Chapter 28 metodos numericos para ingenieros

8/10/2019 Chapter 28 metodos numericos para ingenieros

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CHAPTER 28

28.1The solution with the 2nd-order RK (Heun without corrector) can be laid out as

t c k1 c k2

0 10 2 30 1 1.510 25 1.25 37.5 0.625 0.9375

20 34.375 0.78125 42.1875 0.390625 0.585938

30 40.23438 0.488281 45.11719 0.244141 0.366211

40 43.89648 0.305176 46.94824 0.152588 0.228882

50 46.1853 0.190735 48.09265 0.095367 0.143051

For the 4th-order RK, the solution is

t c k1 cmid k2 cmid k3 cend k4

0 10 2 20 1.5 17.5 1.625 26.25 1.1875 1.57291710 25.72917 1.213542 31.79688 0.910156 30.27995 0.986003 35.58919 0.72054 0.9544

20 35.27317 0.736342 38.95487 0.552256 38.03445 0.598278 41.25594 0.437203 0.579102

30 41.06419 0.446791 43.29814 0.335093 42.73965 0.363017 44.69436 0.265282 0.351382

40 44.57801 0.2711 45.93351 0.203325 45.59463 0.220268 46.78069 0.160965 0.213208

50 46.71009 0.164495 47.53257 0.123371 47.32695 0.133652 48.04662 0.097669 0.129369

A plot of both solutions alon with the anal!tical result is displa!ed below"

28.2The #ass-balance e$uations can be written as

1%&'%1&'% 11 ++= cc

dt

dc

212 2'%2'% ccdt

dc

=

42'%%'% 2 += cc

dt

dc

PROPRIETARY MATERIAL' * The +craw-Hill o#panies, .nc' All rihts reser/ed' 0o part of this +anual#a! be displa!ed, reproduced or distributed in an! for# or b! an! #eans, without the prior written per#ission of the

publisher, or used be!ond the li#ited distribution to teachers and educators per#itted b! +craw-Hill for their

indi/idual course preparation' .f !ou are a student usin this +anual, !ou are usin it without per#ission'


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2

44 %'%12'%%'% ccc

dt

dc+=

21 %&'%%2'%%4'% ccc

dt

dc+=

3elected solution results (uler5s #ethod) are displa!ed below, alon with a plot of the

results'

t c1 c2 c3 c4 c5 dc1/dt dc 2/dt dc 3/dt dc 4/dt dc 5/dt

0 0.0000 0.0000 0.0000 0.0000 0.0000 1.0000 0.0000 4.0000 0.0000 0.00001 1.0000 0.0000 4.0000 0.0000 0.0000 1.0800 0.2000 3.0000 0.3500 0.0400

2 2.0800 0.2000 7.0000 0.3500 0.0400 1.0872 0.3760 2.2600 0.5703 0.0848

3 3.1672 0.5760 9.2600 0.9203 0.1248 1.0488 0.5182 1.7138 0.6999 0.1307

4 4.2160 1.0942 10.9738 1.6201 0.2555 0.9839 0.6244 1.3113 0.7673 0.1752

5 5.1999 1.7186 12.2851 2.3874 0.4307 0.9051 0.6963 1.0147 0.7927 0.2165

76 13.2416 13.2395 18.6473 16.8515 12.9430 0.0002 0.0004 0.0001 0.0106 0.0179

77 13.2418 13.2399 18.6475 16.8621 12.9609 0.0002 0.0004 0.0001 0.0099 0.0168

78 13.2419 13.2403 18.6476 16.8720 12.9777 0.0001 0.0003 0.0001 0.0093 0.0158

79 13.2421 13.2406 18.6477 16.8813 12.9935 0.0001 0.0003 0.0001 0.0088 0.0149

80 13.2422 13.2409 18.6478 16.8901 13.0084 0.0001 0.0003 0.0001 0.0082 0.0140

Finall!, +AT6A7 can be used to deter#ine the eien/alues and eien/ectors"

>> a=[.16 -.06 0 0 0;-.2 .2 0 0 0;0 -.05 .25 0 0;0 0 -.0875 .125 -.0375;-.04

-.02 0 0 .06]

a =

0.1600 -0.0600 0 0 0

-0.2000 0.2000 0 0 0

0 -0.0500 0.2500 0 0

0 0 -0.0875 0.1250 -0.0375

-0.0400 -0.0200 0 0 0.0600

>> [v,d]=eig(a)





