Transcript Slayt 1

Solution of Nonlinear Equation
Dr. Asaf Varol
[email protected]
1
Introduction
•
Solutions to algebraic, differential, and integral equations relating
certain physical parameters for a given problem
• Algebraic equations divided into two categories: (1) Linear and (2)
Nonlinear
• Linear: an equation which contains the first power of the unknown,
explicitly or implicitly, indicating a linear variation of the unknown
parameter as a function of the other known parameters.
• Example: distance, s, traveled by an object which was originally at a
location s = s0 at time t = 0, and moving at constant speed, v, is
related to time by
s = s0 + vt
• which is a linear relationship between s and v , and also between s
and t since v is assumed not changing with time.
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Introduction (Cont’d)
•
Nonlinear: an equation involving variables which contains at least one
term with an exponent other than unity and/or any explicit or implicit nonlinear function of that particular parameter which we want to solve for.
•
Example: Suppose that dv/dt = constant = a  0, i.e. there is a constant acceleration
(or a body force such as gravity) acting on the object, then the relation between s and
t becomes
s = s0 + v0t + (a/2)t2
•
where v0 is the initial speed. This equation has a nonlinear relationship between the
distance, s, and the time, t, because it involves the square of the parameter t.
•
Example: the following relation between the angle, , of a pendulum at time, t, is a
nonlinear relation.
 = 0cos(t)
where  = (g/L)1/2 is the frequency of the oscillations , 0 is the initial starting
angle , g is the acceleration of the gravity, and L is the length of the connecting rod.
This equation is nonlinear because the trigonometric function cos() is a nonlinear
function.
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Introduction (Cont’d)
•
Case for an object with changing acceleration as a result of drag
ds
v
dt
; s( 0)  s 0
dv
 b0  cv
dt
(2.1.4a)
; v 0   v0
(2.1.4b)
where b0 is a body force per unit mass such as acceleration of gravity ,c is a drag coefficient which can be taken as
constant. The solution to the Eqs (2.1.4a,b) is given by
1
v  v 0   b 0  cv 0  1  e  ct
(2.1.5a)
c
b
1
s  s0  0 t  2  b0  cv 0 1  e ct 
(2.1.5b)
c
c


v  100  1981  e01t 
s  98t  19801  e0.1t 
Find the velocity and the time at which the distance s = 0; (Note that t = 0, s = 0, v = 100 m/s is the
trivial solution and we are not looking for this trivial solution).
The equation to be solved first is


s  0  1980 1  e  0.1t  98t  0
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Graphical Method and Scanning
for t=-10:30
ft=(1-exp(-0.1*t)-0.05*t)
plot(t,ft,'r*')
hold on
grid on
end
xlabel('t(sec)')
ylabel('F(t)')
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Graphical Method and Scanning (Cont’d)
for t=-10:30
ft1=exp(-0.1*t)
ft2=1-0.05*t
plot(t,ft1,'r*',t,ft2,'b+')
hold on
end
text(10,2.5,'* f(t)=exp(-0.1t)')
text(10.,2.2,'+ f(t)=1-0.05t')
xlabel('t(sec)')
ylabel('F(t)')
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Example E.2.1
Problem: Find an approximate root by scanning of the function
F ( t )  t  40.8331  e 0.1t 
Solution: Let t = 10 seconds and scan starting from t = 0. From Table 2.1 we detect that t = 40 is an
approximate root. The exact root is somewhere in the interval (40, 50) because the function changes
sign in this interval.
Table 2.1 Values of the function F ( t )  t  40.8331  e
t
0
10
20
30
40
50
0.1t

f(t)
0.0
-15.81
-15.31
-8.80
-0.085
9.44
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Example E.2.2
Problem: Find two smallest positive roots of the nonlinear equation
F(  )     1 exp  0.5   sin100   0
by scanning.
Table E2.2 Scanning ofF(  ) of Example E.2.2 using various intervals
(i) =1.0

0
1
2
3
4
5
6
7
8
9
10
11
12
F()
-1.000
-0.506
-0.505
-0.554
-0.445
-0.139
root1(5.5)
0.293
0.725
1.022
1.087
0.888
0.469
root2(11.5)
-0.061
(ii) =0.1 (iii) =0.01

