BVP of ODE

(1)

BVP of ODE

¹

Boundary-Value Problems for Ordinary Differential Equations

NTNU

Tsung-Min Hwang December 20, 2003

Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

(2)

BVP of ODE

²

1 – Mathematical Theories . . . 4

2 – Finite Difference Method For Linear Problems . . . 15

2.1 – The Finite Difference Formulation . . . 15

2.2 – Convergence Analysis . . . 19

3 – Shooting Methods . . . 22

(3)

BVP of ODE

³

The two-point boundary-value problems (BVP) considered in this chapter involve a

second-order differential equation together with boundary condition in the following form:







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y(b) = β

(1)

The numerical procedures for finding approximate solutions to the initial-value problems can not be adapted to the solution of this problem since the specification of conditions involve two different points,

x = a

^and

x = b

. New techniques are introduced in this chapter for handling problems (1) in which the conditions imposed are of a boundary-value rather than an initial-value type.

(4)

BVP of ODE

³







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y(b) = β

(1)

x = a

^and

x = b

^.

New techniques are introduced in this chapter for handling problems (1) in which the conditions imposed are of a boundary-value rather than an initial-value type.

(5)

BVP of ODE

³







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y(b) = β

(1)

x = a

^and

x = b

. New techniques are introduced in this chapter for handling problems (1) in which the conditions imposed are of a boundary-value rather than an initial-value type.

(6)

BVP of ODE

⁴

1 – Mathematical Theories

Before considering numerical methods, a few mathematical theories about the two-point boundary-value problem (1), such as the existence and uniqueness of solution, shall be discussed in this section.

Theorem 1 Suppose that

f

in (1) is continuous on the set

D = {(x, y, y

⁰

)|a ≤ x ≤ b, −∞ < y < ∞, −∞ < y

⁰

< ∞} ,

and that

∂f

∂y

^and

∂f

∂y

⁰ are also continuous on

D

^{. If}

1.

∂f

∂y (x, y, y

⁰

) > 0

^{for all}

(x, y, y

⁰

) ∈ D

^{, and}

2. a constant

M

exists, with

∂f

∂y

⁰

(x, y, y

⁰

)

≤ M

^,

∀ (x, y, y

⁰

) ∈ D

^,

then (1) has a unique solution.

(7)

BVP of ODE

⁵

When the function

f (x, y, y

⁰

)

has the special form

f (x, y, y

⁰

) = p(x)y

⁰

+ q(x)y + r(x),

the differential equation become a so-called linear problem. The previous theorem can be simplified for this case.

Corollary 1 If the linear two-point boundary-value problem







y

⁰⁰

= p(x)y

⁰

+ q(x)y + r(x) y(a) = α, y(b) = β

satisfies

1.

p(x), q(x)

^{, and}

r(x)

are continuous on

[a, b]

^{, and}

2.

q(x) > 0

^on

[a, b]

^,

then the problem has a unique solution.

(8)

BVP of ODE

⁶

Many theories and application models consider the boundary-value problem in a special form as follows.







y

⁰⁰

= f (x, y)

y(0) = 0, y(1) = 0

We will show that this simple form can be derived from the original problem by simple

techniques such as changes of variables and linear transformation. To do this, we begin by changing the variable. Suppose that the original problem is







y

⁰⁰

= f (x, y)

y(a) = α, y(b) = β

⁽²⁾

where

y = y(x)

^{. Now let}

λ = b − a

and define a new variable

t = x − a

b − a = 1

λ (x − a).

(9)

BVP of ODE

⁷

That is,

x = a + λt

. Notice that

t = 0

corresponds to

x = a

^{, and}

t = 1

corresponds to

x = b

. Then we define

z(t) = y(a + λt) = y(x)

with

λ = b − a

. This gives

z

⁰

(t) = d

dt z(t) = d

dt y(a + λt) = d

dx y(x) d

dt (a + λt)

= λy

⁰

(x)

and, analogously,

z

⁰⁰

(t) = d

dt z

⁰

(t) = λ

²

y

⁰⁰

(x) = λ

²

f (x, y(x)) = λ

²

f (a + λt, z(t)).