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v =

0 0 0 -0.1039 0.2604

0 0 0 -0.1582 -0.5701

0 0.8192 0 -0.0436 0.6892

1.0000 -0.5735 0.4997 0.4956 -0.3635

0 0 0.8662 0.8466 0.0043

d =

0.1250 0 0 0 0

0 0.2500 0 0 0

0 0 0.0600 0 0

0 0 0 0.0686 0

0 0 0 0 0.2914

28.33ubstitutin the para#eters into the differential e$uation i/es

21'%%'%'14 cc

dt

dc=

The #id-point #ethod can be applied with the result"

The results are approachin a stead!-state /alue of 8'&2&'

Challenge question:

The stead! state for# (i'e', dc9dt: %) of the e$uation is % : 14' ; %'%c; %'1c2,which can be sol/ed for 8'&2& and ;1%'141' Thus, there is a neati/e root'

.f we put in the initial y/alue as 1%'141 (or hiher precision), the solution will sta! at the

neati/e root for awhile until roundoff errors #a! lead to the solution oin unstable' .f we

use an initial /alue less than the neati/e root, the solution will also o unstable'

Howe/er, if we pic< a /alue that is slihtl! hiher (as per #achine precision), it will ra/itatetowards the positi/e root' For e=a#ple if we use 1%'14





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4

This


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.n the followin plot, the inflow concentration (thin line) and the outflow concentration

(hea/! line) are both shown'

28.The s!ste# is as depicted below"

Qin = 0.6 m

3

/hr

Qevap = 0.025 m3/hr

s0 = 8000 g/m3

V= 1 m3

Qout = 0.6 m

3

/hr

!a"The #ass of water in the tan< can be #odeled with a si#ple #ass balance

%2'%%2'%&'%&'%e/apoutin === QQQdt

dV

?ith the initial condition that V: 1 #at t: %, this e$uation can be interated to !ield,

tV %2'%1=

Thus, the ti#e to e#pt! the tan< (M: %) can be calculated as t: 19%'%2 : 4% hrs'

!#"The concentration in the tan< o/er this period can be co#puted in se/eral wa!s' The

si#plest is to co#pute the #ass of salt in the tan< o/er ti#e b! sol/in the followin

differential e$uation"

sQsQdt

dmoutinin = (1)

where m: the #ass of salt in the tan


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&

t

m

V

ms

%2'%1==

(2)

Thus, we can interate $' 1 to deter#ine the #ass at each ti#e step and then use $' 2 to

deter#ine the correspondin salt concentration' The first few steps of the solution of the @

usin uler5s #ethod are tabulated below' .n addition, a raph of the entire solution is also

displa!ed' 0ote that the concentration as!#ptoticall! approaches a constant /alue of

,4'2& 9#as the solution e/ol/es'

t Vol m s dm/dt

0 1 8000 8000 0

0.5 0.9875 8000 8101.266 -60.7595

1 0.975 7969.62 8173.969 -104.382

1.5 0.9625 7917.429 8225.901 -135.54

2 0.95 7849.659 8262.799 -157.679

2.5 0.9375 7770.819 8288.874 -173.324

8000

8100

8200

8300

8400

0 10 20 30 40 50

0

0.4

0.8

1.2V

V

s s

Reconi>e that a sinularit! occurs at t: 4%, because the tan< would be totall! e#pt! at this

point'

28.$A heat balance for the sphere can be written as

dH

dthA T T a= ( )

The heat ain can be transfor#ed into a /olu#e loss b! considerin the latent heat of fusion'

dV

dt

hA

L T Tfa= ( ) (1)

where: densit! %'81


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V r A r =4

4

2 : (2)

These can be co#bined with (1) to !ield,

)(44

92

a

f

TTL

Vh

dt

dV

=

where T: %o' This e$uation can be interated alon with the initial condition,

% #%'%%%11:)%'%(

4=V

to !ield the resultin /olu#e as a function of ti#e' This result can be con/erted into dia#eter

usin (2)' 7oth results are shown on the followin plot which indicates that it ta


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0

20

40

60

0 0.005 0.01 0.015 0.02

cbcb

cccc

caca

28.83e/eral #ethods could be used to obtain a solution for this proble# (e'', finite-difference,

shootin #ethod, finite-ele#ent)' The finite-difference approach is straihtforward"

