Initial-Value Problems for Ordinary Differential Equations

Initial-Value Problems for Ordinary Differential Equations

Tsung-Min Hwang

Department of Mathematics National Taiwan Normal University, Taiwan

August 27, 2005

Outline

1 Euler’s Method Algorithm Error analysis

2 Higher-order Taylor methods Taylor methods

Outline

1 Euler’s Method Algorithm Error analysis

2 Higher-order Taylor methods Taylor methods

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

t_i =a + ih, ∀ i = 0, 1, 2, . . . , n.

Recall the Taylor’s Theorem

y (t_i+1) = y (t_i) + (t_i+1− t_i)y⁰(t_i) +(t_i+1− t_i)² 2 y⁰⁰(ξ_i)

= y (t_i) +hy⁰(t_i) +h²

2y⁰⁰(ξ_i) (1)

= y (t_i) +hf (t_i,y (t_i)) + h² 2 y⁰⁰(ξi) for some ξ_i ∈ (ti,t_i+1).

Algorithm

We have the formulation of Euler’s method t_{k +1} = t_k+h,

y_{k +1} = y_k +hf (t_k,y_k), 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 .

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

Error analysis

Lemma

0 ≤ (1 + x )^m ≤ e^mx, ∀x ≥ −1, m > 0.

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

2x²e^ξ, where ξ is between x and zero. Thus

0 ≤ 1 + x ≤ 1 + x + 1

2x²e^ξ=e^x

⇒ 0 ≤ (1 + x)^m≤ e^mx

Error analysis

Lemma

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

then

a_i+1≤ e^(i+1)s

 a₀+ t

s



− t s.

Error analysis

Proof:

a_i+1 ≤ (1 + s)a_i+t

≤ (1 + s) [(1 + s)ai−1+t] + t

≤ (1 + s) {(1 + s) [(1 + s)a_i−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

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) −y_i| ≤ hM 2L

h

e^L(ti−a)− 1i

, ∀i = 0, 1, . . . , n.

Error analysis

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

For i = 0, 1, . . . , n − 1,

y (t_i+1) =y (t_i) +hf (t_i,y (t_i)) + h² 2y⁰⁰(ξ_i) and

y_i+1=y_i+hf (t_i,y_i).

Consequently,

y (t_i+1) −y_i+1=y (t_i) −y_i+h [f (t_i,y (t_i)) −f (t_i,y_i)] + h² 2y⁰⁰(ξ_i) and

|y (t_i+1) −y_i+1| ≤ |y (t_i) −y_i| + h|f (t_i,y (t_i)) −f (t_i,y_i)| + h²

2 |y⁰⁰(ξ_i)|

≤ (1 + hL)|y (t_i) −y_i| +h²M 2

Error analysis

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

|y (t_i+1) −y_i+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 = t_i+1− t0=t_i+1− a.

Taylor methods

Definition (Local truncation error) The difference method

y₀ = α,

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

τ_i+1(h) = y (t_i+1) − [y (t_i) +hφ(t_i,y (t_i))]

h

= y (t_i+1) −y (t_i)

h − φ(t_i,y (t_i)),

∀ i = 0, 1, . . . , n − 1.

Taylor methods

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

τ_i+1(h) = y (t_i+1) −y (t_i)

h − f (t_i,y (t_i))

= y(t_i) +hy⁰(t_i) +h²y⁰⁰(ξ_i) − y(t_i)

h − f (ti,y (t_i))

= h

2y⁰⁰(ξi) for some ξ_i ∈ (ti,t_i+1).

If |y⁰⁰(t)| ≤ M ∀ t ∈ [a, b], then

|τ_i+1(h)| ≤ h 2M,

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

Taylor methods

To improve the convergence of difference methods, one way is selected difference-equations in such that their local truncation errors are O(h^p)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 t_i,

y (t_i+1) = y (t_i) +hy⁰(t_i) +h²

2 y⁰⁰(t_i) + · · · + hⁿ

n!y⁽ⁿ⁾(t_i) + hⁿ⁺¹

(n + 1)!y⁽ⁿ⁺¹⁾(ξ_i) for some ξ_i ∈ (t_i,t_i+1). Since

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

...

y^{(k )}(t) = f^{(k −1)}(t, y ),

Taylor methods

we get

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

2 f⁰(t_i,y (t_i)) + · · · (2) +hⁿ

n!f⁽ⁿ⁻¹⁾(t_i,y (t_i)) + hⁿ⁺¹

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

y₀ = α,

y_i+1 = y_i+hT⁽ⁿ⁾(t_i,y_i), ∀i = 0, 1, . . . , n − 1, where

T⁽ⁿ⁾(t_i,y_i) =f (t_i,y_i) +h

2f⁰(t_i,y_i) + · · · +hⁿ⁻¹

n! f⁽ⁿ⁻¹⁾(t_i,y_i).

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.

Taylor methods

So

T⁽²⁾(t_i,y_i) = f (t_i,y_i) +h

2f⁰(t_i,y_i)

= y_i− t_i²+1 + h 2



y_i− t_i²− 2ti+1

=

 1 + h

2



y_i− t_i²+1

− hti

and

T⁽⁴⁾(t_i,y_i) = f (t_i,y_i) +h

2f⁰(t_i,y_i) +h²

6 f⁰⁰(t_i,y_i) + h³

24f⁰⁰⁰(t_i,y_i)

= y_i − t₁²+1 + h 2



y_i− t_i²− 2t_i+1



+h² 6



y_i− t_i²− 2ti− 1 + h³

24



y_i− t_i²− 2ti− 1

Taylor methods

That is T⁽⁴⁾(t_i,y_i) =

 1 + h

2 +h² 6 + h³

24



(y_i− t_i²) −

 1 + h

3 + h² 12

 ht_i +1 + h

2 −h² 6 − h³

24.

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

y_i+1 = y_i+h



1 +h 2



y_i− t_i²+1

− ht_i

 and

y₀ = 0.5, y_i+1 = y_i +h



1 + h 2+ h²

6 + h³ 24



(y_i− t_i²) −

 1 + h

3 + h² 12

 ht_i +1 + h

2− h² 6 − h³

24

 .

Taylor methods

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

• The second-order method:

y₀ = 0.5, y_i+1 = y_i+0.2



1 + 0.2 2



y_i− 0.04i²+1

− 0.04i



= 1.22y_i− 0.0088i²− 0.008i + 0.22.

• The fourth-order method:

y_i+1 = y_i +0.2



1 + 0.2

2 +0.04

6 +0.008 24



y_i− 0.04i²

−



1 +0.2

3 +0.04 12



(0.04i) + 1 +0.2

2 −0.04

6 −0.008 24



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

Taylor methods

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

• The fourth-order results are vastly superior.

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 (t_i+1) − y (ti) −h



f (t_i,y (t_i)) +h

2f⁰(t_i,y (t_i)) + · · · + hⁿ⁻¹

n! f⁽ⁿ⁻¹⁾(t_i,y (t_i))



= hⁿ⁺¹

(n + 1)!f⁽ⁿ⁾(ξ_i,y (ξ_i)) for some ξ_i in (t_i,t_i+1). So the local truncation error is

τ_i+1(h) = y (t_i+1) −y (t_i)

h − T⁽ⁿ⁾(t_i,y (t_i)) = hⁿ

(n + 1)!f⁽ⁿ⁾(ξ_i,y (ξ_i)).

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