Likewise the boundary conditions are changed to

z(0) = y(a) = α

^and

z(1) = y(b) = β.

(10)

BVP of ODE

⁸

With all these together, the problem (2) is transformed into







z

⁰⁰

(t) = λ

²

f (a + λt, z(t))

z(0) = α, z(1) = β

⁽³⁾

Thus, if

y(x)

is a solution for (2), then

z(t) = y(a + λt)

is a solution for the boundary-value problem (3). Conversely, if

z(t)

is a solution for (3), then

y(x) = z(

_λ¹

(x − a))

is a solution for (2).

Example 1 Simplify the boundary conditions of the following equation by use of changing variables.







y

⁰⁰

= sin(xy) + y

²

y(1) = 3, y(4) = 7

Solution: In this problem

a = 1, b = 4

^{, hence}

λ = 3

. Now define the new variable

t =

¹₃

(x − 1)

^{, hence}

x = 1 + 3t

^{, and let}

z(t) = y(x) = y(1 + 3t)

^{. Then}

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BVP of ODE

⁹

λ

²

f (a + λt, z) = 9 sin(1 + 3t)z + z

²

,

and the original equation is reduced to







z

⁰⁰

(t) = 9 sin((1 + 3t)z) + 9z

²

z(0) = 3, z(1) = 7

To reduce a two-point boundary-value problem







z

⁰⁰

(t) = g(t, z)

z(0) = α, z(1) = β

to a homogeneous system, let

u(t) = z(t) − [α + (β − α)t]

then

u

⁰⁰

(t) = z

⁰⁰

(t)

^{, and}

u(0) = z(0) − α = 0

^and

u(1) = z(1) − β = 0

(12)

BVP of ODE

¹⁰

Moreover,

g(t, z) = g(t, u + α + (β − α)t) ≡ h(t, u).

The system is now transformed into







u

⁰⁰

(t) = h(t, u)

u(0) = 0, u(1) = 0

Example 2 Reduce the system







z

⁰⁰

= [5z − 10t + 35 + sin(3z − 6t + 21)]e

^t

z(0) = −7, z(1) = −5

to a homogeneous problem by linear transformation technique.

Solution: Let

u(t) = z(t) − [−7 + (−5 + 7)t] = z(t) − 2t + 7.

(13)

BVP of ODE

¹¹

Then

z(t) = u(t) + 2t − 7

^{, and}

u

⁰⁰

= z

⁰⁰

= [5z − 10t + 35 + sin(3z − 6t + 21)]e

^t

= [5(u + 2t − 7) − 10t + 35 + sin(3(u + 2t − 7) − 6t + 21)]e

^t

= [5u + sin(3u)]e

^t

The system is transformed to







u

⁰⁰

(t) = [5u + sin(3u)]e

^t

u(0) = u(1) = 0

Example 3 Reduce the following two-point boundary-value problem







y

⁰⁰

= y

²

+ 3 − x

²

+ xy y(3) = 7, y(5) = 9

to a homogeneous system.

(14)

BVP of ODE

¹²

Solution: In the original system,

a = 3, b = 5, α = 7, β = 9

^{. Let}

λ = b − a = 2

^and

define a new variable

t = 1

2 (x − 3) =⇒ x = 2t + 3.

Let the function

z(t) = y(x) = y(2t + 3)

^{. Then}

z

⁰⁰

(t) = λ

²

y

⁰⁰

(2t + 3) = λ

²

f (2t + 3, u)

= 4[z

²

+ 3 − (2t + 3)

²

+ (2t + 3)z]

= 4[z

²

+ 3z + 2tz − 4t

²

− 12t − 6]

The original problem is first transformed into







z

⁰⁰

(t) = 4[z

²

+ 3z + 2tz − 4t

²

− 12t − 6]

z(0) = 7, z(1) = 9

Next let

u(t) = z(t) − [7 + 2t],

or equivalently,

z(t) = u(t) + 2t + 7.

(15)

BVP of ODE

¹³

Then

u

⁰⁰

(t) = 4[(z + 2t + 7)

²

+ 3(u + 2t + 7) + 2t(u + 2t + 7) − 4t

²

− 12t − 6]

= 4[u

²

+ 6tu + 17u + 4t

²

+ 36t + 64].

