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Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005
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Page 1: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Chapter 6: Numerical Methods for OrdinaryDifferential Equations

November 15, 2005

Page 2: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Outline

1 6.1 The Initial Value Problem: Background

2 6.2 Euler’s Method

3 6.4 Variants of Euler’s Method

4 Taylor Series Method (not in the textbook)

5 6.5 Single Step Methods: Runge-Kutta

6 6.8 Applications to Systems of Equations

Page 3: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Differential Equations - Review

• For some simple differential equations, it is possible to finda solution in the form of an elementary function.

• The simplest possible equation is of the form

y ′(x) = g(x)

whose solution is

y(x) =

∫g(x)dx + C (C − constant )

Page 4: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleThe general solution of the equation

y ′(x) = sin x

isy(x) = − cos x + C

If we specify a point on the curve y(x), e.g.

y(π

3) = 2

thenC = 2.5

and we have a unique solution:

y(x) = 2.5− cos x

Page 5: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• The more general type of the first-order equation is

y ′(x) = f (x , y(x))

when y ′ depends on both x and y .• As before, if certain general requirements are imposed on

the function of two variables f , we have a general solutionwhich depends on one constant C

y(x) = h(x) + C

• Again, if we have the initial condition

y(x0) = y0

which tells us that a certain point lies on the curve, thesolution is again unique.

Page 6: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Initial Value Problem

Initial Value Problem: A first-order differential equation togetherwith an initial condition

y ′(x) = f (x , y(x)), x ≥ x0

y(x0) = y0

TheoremLet f (x , z) and ∂f (x , z)/∂z be continuous functions of bothvariables x and z in some neighbourhood of the point (x0, y0).Then there is a unique solution to the initial value problem

y ′(x) = f (x , y(x))

y(x0) = y0

in some neighbourhood of the point x0.

Page 7: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleConsider the initial value problem

y ′ = 2xy2, y(0) = 1

Then,

f (x , z) = 2xz2,∂f (x , z)

∂z= 4xz

and both these functions are continuous in the neighbourhoodof (0, 1), so this initial value problem has a unique solutionaround x0 = 0.It can be shown that the solution is

y(x) =1

1− x2 , −1 < x < 1

Page 8: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

DefinitionWe say that a function f (x , z) is Lipschitz continuous in variablez on a rectangle (a, b)× (c, d) if there is a constant K such that

|f (x , z1)− f (x , z2)| ≤ K |z1 − z2|

for all points z1, z2 in (c, d).

Page 9: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleConsider the initial value problem

y ′ = −(1 + x2)y + sin x , y(0) = 1

on the rectangle (−1, 1)× (0, 2) Since

|f (x , z1)−f (x , z2)| = |−(x2+1)(z1−z2)| = |x2+1||z1−z2| ≤ 2|z1−z2|

The reason for the last inequality is that x ∈ [−1, 1] so

x2 + 1 < 2

Therefore, K = 2.

Page 10: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• Every continuous function satisfies the Lipschitz condition,for some K , on a closed rectangle.

• The converse is not necessarily true; there are functionswhich satisfy the Lipschitz condition but are not continuous.

TheoremIf f (x , z) is a function continuous around the point (x0, y0) andwhich satisfies the Lipschitz condition in a rectangle around(x0, y0), then the initial value problem

y ′(x) = f (x , y(x))

y(x0) = y0

has a unique solution around x0.

Page 11: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• If f (x , z) satisfies the Lipschitz condition, not only will wehave a unique solution y(x) to the initial value problem, butthat solution will also be stable

• Stability of the initial value problem means that, if wechange the initial condition y(x0) = y0 by a small amount,the solution to the initial value problem will be relativelyclose to the solution of the original value problem.

