3 – Shooting Methods

BVP of ODE

²²

BVP of ODE

²²

3 – Shooting Methods

We consider solving the following 2-point boundary-value problem:







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y(b) = β

⁽¹²⁾

The idea of shooting method for (12) is to solve a related initial-value problem with a guess for

y

⁰

(a)

^{, say}

z

^.

The corresponding IVP







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y

⁰

(a) = z

⁽¹³⁾

can then be solved by, for example, Runge-Kutta method. We denote this approximate

solution

y

z and hope

y

z

(b) = β

. If not, we use another guess for

y

⁰

(a)

, and try to solve an altered IVP (13) again. This process is repeated and can be done systematically.

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

BVP of ODE

²²

3 – Shooting Methods

We consider solving the following 2-point boundary-value problem:







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y(b) = β

⁽¹²⁾

The idea of shooting method for (12) is to solve a related initial-value problem with a guess for

y

⁰

(a)

^{, say}

z

. The corresponding IVP







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y

⁰

(a) = z

⁽¹³⁾

can then be solved by, for example, Runge-Kutta method.

We denote this approximate

solution

y

z and hope

y

z

(b) = β

. If not, we use another guess for

y

⁰

(a)

, and try to solve an altered IVP (13) again. This process is repeated and can be done systematically.

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

BVP of ODE

²²

3 – Shooting Methods

We consider solving the following 2-point boundary-value problem:







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y(b) = β

⁽¹²⁾

The idea of shooting method for (12) is to solve a related initial-value problem with a guess for

y

⁰

(a)

^{, say}

z

. The corresponding IVP







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y

⁰

(a) = z

⁽¹³⁾

can then be solved by, for example, Runge-Kutta method. We denote this approximate solution

y

z and hope

y

z

(b) = β

^.

If not, we use another guess for

y

⁰

(a)

, and try to solve an altered IVP (13) again. This process is repeated and can be done systematically.

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

BVP of ODE

²²

3 – Shooting Methods

We consider solving the following 2-point boundary-value problem:







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y(b) = β

⁽¹²⁾

The idea of shooting method for (12) is to solve a related initial-value problem with a guess for

y

⁰

(a)

^{, say}

z

. The corresponding IVP







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y

⁰

(a) = z

⁽¹³⁾

can then be solved by, for example, Runge-Kutta method. We denote this approximate

solution

y

z and hope

y

z

(b) = β

. If not, we use another guess for

y

⁰

(a)

, and try to solve an altered IVP (13) again.

This process is repeated and can be done systematically. Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE

²²

3 – Shooting Methods

We consider solving the following 2-point boundary-value problem:







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y(b) = β

⁽¹²⁾

The idea of shooting method for (12) is to solve a related initial-value problem with a guess for

y

⁰

(a)

^{, say}

z

. The corresponding IVP







y

⁰⁰

= f (x, y, y

⁰

)

y(a) = α, y

⁰

(a) = z

⁽¹³⁾

can then be solved by, for example, Runge-Kutta method. We denote this approximate

solution

y

z and hope

y

z

(b) = β

. If not, we use another guess for

y

⁰

(a)

, and try to solve an altered IVP (13) again. This process is repeated and can be done systematically.

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

BVP of ODE

²³

☞

Objective: select

z

^{, so that}

y

z

(b) = β

^.

Let

φ(z) = y

z

(b) − β.

Now our objective is simply to solve the equation

φ(z) = 0

. Hence secant method can be used.

☞

How to compute

z

^?

Suppose we have solutions

y

z₁

, y

z₂ with guesses

z

₁

, z

₂ ^{and obtain}

φ(z

₁

)

^and

φ(z

₂

)

^.

If these guesses can not generate satisfactory solutions, we can obtain another guess

z

₃ by the secant method

z

₃

= z

₂

− φ(z

₂

) z

₂

− z

₁

φ(z

₂

) − φ(z

₁

) .

In general

z

k+1

= z

k

− φ(z

k

) z

k

− z

k−1

φ(z

k

) − φ(z

k−1

) .

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

BVP of ODE

²³

☞

Objective: select

z

^{, so that}

y

z

(b) = β

^.

Let

φ(z) = y

z

(b) − β.

Now our objective is simply to solve the equation

φ(z) = 0

. Hence secant method can be used.

