acsl, postech1 matlab 입문 chapter 8 numerical calculus and differential equations

ACSL, POSTECH 1

MATLAB 입문

CHAPTER 8 Numerical calculus and differential equations

Acsl, PostechMATLAB 입문 : Chapter 8. Numerical calculus and differential equations

2

Review of integration and differentiation

Engineering applications

Acceleration and velocity:

Velocity and distance:

Capacitor voltage and current:

Work expended:

b

adxxfA )(

A

a bx

f(x)

b

aaxtvbx )()()(

b

vdttabv0

)0()()(

b

aaQdtti

Cbv )()(

1)(

d

d

kxdxW

dxxfW

0

0)(


3

Integrals

Properties of integrals

Definite integrals

Indefinite integrals

b

a

b

a

b

adxxgddxxfcdxxdgxcf )()()]()([

b

a

c

a

b

cdxxfdxxfdxxf )()()(

111

111

n

a

n

b

ax

bx

n

xdxx

nnnb

a

n

abax

bxxdx

x

b

alnlnln

1

2]cos2[cos2

cossin2

x

xxxdx

xxdx sincos

1n

This week's content handles definite

integrals only

the answers are always numeric

see chapter 9


4

Improper integrals and singularities

Improper integrals

singularity

|1|ln1

1

xdxx

001lnlim

1

1lim h

hxdx

x

|)1|ln|1|lim(ln h

)1ln(lim h

h 1- h 1-

h 1-

h 1-


5

Derivatives

dxxfxg )()(dx

dgxf )(

dx

dfxg

dx

dgxf

dx

dh)()( )()()( xgxfxh

)(/)()( xgxfxh 2

)()(

gdxdg

xfdxdf

xg

dx

dh

)(yfz )(xgy dx

dy

dy

dz

dx

dz

xdx

xd

xdx

xdxdx

xd

nxdx

dx nn

sincos

cossin

1ln

1

xxxxdx

xxdsin2cos

)sin( 22

xxdx

dyy

dx

xdcossin22

)(sin2

Integration differentiation

example

: product rule

: quotient rule

: chain rule

1) 2)

3)


6

Numerical integrationRectangular integration trapezoidal integration

Numerical integration functions

Command Description

quad (‘function’,a,b,tol) Uses an adaptive Simpson’s rule to compute the integral of the function ‘function’ with a as the lower integration limit and b as the upper limit. The parameter tol is optional. Tol indicates the specified error tolerance.

quadl (‘function’, a,b,tol) Uses Lobatto quadrature to compute the integral of the function ‘function’. The rest of the syntax is identical to quad.

trapz (x,y) Uses trapezoidal integration to compute the integral of y with respect to x, where the array y contains the function values at the points contained in the array x.

y

xa b

y=f(x) y=f(x)

y

xa b

··· ···

··· ···

0sin dx

2cos0coscossin00

xdx (exact

solution)

(use of the trapz function)


7

Rectangular, Trapezoidal, Simpson Rules

Rectangular rule Simplest, intuitive wi = 1 Error=O(h) Mostly not enough!

Trapezoidal rule Two point formula, h=b-a Linear interpolating polynomial

Simpson’s Rule Three point formula, h=(b-a)/2 Quadratic interpolating polynomial Lucky cancellation due to right-left symmetry

Significantly superior to Trapezoidal rule Problem for large intervals -> Use the extended formula

)(2/)( 310 hOffhI

x0 =a xN=b

f(x)

x1 x2

f(x1)f(x2)

1

0

N

iifhI

x0 =a x1=b

f(x) f0 f1

)()3

1

3

4

3

1( 5

210 hOfffhI


Matlab's Numerical Integration Functions trapz(x,y) When you have a vector of data

trapezoidal integration method not as accurate as Simpson's method very simple to use

quad('fun',a,b,tol) When you have a function adaptive Simpson’s rule integral of ’fun’ from a to b tol is the desired error tolerance and is optional

quad1('fun',a,b,tol) Alternative to quad adaptive Lobatto integration integral of ’fun’ from a to b tol is the desired error tolerance and is optional 8


9

Comparing the 3 integration methods

The answer is A1=A2=0.6667, A3=0.6665

test case:

Trapezoid Method:

x=linspace(0,pi, 10);

y=sin(x);

trapz(x,y);

2cos0coscossin00

xdx

(exact solution)

(

Simpson's Method:

quad('sin',0,pi);

Lobatto's Method

quadl('sin',0,pi);


Integration near a Singularity

Trapezoid:

x=[0:0.01:1];

y=sqrt(x);

A1=trapz(x,y);

Simpson:

A2=quad('sqrt',0,1);

Lobatto:

A3=quadl('sqrt',0,1);

10

xdxdy

xy

/5.0/

(The slope has a singularity at x=0)


Using quad( ) on homemade functions

11

1) Create a function file:

function c2 = cossq(x)

% cosine squared function.

c2 = cos(x.^2);

Note that we must use array exponentiation.

