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# Initial-Value Problems for Ordinary Differential Equations

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### Initial-Value Problems for OrdinaryDifferential Equations

Tsung-Min Hwang

Department of Mathematics National Taiwan Normal University, Taiwan

August 27, 2005

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### Outline

1 Euler’s Method Algorithm Error analysis

2 Higher-order Taylor methods Taylor methods

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### Outline

1 Euler’s Method Algorithm Error analysis

2 Higher-order Taylor methods Taylor methods

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Algorithm

Obtain an approximation to the initial-value problem dy

dt =f (t, y ), a ≤ t ≤ b, y (a) = α.

Subdivide [a, b] into n subintervals of equal length h = (b − a)/n with mesh points {t0,t1, . . . ,tn} where

ti =a + ih, ∀ i = 0, 1, 2, . . . , n.

Recall the Taylor’s Theorem

y (ti+1) = y (ti) + (ti+1− ti)y0(ti) +(ti+1− ti)2 2 y00i)

= y (ti) +hy0(ti) +h2

2y00i) (1)

= y (ti) +hf (ti,y (ti)) + h2 2 y00i) for some ξi ∈ (ti,ti+1).

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Algorithm

We have the formulation of Euler’s method tk +1 = tk+h,

yk +1 = yk +hf (tk,yk), y0= α.

Example

Use Euler’s method to integrate dy

dx =x − 2y , y (0) = 1.

The exact solution is y = 1

4 h

2x − 1 + 5e−2xi .

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Algorithm

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x

y

Example 1 for Euler method

Exact h = 0.2 h = 0.1 h = 0.05

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Error analysis

Lemma

0 ≤ (1 + x )m ≤ emx, ∀x ≥ −1, m > 0.

Proof: Applying Taylor’s Theorem, ex =1 + x + 1

2x2eξ, where ξ is between x and zero. Thus

0 ≤ 1 + x ≤ 1 + x + 1

2x2eξ=ex

⇒ 0 ≤ (1 + x)m≤ emx

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Error analysis

Lemma

If s, t ∈ R+, {ai}ki=0is a sequence satisfying a0≥ −t/s, and ai+1 ≤ (1 + s)ai+t, ∀ i = 0, 1, . . . , k ,

then

ai+1≤ e(i+1)s

 a0+ t

s



− t s.

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Error analysis

Proof:

ai+1 ≤ (1 + s)ai+t

≤ (1 + s) [(1 + s)ai−1+t] + t

≤ (1 + s) {(1 + s) [(1 + s)ai−2+t] + t} + t ...

≤ (1 + s)i+1a0+h

1 + (1 + s) + (1 + s)2+ · · · (1 + s)ii t

= (1 + s)i+1a0+1 − (1 + s)i+1 1 − (1 + s) t

= (1 + s)i+1

 a0+ t

s



− t s

≤ e(i+1)s

 a0+ t

s



− t s

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Error analysis

Theorem

Suppose f ∈ C(D) and satisfies a Lipschitz condition with constant L on

D = {(t, y )|a ≤ t ≤ b, −∞ < y < ∞}

and ∃ M with

|y00(t)| ≤ M, ∀ t ∈ [a, b].

Let y (t) denote the unique solution to (IVP) y0 =f (t, y ), a ≤ t ≤ b, y (a) = α,

and y0,y1, . . . ,ynbe the approximations generated by Euler’s method. Then

|y (ti) −yi| ≤ hM 2L

h

eL(ti−a)− 1i

, ∀i = 0, 1, . . . , n.

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Error analysis

Proof: Since y (t0) =y0= α, it is true for i = 0.

For i = 0, 1, . . . , n − 1,

y (ti+1) =y (ti) +hf (ti,y (ti)) + h2 2y00i) and

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

Consequently,

y (ti+1) −yi+1=y (ti) −yi+h [f (ti,y (ti)) −f (ti,yi)] + h2 2y00i) and

|y (ti+1) −yi+1| ≤ |y (ti) −yi| + h|f (ti,y (ti)) −f (ti,yi)| + h2

2 |y00i)|

≤ (1 + hL)|y (ti) −yi| +h2M 2

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Error analysis

Referring to previous lemma and letting s = hL, t = h2M/2 and aj = |y (tj) −yj| ∀ j = 0, 1, . . . , n, we see that

|y (ti+1) −yi+1| ≤ e(i+1)hL



|y (t0) −y0| +h2M 2hL



−h2M 2hL.

= hM

2L



e(i+1)hL− 1

= hM 2L



e(ti+1−a)L− 1 since (i + 1)h = ti+1− t0=ti+1− a.

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Taylor methods

Definition (Local truncation error) The difference method

y0 = α,

yi+1 = yi+hφ(ti,yi), ∀i = 0, 1, . . . , n − 1, has local truncation error

τi+1(h) = y (ti+1) − [y (ti) +hφ(ti,y (ti))]

h

= y (ti+1) −y (ti)

h − φ(ti,y (ti)),

∀ i = 0, 1, . . . , n − 1.