☞

How to compute

z

^?

Suppose we have solutions

y

z₁

, y

z₂ with guesses

z

₁

, z

₂ ^{and obtain}

φ(z

₁

)

^and

φ(z

₂

)

^.

If these guesses can not generate satisfactory solutions, we can obtain another guess

z

₃ by the secant method

z

₃

= z

₂

− φ(z

₂

) z

₂

− z

₁

φ(z

₂

) − φ(z

₁

) .

In general

z

k+1

= z

k

− φ(z

k

) z

k

− z

k−1

φ(z

k

) − φ(z

k−1

) .

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

BVP of ODE

²³

☞

Objective: select

z

^{, so that}

y

z

(b) = β

^.

Let

φ(z) = y

z

(b) − β.

Now our objective is simply to solve the equation

φ(z) = 0

^.

Hence secant method can be used.

☞

How to compute

z

^?

Suppose we have solutions

y

z₁

, y

z₂ with guesses

z

₁

, z

₂ ^{and obtain}

φ(z

₁

)

^and

φ(z

₂

)

^.

If these guesses can not generate satisfactory solutions, we can obtain another guess

z

₃ by the secant method

z

₃

= z

₂

− φ(z

₂

) z

₂

− z

₁

φ(z

₂

) − φ(z

₁

) .

In general

z

k+1

= z

k

− φ(z

k

) z

k

− z

k−1

φ(z

k

) − φ(z

k−1

) .

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

BVP of ODE

²³

☞

Objective: select

z

^{, so that}

y

z

(b) = β

^.

Let

φ(z) = y

z

(b) − β.

Now our objective is simply to solve the equation

φ(z) = 0

. Hence secant method can be used.

☞

How to compute

z

^?

Suppose we have solutions

y

z₁

, y

z₂ with guesses

z

₁

, z

₂ ^{and obtain}

φ(z

₁

)

^and

φ(z

₂

)

^.

If these guesses can not generate satisfactory solutions, we can obtain another guess

z

₃ by the secant method

z

₃

= z

₂

− φ(z

₂

) z

₂

− z

₁

φ(z

₂

) − φ(z

₁

) .

In general

z

k+1

= z

k

− φ(z

k

) z

k

− z

k−1

φ(z

k

) − φ(z

k−1

) .

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

BVP of ODE

²³

☞

Objective: select

z

^{, so that}

y

z

(b) = β

^.

Let

φ(z) = y

z

(b) − β.

Now our objective is simply to solve the equation

φ(z) = 0

. Hence secant method can be used.

☞

How to compute

z

^?

Suppose we have solutions

y

z₁

, y

z₂ with guesses

z

₁

, z

₂ ^{and obtain}

φ(z

₁

)

^and

φ(z

₂

)

^.

If these guesses can not generate satisfactory solutions, we can obtain another guess

z

₃ by the secant method

z

₃

= z

₂

− φ(z

₂

) z

₂

− z

₁

φ(z

₂

) − φ(z

₁

) .

In general

z

k+1

= z

k

− φ(z

k

) z

k

− z

k−1

φ(z

k

) − φ(z

k−1

) .

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

BVP of ODE

²³

☞

Objective: select

z

^{, so that}

y

z

(b) = β

^.

Let

φ(z) = y

z

(b) − β.

Now our objective is simply to solve the equation

φ(z) = 0

. Hence secant method can be used.

☞

How to compute

z

^?

Suppose we have solutions

y

z₁

, y

z₂ with guesses

z

₁

, z

₂ ^{and obtain}

φ(z

₁

)

^and

φ(z

₂

)

^.

If these guesses can not generate satisfactory solutions, we can obtain another guess

z

₃ by the secant method

z

₃

= z

₂

− φ(z

₂

) z

₂

− z

₁

φ(z

₂

) − φ(z

₁

) .

In general

z

k+1

= z

k

− φ(z

k

) z

k

− z

k−1

φ(z

k

) − φ(z

k−1

) .

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

BVP of ODE

²³

☞

Objective: select

z

^{, so that}

y

z

(b) = β

^.

Let

φ(z) = y

z

(b) − β.

Now our objective is simply to solve the equation

φ(z) = 0

. Hence secant method can be used.

☞

How to compute

z

^?