The original problem is transformed into the following homogeneous system







u

⁰⁰

(t) = 4[u

²

+ 6tu + 17u + 4t

²

+ 36t + 64]

u(0) = u(1) = 0

Theorem 2 The boundary-value problem







y

⁰⁰

= f (x, y)

y(0) = 0, y(1) = 0

has a unique solution if

∂f

∂y

is continuous, non-negative, and bounded in the strip

0 ≤ x ≤ 1

^and

−∞ < y < ∞

^.

(16)

BVP of ODE

¹⁴

Theorem 3 If

f

is a continuous function of

(s, t)

in the domain

0 ≤ s ≤ 1

^and

−∞ < t < ∞

^{such that}

|f (s, t

₁

) − f (s, t

₂

)| ≤ K|t

₁

− t

₂

|, (K < 8).

Then the boundary-value problem







y

⁰⁰

= f (x, y)

y(0) = 0, y(1) = 0

has a unique solution in

C[0, 1]

^.

(17)

BVP of ODE

¹⁵

2 – Finite Difference Method For Linear Problems

We consider finite difference method for solving the linear two-point boundary-value problem of the form







y

⁰⁰

= p(x)y

⁰

+ q(x)y + r(x)

y(a) = α, y(b) = β.

⁽⁴⁾

Methods involving finite differences for solving boundary-value problems replace each of the derivatives in the differential equation by an appropriate difference-quotient approximation.

2.1 – The Finite Difference Formulation

First, partition the interval

[a, b]

^into

n

equally-spaced subintervals by points

a = x

₀

< x

₁

< . . . < x

n

< x

n

= b

. Each mesh point

x

i can be computed by

x

i

= a + i ∗ h, i = 0, 1, . . . , n,

^with

h = b − a n

where

h

is called the mesh size. Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

(18)

BVP of ODE

¹⁵

2 – Finite Difference Method For Linear Problems







y

⁰⁰

= p(x)y

⁰

+ q(x)y + r(x)

y(a) = α, y(b) = β.

⁽⁴⁾

2.1 – The Finite Difference Formulation

[a, b]

^into

n

a = x

₀

< x

₁

< . . . < x

n

< x

n

= b

. Each mesh point

x

i

= a + i ∗ h, i = 0, 1, . . . , n,

^with

h = b − a n

where

h

(19)

BVP of ODE

¹⁵

2 – Finite Difference Method For Linear Problems







y

⁰⁰

= p(x)y

⁰

+ q(x)y + r(x)

y(a) = α, y(b) = β.

⁽⁴⁾

2.1 – The Finite Difference Formulation

[a, b]

^into

n

a = x

₀

< x

₁

< . . . < x

n

< x

n

= b

. Each mesh point

x

i

= a + i ∗ h, i = 0, 1, . . . , n,

^with

h = b − a n

where

h

(20)

BVP of ODE

¹⁵

2 – Finite Difference Method For Linear Problems







y

⁰⁰

= p(x)y

⁰

+ q(x)y + r(x)

y(a) = α, y(b) = β.

⁽⁴⁾

2.1 – The Finite Difference Formulation

[a, b]

^into

n

a = x

₀

< x

₁

< . . . < x

n

< x

n

= b

^.

Each mesh point

x

i

= a + i ∗ h, i = 0, 1, . . . , n,

^with

h = b − a n

where

h

(21)

BVP of ODE

¹⁵

2 – Finite Difference Method For Linear Problems







y

⁰⁰

= p(x)y

⁰

+ q(x)y + r(x)

y(a) = α, y(b) = β.

⁽⁴⁾

2.1 – The Finite Difference Formulation

[a, b]

^into

n

a = x

₀

< x

₁

< . . . < x

n

< x

n

= b

. Each mesh point

x

i

= a + i ∗ h, i = 0, 1, . . . , n,

^with

h = b − a n

where

h

is called the mesh size.

(22)

BVP of ODE

¹⁶

At the interior mesh points,

x

i, for

i = 1, 2, . . . , n − 1

, the differential equation to be approximated satisfies

y

⁰⁰

(x

i

) = p(x

i

)y

⁰

(x

i

) + q(x

i

)y(x

i

) + r(x

i

).

