Download PART 7 Ordinary Differential Equations ODEs

PART 7
Ordinary Differential Equations
ODEs
Ordinary Differential Equations
Part 7
• Equations which are composed of an unknown
function and its derivatives are called
differential equations.
dv
c
g v
dt
m
v - dependent variable
t - independent variable
• Differential equations play a fundamental role in
engineering because many physical phenomena
are best formulated mathematically in terms of
their rate of change.
Ordinary Differential Equations
• When a function involves one dependent
variable, the equation is called an ordinary
differential equation (ODE).
•A partial differential equation (PDE) involves
two or more independent variables.
Ordinary Differential Equations
Differential equations are also classified as to
their order:
1. A first order equation includes a first derivative as
its highest derivative.
- Linear
1st
order ODE
dy
   y  f (x )
dx
dy
- Non-Linear order ODE
 f ( x, y )
dx
Where f(x,y) is nonlinear
1st
Ordinary Differential Equations
2. A second order equation includes a second
derivative.
2
d y
dy
nd
 p  Qy  f (x )
- Linear 2 order ODE
2
dx
dx
- Non-Linear 2nd order ODE
2
d y
dy
 p ( x, y )  Q ( x, y )  y  f ( x )
2
dx
dx
• Higher order equations can be reduced to a system
of first order equations, by redefining a variable.
Ordinary Differential Equations
Runge-Kutta Methods
This chapter is devoted to solving ODE of the
form:
dy
 f ( x, y )
dx
• Euler’s Method
dy
dy  f ( x, y )
dx  f ( x, y )
dx
solution :
Solution
Solution :
yi 1  yi  f ( xi , yi )  h
yi 1  yi  f ( xi , yi )  h
Runge-Kutta Methods
dy
 f ( x, y )
dx
Solution :
dy
yi 1  yi f (fx(,xyi ,)yi )  h
dx
Solution :
yi 1  yi  f ( xi , yi )  h
Euler’s Method: Example
Obtain a solution between x = 0 to x = 4
with a step size of 0.5 for: dy
3
dx
 2 x  12 x 2  20 x  8.5
Initial conditions are: x = 0 to y = 1
ulerSolution:
Solution : yi 1  yi  f ( xi , yi )  h
y (0.5)  y (0 )  f (0,1).(0.5)
 1.0  8.5 x0.5  5.25
y (1.0 )  y (0.5)  f (0.5,5.25).(0.5)
 5.25  ( 2(0.5)3  12(0.5) 2  20(0.5)  8.5).(0.5)  5.875
y ( 2.0 )  y (1.0 )  f (1.0,5.875).(0.5)
 5.25  ( 2(1.0 )3  12(1.0 ) 2  20(1.0 )  8.5).(0.5)  5.125
Euler’s Method: Example
• Although the computation captures the general trend
solution, the error is considerable.
• This error can be reduced by using a smaller step size.
Improvements of Euler’s method
• A fundamental source of error in Euler’s
method is that the derivative at the beginning
of the interval is assumed to apply across the
entire interval.
• Simple modifications are available:
– Heun’s Method
– The Midpoint Method
– Ralston’s Method
Runge-Kutta Methods
• Runge-Kutta methods achieve the accuracy of a Taylor series
approach without requiring the calculation of higher derivatives.
y i 1  y i   (x i , y i , h )h
  a1k 1  a2 k 2 
 an k n
a ' s  constants
k 1  f (x i , y i )
Increment function
(representative slope
over the interval)
k 2  f (x i  p1h , y i  q11k 1h )
k 3  f (x i  p 3 h , y i  q 21k 1h  q 22 k 2 h )
k n  f (x i  p n 1h , y i  q n 1k 1h  q n 1,2 k 2 h 
p ' s and q ' s are constants
 q n 1,n 1k n 1h )
Runge-Kutta Methods
•
Various types of RK methods can be devised by employing
different number of terms in the increment function as
specified by n.
1. First order RK method with n=1 is Euler’s method.
–
Error is proportional to O(h)
2. Second order RK methods:
- Error is proportional to O(h2)
yi 1  yi  (a1k1  a2 k 2 )h
k1  f ( x i , yi )
k 2  f ( xi  p1h, yi  q11k1h)
•
Values of a1, a2, p1, and q11 are evaluated by setting the
second order equation to Taylor series expansion to the
second order term.
Runge-Kutta Methods
• Three equations to evaluate the four unknown constants
are derived:
a1  a2  1
1
a 2 p1 
2
1
a2 q11 
2
A value is assumed for one of the
unknowns to solve for the other
three.
Runge-Kutta Methods
1
1
a1  a2  1, a 2 p1  , a2 q11 
2
2
• We can choose an infinite number of values for a2,there
are an infinite number of second-order RK methods.
• Every version would yield exactly the same results if the
solution to ODE were quadratic, linear, or a constant.
• However, they yield different results if the solution is
more complicated (typically the case).
 f ( x, y )
dxSolution :
Runge-KuttaSolution
Methods
: 1
2
yi 1  yi  ( k1  k 2 )  h
Three
1 3are: 1 3
dy of the most commonly used methods
 f ( x, y )
y i  1  yi  ( k1  k 2 )  h
yi 1  yi  (a1k1  a2 k 2 )h k1  f ( xi , 2yi ) 2
dx
dy
f ( x, y )
: , y ) with a SinglekCorrector
• Huen
Method
(ai )2=1/2)
kSolution