TheoremIf f (x , z) satisfies the Lipschitz condition with the constant K inz, then the solution to

y ′(x) = f (x , y(x))

y(x0) = y0

is stable. Namely, if z(t) is the solution to the same problemwith the initial condition z(x0) = z0, then

|y(x)− z(x)| ≤ eK (x−x0)|y0 − z0|

Page 12: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleIn our previous example,

y ′ = −(1 + x2)y + sin x , y(0) = 1

we saw that K = 2.Suppose we change the initial condition to

z(0) = 1.02

Then,

|y(x)− z(x)| ≤ e2(x−0)|1− 1.02| = 0.02e2x

Page 13: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Other Types of Problems

There are other types of problems from the theory of differentialequations that can be reduced to the initial value problems:

1 Systems of first-order equations2 Second-order equations

Page 14: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

System of Two First-Order Equations

We can consider a system of two differential equations

y ′1(x) = f1(x , y1(x), y2(x)), y1(x0) = y1,0

y ′2(x) = f2(x , y1(x), y2(x)), y2(x0) = y2,0

• One can similarly define a system of n equations in nfunctions y1, y2, . . . yn.

Page 15: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleConsider the system

y ′1 = −4y1 + y2, y1(0) = 1

y ′2 = y1 − 4y2, y2(0) = 0

This system can be rewritten, in the vector/matrix notation, as:[y1y2

]′=

[−4 11 −4

]·[

y1y2

]so we can think of it as

y ′ = Ay

where A is a 2× 2-matrix.It can be shown that the solution of this system is

y1 =12

e−3x +12

e−5x

y2 =12

e−3x − 12

e−5x

Page 16: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Second-Order Equations WithConstant Coefficients

If we are given a second-order equation

y ′′ + ay ′ + by = g(x), y(x0) = y0, y ′(x0) = y0

where a, b are constants, we can reduce it to a system of twoequations in two functions.

1 Introduce the new functions

w1(x) = y(x), w2(x) = y ′(x)

2 Thenw ′

1 = w2

and from the original equation:

w ′2 + aw2 + bw2 = g(x) ⇒ w ′

2 = −aw2 − bw1 + g(x)

3 The conditions become

w1(x0) = y0, w2(x0) = y0

Page 17: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleConsider the second-order problem

y ′′ + 3y ′ + y = 0, y(0) = 1, y ′(0) = 0

Introduce the substitution

w1 = y , w2 = y ′

Then,

w ′1 = w2

w ′2 = −3w2 − w1

and the conditions are

w1(0) = 1, w2(0) = 0

Page 18: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Euler’s Method

• Euler’s Method is the simplest numerical method forsolving initial value problems.

• It is not a very efficient numerical method but, as we willsee, most other numerical methods are based on it.

Page 19: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• Suppose we are given an initial value problem

y ′(x) = f (x , y(x)), x0 ≤ x ≤ by(x0) = y0

• Numerical methods for solving such a problem will give usapproximate values of the solution y(x) at a discrete set ofnodes

x0 < x1 < x2 < . . . < xN = b

• To make things simpler, we will assume that these nodesare equally spaced

xn = x0 + n · h, n = 0, 1, 2, . . . N

Page 20: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• If y(x) is an approximate solution to the problem, let usintroduce the notation

y(xn) = yn, n = 0, 1, 2, . . . , N

• Recall that we used the following approximation to thederivative

y ′(x) ≈ 1h

[y(x + h)− y(x)]

• Since, in the initial value problem,

y ′(xn) = f (xn, y(xn)) = f (xn, yn)

when we apply the approximation to the derivative, wehave

1h

[y(xn+1)− y(xn)] ≈ f (xn, y(xn))

y(xn+1) ≈ y(xn) + hf (xn, yn)

Page 21: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• From the last formula, we get the recursive formula forEuler’s Method:

yn+1 = yn + hf (xn, yn)

• For the initial value y0, we take the value of the function

f (x0) = y0

from the statement of the initial value problem.• Geometrically, the tangent line to the solution curve y(x) at

xn has the slope

y ′(xn) = f (xn, y(xn))

and we use this tangent line to approximate the graph ofthe solution to the right of the point xn+1

Page 22: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Figure: Euler’s Method

Page 23: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleUse Euler’s Method with the step h = 0.2 to find theapproximate value of the solution to the initial value problem

y ′(x) =y(x) + x2 − 2

x + 1, y(0) = 2

at x = 1Euler’s Method for this equation is

yn+1 = yn +h(yn + x2

n − 2)

xn + 1

with y0 = 2 and x0 = 0 and the step h = 0.2The values at the nodes between x0 = 0 and x5 = 1 are shownin the table.

