Ordinary Differential Equations Part 7
Transcript of Ordinary Differential Equations Part 7
Introduction
• In Chapter 1, we have Newton's Second Law of falling parachutist;
---------- (PT7.1)
where; g = gravitational constant
m = mass
c = drag coefficient
dv/dt = unknown function/differential equation
(rate equation)
• Equation (PT7.1) is called a rate equation because the rate of
change of a variable i.e. v = dependent variable and t =
independent variable
• When function involves one independent variable, the equation is
called Ordinary Differential Equation (ODE).
ordinary differential equations 2
vm
cg
dt
dv
• A partial differential equation (or PDE) involves two or more
independent variables.
• The fundamental laws of physics, mechanics, electricity, and
thermodynamics are usually based on empirical (experimentally)
observations that explain variations in physical properties and
states of systems.
• However, as described in previous chapter, many of differential
equations of practical significant cannot be solved using analytical
methods of calculus.
• Thus, this chapter covers important methods (Numerical
approach) for solving all engineering applications.
3 ordinary differential equations
ordinary differential equations 4
Law Mathematical expression Variables & Parameters
Newton’s 2nd
law
of motion
Fourier’s heat
law
Fick’s law of
Diffusion
Faraday’s law
dv
dt
F
m
q k 'dT
dx
J Ddc
dx
VL Ldi
dt
V=velocity F=force m=mass q=heat flux k’=thermal conductivity T=temp. J=mass flux D=diffusion coefficient c=concentration VL=voltage drop L=inductance i=current
Table PT7.1 : Examples of fundamental laws in terms of rate of change of
variables (t = time, x = position)
5 ordinary differential equations
Figure PT7.2 : The sequence of events in the application of ODEs for
engineering problem solving e.g. a falling parachutist
• In this chapter we will focus on one step method which consists of:
Euler’s method
Improvement of Euler method
a) Huen’s method
b) The midpoint (or improved Polygon) method
Runge-Kutta method (2nd, 3rd, 4th and higher order)
One step method
• To solve ODE's in the form of;
dy/dx = f(x,y)
• In previous chapter, we used a numerical method to solve such an
equation for the velocity of the falling parachutist.
6 ordinary differential equations
• In general form;
New value = old value + slope x (step size)
• or, in mathematical terms,
yi+1 = yi + h ------ (25.1)
where, yi+1 = new value
yi = old value
h = step size
= slope; extrapolation from yi to yi+1 over a distance
h (Fig. 25.1)
• This formula can be applied step by step to compute the final
results.
ordinary differential equations 7
Step size = h
xi xi+1
y
x
hyy ii 1
Slope =
Figure 25.1: Graphical depiction of a one-step method
8 ordinary differential equations
25.1 EULER’s METHOD
• This method follows the approximation of Taylor’s expansion.
• The slope at the beginning of the interval is taken as an
approximation of the average slope over the whole interval. The first
derivative provides a direct estimate of the slope at xi.
• Slope at xi;
• Subs eq a) into eq 25.1 yield;
• Eq 25.2 is called as Euler’s Method or point-slope method.
hyxfyyiiii),(
1
9 ordinary differential equations
),( ii yxf
= differential equation evaluated at xi and yi
size step
),(
h
yxfii
----- a
----- 25.2
A new value of y is predicted using the slope (equal to the first
derivative at the original value of x) to extrapolate linearly over the
step size h
10 ordinary differential equations
Fig. 25.2: Euler’s Method
Example 1: Use Euler’s method to numerically integrate:
11 ordinary differential equations
5.820122 23 xxxdx
dy
from x = 0 to x = 4 with a step size of 0.5. Initial condition at x = 0 is
y = 1. Exact solution is given as;
y = -0.5x4 + 4x3 – 10x2 + 8.5x + 1
Example 1: Use Euler’s method to numerically integrate:
12 ordinary differential equations
5.820122 23 xxxdx
dy
from x = 0 to x = 4 with a step size of 0.5. Initial condition at x = 0 is
y = 1. Exact solution is given as;