⁽⁵⁾

The central finite difference formulae

y

⁰

(x

i

) = y(x

i+1

) − y(x

i−1

)

2h − h

²

6 y

⁽³⁾

(η

i

),

⁽⁶⁾

for some

η

i in the interval

(x

i−1

, x

i+1

)

^{, and}

y

⁰⁰

(x

i

) = y(x

i+1

) − 2y(x

i

) + y(x

i−1

)

h

²

− h

²

12 y

⁽⁴⁾

(ξ

i

),

⁽⁷⁾

for some

ξ

i in the interval

(x

i−1

, x

i+1

)

, can be derived from Taylor’s theorem by expanding

y

^about

x

i.

Let

u

i denote the approximate value of

y

i

= y(x

i

)

^{. If}

y ∈ C

⁴

[a, b]

, then a finite difference method with truncation error of order

O(h

²

)

can be obtained by using the approximations Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

(23)

BVP of ODE

¹⁶

x

i, for

i = 1, 2, . . . , n − 1

y

⁰⁰

(x

i

) = p(x

i

)y

⁰

(x

i

) + q(x

i

)y(x

i

) + r(x

i

).

⁽⁵⁾

y

⁰

(x

i

) = y(x

i+1

) − y(x

i−1

)

2h − h

²

6 y

⁽³⁾

(η

i

),

⁽⁶⁾

for some

η

i in the interval

(x

i−1

, x

i+1

)

^,

and

y

⁰⁰

(x

i

) = y(x

i+1

) − 2y(x

i

) + y(x

i−1

)

h

²

− h

²

12 y

⁽⁴⁾

(ξ

i

),

⁽⁷⁾

for some

ξ

i in the interval

(x

i−1

, x

i+1

)

y

^about

x

i.

Let

u

y

i

= y(x

i

)

^{. If}

y ∈ C

⁴

[a, b]

O(h

²

)

(24)

BVP of ODE

¹⁶

x

i, for

i = 1, 2, . . . , n − 1

y

⁰⁰

(x

i

) = p(x

i

)y

⁰

(x

i

) + q(x

i

)y(x

i

) + r(x

i

).

⁽⁵⁾

y

⁰

(x

i

) = y(x

i+1

) − y(x

i−1

)

2h − h

²

6 y

⁽³⁾

(η

i

),

⁽⁶⁾

for some

η

i in the interval

(x

i−1

, x

i+1

)

^{, and}

y

⁰⁰

(x

i

) = y(x

i+1

) − 2y(x

i

) + y(x

i−1

)

h

²

− h

²

12 y

⁽⁴⁾

(ξ

i

),

⁽⁷⁾

for some

ξ

i in the interval

(x

i−1

, x

i+1

)

y

^about

x

i.

Let

u

y

i

= y(x

i

)

^{. If}

y ∈ C

⁴

[a, b]

O(h

²

)

(25)

BVP of ODE

¹⁶

x

i, for

i = 1, 2, . . . , n − 1

y

⁰⁰

(x

i

) = p(x

i

)y

⁰

(x

i

) + q(x

i

)y(x

i

) + r(x

i

).

⁽⁵⁾

y

⁰

(x

i

) = y(x

i+1

) − y(x

i−1

)

2h − h

²

6 y

⁽³⁾

(η

i

),

⁽⁶⁾

for some

η

i in the interval

(x

i−1

, x

i+1

)

^{, and}

y

⁰⁰

(x

i

) = y(x

i+1

) − 2y(x

i

) + y(x

i−1

)

h

²

− h

²

12 y

⁽⁴⁾

(ξ

i

),

⁽⁷⁾

for some

ξ

i in the interval

(x

i−1

, x

i+1

)

y

^about

x

i.

Let

u

y

i

= y(x

i

)

^.

If

y ∈ C

⁴

[a, b]

O(h

²

)

(26)

BVP of ODE

¹⁶

x

i, for

i = 1, 2, . . . , n − 1

y

⁰⁰

(x

i

) = p(x

i

)y

⁰

(x

i

) + q(x

i

)y(x

i

) + r(x

i

).