Page 24: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

n xn yn0 0.0000 2.00001 0.2000 2.00002 0.4000 2.00673 0.6000 2.03054 0.8000 2.07935 1.0000 2.1592

Therefore,y(1) ≈ 2.1592

Page 25: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• The actual solution to this initial value problem is thefunction

y(x) = x2 + 2x + 2− 2(x + 1) ln(x + 1)

whose value at x = 1 is:

y(1) = 2.2274

• One way to improve the accuracy is to use a finer partitionof the interval [0, 1]

• For example, if we use the step h = 0.1, we get

y(1) ≈ 2.1912

and with h = 0.05, we have

y(1) ≈ 2.2087

Page 26: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

The Backward Euler Method

• In the derivation for the formula for Euler’s Method, we usethe approximation for the derivative

y ′(x) ≈ 1h

[y(x + h)− y(x)]

• We can use the backward difference formula instead:

y ′(x) ≈ 1h

[y(x)− y(x − h)]

• Using this approximation, we get the Backward Euler’sMethod:

yn+1 = yn + hf (xn+1, yn+1)

with the initial approximation y0 = y(x0)

Page 27: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Midpoint Method

If we use the centered difference formula for approximating thederivative

y ′(x) ≈ 12h

[f (x + h)− f (x − h)]

then we get the following version of Euler’s Method, which iscalled the Midpoint Method:

yn+1 = yn−1 + 2hf (xn, yn)

Page 28: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleUse the Midpoint Method with the step h = 0.2 to find theapproximate value of the solution to the initial value problem

y ′(x) =y(x) + x2 − 2

x + 1, y(0) = 2

at x = 1The Midpoint Method for this equation is

yn+1 = yn−1 +2h(yn + x2

n − 2)

xn + 1

with y0 = 2 and x0 = 0 and the step h = 0.2

Page 29: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Since yn+1 requires that yn−1 and yn be already available, wewill use the Euler’s Method to calculate the first iteration, but forfurther values y2, y3, . . . , we can use the Midpoint Rule formula

n xn yn0 0.0000 2.00001 0.2000 2.00002 0.4000 2.00673 0.6000 2.02964 0.8000 2.07545 1.0000 2.1241

Therefore,y(1) ≈ 2.1241

Page 30: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Trapezoidal method

• All the methods we have seen so far (Euler’s Method,Backward Euler’s Method, and the Midpoint Method) arerelatively inefficient.

• The Trapezoidal Method has a higher convergence order;the previous methods are linear in h while the TrapezoidalMethod is quadratic in accuracy.

• We start with the initial value problem

y ′(x) = f (x , y(x))

Page 31: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• We integrate this equation on the interval [xn, xn+1]:

y(xn+1) = y(xn) +

∫ xn+1

xn

f (x , y(x))dx

• Next, we use the Trapezoidal Rule to approximate thisintegral

y(xn+1) ≈ y(xn) +h2

[f (xn, y(xn)) + f (xn+1, y(xn+1))]

• Then, we get the following formula for the TrapezoidalMethod:

yn+1 = yn +h2

[f (xn, yn) + f (xn+1, yn+1)]

with y0 being f (x0).

Page 32: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• Problem: Both the Backward Euler’s Method and theTrapezoidal Method suffer from the same problem which isthat yn+1 appears on both sides of the formula for yn+1.

• Such methods are called implicit unlike Euler’s Method andthe Midpoint Method, for which yn+1 was defined explicitlyin terms of xn and previously computed values of y only.

• If we attempt to use e.g Trapezoidal method in its currentform, then, at each step, we have to use a numericalmethod for solving equations (e.g Newton’s Method) to findyn+1 from that implicit equation.

• This would be too complicated so we will attempt to rectifythe problem in a different way.

Page 33: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• A common approach to resolve this issue is to use thepredictor-corrector idea.