2) The quad function is called as follows:

quad('cossq',0,sqrt(2*pi))

3) The result is 0.6119.


Problem 1 and 5 -- Integrate

Hint: position is the integral of velocity*dt Hint: Work is the integral of force*dx

12


Problem 11

First find V(h=4) which is the Volume of the full cup Hint: Flow = dV/dt, so the integral of flow (dV/dt) from 0 to t = V(t) Set up the integral using trapezoid (for part a) and Simpson (for b) Hint: see next slide for how to make a vector of integrals with changing b values So you can calculate vector V(t), i.e. volume over time for a given flow rate

V = quad(....., 0, t) where t is a vector of times and then plot and determine graphically when V crosses full cup 13


To obtain a vector of integration results:

For example, sin(x) is the integral of the cosine(z) from z=0 to z=x

In matlab we can prove this by calculating the quad integral in a loop:for k=1:101

x(k)= (k-1)* pi/100; % x vector will go from 0 to pi

sine(k)=quad('cos',0,x(k)); % this calculates the integral from 0 to x

end

plot(x, sine) % this shows the first half of a sine wave

% calculated by integrating the cosine

14


Problem 14

Hint: The quad function can take a vector of values for the endpoint Set up a function file for current Set up a vector for the time range from 0 to .3 seconds Then find v = quad('current',0,t) will give a vector result and plot v

15


16

Numerical differentiation

x

yy

xx

yymB

23

23

23

x1 x2 x3

y1

y2

y3

Δx Δx

y=f(x)

True slope

A

B

C

x

yy

x

yy

x

yymmm

BA

C

22

1

2

132312

x

yy

xx

yymA

12

12

12

x

y

dx

dy

limΔx 0

: backward difference

: central difference

: forward difference


17

The diff function

The diff Function

d= diff (x)

Backward difference & central difference method

example

example x=[5, 7, 12, -20]; d= diff(x)

d = 2 5 -32


Try that: Compare Backward vs Central

Construct function with noise

x=[0:pi/50:pi];

n=length(x);

% y=sin(x) +/- .025 random error

y=sin(x) + .05*(rand(1,51)-0.5);

td=cos(x); % true derivative

Backward difference using diff:

d1= diff(y)./diff(x);

subplot(2,1,1)

plot(x(2:n),td(2:n),x(2:n),d1,'o');

xlabel('x'); ylabel('Derivative');

axis([0 pi -2 2])

title('Backward Difference Estimate')

Central Difference

d2=(y(3:n)-y(1:n-2))./(x(3:n)-x(1:n-2));

subplot(2,1,2)

plot(x(2:n-1),td(2:n-1),x(2:n-1),d2,'o');

xlabel('x'); ylabel('Derivative');

axis([0 pi -2 2])

title('Central Difference Estimate');

18


Problem 7: Derivative problem

Hint: velocity is the derivative of height

19


Problem 17

20


21

Polynomial derivatives -- trivia

b = polyder (p)

p = [a1,a2,…,an]

b = [b1,b2,…,bn-1] b = polyder (p1,p2) [num, den] = polyder(p2,p1)

nnnnn axaxaxaxaxf

12

31

21)(

12

21

1 )1( nnn axanxna

dx

df

12

21

1 nnn bxbxb

Example p1= 5x +2

p2=10x2+4x-3

Result: der2 = [10, 4, -3]

prod = [150, 80, -7]

num = [50, 40, 23]

den = [25, 20, 4]

Command Description

d=diff(x) Returns a vector d containing the differences between adjacent elements in the vector x.

b=polyder(p) Returns a vector b containing the coefficients of the derivative of the polynomial represented by the vector p.

b=polyder(p1,p2) Returns a vector b containing the coefficients of the polynomial that is the derivative of the product of the polynomials represented by p1 and p2. cf. equivalent comment: b=polyder(conv(p1,p2))

[num, den]= polyder(p2,p1)

Returns the vector num and den containing the coefficients of the numerator and denominator polynomials of the derivative of the quotient p2/p1, where p1 and p2 are polynomials.