⁽⁵⁾

y

⁰

(x

i

) = y(x

i+1

) − y(x

i−1

)

2h − h

²

6 y

⁽³⁾

(η

i

),

⁽⁶⁾

for some

η

i in the interval

(x

i−1

, x

i+1

)

^{, and}

y

⁰⁰

(x

i

) = y(x

i+1

) − 2y(x

i

) + y(x

i−1

)

h

²

− h

²

12 y

⁽⁴⁾

(ξ

i

),

⁽⁷⁾

for some

ξ

i in the interval

(x

i−1

, x

i+1

)

y

^about

x

i.

Let

u

y

i

= y(x

i

)

^{. If}

y ∈ C

⁴

[a, b]

O(h

²

)

can be obtained by using the approximations

(27)

BVP of ODE

¹⁷

y

⁰

(x

i

) ≈ u

i+1

− u

i−1

2h

^and

y

⁰⁰

(x

i

) ≈ u

i+1

− 2u

i

+ u

i−1

h

²

for

y

⁰

(x

i

)

^and

y

⁰⁰

(x

i

)

, respectively.

Furthermore, let

p

i

= p(x

i

), q

i

= q(x

i

), r

i

= r(x

i

).

The discrete version of equation (4) is then

u

i+1

− 2u

i

+ u

i−1

h

²

= p

i

u

i+1

− u

i−1

2h + q

i

u

i

+ r

i

, i = 1, 2, . . . , n − 1,

⁽⁸⁾

together with boundary conditions

u

₀

= α

^and

u

n

= β

. Equation (8) can be written in the

form

1 + h 2 p

i

u

i−1

− 2 + h

²

q

i

u

i

+

1 − h 2 p

i

u

i+1

= h

²

r

i

,

⁽⁹⁾

for

i = 1, 2, . . . , n − 1

^{. In (8),}

u

₁

, u

₂

, . . . , u

n−1 are the unknown, and there are

n − 1

linear equations to be solved. The resulting system of linear equations can be expressed in the matrix form

Au = f,

⁽¹⁰⁾

(28)

BVP of ODE

¹⁷

y

⁰

(x

i

) ≈ u

i+1

− u

i−1

2h

^and

y

⁰⁰

(x

i

) ≈ u

i+1

− 2u

i

+ u

i−1

h

²

for

y

⁰

(x

i

)

^and

y

⁰⁰

(x

i

)

, respectively. Furthermore, let

p

i

= p(x

i

), q

i

= q(x

i

), r

i

= r(x

i

).

u

i+1

− 2u

i

+ u

i−1

h

²

= p

i

u

i+1

− u

i−1

2h + q

i

u

i

+ r

i

, i = 1, 2, . . . , n − 1,

⁽⁸⁾

u

₀

= α

^and

u

n

= β

form

1 + h 2 p

i

u

i−1

− 2 + h

²

q

i

u

i

+

1 − h 2 p

i

u

i+1

= h

²

r

i

,

⁽⁹⁾

for

i = 1, 2, . . . , n − 1

^{. In (8),}

u

₁

, u

₂

, . . . , u

n − 1

Au = f,

⁽¹⁰⁾

(29)

BVP of ODE

¹⁷

y

⁰

(x

i

) ≈ u

i+1

− u

i−1

2h

^and

y

⁰⁰

(x

i

) ≈ u

i+1

− 2u

i

+ u

i−1

h

²

for

y

⁰

(x

i

)

^and

y

⁰⁰

(x

i

)

p

i

= p(x

i

), q

i

= q(x

i

), r

i

= r(x

i

).

u

i+1

− 2u

i

+ u

i−1

h

²

= p

i

u

i+1

− u

i−1

2h + q

i

u

i

+ r

i

, i = 1, 2, . . . , n − 1,

⁽⁸⁾

u

₀

= α

^and

u

n

= β

^.