• First, we “predict” the value of yn+1 to be used on theright-hand side, inside the function f (xn+1, yn+1) usingsome explicit formula, e.g. by using Euler’s Method.

• Then, we use this “predicted” value yn+1 and “correct” itusing the Trapezoidal Rule.

Page 34: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Trapezoidal Predictor-CorrectorMethod

1 Predictor Step:

yn+1 = yn + hf (xn, yn)

[Use Euler’s Method to predict the value of yn+1]2 Corrector Step:

yn+1 = yn +h2

[f (xn+1, yn+1) + f (xn, yn)]

[Correct the calculated value from the first step using theTrapezoidal Method]

Page 35: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleConsider the initial value problem

y ′ = −y ln y , y(0) =12

Use the Trapezoidal predictor-Corrector method to approximatethe solution at x = 1 using the step h = 0.25.

n xn yn yn0 0.00000 0.500001 0.25000 0.58664 0.582432 0.50000 0.66114 0.655983 0.75000 0.72513 0.719694 1.00000 0.77887 0.77361

So,y(1) ≈ 0.77361

Page 36: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Taylor Series method

• Again, suppose we are given an initial value problem

y ′(x) = f (x , y(x)), y(x0) = y0

• We can try to approximate the solution function by itsTaylor polynomial of degree n around the initial point x0

y(x) ≈ y(x0) + y ′(x0)(x − x0) +y ′′(x0)

2!(x − x0)

2+

+ . . . +y (n)(x0)

n!(x − x0)

n

Page 37: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• If we introduce the substitution h = x − x0, this becomes

y(x0 + h) ≈ y(x0) + y ′(x0)h +y ′′(x0)

2!h2 + . . . +

y (n)(x0)

n!hn

• The coefficient y(x0) = y0 is given in the initial valueproblem, while we can determine the other coefficients bydifferentiating the right hand side of the equation

f (x , y(x))

repeatedly (using the Chain Rule)

Page 38: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleConsider the initial value problem

y ′ = −2x − y , y(0) = −1

and suppose we want to find, using the Taylor polynomial ofdegree four, the approximate value of y(0.4).Solution: We expand y(x) into the fourth degree Taylorpolynomial around x0 = 0:

y(h) ≈ y(0) + y ′(0)h +y ′′(0)

2!h2 +

y ′′′(0)

3!h3 +

y (4)(0)

4!h4

Since y(0) = −1, we have:

y(h) ≈ −1 + y ′(0)h +y ′′(0)

2!h2 +

y ′′′(0)

3!h3 +

y (4)(0)

4!h4

Next, we calculate y ′(0):

y ′(x) = −2x − y(x) ⇒ y ′(0) = −2 · 0− (−1) = 1

Page 39: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

So,

y(h) ≈ −1 + 1 · h +y ′′(0)

2!h2 +

y ′′′(0)

3!h3 +

y (4)(0)

4!h4

Then, we calculate y ′′(0):

y ′′(x) = −2− y ′(x) ⇒ y ′′(0) = −2− y ′(0) = −2− 1 = −3

Similarly, we calculate y ′′′(0) and y (4)(0):

y ′′′(x) = −y ′′(x) ⇒ y ′′′(0) = −y ′′(0) = 3

y (4)(x) = −y ′′′(x) ⇒ y (4)(0) = −y ′′′(0) = −3

So, the approximation to the solution around x0 = 0 is:

y(h) ≈ −1 + h − 32

h2 +12

h3 − 18

h4

and the approximate value of y(0.4) is:

y(0.4) ≈ −0.8112

Page 40: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

The actual solution to this initial value problem is

y(x) = −3e−x − 2x + 2

and the actual value of the solution at x = 0.4 is

y(0.4) = −0.81096

The estimate for the upper bound of the error is the same onethat we had for approximating the function by its fourth degreeTaylor polynomial on the interval [0, 0.4] which is

15!

(0.4)5 max[0,0.4]

|f (5)(x)|

Page 41: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Runge-Kutta Methods

• Runge-Kutta family of methods is among the most popularmethods for solving initial value problems.