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Taylor methods

For example, the local truncation error in Euler’s method at ith step is

τi+1(h) = y (ti+1) −y (ti)

h − f (ti,y (ti))

= y(ti) +hy0(ti) +h2y00i) − y(ti)

h − f (ti,y (ti))

= h

2y00i) for some ξi ∈ (ti,ti+1).

If |y00(t)| ≤ M ∀ t ∈ [a, b], then

i+1(h)| ≤ h 2M,

so the local truncation error in Euler’s method is O(h).

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Taylor methods

To improve the convergence of difference methods, one way is selected difference-equations in such that their local truncation errors are O(hp)for as large a value of p as possible.

Suppose the solution y to (IVP) has (n + 1) continuous derivatives. Consider the nth Taylor polynomial of y (t) at ti,

y (ti+1) = y (ti) +hy0(ti) +h2

2 y00(ti) + · · · + hn

n!y(n)(ti) + hn+1

(n + 1)!y(n+1)i) for some ξi ∈ (ti,ti+1). Since

y0(t) = f (t, y ), y00(t) = f0(t, y ),

...

y(k )(t) = f(k −1)(t, y ),

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Taylor methods

we get

y (ti+1) = y (ti) +hf (ti,y (ti)) + h2

2 f0(ti,y (ti)) + · · · (2) +hn

n!f(n−1)(ti,y (ti)) + hn+1

(n + 1)!f(n)i,y (ξi)). (3) Taylor method of order n

y0 = α,

yi+1 = yi+hT(n)(ti,yi), ∀i = 0, 1, . . . , n − 1, where

T(n)(ti,yi) =f (ti,yi) +h

2f0(ti,yi) + · · · +hn−1

n! f(n−1)(ti,yi).

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Taylor methods

Example

y0 =y − t2+1, 0 ≤ t ≤ 2, y (0) = 0.5.

Consider Taylor’s method of order two and four.

f (t, y ) = y − t2+1, f0(t, y ) = d

dt



y − t2+1

=y0− 2t = y − t2+1 − 2t, f00(t, y ) = d

dt



y − t2+1 − 2t

=y0− 2t − 2

= y − t2+1 − 2t − 2 = y − t2− 2t − 1, f000(t, y ) = d

dt



y − t2− 2t − 1

=y0− 2t − 2

= y − t2+1 − 2t − 2 = y − t2− 2t − 1.

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Taylor methods

So

T(2)(ti,yi) = f (ti,yi) +h

2f0(ti,yi)

= yi− ti2+1 + h 2



yi− ti2− 2ti+1

=

 1 + h

2



yi− ti2+1

− hti

and

T(4)(ti,yi) = f (ti,yi) +h

2f0(ti,yi) +h2

6 f00(ti,yi) + h3

24f000(ti,yi)

= yi − t12+1 + h 2



yi− ti2− 2ti+1



+h2 6



yi− ti2− 2ti− 1 + h3

24



yi− ti2− 2ti− 1

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Taylor methods

That is T(4)(ti,yi) =

 1 + h

2 +h2 6 + h3

24



(yi− ti2) −

 1 + h

3 + h2 12

 hti +1 + h

2 −h2 6 − h3

24.

The Taylor methods of orders two and four are, consequently, y0 = 0.5,

yi+1 = yi+h



1 +h 2



yi− ti2+1

− hti

 and

y0 = 0.5, yi+1 = yi +h



1 + h 2+ h2

6 + h3 24



(yi− ti2) −

 1 + h

3 + h2 12

 hti +1 + h

2− h2 6 − h3

24

 .

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Taylor methods

If h = 0.2, then n = 10 and ti =0.2i ∀ i = 1, 2, . . . , 10.

• The second-order method:

y0 = 0.5, yi+1 = yi+0.2



1 + 0.2 2



yi− 0.04i2+1

− 0.04i



= 1.22yi− 0.0088i2− 0.008i + 0.22.

• The fourth-order method:

yi+1 = yi +0.2



1 + 0.2

2 +0.04

6 +0.008 24



yi− 0.04i2



1 +0.2

3 +0.04 12



(0.04i) + 1 +0.2

2 −0.04

6 −0.008 24



= 1.2214yi − 0.008856i2− 0.00856i + 0.2186.

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Taylor methods

• Exact solution y (t) = (t + 1)2− 0.5et.

• The fourth-order results are vastly superior.

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Taylor methods

Theorem

If y ∈ Cn+1[a, b], then the local truncation error of Taylor’s method of order n is O(hn).

Proof: From Eq. (3), we have y (ti+1) − y (ti) −h



f (ti,y (ti)) +h

2f0(ti,y (ti)) + · · · + hn−1

n! f(n−1)(ti,y (ti))



= hn+1

(n + 1)!f(n)i,y (ξi)) for some ξi in (ti,ti+1). So the local truncation error is

τi+1(h) = y (ti+1) −y (ti)

h − T(n)(ti,y (ti)) = hn

(n + 1)!f(n)i,y (ξi)).

Since y ∈ Cn+1[a, b], we have y(n+1)(t) = f(n)(t, y (t)) bounded on [a, b] and τi =O(hn), ∀i = 1, 2, . . . , N.

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