y = -0.5x4 + 4x3 – 10x2 + 8.5x + 1 Solution:
From Euler’s method equation;
1. Given; h = 0.5, x = 0 to 4, y(0) = 1
2. At x=0, y0 = 1
x=0.5, y0.5 = y0 + f (0,1) (0.5) = 1 + f (0,1) (0.5)
hyxfyy iiii ),(1
hyxfyy iiii ),(1
x yEuler
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
• Given;
• Therefore: f(0,1) = -2(0)3 + 12 (0)2 + 20(0) + 8.5 = 8.5
• From point 2,
y0.5 = 1 + f (0,1) (0.5) = 1 + (8.5) (0.5) = 5.25
Thus; y(0.5) = 1.0 + 8.5(0.5) = 5.25
ordinary differential equations 13
5.820122 23 xxxdx
dy
%1.63%100x21875.3
03125.2
03125.225.521875.3
iserror true thethus,
21875.31)5.0(5.8)5.0(10)5.0(4)5.0(5.0
:is 0.5at xy newfor solution True
t
234
tE
y
3. At x = 1.0, y1 = y0.5 + f (0.5,5.25) (0.5) = 5.25 + f (0.5,5.25) (0.5)
• Therefore: f(0.5,5.25) = -2(0.5)3 + 12 (0.5)2 + 20(0.5) + 8.5 = 1.25
y1 = 5.25 + f(0.5, 5.25) (0.5)
Thus; y(1) = 5.25 + (1.25)(0.5) = 5.875
ordinary differential equations 14
hyxfyy iiii ),(1
%8.95
875.2875.50.3
iserror true thethus,
0.31)1(5.8)1(10)1(4)1(5.0
is; 1at x solution The
t
234
tE
y
ordinary differential equations 15
Table 25.1: Comparison true and approximation values
x ytrue yEuler t% Global t% Local 0.0 1.000 1.000 0.5 3.21875 5.250 -63.1 -63.1 1.0 3.000 5.875 -95.8 -128.0 1.5 2.21875 5.125 131.0 -1.41 2.0 2.000 4.500 -125.0 20.5 2.5 2.71875 4.7500 -74.7 17.3 3.0 4.000 5.8750 46.9 4.0 3.5 4.71875 7.1250 -51.0 -11.3 4.0 3.000 7.000 -133.3 -53.0
• The results for all data are compiled in table 25.1 and figure 25.3.
• Although the computation captures the general trend of the true
solution, the error is considerable.
• As discussed in the next section, this error can be reduced by using
a smaller step size
Fig. 25.3: Comparison of curve true solution with numerical solution
using Euler’s Method
ordinary differential equations 16
Fig. 25.4: Comparison of curve true solution with numerical
solution using Euler’s Method
ordinary differential equations 17
25.2 Improvements of Euler’s method
• Can be divided into 2 types;
a. Huen's Method
b. Mid-point Method (Improve Polygon Method)
25.2.1 Huen’s Method
• To improve the estimate of the slope.
• Involves the determination of two derivatives for the interval (h) at;
i. Initial Point
ii. End Point
• Then the two derivatives are averaged to obtain an improved
estimate of the slope for the entire interval, h.
ordinary differential equations 18
• Recall Euler's Method; slop at beginning an interval;
y'i = (f (xi , yi ) -------- 25.12
• Used to extrapolate linearly to yi+1;
Note: for Euler's method we would stop at this point;
• However in Huen's Method yoi+1 is an intermediate prediction.
• Thus, equation 25.13 is called as Predictor equation and this
equation allowing the calculation of an estimated slope at the end of
interval:
hyxfy iii ),( 10
1
'
1
19 ordinary differential equations
13.25 ),(1
hyxfyyiii
o
i
----- 25.14
• Thus an average slope of eqs 25.12 and 25.14 for the interval is
given as;
• This average slope is the used to extrapolate linearly from yi to yi+1
using Euler’s method:
which is called a corrector equation.
• Fig. 25.9a shows the predictor approach (equation 25.13), and Fig.
25.9b shows the corrector equation (equation 25.16).
2
),(),(
2'
0
11
'
1
'
iiiiii yxfyxfyy
y
hyxfyxf
yy iiiiii
2
),(),( 0
111
20 ordinary differential equations
----- 25.16
Fig. 25.9: (a) Predictor of Huen’s Method and (b) Corrector of Huen’s
Method
ordinary differential equations 21
Termination Criterion:
ordinary differential equations 22
%1001
1
11
j
i
j
i
j
ia
y
yy
j
iy 1
1
1
j
iy
where; Result from present iteration of corrector
Result from previous iteration of corrector
----- 25.17
Example 2: Use Huen’s method to integrate y’=4e 0.8x–0.5y from x = 0
to x = 4 with a step size of 1.0. The initial condition at x = 0 is y = 2.
Analytical solution for integral equation;
y’=4e 0.8x – 0.5y is;
ordinary differential equations 23
xxx eeey 5.05.08.0 2)(3.1
4
Example 2: Use Huen’s method to integrate y’=4e 0.8x–0.5y from x = 0
to x = 4 with a step size of 1.0. The initial condition at x = 0 is y = 2.