D A A A

xkA

i i i

i

+ + =1 12

2%

3ubstitutin para#eter /alues and collectin ter#s i/es

%1%'1)1%1%(1%'1 1&2&&1& =++ + iii AAxA

Dsin a x: %'2 c# this e$uation can be written for all the interior nodes' The resultin

linear s!ste# can be sol/ed with an approach li


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8

ao

a

TT

TtT

tK

=)(

ln1

3ubstitutin the /alues !ields

hr94882&4'%2%'28

2%'2ln2

1=

=K

The ti#e of death can then be co#puted as

hr1&&'12%'28

2%ln

4882&4'%

1)(ln

1=

=

=

ao

add

TT

TtT

Kt

Thus, the person died 1'1&& hrs prior to bein disco/ered' The non-self-startin Heun !ielded

the followin ti#e series of te#perature' For con/enience, we ha/e redefined the ti#e of

death as t: %'"

20

30

40

0 1 2 3 4

bodydiscoveredtime ofdet! seco"dtem# redi"$

28.1'The classical 4thorder RK #ethod !ields





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1%

28.11The classical 4thorder RK #ethod !ields

t C Te

0 1 15

0.0625 0.941261 60.796750.125 0.885808 82.90298

0.1875 0.83356 92.37647

0.25 0.784373 95.1878

0.3125 0.738086 94.54859

0.375 0.694535 92.18087

0.4375 0.653562 89.00406

0.5 0.615016 85.50575

0.5625 0.578753 81.94143

0.625 0.544638 78.4421

0.6875 0.512542 75.07218

0.75 0.482346 71.86061

0.8125 0.453936 68.81760.875 0.427205 65.94362

0.9375 0.402055 63.23421

1 0.378391 60.68253

28.12.n +AT6A7, the first step is to set up a function to hold the differential e$uations"

funci!n dc = dcd"iff(, c)

dc = [-0.013#c(1)-1000#c(1)#c(3);-2500#c(2)#c(3);-0.013#c(1)-

1000#c(1)#c(3)-2500#c(2)#c(3)];

Then, an @ sol/er li> "$an=[0,50];

>> %0=[1,1,0];

>> [,%]=!de45(&dcd"iff,"$an,%0);





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11

.f this is done, the solution will ta> [,%]=!de23"(&dcd"iff,"$an,%0);

.n either case, a plot of the results can be de/eloped as

>> $'!(,%)

28.13 !a"The first few steps of uler5s #ethod are shown in the followin table

t x y0 2 1

0.1 2.12 0.98

0.2 2.249744 0.963928

0.3 2.389598 0.951871

0.4 2.539874 0.943959

0.5 2.700807 0.940369

A plot of the entire si#ulation is shown below"





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12

0otice that because the uler #ethod is lower order, the pea


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1

!("The first few steps of the 4th-order RK #ethod is shown in the followin table

t x y

0 2 10.1 2.124862 0.982012

0.2 2.259682 0.968111

0.3 2.404758 0.958421

0.4 2.560303 0.9531160.5 2.726403 0.952425

The results are $uite close to those obtained with the Heun #ethod in part (b)' .n fact, both

the ti#e series and state-space plots are indistinuishable fro# each other'

!)".n order to sol/e the proble# in +AT6A7, a function is first de/eloped to hold the

@s,

funci!n %$=$ed$e%(,%)

%$=[1.2#%(1)-0.6#%(1)#%(2);-0.8#%(2)0.3#%(1)#%(2)];


>> %0=[2,1];

>> [,%]=!de45(&$ed$e%,"$an,%0);

>> $'!(,%)





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14

The state-space plot can be enerated as

>> $'!(%(*,1),%(*,2))

28.14Dsin the step si>e of %'1, !a"and !#"both i/e unstable results'

!("The 4th-order RK #ethod !ields a stable solution' The first few /alues are shown in the

followin table' A ti#e series plot of the results is also shown below' 0otice how after about t

: &, this solution di/eres fro# the double precision /ersion in Fi' 2'8'

t x y z

0 5 5 50.1 9.781469 17.07946 10.43947

0.2 17.70297 20.8741 35.89687

0.3 10.81088 -2.52924 39.30745

0.4 0.549577 -5.54419 28.07462

0.5 -3.16461 -5.84129 22.36889





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!)".n order to sol/e the proble# in +AT6A7, a function is first de/eloped to hold the

@s,

funci!n %$='!en+(,%)

%$=[-10#%(1)10#%(2);28#%(1)-%(2)-%(1)#%(3);-

2.666667#%(3)%(1)#%(2)];


>> %0=[5,5,5];

>> [,%]=!de45(&'!en+,"$an,%0);

>> $'!(,%)

28.1The second-order e$uation can be e=pressed as a pair of first-order e$uations,

dydz

w

dw

dz

f

EIL z

=

= 2

2( )

?e used uler5s #ethod with h: 1 to obtain the solution"





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1&

z y w dy /dz dw /dz

0 0 0 0 0.004321 0 0.00432 0.00432 0.004037

2 0.00432 0.008357 0.008357 0.003763

3 0.012677 0.01212 0.01212 0.003499

4 0.024797 0.015619 0.015619 0.003245

5 0.040416 0.018864 0.018864 0.003

26 0.8112 0.04524 0.04524 7.68E-05

27 0.85644 0.045317 0.045317 4.32E-05

28 0.901757 0.04536 0.04536 1.92E-05

29 0.947117 0.045379 0.045379 4.8E-06

30 0.992496 0.045384 0.045384 0

y

z

0

10

20

30

%1 0 1

28.1$The second-order e$uation can be e=pressed as a pair of first-order e$uations,

2%92

)(2)(

2%% zL

EIz

ze

dz

dww

dz

dy z

+==

?e used uler5s #ethod with h: 1 to obtain the solution"

z f&z' y w dy /dz dw /dz

0 0 0 0 0 0

1 31.18357 0 0 0 0.0020982 50.0099 0 0.002098 0.002098 0.003137

3 61.40481 0.002098 0.005235 0.005235 0.003581

4 68.08252 0.007333 0.008816 0.008816 0.003682

5 71.65313 0.016148 0.012498 0.012498 0.003583





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26 29.63907 0.700979 0.040713 0.040713 3.79E-05

27 27.89419 0.741693 0.040751 0.040751 2.01E-05

28 26.24164 0.782444 0.040771 0.040771 8.4E-06

29 24.67818 0.823216 0.04078 0.04078 1.97E-06

30 23.20033 0.863995 0.040782 0.040782 0

y

z

10

20

30

%1 0 1

28.1%This proble# was sol/ed usin the =cel spreadsheet in a fashion si#ilar to the last

e=a#ple in 3ec' 2'1' ?e set up uler5s #ethod to sol/e the @s usin uesses for the

diffusion coefficients' Then we for#ed a colu#n containin the s$uared residuals between

our predictions and the #easured /alues' AdCustin the diffusion coefficients with the 3ol/er

tool #ini#i>ed the su# of the s$uares' At first, we assu#ed that the diffusion coefficients

were >ero' For this case the 3ol/er did not con/ere on a credible answer' ?e then #adeuesses of 11%&for both' This #anitude was based on the fact that the /olu#es were of this

order of #anitude' The resultin si#ulation did not fit the data /er! well, but was #uchbetter than when we had uessed >ero' ?hen we used 3ol/er, it con/ered onE12:

1'241%&and E1: 2'2&41%&which corresponded to a su# of the s$uares of residuals of

'4' 3o#e of the uler calculations are displa!ed below alon with a plot of the fit'

t c1 c2 c3 dc1/dt dc 2/dt dc 3/dt

0 0 0 100 22.63911 0 -45.2782

0.1 2.2639108 0 95.47218 19.91464 0.351651 -42.203

0.2 4.2553743 0.035165 91.25187 17.46867 0.655521 -39.3905

0.3 6.0022409 0.100717 87.31283 15.27375 0.916677 -36.816

0.4 7.5296162 0.192385 83.63123 13.30513 1.139684 -34.45750.5 8.8601294 0.306353 80.18548 11.54045 1.328649 -32.2948





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1

.t should be noted that we #ade up the E#easure#ents for this proble# usin the 4 th-order

RK #ethod with /alues for diffusi/e #i=in of E12: 11%&and E1: 21%

&' ?e then used a

rando# nu#ber enerator to add so#e error to this Edata'

28.18The Heun #ethod without corrector can be used to co#pute

t p k1 pend k2

0 6000 450 6225 466.875 458.43750.5 6229.219 467.1914 6462.814 484.7111 475.9512

1 6467.194 485.0396 6709.714 503.2286 494.1341

1.5 6714.261 503.5696 6966.046 522.4535 513.0115

2 6970.767 522.8075 7232.171 542.4128 532.6102

18 23137.43 1735.308 24005.09 1800.382 1767.845

18.5 24021.36 1801.602 24922.16 1869.162 1835.382

19 24939.05 1870.429 25874.26 1940.57 1905.49919.5 25891.8 1941.885 26862.74 2014.705 1978.295

20 26880.94 2016.071 27888.98 2091.673 2053.872

The results can be plotted' .n addition, linear reression can be used to fit a straiht line tolnp/ersus tto i/e lnp: '&88 G %'%t' Thus, as would be e=pected fro# a first-order

#odel, the slope is e$ual to the rowth rate of the population'





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18

0

5000

10000

15000

20000

0 10 20

8.4

8.8

(.2

(.6

10

10.4

0 10 20

p lnp

28.1&The Heun #ethod can be used to co#pute

t p k1 pend k2 0 6000 840 6420 871.836 855.918

0.5 6427.959 872.4052 6864.162 901.6652 887.0352

1 6871.477 902.1234 7322.538 928.312 915.2177

1.5 7329.085 928.6622 7793.417 951.3099 939.986

18 18798.01 225.95 18910.99 205.9432 215.9466

18.5 18905.98 206.8344 19009.4 188.3068 197.5706

19 19004.77 189.1412 19099.34 172.02 180.5806

19.5 19095.06 172.7988 19181.46 157.008 164.9034

20 19177.51 157.7327 19256.38 143.1946 150.4637

The results can be plotted'

The cur/e is s-shaped' This shape occurs because initiall! the population is increasin

e=ponentiall! since pis #uch less than p#a=' Howe/er, as papproaches p#a=, the rowth rate

decreases and the population le/els off'





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2%

28.2' !a"0onlinear reression (e'', usin the =cel sol/er option) can be used to #ini#i>e the

su# of the s$uares of the residuals between the data and the si#ulation' The resultin

esti#ates are" a: %'22, b: %'%121, c: %'2244, and d: %'%%%28' The fit is"

0

400

800

1200

1(60 1((0 2020

#oose

wol/es

!#" The results in state space are,

)o*ves

0

100

200

300

0 500 1000 1500

+oose!("

0

400

800

1200

1(60 1((0 2020





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21

)o*ves

+oose

0

100

200

300

0 500 1000 1500

!)"

0

400

800

1200

1(60 1((0 2020

)o*ves

+oose

0

100

200

300

0 500 1000 1500

28.21Main Program:

anging "aic ca'e - /=/()

aa!'ic "!'ui!n /=/()

ni" (',f,")

/ = /!(1"in($i2#') nde$enden aia'e , "="a , f=end

iniia' c!ndii!n" [%(1)=ca'e %-c!!dinae, %(2)=ca'e "'!$e];

e"=0.5e-7;

"$an=[0,200];

ic=[0 0];

g'!a' /!

/!=0.0025;

[,%]=!de45(&"'$,"$an,ic);





22/43

22

%f(1)=%('eng());

/!(1)=/!;

ea(1)=1;

/!=0.002;

[,%]=!de45(&"'$,"$an,ic);

%f(2)=%('eng());

/!(2)=/!;

ea(2)=a"( (%f(2)-%f(1))%f(2) );

f! :=3*10

/!(:)=/!(:-1)(/!(:-1)-/!(:-2))(%f(:-1)-%f(:-2))#(50-%f(:-1));

/!=/!(:);

[,%]=!de45(&"'$,"$an,ic);

%f(:)=%('eng());

ea(:)=a"( (%f(:)-%f(:-1))%f(:) );

if (ea(:)=e")

$'!(,%(*,1)); gid;

'ae'(


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2

)( xLEI

P

dx

dz

zdx

dy

=

=

These e$uations can be si#pl! interated with an! of the one-step techni$ues fro# this part

of the boo


24/43

24

EI

wx

EI

wLx

dx

dz

zdx

dy

22

2

=

=

These can be sol/ed usin two uesses for the initial condition of z' For our cases we used

z(%) : ;11%;& and z(%) : ;21%;&' ?e sol/ed the @s with uler5s #ethod usin a step

si>e of %'2' The results are

z(0 1E6 2E6

y(10 6.48275E-06 3.51725E-06

These /alues can then be used to deri/e the correct initial condition that corresponds with

y(1%) : %' Dsin linear interpolation, the result is ;1'&421%;&' Dsin this /alue, the results

of the final shot can be tabulated alon with the anal!tical solution"

x y z dydx dzdx yanal

0 0 -1.6483E-06 -1.6483E-06 0 0

0.25 -4.1207E-07 -1.6483E-06 -1.6483E-06 5.0781E-08 -4.3349E-07

0.5 -8.2414E-07 -1.6356E-06 -1.6356E-06 9.8958E-08 -8.6382E-07

0.75 -1.2330E-06 -1.6108E-06 -1.6108E-06 1.4453E-07 -1.2880E-06

1 -1.6357E-06 -1.5747E-06 -1.5747E-06 1.8750E-07 -1.7031E-06

9 -1.7607E-06 1.7013E-06 1.7013E-06 1.8750E-07 -1.7031E-06

9.25 -1.3354E-06 1.7482E-06 1.7482E-06 1.4453E-07 -1.2880E-06

9.5 -8.9836E-07 1.7843E-06 1.7843E-06 9.8958E-08 -8.6382E-07

9.75 -4.5227E-07 1.8091E-06 1.8091E-06 5.0781E-08 -4.3349E-07

10 7.4115E-22 1.8218E-06 1.8218E-06 0.0000E+00 0.0000E+00

0ote that althouh there is a discrepanc!, the results are fairl! close' .n particular, notice that

the end displace#ent is effecti/el! >ero' As shown below, the close aree#ent between the

approaches can be seen when the nu#erical and anal!tical solutions are plotted on the sa#e

raph' .n fact, if a hiher-order approach li


25/43

2

28.24 *E++ata: In the ,i+st -+inting a /inus sign shoul) #e in(lu)e) in the )i,,e+ential

equation:

)(2)(4

2

ehhA

d

dt

dh+=

Late+ -+intings shoul) ha0e the (o++e(t equation.

This proble# can be approached in a nu#ber of wa!s' The si#plest wa! is to fit the area-

depth data with pol!no#ial reression' A fifth-order pol!no#ial with a >ero intercept !ields a

perfect fit"

This pol!no#ial can then be substituted into the differential e$uation to !ield

)(2)114'1224G2'%G8'24111'881G1%'%%4(4 24

2

ehhhhhh

d

dt

dh+

=

This e$uation can then be interated nu#ericall!' This is a little tric

occurs as the laero' Therefore, the software to sol/e this proble#

should be desined to ter#inate Cust prior to this occurrin' For e=a#ple, the software can be

desined to ter#inate when a neati/e area is detected' As displa!ed below, the resultsindicate that the reser/oir will e#pt! in a little o/er &,%% s'





26/43

2&

28.2 The para#eters can be substituted into force balance e$uations to i/e

( )( )

( ) %22'%22'%%1'%42'%24'%

%2'%4'%

2

2

2

2

1

212

=+=+=

!!

!!!

!!

A +AT6A7 session can be conducted to e/aluate the eien/alues and eien/ectors as

>> = [0.450 -0.200 0.000;-0.240 0.420 -0.180;0.000 -0.225 0.225];

>> [v,d] = eig()

v =

-0.5879 -0.6344 0.2913

0.7307 -0.3506 0.5725

-0.3471 0.6890 0.7664

d =

0.6986 0 0

0 0.3395 0

0 0 0.0569

Therefore, the eien/alues are %'&8&, %'8 and %'%&8' The correspondin eien/ectors

are (nor#ali>in so that the a#plitude for the third floor is one),

1

4&888'%%%8'%

1%'%82%'%

11%1&'2

&84'1

A raph can be #ade showin the three #odes





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2

%2 %1 0 1 2%2 %1 0 1 2 %1 0 1 0 1

+ode 1 +ode 2 +ode 3

28.2$The second-order e$uation can be e=pressed as a pair of first-order e$uations,

t"idt

di

idt

d"

%'1sin4%2'% +=

=

The para#eters can be substituted and the s!ste# sol/ed with the 4 th-order RK #ethod in

double-precision with h: %'1' A table showin the first few steps and a raph of the entiresolution are shown below'

t i q k11 k12 imid qmid k21 k22 imid qmid k31 k32 iend qend k41 k42 phi1 phi2

% %'%%%% %'%%%% %'%%%% %'%%%% %'%%%% %'%%%% %'%84 %'%%%% %'%%4 %'%%%% %'%8 %'%%4 %'%%8 %'%%% %'18 %'%%8 %'%828 %'%%1

%'1 %'%%8 %'%%% %'14 %'%%8 %'%1 %'%%% %'24 %'%1 %'%2% %'%%12 %'214 %'%2% %'%&4 %'%%2& %'42 %'%&4 %'21 %'%214

%'2 %'%&4 %'%%2 %'4 %'%&4 %'%42 %'%%4 %'424 %'%42 %'%% %'%%2 %'42 %'%% %'%8 %'%% %'482 %'%8 %'428% %'%&

%' %'%8 %'%%1 %'48 %'%8 %'1%42 %'%121 %'% %'1%42 %'1%2 %'%1 %'% %'1%2 %'14& %'%1 %'&%1 %'14& %'& %'1%&1

%'4 %'14 %'%1 %'&%21 %'14 %'1&4 %'%2 %'&88 %'1&4 %'1&& %'%2% %'& %'1&& %'181 %'%4 %'& %'181 %'&4& %'1&8

28.2%The second-order e$uation can be e=pressed as a pair of first-order e$uations,





28/43

2

#L

"i

L

$

dt

di

idt

d"

=

=

The para#eters can be substituted and the s!ste# sol/ed with the 4

th

-order RK #ethodwith h: %'%%' A table showin the first few steps and a raph of the entire solution are

shown below'

t i q k11 k12 imid qmid k21 k22 imid qmid k31 k32 iend qend k41 k42 phi1 phi

%'%%% -'22 %'1%% -1&'1 -'22 -'&88 %'%82 -14&'&% -'&88 -'&4 %'%81 -14'%2 -'&4 -4'%% %'%2 -12'4 -4'%% -14'& -'&&

%'%% -4'%1% %'%2 -12'2& -4'%1% -4'1 %'%2 -1%%'1 -4'1 -4'2&% %'%1 -88'1 -4'2&% -4'%& %'%&% -'% -4'%& -88'&% -4'2

%'%1% -4'% %'%&% -'488 -4'% -4'&8& %'%48 -1'% -4'&8& -4'& %'%48 -%'4 -4'& -4'&1 %'% -2&'&1 -4'&1 -%'8& -4'&

%'%1 -4'& %'% -2&'8& -4'& -4'2 %'%2 -1'82 -4'2 -4'& %'%2 -2'21 -4'& -4'4 %'%1 21'8 -4'4 -2'21 -4'

%'%2% -4'4 %'%1 21'84 -4'4 -4'2% %'%%1 44'82 -4'2% -4'&&1 %'%%1 44'% -4'&&1 -4' -%'%1% &&'%% -4' 44'2& -4'&

it

%4

0

4

0 0.1 0.2 0.3 0.4 0.5

it

%4

0

4

0 0.1 0.2 0.3 0.4 0.5

q

t

%0.1

0

0.1

0 0.1 0.2 0.3 0.4 0.5

q

t

%0.1

0

0.1

0 0.1 0.2 0.3 0.4 0.5

28.28The e$uation can be sol/ed anal!ticall! as





29/43

28

tei

tL$i

#

#tL$i

dtL

$

i

di

iL

$

dt

di

2'%

)9()'%9ln(

'%ln

)9(ln

=

=

=

+=

=

=

The nu#erical solution can be obtained b! e=pressin the e$uation as

idtdi '1=

and usin uler5s #ethod with h: %'% to sol/e for the current' 3o#e selected /alues areshown, alon with a plot of both the anal!tical and nu#erical result' A better #atch would be

obtained b! usin a s#aller step or a hiher-order #ethod'

t analytical Euler di /dt

0 0.500000 0.500000 -0.7500000.05 0.463872 0.462500 -0.693750

0.1 0.430354 0.427813 -0.641719

0.15 0.399258 0.395727 -0.593590

0.2 0.370409 0.366047 -0.5490710.25 0.343645 0.338594 -0.507890

28.2&The nu#erical solution can be obtained b! e=pressin the e$uation as

)('1

iidt

di=





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%

and usin uler5s #ethod with h: %'% to sol/e for the current' 3o#e selected /alues are

shown, alon with a plot of the nu#erical result' 0ote that the table and plot also show the

anal!tical solution for the linear case co#puted in Jrob' 2'18'

t analytical Euler di /dt

0 0.500000 0.500000 -0.562500

0.05 0.463872 0.471875 -0.5502070.1 0.430354 0.444365 -0.534931

0.15 0.399258 0.417618 -0.517175

0.2 0.370409 0.391759 -0.497451

0.25 0.343645 0.366887 -0.476253

28.3'Dsin an approach si#ilar to 3ec' 2', the s!ste# can be e=pressed in #atri= for# as

