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### Transcript of Numerical Diﬀerential Equations: dattab/MATH435.2013/NUMERICAL_ODE.IVP.pdfNumerical Diﬀerential...

• Chapter 11

Numerical Differential Equations:

IVP

**** 4/16/13 EC

(Incomplete)

11.1 Initial Value Problem for Ordinary Differential Equations

We consider the problem of numerically solving a differential equation of the form

dy

dt= f(t, y), a t b, y(a) = (given).

Such a problem is called the Initial Value Problem or in short IVP, because the initial value

of the solution y(a) = is given.

Since there are infinitely many values between a and b, we will only be concerned here to find

approximations of the solution y(t) at several specified values of t in [a, b], rather than finding

y(t) at every value between a and b.

The following strategy will be used:

Divide [a, b] into N equal subintervals each of length h:

t0 = a < t1 < t2 < < tN = b.

491

• 492 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP

a = t0 t1 t2 tN = b

Set h =b a

N(step size)

Compute approximations yi to y(ti) at t = ti; that is, given y(0) = y0, compute, recur-

sively, y1, y2, . . . , yn.

y(ti) = Exact value of y(t) at t = ti.

yi = An approximation of y(ti) at t = ti.

The Initial Value Problem

Given

(i) y = f(y, t), a t b

(ii) The initial value y(a) = y(t0) = y0 =

(iii) The step-size h.

Find yi (approximate value of y(ti)), i = 1, . . . , N , where N =bah

.

We will briefly describe here the following well-known numerical methods for solving the IVP:

The Euler and Modified Euler Method (Taylor Method of order 1)

The Higher-order Taylor Methods

The Runge-Kutta Methods

The Predictor-Corrector Methods

Methods for Systems of IVP Differential Equations

We will also discuss the error behavior and convergence of these methods. However, before doing

so, we state a result without proof, in the following section on the existence and uniqueness

of the solution for the IVP. The proof can be found in most books on ordinary differential

equations.

• 11.1. INITIAL VALUE PROBLEM FOR ORDINARY DIFFERENTIAL EQUATIONS 493

11.1.1 Existence and Uniqueness of the Solution for the IVP

Theorem 11.1 (Existence and Uniqueness Theorem for the IVP). Suppose

(i) f(t, y) is continuous on the domain defined by R = {a t b, < y < },

and

(ii) the following inequality is satisfied:

|f(t, y) f(t, y)| L|y y|,

whenever (t, y) and (t, y) R, and L is constant.

Then the IVP

y = f(t, y), y(a)

has a unique solution.

Definition 11.2. The condition |f(t, y) f(t, y)| L|y y| is called the Lipschitz Con-

dition. The number L is called a Lipschitz Constant.

Definition 11.3. A set S is said to be convex if whenever (t1, y1) and (t2, y2) belong to S,

the point ((1 )t1 + t2, (1 )y1 + y2) also belongs to S for each , 0 1.

11.1.2 Simplification of the Lipschitz Condition for the Convex Domain

Suppose that the domain R in Theorem 11.1 is convex set, then the The IVP has a unique

solution if

f

y(t, y)

L for all (t, y) R.

11.1.3 Liptischitz Condition and Well-Posedness

Definition 11.4. An IVP is said to be well-posed if a small perturbation in the data of the

problem leads to only a small change in the solution.

Since numerical computation may introduce some perturbations to the problem, it is important

that the problem that is to be solved is well-posed. Fortunately, the Lipschitz condition is a

sufficient condition for the IVP problem to be well-posed.

Theorem 11.5 (Well-Posedness of the IVP problem). If f(t, y) satisfies the Lipschitz

condition, then the IVP is well-posed.

• 494 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP

11.2 The Euler Method

One of the simplest methods for solving the IVP is the classical Euler method.

The method is derived from the Taylor Series expansion of the function y(t).

Recall that the function y(t) has the following Taylor series expansion of order n at t = ti+1:

y(ti+1) = y(ti) + (ti+1 ti)y(ti) +

(ti+1 ti)

2!

2

y(ti)

+ +(ti+1 ti)

n!

n

y(n)(ti) +(ti+1 ti)

(n+ 1)!

n+1

yn+1(i), where i is in (ti, ti+1).

Substitute h = ti+1 ti. Then

Taylor Series Expansion of y(t) of order n at t = ti+1

y(ti+1) = y(ti) + hy(ti) +

h2

2!y(ti) + +

hn

n!y(n)(ti) +

hn+1

(n + 1)!y(n+1)(i).

For n = 1, this formula reduces to

y(ti+1) = y(ti) + hy(ti) +

h2

2!y(i).

The term h2

2! y(2)(i) is called the remainder term.

Neglecting the remainder term, we can write the above equation as:

y(ti+1) y(ti) + hy(ti)

Using our notations, established earlier, we then have

yi+1 = yi + hf(ti, yi),

since yi = f(ti, yi)

We can then resursively generate the successive approximations yi, y2, . . . , yN to y(t1), y(t2), . . .

and y(tN ) as follows:

• 11.2. THE EULER METHOD 495

y0 = (given)

y1 = y0 + hf(t0, y0)

y2 = y1 + hf(t1, y1)...

yN = yN1 + hf(tN1, yN1)

.