f
(
x
dx 1  f ( xi , y
1
i
i
3
3
k 2 : f ( xi  h, yi  k1h )
dy
1
1
Solution
ky2i 1f y(fix(i x(, p
yk1)h1 ,yi k2q)11 hk1hk) 2dy f ( xi  h4), ( yi  k14h ) 
2
2
yi 1  yi k 2f(hx, y )
dx
dx
k1  f ( xi , yi )
k12
f ( xi , yi )
• The
Midpoint
= 1)
Solution
: Method (a
Solution :
k 2  f ( xi  h ), ( yi  k1h ) 
1
1
dy
yi 1 f ( yx,i y) k 2  h
k 2  f ( xi  h, yi 1
 k1h2)
yi 1 2yi  ( k21  k 2 )  h
dx
3
3
k1  f ( xi , yi )
Solution : Method (a2= 2/3) k1  f ( xi , yi )
• Raltson’s
1 2
1
1
3
3
yki 21 
 yfi (x(i k1 h, ky2i ) h k1hk)2  f ( xi  h, yi  k1h )
2
3 2 3
4
4
Runge-Kutta Methods
y
• Heun’s Method:
Involves the determination
dydy
of
two derivatives for the

f
(
x
,
y
)
interval
, y )initial point
dxdx  fat( xthe
and the end point.
f(xi,yi)
xi
Solution
Solution ::
11 1 1
yiyi11 yyii (( k1k1  k2 )k 2h)  h
22 2 2
k

f
(
x
,
y
)
1
i
i
k  f (x , y )
i
x
xi+h
y
ea
k 2  f ( xi  h ), ( yi  k1h ) 
1
f(xi+h,yi+k1h)
i
Slope: 0.5(k1+k2)
k 2  f ( xi  h ), ( yi  k1h ) 
xi
x
xi+h
Runge-Kutta Methods
y
• Midpoint Method:
Uses
dy Euler’s method to predict
 off (yxat
, ythe
) midpoint of
ady
value
dx f ( x, y )
the
dxinterval:
Solution :
Solution
yi 1  yi:  k 2  h
yki 1 fy(i x,ky2 ) h
1
i
f(xi,yi)
xi
f(xi+h/2,yi+k1h/2)
x
xi+h/2
y
i
k1  f ( xi , yi )1
1
k 2  f ( xi 1 h, yi 1 k1h )
k 2  f ( xi  2h, yi  2k1h )
2
2
Chapter 25
ea
Slope: k2
xi
x
xi+h
dy
Runge-Kutta
Methods
dy  f ( x, y )
dx  f ( x, y )
•dx
Ralston’s Method:
Solution :
Solution :
1
2
yi 1  yi  ( 1k1  2k 2 )  h
yi 1  yi  (3 k1 3 k 2 )  h
3
k1  f ( xi , yi )3
k1  f ( xi , y3i )
3
k 2  f ( xi  h, yi  k1h )
3
3
4
4
k  f ( x  h, y  k h )
2
i
4
i
4
y
f(xi,yi)
xi
f(xi+ 3/4 h,
yi+3/4k1h)
xi+3/4h
x
y
ea
1
Slope:
(1/3k1+2/3k2)
Chapter 25
xi
xi+h
x
Chapter 25
22
Runge-Kutta Methods
3. Third order RK methods
1
yi 1  yi  (k1  4k 2  k3 ) h
16
yi 1  yi  (k1  4k 2  k3 )h
6
where
k where
 f (x , y )
1
i
i
k1  f ( x i , yi )1
1
k 2  f ( xi  h, yi  k1h)
12
1 2
k 2  f ( xi  h, yi  k1h)
k3  f ( xi  2h, yi  k21h  2k 2 h)
k3  f ( xi  h, yi  k1h  2k 2 h)
Chapter 25
Runge-Kutta Methods
4. Fourth order RK methods
1
yi 1  yi  (k1  2k 2  2k3  k 4 )h
6
where
k1  f ( x i , yi )
1
1
k 2  f ( xi  h, yi  k1h)
2
2
1
1
k3  f ( xi  h, yi  k 2 h)
2
2
k3  f ( xi  h, yi  k3 h)
Chapter 25
Comparison of Runge-Kutta Methods
Use first to fourth order RK methods to solve the equation from
x = 0 to x = 4
Initial condition y(0) = 2, exact answer of y(4) = 75.33896
f ( x, y)  4e
0.8 x
 0.5 y
Chapter 25