• Recall the Trapezoidal Predictor-Corrector Method:

yn+1 = yn + hf (xn, yn)

yn+1 = yn +h2

[f (xn+1, yn+1) + f (xn, yn)]

• Substituting the predictor equation into the correctorequation, we get

yn+1 = yn +h2

[f (xn+1, yn + hf (xn, yn)) + f (xn, yn)]

Page 42: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• The geometric meaning of this method is that, in order todefine yn+1 from yn, we advance along a straight linewhose slope is the average of the two slopes

f (xn, yn), and f (xn+1, yn+1)

• The second-order Runge-Kutta methods are based on thegeneralization of this formula.

• We may want to use different averaging of slopes, and getthe more general method:

yn+1 = yn + c1hf (xn, yn) + c2hf (xn + αh, yn + βhf (xn, yn))

where c1, c2, α, and β are parameters that we want todetermine in some way.

Page 43: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• Using Taylor series for functions of two variables, it can beshown that these parameters are bound by the equations

0 = 1− c1 − c2

0 =12− c2α

0 =12− c2β

• Since this is a system of three equations in four variables,there are infinitely many solutions. In fact, we can choosethe value of one parameter, for example c1, in an arbitraryway.

Page 44: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• If we choose c1 = 12 , then

c2 =12, α = β = 1

and we get the Trapezoidal Predictor-Corrector Method:

yn+1 = yn +h2

[f (xn+1, yn + hf (xn, yn)) + f (xn, yn)]

• If c1 = 14 , then

c2 =34, α = β =

23

and we get

yn+1 = yn +14

[f (xn, yn) + 3f

(xn +

23

h, yn +23

hf (xn, yn)

)]which is called the Heun’s Method

Page 45: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Fourth-Order Runge-Kutta Method

• The most commonly used Runge-Kutta method is thefourth-order method, which we will call RK4 for short.

• It is derived in a similar fashion to the methods we havejust described, but using four slopes instead of two.

k1 = hf (xn, yn)

k2 = hf (xn +12

h, yn +12

k1)

k3 = hf (xn +12

h, yn +12

k2)

k4 = hf (xn + h, yn + k3)

yn+1 = yn +16(k1 + 2k2 + 2k3 + k4)

Page 46: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleConsider the initial value problem

y ′ = −y ln y , y(0) =12

We will find the approximate value y(0.3) using h = 0.1 and theHeun’s Method.Solution: The formula is

yn+1 = yn +14

[f (xn, yn) + 3f

(xn +

23

h, yn +23

hf (xn, yn)

)]The table of values is

n xn yn0 0.00000 0.000001 0.10000 0.535142 0.20000 0.569193 0.30000 0.60194

Page 47: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleConsider the initial value problem

y ′ = −y ln y , y(0) =12

We will find the approximate value y(0.3) using h = 0.1 and theRK4 Method.Solution: The formulas are

k1 = −hyn ln(yn)

k2 = −h(yn +12

k1) ln(yn +12

k1)

k3 = −h(yn +12

k2) ln(yn +12

k2)

k4 = −h(yn + k3) ln(yn + k3)

yn+1 = yn +16(k1 + 2k2 + 2k3 + k4)

Page 48: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

The table of values is then

n xn yn k1 k2 k3 k40 0.0000 0.5000 0.0346 0.0340 0.0341 0.03441 0.1000 0.5340 0.0335 0.0328 0.0328 0.03212 0.2000 0.5668 0.0321 0.0314 0.0314 0.03073 0.3000 0.5982

Page 49: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Implementation and Examples

• Suppose we are given a system of two first-orderdifferential equations with initial conditions:

y ′1 = f1(x , y1, y2), y1(0) = y10

y ′2 = f2(x , y1, y2), y2(0) = y20

• All the numerical methods for solving initial value problemswe have studied so far can be applied to such a system ofequations.

