Initial valueproblems

Initial Value Problems
Mohammad Tawfik
#WikiCourses
http://WikiCourses.WikiSpaces.com
Mohammad Tawfik

Mohammad Tawfik
#WikiCourses
Objectives
• Understand the applications of initial-value
problems
• Be able to apply the Euler method for
solving initial value problems
• Be able to apply the Runge-Kutta method
for solving initial value problem

Mohammad Tawfik
#WikiCourses
Example Problem
dt
dv
mmaF 
cvmgFFF UD 
cvmgvm
dt
dv
m  
m
cvmg
v

    mct
e
c
mg
tv /
1 


Mohammad Tawfik
#WikiCourses
Exact Solution

Mohammad Tawfik
#WikiCourses
Approximate Solution
12
12
tt
vv
t
v
dt
dv






m
cvmg
tt
vv 



12
12
m
cvmg
tt
vv 1
12
12 




Mohammad Tawfik
#WikiCourses
Approximate Solution
 
m
cvmg
ttvv 1
1212



Mohammad Tawfik
#WikiCourses
Euler Method
• Given the differential
equation:
• We may write:
• Giving:
 tyf
dt
dy
,
 tyf
t
yy
t
y
dt
dy ttt
,





 
 tytfyy ttt ,

Mohammad Tawfik
#WikiCourses
Example
equation:
• The exact solution is:
• At t=0, y=2
• Find y(4) using Euler
method with step
t=1
ye
dt
dy t
5.04 8.0

  tt
eety 5.08.0
08.108.3 


Mohammad Tawfik
#WikiCourses
Solution
ye
dt
dy t
5.04 8.0
  yetyy t
tttt 5.04 8.0
 
 
  5142
5.04 0
0
01

 yeyy  
  4.115.245
5.04
8.0
1
8.0
12


e
yeyy
 
  5.254.11*5.044.11
5.04
6.1
2
2*8.0
23


e
yeyy
 
  8.565.25*5.045.25
5.04
4.2
3
3*8.0
34


e
yeyy

Mohammad Tawfik
#WikiCourses
Convergence!
0
10
20
30
40
50
60
70
80
0 0.5 1 1.5 2 2.5 3 3.5 4
Time
y(t)
Exact
Dt=1.0
Dt=0.5
Dt=0.1

Mohammad Tawfik
#WikiCourses
Runge-Kutta Methods
• The Runge-Kutta methods achieves the
Taylor series accuracy
• Many forms of the method are available;
we will use 2nd order and 3rd order
methods only

Mohammad Tawfik
#WikiCourses
2nd Order Runge-Kutta method
• For the DE:
• The 2nd order R.K. solution
is:
• Where:
• Note: in the textbook there
are 3 different methods, we
are using the first!
 tyf
dt
dy
,
 21
2
kk
t
yy ttt 


 tyfk t ,1 
 tttkyfk t  ,12

Mohammad Tawfik
#WikiCourses
Example
equation:
• The exact solution is:
• At t=0, y=2
• Find y(4) using 2nd
order R.K. method
with step t=1
ye
dt
dy t
5.04 8.0

  tt
eety 5.08.0
08.108.3 


Mohammad Tawfik
#WikiCourses
Solution
• At t=0
32*5.04 0
1  ek
 
  4.61*325.04 108.0
2  
ek
  7.6
2
2101 

 kk
t
yy
• Repeat for all t

Mohammad Tawfik
#WikiCourses
Solution
yk2k1t
20
6.7010826.40216431
16.3197813.685785.5516232
37.1992530.106711.652243
83.3377766.7839625.493084
yk2k1t
20
3.8043254.21729930.5
6.3165385.9837174.0651361
9.9240688.6862255.7438951.5
15.196312.770498.3184342
22.9759418.9045812.213982.5
34.5149228.0876718.068263
51.6768141.8123226.835253.5
77.2385262.3066739.940184

Mohammad Tawfik
#WikiCourses
Convergence
0
10
20
30
40
50
60
70
80
90
0 0.5 1 1.5 2 2.5 3 3.5 4
Time
y(t)
Exact
Dt=1.0
Dt=0.5
Dt=0.1

Mohammad Tawfik
#WikiCourses
3rd Order Runge-Kutta method
• For the DE:
• The 3rd order R.K.
solution is:
• Where:
 tyf
dt
dy
,
 321 4
6
kkk
t
yy ttt 


 tyfk t ,1 





 



2
,
2
1
2
t
t
tk
yfk t
 tttktkyfk t  ,2 213

Mohammad Tawfik
#WikiCourses
Homework #4
• Solve:
• Given y(0)=1
1. Analytically
2. Using Euler method until t=2, with t=0.5
3. Repeat part 2 using 2nd order RK method
4. Repeat part 2 using 3rd order RK method
5. Repeat parts 2 through 4 using t=0.25
6. Compare results of all parts above
yyt
dt
dy
2.12


Initial valueproblems

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Initial valueproblems