Equation (8) can be written in the

form

1 + h 2 p

i

u

i−1

− 2 + h

²

q

i

u

i

+

1 − h 2 p

i

u

i+1

= h

²

r

i

,

⁽⁹⁾

for

i = 1, 2, . . . , n − 1

^{. In (8),}

u

₁

, u

₂

, . . . , u

n − 1

Au = f,

⁽¹⁰⁾

(30)

BVP of ODE

¹⁷

y

⁰

(x

i

) ≈ u

i+1

− u

i−1

2h

^and

y

⁰⁰

(x

i

) ≈ u

i+1

− 2u

i

+ u

i−1

h

²

for

y

⁰

(x

i

)

^and

y

⁰⁰

(x

i

)

p

i

= p(x

i

), q

i

= q(x

i

), r

i

= r(x

i

).

u

i+1

− 2u

i

+ u

i−1

h

²

= p

i

u

i+1

− u

i−1

2h + q

i

u

i

+ r

i

, i = 1, 2, . . . , n − 1,

⁽⁸⁾

u

₀

= α

^and

u

n

= β

form

1 + h 2 p

i

u

i−1

− 2 + h

²

q

i

u

i

+

1 − h 2 p

i

u

i+1

= h

²

r

i

,

⁽⁹⁾

for

i = 1, 2, . . . , n − 1

^.

In (8),

u

₁

, u

₂

, . . . , u

n − 1

Au = f,

⁽¹⁰⁾

(31)

BVP of ODE

¹⁷

y

⁰

(x

i

) ≈ u

i+1

− u

i−1

2h

^and

y

⁰⁰

(x

i

) ≈ u

i+1

− 2u

i

+ u

i−1

h

²

for

y

⁰

(x

i

)

^and

y

⁰⁰

(x

i

)

p

i

= p(x

i

), q

i

= q(x

i

), r

i

= r(x

i

).

u

i+1

− 2u

i

+ u

i−1

h

²

= p

i

u

i+1

− u

i−1

2h + q

i

u

i

+ r

i

, i = 1, 2, . . . , n − 1,

⁽⁸⁾

u

₀

= α

^and

u

n

= β

form

1 + h 2 p

i

u

i−1

− 2 + h

²

q

i

u

i

+

1 − h 2 p

i

u

i+1

= h

²

r

i

,

⁽⁹⁾

for

i = 1, 2, . . . , n − 1

^{. In (8),}

u

₁

, u

₂

, . . . , u

n − 1

linear equations to be solved.

The resulting system of linear equations can be expressed in the matrix form

Au = f,

⁽¹⁰⁾

(32)

BVP of ODE

¹⁷

y

⁰

(x

i

) ≈ u

i+1

− u

i−1

2h

^and

y

⁰⁰

(x

i

) ≈ u

i+1

− 2u

i

+ u

i−1

h

²

for

y

⁰

(x

i

)

^and

y

⁰⁰

(x

i

)

p

i

= p(x

i

), q

i

= q(x

i

), r

i

= r(x

i

).

u

i+1

− 2u

i

+ u

i−1

h

²

= p

i

u

i+1

− u

i−1

2h + q

i

u

i

+ r

i

, i = 1, 2, . . . , n − 1,

⁽⁸⁾

u

₀

= α

^and

u

n

= β

form

1 + h 2 p

i

u

i−1

− 2 + h

²

q

i

u

i

+

1 − h 2 p

i

u

i+1

= h

²

r

i

,

⁽⁹⁾

for

i = 1, 2, . . . , n − 1

^{. In (8),}

u

₁

, u

₂

, . . . , u

n − 1

Au = f,

⁽¹⁰⁾

(33)

BVP of ODE

¹⁸

where A =







−2 − h²q₁ 1 −

h 2 p₁ 1 + ^h

2p₂ −2 − h²q₂ 1 −

h 2p₂

. .. . .. . ..

1 + ^h

2pⁿ₋₂ −2 − h²qⁿ₋₂ 1 −

h

2pⁿ₋₂ 1 + ^h

2pⁿ₋₁ −2 − h²qⁿ₋₁





 ,

u =







u

₁

u

₂

.. .

u

n−2

u

n−1







,

^and

f =







h

²

r

₁

− 1 +

^h₂

p

₁

α h

²

r

₂

.. .

h

²

r

n−2

h

²

r

n−1

− 1 −

^h₂

p

n−1

β







(34)

BVP of ODE

¹⁹

Since the matrix

A

is tridiagonal, this system can be solved by a special Gaussian elimination in

O(n

²

)

^flops.

p(x), q(x)

^{, and}

r(x)

in (4) are continuous on

[a, b]

^{, and}

q(x) > 0

^on

[a, b]

. Then (10) has a unique solution provided that

h < 2/L

^{, where}

L = max

_a≤x≤b

|p(x)|

^.

