Shooting Method - Numerical Methods - Lecture Slides, Slides of Mathematical Methods for Numerical Analysis and Optimization

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Shooting Method

Shooting Method

The shooting method uses the methods used in solving initial value problems. This is done by assuming initial values that would have been given if the ordinary differential equation were a initial value problem. The boundary value obtained is compared with the actual boundary value. Using trial and error or some scientific approach, one tries to get as close to the boundary value as possible.

Solution

Two first order differential equations are given as

= w , u ( ) 5 = 0. 0038371 dr

du

w ( ) notknown r

u r

w dr

dw (^) , 5 =− + 2 =

Let us assume

( ) ( ) ( )^ ( )^0. 00026538 8 5

5 5 8 5 =− −

= ≈ u − u dr

w du

To set up initial value problem

= w = f 1 ( r , u , w ), u ( ) 5 = 0. 0038371 dr

du

= − + 2 = f 2 ( r , u , w ), w ( ) 5 =− 0. 00026538 r

u r

w dr

dw

Solution Cont

Using Euler’s method,

ui + 1 = ui + f 1 ( r i , ui , wi ) h

wi + 1 = wi + f 2 ( ri , ui , wi ) h

Let us consider 4 segments between the two boundaries, and then,

r = 5 r = 8

1. 75 4

8 5

− h =

Solution Cont

For i^ =^1 ,^ r 1 = r 0 + h =^5 +^0.^75 =^5.^75 , u 1 = 0. 0036741 , w 1 =− 0. 00010940

1

2 1 1 1 1 1

f

u u f r u w h

1. 000011769

2. 00010938 0. 00013015 0. 75

3. 00010938 5. 75 , 0. 0036741 , 0. 00010938 0. 75

, ,

2

2 1 2 1 1 1

= −

=− +

=− + −

= + f

w w f r u w h

Solution Cont

For (^) i = 2 , r 2 = r 1 + h = 5. 75 + 0. 75 = 6. (^5) u 2 = 0. 0035920 , w 2 =− 0. 000011785

1

3 2 1 2 2 2

f

u u f r u w h

2

3 2 2 2 2 2

f

w w f r u w h

Solution Cont

Let us assume a new value for ( ) 5 dr

du

( ) ( )

( ) ( ) 2 ( 0. 00026538 ) 0. 00053076 8 5

u u dr

du w

Using (^) h = 0. 75 and Euler’s method, we get

u ( ) 8 ≈ u 4 = 0. 0029665 "

While the given value of this boundary condition is

u ( ) 8 ≈ u 4 = 0. 0030770

Solution Cont

Using linear interpolation on the obtained data for the two assumed values of

dr

du (^) we get

u^ ( ) 8 = 0. 00030770

( )

( ) ( 0. 0030770 0. 0036232 ) ( 0. 00026538 )

1. 0029645 0. 0036232

dr

du

Using h = 0. 75 and repeating the Euler’s method with (^) w ( 5 )=− 0. 00048611

u^ ( ) 8 ≈ u 4 = 0. 0030769

Comparisons of different initial guesses

3.0E-

3.2E-

3.4E-

3.6E-

3.8E-

4.0E-

(^5 6) Radial Location, r (in) 7 8

Radial Displacement,

u^ (in)

du/dr = -0.

du/dr = -0.

du/d r= -0.

Exact

Comparison of Euler and Runge-Kutta

Results with exact results Table 1 Comparison of Euler and Runge-Kutta results with exact results. r (in) Exact (in) Euler (in) Runge-Kutta(in)

• 1. • 3.8731×10−
     • 3.5567×10−
     • 3.3366×10−
     • 3.1829×10−
     • 3.0770×10−
       • 3.8731×10−
       • 3.5085×10−
       • 3.2858×10−
       • 3.1518×10−
       • 3.0770×10− - 0. - 1. - 1.
         • 9.8967×10−
         • 1.9500×10− - 3.8731×10− - 3.5554×10− - 3.3341×10− - 3.1792×10− - 3.0723×10− - 0. - 3.5824×10− - 7.4037×10− - 1.1612×10− - 1.5168×10−