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# 3 – Shooting Methods

22

## BVP of ODE

22

### 3 – Shooting Methods

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

00

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

0

, say

### z

.

The corresponding IVP

00

0

0

### (a) = z

(13)

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

solution

z and hope

z

### (b) = β

. If not, we use another guess for

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:

00

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

0

, say

### z

. The corresponding IVP

00

0

0

### (a) = z

(13)

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

We denote this approximate

solution

z and hope

z

### (b) = β

. If not, we use another guess for

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:

00

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

0

, say

### z

. The corresponding IVP

00

0

0

### (a) = z

(13)

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

z and hope

z

### (b) = β

.

If not, we use another guess for

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:

00

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

0

, say

### z

. The corresponding IVP

00

0

0

### (a) = z

(13)

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

solution

z and hope

z

### (b) = β

. If not, we use another guess for

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:

00

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

0

, say

### z

. The corresponding IVP

00

0

0

### (a) = z

(13)

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

solution

z and hope

z

### (b) = β

. If not, we use another guess for

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

, so that

z

.

Let

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

z1

z2 with guesses

1

2 and obtain

1

and

2

### )

.

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

### z

3 by the secant method

3

2

2

2

1

2

1

In general

k+1

k

k

k

k−1

k

k−1

### ) .

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

## BVP of ODE

23

### ☞

Objective: select

, so that

z

.

Let

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

z1

z2 with guesses

1

2 and obtain

1

and

2

### )

.

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

### z

3 by the secant method

3

2

2

2

1

2

1

In general

k+1

k

k

k

k−1

k

k−1

### ) .

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

## BVP of ODE

23

### ☞

Objective: select

, so that

z

.

Let

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

z1

z2 with guesses

1

2 and obtain

1

and

2

### )

.

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

### z

3 by the secant method

3

2

2

2

1

2

1

In general

k+1

k

k

k

k−1

k

k−1

### ) .

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

## BVP of ODE

23

### ☞

Objective: select

, so that

z

.

Let

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

z1

z2 with guesses

1

2 and obtain

1

and

2

### )

.

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

### z

3 by the secant method

3

2

2

2

1

2

1

In general

k+1

k

k

k

k−1

k

k−1

### ) .

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

## BVP of ODE

23

### ☞

Objective: select

, so that

z

.

Let

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

z1

z2 with guesses

1

2 and obtain

1

and

2

### )

.

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

### z

3 by the secant method

3

2

2

2

1

2

1

In general

k+1

k

k

k

k−1

k

k−1

### ) .

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

## BVP of ODE

23

### ☞

Objective: select

, so that

z

.

Let

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

z1

z2 with guesses

1

2 and obtain

1

and

2

### )

.

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

### z

3 by the secant method

3

2

2

2

1

2

1

In general

k+1

k

k

k

k−1

k

k−1

### ) .

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

## BVP of ODE

23

### ☞

Objective: select

, so that

z

.

Let

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

z1

z2 with guesses

1

2 and obtain

1

and

2

### )

.

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

### z

3 by the secant method

3

2

2

2

1

2

1

In general

k+1

k

k

k

k−1

k

k−1

### ) .

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

## BVP of ODE

23

### ☞

Objective: select

, so that

z

.

Let

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

z1

z2 with guesses

1

2 and obtain

1

and

2

### )

.

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

### z

3 by the secant method

3

2

2

2

1

2

1

In general

k+1

k

k

k

k−1

k

k−1

### ) .

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

## BVP of ODE

24

Special BVP:

00

0

(14)

where

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

are continuous in

### [a, b]

.

Suppose we have solved (14) twice with initial guesses

1 and

### z

2, and obtain approximate solutions

1 and

2, hence

100

1

10

1

10

1 and

200

2

20

2

20

2

Now let

1

2

### (x)

for some parameter

, we can show

00

### = u + vy + wy

0

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

## BVP of ODE

24

Special BVP:

00

0

(14)

where

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

are continuous in

### [a, b]

.

Suppose we have solved (14) twice with initial guesses

1 and

### z

2, and obtain approximate solutions

1 and

2,

hence

100

1

10

1

10

1 and

200

2

20

2

20

2

Now let

1

2

### (x)

for some parameter

, we can show

00

### = u + vy + wy

0

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

## BVP of ODE

24

Special BVP:

00

0

(14)

where

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

are continuous in

### [a, b]

.

Suppose we have solved (14) twice with initial guesses

1 and

### z

2, and obtain approximate solutions

1 and

2, hence

100

1

10

1

10

1 and

200

2

20

2

20

2

Now let

1

2

### (x)

for some parameter

, we can show

00

### = u + vy + wy

0

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

## BVP of ODE

24

Special BVP:

00

0

(14)

where

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

are continuous in

### [a, b]

.

Suppose we have solved (14) twice with initial guesses

1 and

### z

2, and obtain approximate solutions

1 and

2, hence

100

1

10

1

10

1 and

200

2

20

2

20

2

Now let

1

2

### (x)

for some parameter

,

we can show

00

### = u + vy + wy

0

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

## BVP of ODE

24

Special BVP:

00

0

(14)

where

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

are continuous in

### [a, b]

.

Suppose we have solved (14) twice with initial guesses

1 and

### z

2, and obtain approximate solutions

1 and

2, hence

100

1

10

1

10

1 and

200

2

20

2

20

2

Now let

1

2

### (x)

for some parameter

, we can show

00

### = u + vy + wy

0

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

## BVP of ODE

25

and

1

2

We can select

so that

.

1

2

1

2

2

2

1

2

### (b))

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

00

0

0

and

00

0

0

### (a) = 1

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

## BVP of ODE

25

and

1

2

We can select

so that

.

1

2

1

2

2

2

1

2

### (b))

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

00

0

0

and

00

0

0

### (a) = 1

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

## BVP of ODE

25

and

1

2

We can select

so that

.

1

2

1

2

2

2

1

2

### (b))

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

00

0

0

and

00

0

0

### (a) = 1

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

## BVP of ODE

25

and

1

2

We can select

so that

.

1

2

1

2

2

2

1

2

### (b))

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

00

0

0

and

00

0

0

### (a) = 1

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

## BVP of ODE

25

and

1

2

We can select

so that

.

1

2

1

2

2

2

1

2

### (b))

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

00

0

0

and

00

0

0

### (a) = 1

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