Analytical solution for integral equation;
y’=4e 0.8x – 0.5y is;
Solution
ordinary differential equations 24
xxx eeey 5.05.08.0 2)(3.1
4
This formula can be used to calculate true value/solution as shown in table
25.2;
(1 iteration each step) (15 iteration each step)
x ytrue yHeun t% yHeun t% 0 2.000 2.000 0.00 2.000 0.00 1 6.1946314 6.7010819 8.18 6.3608655 2.68 2 14.8439219 16.3197819 9.94 15.3022367 3.09 3 33.6771718 37.1992489 10.46 34.7432761 3.17 4 75.3389626 83.3377674 10.62 77.7350962 3.18
To calculate yHeun ; as Numerical Approach;
Given initial value; xi = 0, yi = 2
Predictor (eq. 25.13): yoi+1 = yi + f(xi , yi )h
= 2 + (1)
Step 1: f(xi , yi ) = f(0 , 2) = slope at point (xi , yi )
y’=4e 0.8x–0.5y
= 4e 0.8(0) – 0.5(2)
= 4-1 = 3
Step 2: predictor (eq 25.13);
yoi+1 = yi + f(xi , yi )h
= 2 + (1)
yoi+1 = 2 + 3(1) = 5
ordinary differential equations 25
Thus, next point: xi+1 = 1 , yi+1 = 5
Step 3 : f(xi+1 , yo
i+1 )= f(1 , 5) = slope at point (xi+1 ,yo
i+1)
y’=4e 0.8x–0.5y
y’=4e 0.8(1)–0.5(5) = 6.402164
Step 4 : average slope
Step 5 : Corrector (eq 25.16);
ordinary differential equations 26
2
),(),('
11
io
iii yxfyxfy
701082.42
)402164.6()3('
y
hyxfyxf
yyi
o
iii
ii2
),(),( 11
1
Thus;
Step 6 : repeat step 3 through step 5
Step 3: f(xi+1 , yi+1) = f(1 , 6.701082)
= 4e 0.8(1) - 0.5 (6.701082)
= 5.551622714
ordinary differential equations 27
701082.6
)1(701082.42
1
hyy
or
o
%18.8
%1001946314.6
701082.61946314.6%
t
195.6
2)(3.1
4
2)(3.1
4
)1(5.0)1(5.0)1(8.0
5.05.08.0
eee
eeey xxx
Step 4 : slope average;
Step 5: corrector equation;
y1 = yo + h
= 2 + 4.275811357(1)
= 6.275811357
Thus;
ordinary differential equations 28
275811357.42
551622714.53'
y
%31.1
%1001946314.6
275811.61946314.6%
t
• Again, repeat step 3 to step 5:
• t% increasing when iteration increase. It happen especially for
large step size. For small step size, the iteration coverage on a
single value.
• For our case, if step 3 to 5 are repeated at iteration 15; we get y1 =
6.3608655 and εt% = 2.68
ordinary differential equations 29
03.3%
;
382129.6
)1(2
)1(2758.6(5.0432
)1(8.0
1
t
thus
ey
Fig. 25.11: Comparison of true solution with numerical solution using
Euler's and Heun's method for y' = -2x3+12x2 -20x +8.5
ordinary differential equations 30
25.2.2 Midpoint Method (Improved Polygon Method)
• Another simple modification of Euler’s method.
• This technique uses Euler's method to predict a value of y at
midpoint of interval as shown in Fig. 25.12 b).
• Thus the new value of y;
• Then, this predicted value is used to calculate a slope at the
midpoint:
which is assumed to represent a valid approximation of the average
slope for the entire interval.
ordinary differential equations 31
2),(
2
1
hyxfyy iii
i
yi1
2
f (xi1
2
,yi1
2
)
----- 25.25
----- 25.26
Fig. 25.12 : Mid-point method for equations 25.24 and 25.27
ordinary differential equations 32
Note: because yi+1
is not both sides,
the corrector (eq
25.16 cannot be
applied iteratively
to improve the
solution
• This slope is then used to extrapolate linearly from xi to xi+1 (Fig.
25.12).
Example 3: Use Midpoint method to solve this equation;
for range x = 0 to 2 and h = 0.5 where y(0) = 1.
ordinary differential equations 33
27.25 ),(2
1
2
11
hyxfyyii
ii
y yx2 1.2y
Solution:
At x = 0, y = 1
At x = 0.5
(a)
(b) slope:
ordinary differential equations 34
yi1
2
y i f (x i,y i)h
2
yi1
2
1 f (0,1)0.5
2
f (0,1) (1)(0)2 1.2(1) 1.2
yi1
2
1 (1.2)0.5
2 0.7
yi1
2
f (xi1
2
,yi1
2
) f (0.25,0.7)
f (0.25,0.7) (0.7)(0.25)2 1.2(0.7) 0.79625