These iterations are known as Eulers iterations and the method is the historical Eulers

method.

Figure 11.1: Geometrical Interpretation

y(tN )

= y(b)

y(t2)

y(t1)

y(t0)

= y(a) =

a t1 t2 tN1 b = tN

• 496 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP

Algorithm 11.6 (Eulers Method for IVP).

Input: (i) The function f(t, y)

(ii) The end points of the interval [a, b]: a and b

(iii) The initial value: = y(t0) = y(a)

(iv) The step size: h

Output:Approximations yi+1 of y(ti + 1), i = 0, 1, . . . , N 1.

Step 1. Initialization: Set t0 = a, y0 = y(t0) = y(a) = and N =bah

.

Step 2. For i = 0, 1, . . . , N 1 do

Compute yi+1 = yi + hf(ti, yi)

End

Example 11.7

Using Eulers method, solve numerically y = t2 + 5, 0 t 1, with h = 0.25.

Input Data:

f(t, y) = t2 + 5

t0 = a = 0, t1 = t0 + h = 0.25, t2 = t1 + h = 0.50

t3 = t2 + h = 0.75, t4 = b = t3 + h = 1.

y0 = y(t0) = y(0) = 0.

Find:

y1 = approximation of y(t1) = y(0.25)

y2 = approximation of y(t2) = y(0.50)

y3 = approximation of y(t3) = y(0.75)

y4 = approximation of y(t4) = y(1)

The exact solution obtained by direct integration:

y(t) =t3

3+ 5t.

Solution.

i = 0: Compute y1 from y0 (Set i = 0 in Step 2) of Algorithm 11.6:

• 11.2. THE EULER METHOD 497

y1 = y0 + hf(t0, y0) = 0 + 0.25(5)

= 1.25 (Approximate value of y(0.25))

Actual value of y(0.25) = 1.2552.

i = 1: Compute y2 from y1 (set i = 1 in Step 2).

y2 = y1 + hf(t1, y1)

= 1.25 + 0.25(t21 + 5) = 1.25 + 0.25((0.25)2 + 5)

= 2.5156 (Approximate value of y(0.50))

Note: of y(0.5) = 2.5417. (Four significant digits)

i = 2 : Compute y3 from y2 (Set i = 2 in Step 2).

y3 = y2 + hf(t2, y2)

= 2.5156 + 0.25((0.5)2 + 5)

= 3.8281 (Approximate value of y(0.75))

Actual value of y(0.75) = 3.8696.

Etc.

Example 11.8

y = t2 + 5, 0 t 2

y(0) = 0, h = 0.5

Solution.

i = 0 : Compute y1 from y0 (Set i = 0 in Step 2 of Algorithm 11.6).

y1 = y0 + hf(t0, y0) = y(0) + hf(0, 0) = 0 + 0.5 5

= 2.5 (Approximate value of y(0.50))

Note:y(0.50) = 2.5417 (four significant digists)

i = 1 : Compute y2 from y1 (Set i = 1 in Step 2 in Algorithm 11.6).

• 498 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP

y2 = y1 + hf(t1, y1) = 2.5 + 0.5((0.5)2 + 5)

= 5.1250 (Approximate value of y(1))

Note: y(1) = 5.3333 (four significant digits).

i = 3 : Compute y3 from y2 (Set i = 2 in Step 2).

y3 = y2 + hf(t2, y2) = 5.1250 + 0.5(t22 + 5) = 5.1250 + 0.5(1 + 5)

= 8.1250 (Approximate value of y(1.5))

Note: of y(1.5) = 8.6250 (four significant digits).

Etc.

11.2.1 The Errors in Eulers Method

The approximations obtained by a numerical method to solve the IVP are usually subjected to

three types of errors:

Local Truncation Error

Global Truncation Error

Round-off Error

We will not consider the round-off error for our discussions below.

Definition 11.9. The local error is the error made at a single step due to the truncation

of the series used to solve the problem.

Recall that the Euler Method was obtained by truncating the Taylor series

y(ti+1) = y(ti) + hy(ti) +

h2

2y(ti) +

after two terms. Thus, in obtaining Eulers method, the first term neglected is h2

2 y(ti).

The local error in Eulers method is:

• 11.2. THE EULER METHOD 499

ELE =h2

2y

(i), where i lies between ti and ti+1.

In this case, we say that the local error is of order h2, written as O(h2). Note that the local

error ELE converges to zero as h 0. That means, the smaller h is, the better accuracy will be.

Definition 11.10. The global error is the difference between the true solution y(ti) and the

approximate solution yi at t = ti. Thus, Global error = y(ti) yi. Denote this by EGE .

The following theorem shows that the global error, EGE , is of order h.

Theorem 11.11 (Global Error Bound for the Euler Method). (i) Let y(t) be the unique

solution of the IVP: y = f(t, y); y(a) = , where

a t b, < y < .

(ii) Let L and M be two numbers such that

f(t, y)

y

L, and |y(t)| M in [a, b].

Then the global error EGE at t = ti satisfies the following inequality: