[email protected] MTH16_Lec-14_sec_9-4_ODE_SlopeFields_Euler.pptx 1 Bruce Mayer, PE Chabot...

[email protected] • MTH16_Lec-14_sec_9-4_ODE_SlopeFields_Euler.pptx 1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§9.4 ODEAnalytic

s



Review §

Any QUESTIONS About• §9.3 Differential Equation Applications

Any QUESTIONS About HomeWork• §9.3 → HW-15

9.3



§9.4 Learning Goals

Analyze solutions ofdifferential equationsusing slope fields

Use Euler’s methodfor approximating solutions of initialvalue problems



Slope Fields

Recall that indefinite integration, or AntiDifferentiation, is the process of reverting a function from its derivative. • In other words, if we have a derivative, the

AntiDerivative allows us to regain the function before it was differentiated – EXCEPT for the CONSTANT, of course.

Given the derivative dy/dx = f ‘(x) then solving for y (or f(x)), produces the General Solution of a Differential Eqn



Slope Fields

AntiDifferentiation (Separate Variables) Example• Let: • Then Separating the Variables:• Now take the AntiDerivative: • To Produce the General Solution:

This Method Produces an EXACT and SYMBOLIC Solution which is also called an ANALYTICAL Solution

xdx

dy2

dxxdy 2

dxxdy 2

Cxy 2



Slope Fields

Slope Fields, on the other hand, provide a Graphical Method for ODE Solution

Slope, or Direction, fields basically draw slopes at various CoOrdinates for differing values of C.

Example: The Slope Field for ODE x

dx

dy



Slope Fields

slope field describes several different parabolas based on varying values of C

Slope Field Example: create the slope field for the Ordinary Differential Eequation:

Cx

ydxxdyxdx

dy 2

12

y

x

dx

dy



Slope Fields

Note that dy/dx = x/y calculates the slope at any (x,y) CoOrdinate point• At (x,y) = (−2, 2),

dy/dx = −2/2 = −1• At (x,y) = (−2, 1),

dy/dx = −2/1 = −2• At (x,y) = (−2, 0),

dy/dx = −2/0 = UnDef.• And SoOn

Produces OutLine of a HYPERBOLA

x

y

-2

-1

1

2

-2 -1 1 2

x

y



Slope Fields

Of course this Variable Separable ODE can be easily solved analytically

y

x

dx

dy dxxdyy

dxxdyy

Cxy 22

2

1

2

1

Cxy 22

Cyx 22

x

y

-2

-1

1

2

-2 -1 1 2

x

y



Slope Fields Example

For the given slope field, sketch two approximate solutions – one of which is passes through(4,2):• Solve ODE Analytically using

using (4,2) BC

12 xm

2,4

12

1 x

dx

dy dxxdy

1

2

1

dxxdy

1

2

1

C 444

12 2

Cxxy 2

4

1

C 2 24

1 2 xxySoln



Slope Field Identification

C

3xdx

dy

In order to determine a slope field from a

differential equation, we should consider the

following:

i) If isoclines (points with the same slope) are along horizontal lines, then DE depends only on y

ii) Do you know a slope at a particular point?

iii) If we have the same slope along vertical lines, then DE depends only on x

iv) Is the slope field sinusoidal?

v) What x and y values make the slope 0, 1, or undefined?

vi) dy/dx = a(x ± y) has similar slopes along a diagonal.

vii) Can you solve the separable DE?

1. _____

2. _____

3. _____

4. _____

5. _____

6. _____

7. _____

8. _____

Match the correct DE with its graph:2y

dx

dy

xdx

dycos

xdx

dysin

yxdx

dy

22 yxdx

dy

1 yydx

dy

y

x

dx

dy

A B

C

E

G

D

F

H

H

B

F

D

G

E

A



Example Demand Slope Field

Imagine that the change in fraction of a production facility’s inventory that is demanded, D, each period is given by• Where p is the unit price in $k

Draw a slope field to approximate a solution assuming a half-stocked (50%) inventory and $2k per item, and then • Verify the Slope-Field solution using

Separation of Variables.

c




SOLUTION: Calculate some

Slope Values from peDm

dp

dD 1

1100,0 0 emdp

dD

0111,1 1 emdp

dD

068.015.05.0,2 2 emdp

dD




An approximate solution passing through (2,0.5) with slope field on the window 0 < x < 3 and 0 < y < 1

$k/unit p

fr

acti

onal

p

D




Find an exact solution to this differential equation using separation of variables:

Remove absolute-value and then change signs as inventory demanded satisfies: 0≤ D ≤1




Removing ABS Bars

Or Now use Boundary Value ($2k/unit,0.5)

CeCeDp eeDeeCeDpp

11ln 1ln

pp eeC eADDeeD 110with 1

572.05.02

eeA




Graph for

This is VERY SIMILAR to the Slope Field Graph Sketched Before

peepD 572.01



Numerical ODE Solutions Next We’ll “look

under the hood” of NUMERICAL Solutions to ODE’s

The BASIC Game-Plan for even the most Sophisticated Solvers:• Given a STARTING

POINT, y(0)• Use ODE to find dy/dt at t=0

• ESTIMATE y1 as

001

tdt

dytyy



Numerical Solution - 1

Notation Exact Numerical Method (impossible to achieve) by Forward Steps

tntn

)( nn tyy

),( nnn ytff

Number Step n

Length Step Time t

),( ytfdt

dy

Now Consider

yn+1

tn

yn

tn+1

t

Dt



Numerical Solution - 2 The diagram at Left shows

that the relationship between yn, yn+1 and the CHORD slope

yn+1

tn

yn

tn+1

t

Dt

slope chord 1

t

yy nn

The problem with this formula is we canNOT calculate the CHORD slope exactly • We Know Only Δt & yn, but

NOT the NEXT Step yn+1

The AnalystChooses Δt

ChordSlope

Tangent Slope



Numerical Solution -3 However, we can

calculate the TANGENT slope at any point FROM the differential equation itself

The Basic Concept for all numerical methods for solving ODE’s is to use the TANGENT slope, available from the R.H.S. of the ODE, to approximate the chord slope

Recognize dy/dt as the Tangent Slope

),( nntt

n ytfdt

dym

n

),( slopetangent ytf nnt

nn ytfdt

dy

t

yy

n

,1



Euler Method – 1st Order ODE

Solve 1st Order ODE with I.C.

ReArranging

Use: [Chord Slope] [Tangent Slope at start of time step]

),( ytfdt

dy

by )0( nnn ftyy 1

Then Start the “Forward March” with Initial Conditions

byt 00 0 nnt

nn ytfdt

dy

t

yy

n

,1

or1 nt

n ydt

dyty

n



Example Euler Estimate

Consider 1st Order ODE with I.C.

Use The Euler Forward-Step Reln

See Next Slide for the 1st Nine Steps For Δt = 0.1

1 ydt

dy

0)0( y

)1(1 nnn ytyy

ntn

nnn

dt

dyty

ftyy

1

But from ODE

So In This Example:

1 nt

ydt

dy

n



Euler Exmple Calc

n tn yn fn= – yn+1 yn+1= yn+Dt fn

0 0 0.000 1.000 0.100

1 0.1 0.100 0.900 0.190

2 0.2 0.190 0.810 0.271

3 0.3 0.271 0.729 0.344

4 0.4 0.344 0.656 0.410

5 0.5 0.410 0.590 0.469

6 0.6 0.469 0.531 0.522

7 0.7 0.522 0.478 0.570

8 0.8 0.570 0.430 0.613

9 0.9 0.613 0.387 0.651

1.01 tydt

dy

Plot

Slope



Euler vs Analytical

0

0.2

0.4

0.6

0.8

0

0.2

5

0.5

0.7

5 1

1.2

5t

Exact

Numerical

y

tey 1

The Analytical Solution



Analytical Soln

Let u = −y+1 Then

001 tyydt

dy

dudy

dydu

yu

10

1

Sub for y & dy in ODE

udt

du

Separate Variables

dtu

du

Integrate Both Sides

dtu

du 1

Recognize LHS as Natural Log

Ctu ln

Raise “e” to the power of both sides

Ctu ee ln



Analytical Soln

And

001 tyydt

dy

Thus Soln u(t)tKeu

Sub u = 1−y

Now use IC

The Analytical Soln

ttCCt

u

Keeee

ue

ln

tKey 1

1

01 0

K

Ke

tey 11

tey 1



ODE Example: Euler Solution with

∆t = 0.25, y(t=0) = 37 The Solution Table

61.5ln2.4cos9.3 tydt

dy

0 1 2 3 4 5 6 7 8 9 1022

24

26

28

30

32

34

36

38

t

y(t)

by

Eul

er

Euler Solution to dy/dt = 3.9cos(4.2y)-ln(5.1t+6)

n t y dy/dt dely yn+1

0 0 37.0000 -1.7457 -0.4364 36.56361 0.25 36.5636 1.4027 0.3507 36.91432 0.5 36.9143 -1.3492 -0.3373 36.57693 0.75 36.5769 1.2410 0.3103 36.88724 1 36.8872 -1.2264 -0.3066 36.58065 1.25 36.5806 1.0448 0.2612 36.84186 1.5 36.8418 -0.7108 -0.1777 36.66417 1.75 36.6641 1.1868 0.2967 36.96088 2 36.9608 -2.5004 -0.6251 36.33579 2.25 36.3357 -2.6357 -0.6589 35.6768

