for the
and
simplify substantially, and
they are routinely tabulated:
Example 6.17
Table 6.6
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Example 6.17 (con.)
Table 6.6
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Adams-Moulton Methods
If k = -1 then tn-k= tn+1
If k = -1 then tn-k= tn+1
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Example 6.18
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6.6.2 The BDF Family

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BDF: backward difference formula
The BDF Method (con.)
K = -1
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The BDF Method (con.)
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6.7 Stability Issues
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Definition 6.6
A root having multiplicity n = 1 is called a simple root. For example, f (z) = (z-1)(z-2) has a
simple root at z = 1, but g(z) = (z-1)2 has a root of multiplicity 2 at z = 1, which is therefore not
75 a simple root.
Discussion
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Example 6.19
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