2.2 – Convergence Analysis

We shall analyze that when

h

converges to zero, the solution

u

i of the discrete problem (8) converges to the solution

y

i of the original continuous problem (5).

y

i satisfies the following system of equations

y

i+1

− 2y

i

+ y

i−1

h

²

− h

²

12 y

⁽⁴⁾

(ξ

i

) = p

i

y

i+1

− y

i−1

2h − h

²

6 y

⁽³⁾

(η

i

)

+ q

i

y

i

+ r

i

,

(11) for

i = 1, 2, . . . , n − 1

. Subtract (8) from (11) and let

e

i

= y

i

− u

i, the result is

e

_i+1

− 2e

i

+ e

_i−1

h

²

= p

i

e

_i+1

− e

_i−1

2h + q

i

e

i

+ h

²

g

i

, i = 1, 2, . . . , n − 1,

(35)

BVP of ODE

¹⁹

Since the matrix

A

O(n

²

)

^flops.

p(x), q(x)

^{, and}

r(x)

[a, b]

^{, and}

q(x) > 0

^on

[a, b]

h < 2/L

^{, where}

L = max

_a≤x≤b

|p(x)|

^.

2.2 – Convergence Analysis

h

u

y

i+1

− 2y

i

+ y

i−1

h

²

− h

²

12 y

⁽⁴⁾

(ξ

i

) = p

i

y

i+1

− y

i−1

2h − h

²

6 y

⁽³⁾

(η

i

)

+ q

i

y

i

+ r

i

,

(11) for

i = 1, 2, . . . , n − 1

e

i

= y

i

− u

i, the result is

e

_i+1

− 2e

i

+ e

_i−1

h

²

= p

i

e

_i+1

− e

_i−1

2h + q

i

e

i

+ h

²

g

i

, i = 1, 2, . . . , n − 1,

(36)

BVP of ODE

¹⁹

Since the matrix

A

O(n

²

)

^flops.

p(x), q(x)

^{, and}

r(x)

[a, b]

^{, and}

q(x) > 0

^on

[a, b]

h < 2/L

^{, where}

L = max

_a≤x≤b

|p(x)|

^.

2.2 – Convergence Analysis

h

u

y

i+1

− 2y

i

+ y

i−1

h

²

− h

²

12 y

⁽⁴⁾

(ξ

i

) = p

i

y

i+1

− y

i−1

2h − h

²

6 y

⁽³⁾

(η

i

)

+ q

i

y

i

+ r

i

,

(11) for

i = 1, 2, . . . , n − 1

e

i

= y

i

− u

i, the result is

e

_i+1

− 2e

i

+ e

_i−1

h

²

= p

i

e

_i+1

− e

_i−1

2h + q

i

e

i

+ h

²

g

i

, i = 1, 2, . . . , n − 1,

(37)

BVP of ODE

¹⁹

Since the matrix

A

O(n

²

)

^flops.

p(x), q(x)

^{, and}

r(x)

[a, b]

^{, and}

q(x) > 0

^on

[a, b]

h < 2/L

^{, where}

L = max

_a≤x≤b

|p(x)|

^.

2.2 – Convergence Analysis

h

u

y

i+1

− 2y

i

+ y

i−1

h

²

− h

²

12 y

⁽⁴⁾

(ξ

i

) = p

i

y

i+1

− y

i−1

2h − h

²

6 y

⁽³⁾

(η

i

)

+ q

i

y

i

+ r

i

,

(11) for

i = 1, 2, . . . , n − 1

^.

Subtract (8) from (11) and let

e

i

= y

i

− u

i, the result is

e

_i+1

− 2e

i

+ e

_i−1

h

²

= p

i

e

_i+1

− e

_i−1

2h + q

i

e

i

+ h

²

g

i

, i = 1, 2, . . . , n − 1,

(38)

BVP of ODE

¹⁹

Since the matrix

A

O(n

²

)

^flops.

p(x), q(x)

^{, and}

r(x)

[a, b]

^{, and}

q(x) > 0

^on

[a, b]

h < 2/L

^{, where}

L = max

_a≤x≤b

|p(x)|

^.