0.0
0.1
0.2
0.3
F()
-1.000
-1.400
root1(0.15)
0.1890
root2(0.25)
-1.591

0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
(iv) =0.005

0.00
0.005
0.010
F()
-1.000
-0.144
-0.061
-0.814
-1.698
-1.885
-1.192
-0.241
root1(0.075)
0.105
root2(0.085)
-0.458
0.015
0.020
F()
0.000
-0.513
-0.144
root1(
0.0125)
0.020
root2(
0.0175)
-0.061
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Bisection Method
• Bisection is a systematic search technique for finding a
zero of a continuous function. The method is based on
first finding an interval in which a zero is known to occur
because the function has opposite signs at the ends of
the interval, then dividing the interval into two equal
subintervals, and determining which subinterval contains
a zero, and continuing the computations on the
subinterval that contains the zero.
• Suppose that an interval [a, b] has been located which is
known to contain a zero, since the function changes
signs between a and b. The midpoint of the interval is
m=(a+b)/2
and a zero must lie in either [a, m] or [m, b].
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Bisection method
• The appropriate
subinterval is determined
by testing the function to
see whether it changes
sign on [a, m]. If so, the
search continues on that
interval; otherwise, it
continues on the interval
[m, b]. Figure illustrates
the first approximation to
the zero of the function
• Y=f(x)=x2-3, starting with
the interval [1, 2] [2].
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Example
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Example (Cont’d)
• It is
convenient
to keep track
of the
calculations
in tabular
form.
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Matlab program for Bisection Method
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Results
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Bisection Method (sometimes does not work at all)
•
Bisection method frequently converges slowly and sometimes does not work at
all.
Figure 2.2.3 Illustration of a situation where bisection method will not
work F(x)0 in the interval (x1 ,x2 ); F(x)0 in the interval (x3 ,x4 ):in both
intervals there is a root.
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Example E2.2.1
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Example E2.2.3
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Example E2.2.3 (Cont’d)
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False-Position (Regula Falsi) Method
• False-Position (or Regula-Falsi) Method is similar to the
bisection method, with a slight improvement in the
convergence rate.
• Upper and lower bound of the root are used to locate a
usually better estimate.
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False-Position (Regula Falsi) Method (Cont’d)
xr  xl 
 x  x F x 
F x   F x 
u
l
l
u
l
similarly one can show
xr  xu
x  x F x 


Fx   Fx 
u
l
u
u
l
Figure 2.2.4 Illustration of the false-position method
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False-Position Method (Cont’d)
•
However, there are special cases where the false-position method does not give
a better estimate than bisection method.
Figure 2.2.5 Special case for which the false-position method is not better
than the bisection method.
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Matlab program for Regula Falsi
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Results
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Fixed-Point Iteration Method
•
Simplest method for finding roots of nonlinear equations (also known as
the “Simple Iteration Method”)
•
Equations need to be of the form a = g(a)
– Example:
F(x) = x – ln (4 + x) = 0
can be rearranged to
x = ln (4 + x) = f(x)
• General Iteration Procedure:
xi+1 = f(xi)
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Matlab program for fixed point Method
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Diagram
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Results
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Newton-Raphson Method
Newton-Raphson method can be derived using a Taylor series
expansion about the point x = xi
F(xi+1) = F(xi) + F(xi)h + F(xi)h2/2 + F(xi)h3/6 + ...
where xi+1 is the next estimate to the root, and h = xi+1 - xi
• If the next iterate (xi+1) is approximately the root, then F(xi+1)  0
• The Newton-Raphson method is obtained by truncating the
Taylor series at the second term and solving for xi+1
F( x i )
x i 1  x i 
F' ( x i )
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Flow Diagram (Newton Raphson)
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Newton-Raphson Method (Cont’d)
•
Error - the error is determined from the leading truncation error of the
Taylor series.
 F' ' ( x i ) 
Ei1   
( Ei )2

 2 F' ( x i ) 
•
where Ei = xr - xi is the error in the previous iteration and Ei+1 is the error
in the current iteration
Convergence - such that successive errors become smaller, the
convergence of the method is sufficient that
 F' ' ( x i ) 
Abs
Ei   1
 2F' ( x i ) 
•
Since the error in the next iteration is proportional to the square of the
error is the previous iteration, the Newton-Raphson method has
quadratic convergence, and thus is a second-order method
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Matlab Program for Newton-Raphson Method
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Results for the Method of Newton-Raphson
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•
End of Chapter 2a
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References
•
•
•
•
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Celik, Ismail, B., “Introductory Numerical Methods for Engineering Applications”,
Ararat Books & Publishing, LCC., Morgantown, 2001
Fausett, Laurene, V. “Numerical Methods, Algorithms and Applications”, Prentice Hall,
2003 by Pearson Education, Inc., Upper Saddle River, NJ 07458
Rao, Singiresu, S., “Applied Numerical Methods for Engineers and Scientists, 2002
Prentice Hall, Upper Saddle River, NJ 07458
Mathews, John, H.; Fink, Kurtis, D., “Numerical Methods Using MATLAB” Fourth
Edition, 2004 Prentice Hall, Upper Saddle River, NJ 07458
Varol, A., “Sayisal Analiz (Numerical Analysis), in Turkish, Course notes, Firat
University, 2001
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