Numerical differentiation functions


22

Analytical solutions to differential equations(1/6)

Solution by Direct Integration Ordinary differential equation (ODE)

example

Partial differential equation (PDE) -- not covered

2tdt

dy

303

33

0 0

2 tttdttdt

dt

dyt t

3)0()(

3tyty

2

2

)(

)(

dt

ydty

dt

dyty


23


Oscillatory forcing function

A second-order equation

)(tfdt

dy Forcing function example wttf sin)(

w

wtt

w

wtwtdtdt

dt

dy tt cos1

0

cossin

00

w

wtyty

tty

cos1)0()(

0)(

w

wtyty

cos1)0()(

32

2

tdt

yd

4)0(

4

0

3

0 2

2 tdtty

dt

dydt

dt

yd tt

)0(4

4

yt

dt

dy

)0(20

)0(4

)0()(5

0

4

0yt

tdty

tytydt

dt

dy tt

)0()0(20/)( 5 yyttty


24


Substitution method for first-order equations

)(tfydt

dy stAety )( stsAedtdy / 0)( tf

0 stst AesAeydt

dy

Characteristic equation

/1s

AAey 0)0(

/)0()( teyty

Characteristic root

( ),

01s

to find A

Solution, Free response

∵The solution y(t) decays with time if τ>0

τ is called the time constant.

BAety st )(

AyB )0(

AyAety st )0()(

01s MyA )0(

)1()0()( // tt eMeyty

0)( tf 0tsuppose

for

M at t=0 Such a function is the step

function

( ),

Forced response

Natural (by Internal Energy) v.s. Forced Response (by External Energy)

Transient (Dynamics-dependent) v.s. Steady-state Response (External Energy-dependent)


Problem 18 – solve on paper then plot(next week we solve with Matlab)

Use the method in previous slide

25


Problem 21 – solve on paper then plot(next week we solve with Matlab)

Re-arrange terms:

mv' + cv = f , OR

(m/c) v' + v = f/c

now it's in the same form as 2 slides back

26


27


Nonlinear equations

Substitution method for second order equations(1/3)

example

05 yyyy

0sin yy

0 yy

02 ycy stAety )(

0)( 22 stAecs

( )

cs ctct eAeAty 21)( general solution

suppose that 4)0(,6)0(,2 yyc

216)0( AAy

422)0( 21 AAy2,4 21 AA solution

1)


28



02 yy stAety )(

tite ti sincos

titi eAeAty 21)(

tBtBty cossin)( 21

)0(,/)0( 21 yByB

tyty

ty

cos)0(sin)0(

)(

2

P Pf /1

)(tfkyycym stAetytf )(,0)(

tsts eAeAty 2121)(

tsetAAty 1)()( 21

teBteBty atat cossin)( 21

( )

general solution

Euler’s identities

period frequency

2)

3) ( )

0)( 2 stAekcsms

1. Real and distinct: s1 and s2.

2. Real and equal: s1.

3. Complex conjugates: ias


29



1. real, distinct roots: m=1,c=8, k=15.characteristic roots s=-3,-5.

tt eAeAty 52

31)(

2. complex roots: m=1,c=10,k=601

characteristic roots is 245

teBteBty tt 24cos24sin)( 52

51

12/24/2 P

3. unstable case, complex roots: m=1,c=-4,k=20 is 42 characteristic roots

teBteBty tt 4cos4sin)( 22

21

2/4/2 P

4. unstable case, real roots: m=1,c=3,k=-10characteristic roots s=2,-5.

tt eAeAty 52

21)(


Problem 22

Just find the roots to the characteristic equation and determine which

form the free response will take (either 1, 2, 3 or 4 in previous slide)

This alone can be a helpful check on matlab's solutions—if you don't get the right form, you can suspect there is an error somewhere in your solution

30


Next Week:

we learn how to solve ODE's with Matlab

31

acsl, postech1 matlab 입문 chapter 8 numerical calculus and differential equations

Documents

numerical calculus

function function

postech matlab

fx y x

integral of y

trapz function slide

review of integration

integration methods