Suppose we have solutions

y

z₁

, y

z₂ with guesses

z

₁

, z

₂ ^{and obtain}

φ(z

₁

)

^and

φ(z

₂

)

^.

If these guesses can not generate satisfactory solutions, we can obtain another guess

z

₃ by the secant method

z

₃

= z

₂

− φ(z

₂

) z

₂

− z

₁

φ(z

₂

) − φ(z

₁

) .

In general

z

k+1

= z

k

− φ(z

k

) z

k

− z

k−1

φ(z

k

) − φ(z

k−1

) .

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

BVP of ODE

²³

☞

Objective: select

z

^{, so that}

y

z

(b) = β

^.

Let

φ(z) = y

z

(b) − β.

Now our objective is simply to solve the equation

φ(z) = 0

. Hence secant method can be used.

☞

How to compute

z

^?

Suppose we have solutions

y

z₁

, y

z₂ with guesses

z

₁

, z

₂ ^{and obtain}

φ(z

₁

)

^and

φ(z

₂

)

^.

If these guesses can not generate satisfactory solutions, we can obtain another guess

z

₃ by the secant method

z

₃

= z

₂

− φ(z

₂

) z

₂

− z

₁

φ(z

₂

) − φ(z

₁

) .

In general

z

k+1

= z

k

− φ(z

k

) z

k

− z

k−1

φ(z

k

) − φ(z

k−1

) .

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

BVP of ODE

²⁴

☞

Special BVP:







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y(b) = β

⁽¹⁴⁾

where

u(x), v(x), w(x)

are continuous in

[a, b]

^.

Suppose we have solved (14) twice with initial guesses

z

₁ ^and

z

₂, and obtain approximate solutions

y

₁ ^and

y

₂^{, hence}







y

₁⁰⁰

= u + vy

₁

+ wy

₁⁰

y

₁

(a) = α, y

₁⁰

(a) = z

₁ ^and







y

₂⁰⁰

= u + vy

₂

+ wy

₂⁰

y

₂

(a) = α, y

₂⁰

(a) = z

₂

Now let

y(x) = λy

₁

(x) + (1 − λ)y

₂

(x)

for some parameter

λ

, we can show

y

⁰⁰

= u + vy + wy

⁰

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

BVP of ODE

²⁴

☞

Special BVP:







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y(b) = β

⁽¹⁴⁾

where

u(x), v(x), w(x)

are continuous in

[a, b]

^.

Suppose we have solved (14) twice with initial guesses

z

₁ ^and

z

₂, and obtain approximate solutions

y

₁ ^and

y

₂^,

hence







y

₁⁰⁰

= u + vy

₁

+ wy

₁⁰

y

₁

(a) = α, y

₁⁰

(a) = z

₁ ^and







y

₂⁰⁰

= u + vy

₂

+ wy

₂⁰

y

₂

(a) = α, y

₂⁰

(a) = z

₂

Now let

y(x) = λy

₁

(x) + (1 − λ)y

₂

(x)

for some parameter

λ

, we can show

y

⁰⁰

= u + vy + wy

⁰

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

BVP of ODE

²⁴

☞

Special BVP:







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y(b) = β

⁽¹⁴⁾

where

u(x), v(x), w(x)

are continuous in

[a, b]

^.

Suppose we have solved (14) twice with initial guesses

z

₁ ^and

z

₂, and obtain approximate solutions

y

₁ ^and

y

₂^{, hence}







y

₁⁰⁰

= u + vy

₁

+ wy

₁⁰

y

₁

(a) = α, y

₁⁰

(a) = z

₁ ^and







y

₂⁰⁰

= u + vy

₂

+ wy

₂⁰

y

₂

(a) = α, y

₂⁰

(a) = z

₂

Now let

y(x) = λy

₁

(x) + (1 − λ)y

₂

(x)

for some parameter

λ

, we can show

y

⁰⁰

= u + vy + wy

⁰

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

BVP of ODE

²⁴

☞

Special BVP:







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y(b) = β

⁽¹⁴⁾

where

u(x), v(x), w(x)

are continuous in

[a, b]

^.

