**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

z_{1}

### , y

z_{2}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

z_{1}

### , y

z_{2}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

z_{1}

### , y

z_{2}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

z_{1}

### , y

z_{2}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

z_{1}

### , y

z_{2}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

z_{1}

### , y

z_{2}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

z_{1}

### , y

z_{2}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

z_{1}

### , y

z_{2}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

_{1}

^{00}

### = u + vy

_{1}

### + wy

_{1}

^{0}

### y

_{1}

### (a) = α, y

_{1}

^{0}

### (a) = z

_{1}

^{and}

###

###

###

### y

_{2}

^{00}

### = u + vy

_{2}

### + wy

_{2}

^{0}

### y

_{2}

### (a) = α, y

_{2}

^{0}

### (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

_{1}

^{00}

### = u + vy

_{1}

### + wy

_{1}

^{0}

### y

_{1}

### (a) = α, y

_{1}

^{0}

### (a) = z

_{1}

^{and}

###

###

###

### y

_{2}

^{00}

### = u + vy

_{2}

### + wy

_{2}

^{0}

### y

_{2}

### (a) = α, y

_{2}

^{0}

### (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

_{1}

^{00}

### = u + vy

_{1}

### + wy

_{1}

^{0}

### y

_{1}

### (a) = α, y

_{1}

^{0}

### (a) = z

_{1}

^{and}

###

###

###

### y

_{2}

^{00}

### = u + vy

_{2}

### + wy

_{2}

^{0}

### y

_{2}

### (a) = α, y

_{2}

^{0}

### (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

_{1}

^{00}

### = u + vy

_{1}

### + wy

_{1}

^{0}

### y

_{1}

### (a) = α, y

_{1}

^{0}

### (a) = z

_{1}

^{and}

###

###

###

### y

_{2}

^{00}

### = u + vy

_{2}

### + wy

_{2}

^{0}

### y

_{2}

### (a) = α, y

_{2}

^{0}

### (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

_{1}

^{00}

### = u + vy

_{1}

### + wy

_{1}

^{0}

### y

_{1}

### (a) = α, y

_{1}

^{0}

### (a) = z

_{1}

^{and}

###

###

###

### y

_{2}

^{00}

### = u + vy

_{2}

### + wy

_{2}

^{0}

### y

_{2}

### (a) = α, y

_{2}

^{0}

### (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**