2.2 – Convergence Analysis

h

u

y

i+1

− 2y

i

+ y

i−1

h

²

− h

²

12 y

⁽⁴⁾

(ξ

i

) = p

i

y

i+1

− y

i−1

2h − h

²

6 y

⁽³⁾

(η

i

)

+ q

i

y

i

+ r

i

,

(11) for

i = 1, 2, . . . , n − 1

e

i

= y

i

− u

i, the result is

e

_i+1

− 2e

i

+ e

_i−1

h

²

= p

i

e

_i+1

− e

_i−1

2h + q

i

e

i

+ h

²

g

i

, i = 1, 2, . . . , n − 1,

(39)

BVP of ODE

²⁰

where

g

i

= 1

12 y

⁽⁴⁾

(ξ

i

) − 1

6 p

i

y

⁽³⁾

(η

i

).

After collecting terms and multiplying by

h

²^{, we have}

1 + h 2 p

i

e

i−1

− 2 + h

²

q

i

e

i

+

1 − h 2 p

i

e

i+1

= h

⁴

g

i

, i = 1, 2, . . . , n − 1.

Let

e = [e

₁

, e

₂

, . . . , e

_n−1

]

^T ^and

|e

k

| = kek

_∞^{. Then}

2 + h

²

q

k

e

k

=

1 + h 2 p

k

e

k−1

+

1 − h 2 p

k

e

k+1

− h

⁴

g

k

,

and, hence

2 + h

²

q

k

|e

k

| ≤

1 + h 2 p

k

|e

k−1

| +

1 − h 2 p

k

|e

k+1

| + h

⁴

|g

k

|

≤

1 + h 2 p

k

kek

_∞

+

1 − h 2 p

k

kek

_∞

+ h

⁴

kgk

_∞

(40)

BVP of ODE

²⁰

where

g

i

= 1

12 y

⁽⁴⁾

(ξ

i

) − 1

6 p

i

y

⁽³⁾

(η

i

).

h

²^{, we have}

1 + h 2 p

i

e

i−1

− 2 + h

²

q

i

e

i

+

1 − h 2 p

i

e

i+1

= h

⁴

g

i

, i = 1, 2, . . . , n − 1.

Let

e = [e

₁

, e

₂

, . . . , e

_n−1

]

^T ^and

|e

k

| = kek

_∞^{. Then}

2 + h

²

q

k

e

k

=

1 + h 2 p

k

e

k−1

+

1 − h 2 p

k

e

k+1

− h

⁴

g

k

,

and, hence

2 + h

²

q

k

|e

k

| ≤

1 + h 2 p

k

|e

k−1

| +

1 − h 2 p

k

|e

k+1

| + h

⁴

|g

k

|

≤

1 + h 2 p

k

kek

_∞

+

1 − h 2 p

k

kek

_∞

+ h

⁴

kgk

_∞

(41)

BVP of ODE

²⁰

where

g

i

= 1

12 y

⁽⁴⁾

(ξ

i

) − 1

6 p

i

y

⁽³⁾

(η

i

).

h

²^{, we have}

1 + h 2 p

i

e

i−1

− 2 + h

²

q

i

e

i

+

1 − h 2 p

i

e

i+1

= h

⁴

g

i

, i = 1, 2, . . . , n − 1.

Let

e = [e

₁

, e

₂

, . . . , e

_n−1

]

^T ^and

|e

k

| = kek

_∞^.

Then

2 + h

²

q

k

e

k

=

1 + h 2 p

k

e

k−1

+

1 − h 2 p

k

e

k+1

− h

⁴

g

k

,

and, hence

2 + h

²

q

k

|e

k

| ≤

1 + h 2 p

k

|e

k−1

| +

1 − h 2 p

k

|e

k+1

| + h

⁴

|g

k

|

≤

1 + h 2 p

k

kek

_∞

+

1 − h 2 p

k

kek

_∞

+ h

⁴

kgk

_∞