Suppose we have solved (14) twice with initial guesses

z

₁ ^and

z

₂, and obtain approximate solutions

y

₁ ^and

y

₂^{, hence}







y

₁⁰⁰

= u + vy

₁

+ wy

₁⁰

y

₁

(a) = α, y

₁⁰

(a) = z

₁ ^and







y

₂⁰⁰

= u + vy

₂

+ wy

₂⁰

y

₂

(a) = α, y

₂⁰

(a) = z

₂

Now let

y(x) = λy

₁

(x) + (1 − λ)y

₂

(x)

for some parameter

λ

^,

we can show

y

⁰⁰

= u + vy + wy

⁰

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

BVP of ODE

²⁴

☞

Special BVP:







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y(b) = β

⁽¹⁴⁾

where

u(x), v(x), w(x)

are continuous in

[a, b]

^.

Suppose we have solved (14) twice with initial guesses

z

₁ ^and

z

₂, and obtain approximate solutions

y

₁ ^and

y

₂^{, hence}







y

₁⁰⁰

= u + vy

₁

+ wy

₁⁰

y

₁

(a) = α, y

₁⁰

(a) = z

₁ ^and







y

₂⁰⁰

= u + vy

₂

+ wy

₂⁰

y

₂

(a) = α, y

₂⁰

(a) = z

₂

Now let

y(x) = λy

₁

(x) + (1 − λ)y

₂

(x)

for some parameter

λ

, we can show

y

⁰⁰

= u + vy + wy

⁰

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

BVP of ODE

²⁵

and

y(a) = λy

₁

(a) + (1 − λ)y

₂

(a) = α

We can select

λ

^{so that}

y(b) = β

^.

β = y(b) = λy

₁

(b) + (1 − λ)y

₂

(b)

= λ(y

₁

(b) − y

₂

(b)) + y

₂

(b)

⇒ λ = β − y

₂

(b) (y

₁

(b) − y

₂

(b))

In practice, we can solve the following two IVPs (in parallel)







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y

⁰

(a) = 0

and







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y

⁰

(a) = 1

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

BVP of ODE

²⁵

and

y(a) = λy

₁

(a) + (1 − λ)y

₂

(a) = α

We can select

λ

^{so that}

y(b) = β

^.

β = y(b) = λy

₁

(b) + (1 − λ)y

₂

(b)

= λ(y

₁

(b) − y

₂

(b)) + y

₂

(b)

⇒ λ = β − y

₂

(b) (y

₁

(b) − y

₂

(b))

In practice, we can solve the following two IVPs (in parallel)







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y

⁰

(a) = 0

and







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y

⁰

(a) = 1

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

BVP of ODE

²⁵

and

y(a) = λy

₁

(a) + (1 − λ)y

₂

(a) = α

We can select

λ

^{so that}

y(b) = β

^.

β = y(b) = λy

₁

(b) + (1 − λ)y

₂

(b)

= λ(y

₁

(b) − y

₂

(b)) + y

₂

(b)

⇒ λ = β − y

₂

(b) (y

₁

(b) − y

₂

(b))

In practice, we can solve the following two IVPs (in parallel)







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y

⁰

(a) = 0

and







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y

⁰

(a) = 1

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

BVP of ODE

²⁵

and

y(a) = λy

₁

(a) + (1 − λ)y

₂

(a) = α

We can select

λ

^{so that}

y(b) = β

^.

β = y(b) = λy

₁

(b) + (1 − λ)y

₂

(b)

= λ(y

₁

(b) − y

₂

(b)) + y

₂

(b)

⇒ λ = β − y

₂

(b) (y

₁

(b) − y

₂

(b))

In practice, we can solve the following two IVPs (in parallel)







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y

⁰

(a) = 0

and







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y

⁰

(a) = 1

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

BVP of ODE

²⁵

and

y(a) = λy

₁

(a) + (1 − λ)y

₂

(a) = α

We can select

λ

^{so that}

y(b) = β

^.

β = y(b) = λy

₁

(b) + (1 − λ)y

₂

(b)

= λ(y

₁

(b) − y

₂

(b)) + y

₂

(b)

⇒ λ = β − y

₂

(b) (y

₁

(b) − y

₂

(b))

In practice, we can solve the following two IVPs (in parallel)







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y

⁰

(a) = 0

and







y

⁰⁰

= u(x) + v(x)y + w(x)y

⁰

y(a) = α, y

⁰

(a) = 1

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

在文檔中 BVP of ODE (頁 49-73)