Page 50: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

• For instance, Euler’s method would be given by these tworecursion formulas

y1,n+1 = y1,n + hf1(xn, y1,n, y2,n)

y2,n+1 = y2,n + hf2(xn, y1,n, y2,n)

• The Trapezoidal Predictor-Corrector method would use thefollowing formulas:

y1,n+1 = y1,n + hf1(xn, y1,n, y2,n)

y2,n+1 = y2,n + hf2(xn, y1,n, y2,n)

y1,n+1 = y1,n +h2

[f1(xn+1, y1,n+1, y2,n+1) + f1(xn, y1,n, y2,n)]

y2,n+1 = y2,n +h2

[f2(xn+1, y1,n+1, y2,n+1) + f2(xn, y1,n, y2,n)]

Page 51: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleGiven the system

y ′1 = y1 − y1y2 + sin(πx), y1(0) = 2

y ′2 = y1y2 − y2, y2(0) = 1

find the approximate values y1(0.2) and y2(0.2), using Euler’sMethod and the step h = 0.1.Solution: The recursion formulas will be

y1,n+1 = y1,n + h (y1,n − y1,ny2,n + sin(πxn))

y2,n+1 = y2,n + h (y1,ny2,n − y2,n)

Page 52: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

n xn y1,n y2,n0 0.00000 2.00000 1.000001 0.10000 2.00000 1.100002 0.20000 2.01090 1.21000

So,y1(0.2) ≈ 2.01090, y2(0) ≈ 1.21000

Page 53: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

RK4 for Systems of Equations

k1,1 = h f1(xn, y1,n, y2,n)

k2,1 = h f2(xn, y1,n, y2,n)

k1,2 = h f1(xn +12

h, y1,n +12

k1,1, y2,n +12

k2,1)

k2,2 = h f2(xn +12

h, y1,n +12

k1,1, y2,n +12

k2,1)

k3,1 = h f1(xn +12

h, y1,n +12

k2,1, y2,n +12

k2,2)

k3,2 = h f2(xn +12

h, y1,n +12

k1,2, y2,n +12

k2,2)

k4,1 = h f1(xn + h, y1,n + k3,1, y2,n + k3,2)

k4,2 = h f2(xn + h, y1,n + k3,1, y2,n + k3,2)

Page 54: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

y1,n+1 = y1,n +16(k1,1 + 2k2,1 + 2k3,1 + k4,1)

y2,n+1 = y2,n +16(k1,2 + 2k2,2 + 2k3,2 + k4,2)

Page 55: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Taylor Series Method for Systems

• We can also use Taylor Series Method to solve systems ofdifferential equations with initial conditions.

• We approximate both functions y1 and y2 with their Taylorpolynomials, which are of the same degree, around thepoint x = x0.

Page 56: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

ExampleSuppose we want to find approximate values of y1(0.2) andy2(0.2) if

y ′1 = y1 − y1y2 + sin(πx), y1(0) = 2

y ′2 = y1y2 − y2, y2(0) = 1

using Taylor polynomials of degree three.Solution: We find the first three derivatives of both y1 and y2around the point x0 = 0:

y ′1(0) = y1(0)− y1(0)y2(0) + sin(π · 0) = 0

y ′2(0) = y1(0)y2(0)− y2(0) = −1

Page 57: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Since

y ′′1 = y ′

1 − y ′1y2 − y1y ′

2 + π cos(πx)

y ′′2 = y ′

1y2 + y1y ′2 − y ′

2

soy ′′

1 (0) = 2 + π, y ′′2 (0) = −1

Finally,

y ′′′1 = y ′′

1 − y ′′1 y2 − 2y ′

1y ′2 − y1y ′′

2 − π2 sin(πx)

y ′′′2 = y ′′

1 y2 + 2y ′1y ′

2 + y1y ′′2 − y ′′

2

andy ′′′

1 (0) = 2, y ′′′2 (0) = 1 + π

Page 58: Chapter 6: Numerical Methods for Ordinary Differential ...ddelic/Ch6handout.pdf · Chapter 6: Numerical Methods for Ordinary Differential Equations November 15, 2005. Outline ...

Taylor polynomials of degree three are

y1(h) ≈ 2 +2 + π

2h2 +

13

h3

y2(h) ≈ 1− h − 12

h2 +1 + π

6h3

andy1(0.2) ≈ 2.10549, y2(0.2) ≈ 0.78552


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