10 2.5 35.6768 -1.6265 -0.4066 35.270111 2.75 35.2701 0.0722 0.0181 35.288212 3 35.2882 -0.2436 -0.0609 35.227313 3.25 35.2273 0.4430 0.1107 35.338014 3.5 35.3380 -1.1420 -0.2855 35.052615 3.75 35.0526 -0.0139 -0.0035 35.049116 4 35.0491 -0.1072 -0.0268 35.022317 4.25 35.0223 -0.5255 -0.1314 34.890918 4.5 34.8909 -2.6041 -0.6510 34.239919 4.75 34.2399 -1.1497 -0.2874 33.952420 5 33.9524 -3.0108 -0.7527 33.199721 5.25 33.1997 -3.0006 -0.7502 32.449622 5.5 32.4496 -3.0151 -0.7538 31.695823 5.75 31.6958 -2.9862 -0.7466 30.949224 6 30.9492 -3.0384 -0.7596 30.189725 6.25 30.1897 -2.9328 -0.7332 29.456426 6.5 29.4564 -3.1419 -0.7855 28.671027 6.75 28.6710 -2.6916 -0.6729 27.998128 7 27.9981 -3.5484 -0.8871 27.111029 7.25 27.1110 -1.7458 -0.4365 26.674530 7.5 26.6745 -2.8722 -0.7180 25.956531 7.75 25.9565 -2.4562 -0.6141 25.342432 8 25.3424 -0.4717 -0.1179 25.224533 8.25 25.2245 -2.2562 -0.5641 24.660434 8.5 24.6604 -0.0369 -0.0092 24.651235 8.75 24.6512 -0.0977 -0.0244 24.626836 9 24.6268 -0.2699 -0.0675 24.559337 9.25 24.5593 -1.0481 -0.2620 24.297338 9.5 24.2973 -3.9863 -0.9966 23.300739 9.75 23.3007 -0.9318 -0.2329 23.067840 10 23.0678 -1.0551 -0.2638 22.8040



Compare Euler vs. ODE45Euler Solution ODE45 Solution

0 1 2 3 4 5 6 7 8 9 1022

24

26

28

30

32

34

36

38

t

y(t)

by

Eul

er


0 1 2 3 4 5 6 7 8 9 1034.5

35

35.5

36

36.5

37

37.5

T by ODE45

Y b

y O

DE

45

Euler is Much LESS accurate



Compare Again with ∆t = 0.025Euler Solution ODE45 Solution

0 1 2 3 4 5 6 7 8 9 1034.5

35

35.5

36

36.5

37

37.5

T by ODE45

Y b

y O

DE

45

Smaller ∆T greatly improves Result

0 1 2 3 4 5 6 7 8 9 1035.8

36

36.2

36.4

36.6

36.8

37

37.2

t

y(t)

by

Eul

er




MatLAB Code for Euler% Bruce Mayer, PE% ENGR25 * 04Jan11% file = Euler_ODE_Numerical_Example_1201.m%y0= 37;delt = 0.25;t= [0:delt:10]; n = length(t);yp(1) = y0; % vector/array indices MUST start at 1tp(1) = 0;for k = 1:(n-1) % fence-post adjustment to start at 0 dydt = 3.9*cos(4.2*yp(k))^2-log(5.1*tp(k)+6); dydtp(k) = dydt % keep track of tangent slope tp(k+1) = tp(k) + delt; dely = delt*dydt delyp(k) = dely yp(k+1) = yp(k) + dely;endplot(tp,yp, 'LineWidth', 3), grid, xlabel('t'),ylabel('y(t) by Euler'),... title('Euler Solution to dy/dt = 3.9cos(4.2y)-ln(5.1t+6)')



MatLAB Command Window forODE45

>> dydtfcn = @(tf,yf) 3.9*(cos(4.2*yf))^2-log(5.1*tf+6);>> [T,Y] = ode45(dydtfcn,[0 10],[37]);>> plot(T,Y, 'LineWidth', 3), grid, xlabel('T by ODE45'), ylabel('Y by ODE45')



Example Euler Approximation

Use four steps of Δt = 0.1 with Euler’s Method to approximate the solution to

• With I.C.

SOLUTION: Make a table of values, keeping track

of the current values of t and y, the derivative at that point, and the projected next value.

10 ty




Use I.C. to calculate the Initial Slope

Use this slope to Project to the NEW value of yn+1 = yn + Δy:

Then the NEW value for y:

210

11

run

rise1,0, 2

00

mdt

dyyx

2.01.02 ytdt

dytytm

2.12.01001 yyy




Tabulating the remaining Calculations

The table then DEFINES y = f(t) Thus, for example, y(t=0.3) = 1.685

yhdtdy dtdy yy



WhiteBoard Work

Problems From §9.4• P32 Population Extinction

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&uact=8&docid=mwI7LyqTNR2-oM&tbnid=smndac9x-ORZlM:&ved=0CAUQjRw&url=http://warnercnr.colostate.edu/~gwhite/pva/fslide45.htm&ei=V444U_OOGYaayQGA3YDgBA&bvm=bv.63808443,d.aWc&psig=AFQjCNFvu4KGyanUyDOagT_9apqOceJ92w&ust=1396301754935482



All Done for Today

CarlRunge

Carl David Tolmé Runge

Born: 1856 in Bremen, Germany

Died: 1927 in Göttingen, Germany



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

Appendix

–

srsrsr 22

a2 b2

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