1 11 2

%1

2

=

i

i K L

A pac a=[1 -1;-1 2];>> [v,d]=eig(a)

v =

0.8507 -0.5257

0.5257 0.8507

d =

0.3820 0

0 2.6180

Thus, we can see that the eien/alues are : %'2 and 2'&1 or natural fre$uencies of :

%'&19 L# and 1'&19 L# ' The eien/ectors tell us that these correspond to oscillations

that coincide (%'% %'2) and which run counter to each other (%'2 %'%)'





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1

L#

&1'1=

L#

&1'1=

L#

&1'%=

L#

&1'%=

28.31The differential e$uations to be sol/ed are

sin

1

1'8

1

1'8

"nonlinear"linear

==

==

dt

d%

dt

d%

%dt

d%

dt

d

A few steps for the 4th-order RK solution of the nonlinear s!ste# are contained in the

followin table and a plot of both solutions is shown below'

t theta v k 11 k12 theta v k 21 k22 theta v k 31 k32 theta v k 41 k42 1 2

% %' %'%%% %'%%% -&'8 %' -%'% -%'% -&'8 %' -%'% -%'% -&'8& %' -%'%&8 -%'%&8 -&'84 -%'% -&'8&

%'%1 %' -%'%&8 -%'%&8 -&'84 %' -%'1%4 -%'1%4 -&'82 %' -%'1%4 -%'1%4 -&'81 %'4 -%'18 -%'18 -&'82 -%'1%4 -&'81

%'%2 %'4 -%'18 -%'18 -&'82 %' -%'1 -%'1 -&'822 %' -%'1 -%'1 -&'821 %'2 -%'2% -%'2% -&'81 -%'1 -&'821

%'% %'2 -%'2% -%'2% -&'81 %'1 -%'242 -%'242 -&'8% %'1 -%'242 -%'242 -&'8% %'% -%'2 -%'2 -&'8 -%'242 -&'8%

%'%4 %'% -%'2 -%'2 -&'8 %' -%'11 -%'11 -&' %' -%'11 -%'11 -&' %' -%'4& -%'4& -&'& -%'11 -&'

%4

%2

0

2

4

0 2 4 6

*i"%t!et

*i"%v

"o"*i"%t!et

"o"*i"%v


xm

k%

m

c

dt

d%

%dt

dx

=

=





32/43

2

A few steps for the 2nd-order RK solution (Heun without iteration) are shown in the followin

table and plots of displace#ent and /elocit! are shown below'

t x v k 11 k12 x v k21 k22 1 2

0 0.4000 0.0000 0.0000 -416.67 0.4000 -0.4167 -0.4167 -413.19 -0.2083 -414.93

0.001 0.3998 -0.4149 -0.4149 -412.99 0.3994 -0.8279 -0.8279 -409.12 -0.6214 -411.05

0.002 0.3992 -0.8260 -0.8260 -408.92 0.3983 -1.2349 -1.2349 -404.65 -1.0304 -406.79

0.003 0.3981 -1.2328 -1.2328 -404.46 0.3969 -1.6372 -1.6372 -399.80 -1.4350 -402.13

0.004 0.3967 -1.6349 -1.6349 -399.61 0.3951 -2.0345 -2.0345 -394.58 -1.8347 -397.09

0.005 0.3949 -2.0320 -2.0320 -394.39 0.3928 -2.4264 -2.4264 -388.99 -2.2292 -391.69

isplace#ent"

Melocit!"

28.33The differential e$uation to be sol/ed is

)2%(2'% Tdt

dT=

A few steps for the 2nd-order RK solution (Heun without iteration) are shown in the followintable and a plot of te#perature /ersus ti#e is shown below'

t T k11 Tend k21

0 90 -17.5 88.25 -17.0625 -17.28130.1 88.27188 -17.068 86.56508 -16.6413 -16.8546





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0.2 86.58641 -16.6466 84.92175 -16.2304 -16.4385

0.3 84.94256 -16.2356 83.31900 -15.8297 -16.0327

0.4 83.33929 -15.8348 81.75581 -15.4390 -15.6369

0.5 81.77560 -15.4439 80.23121 -15.0578 -15.2509

The te#perature will drop to 4%o in appro=i#atel! #inutes'

28.34The differential e$uation to be sol/ed is

t

t

dt

dQ

'21%%

)2%)('22%(1%%)12('%

=

A few steps for the 2nd-order RK solution (Heun without iteration) are shown in the followin

table and a plot of heat flow /ersus ti#e is shown below'

t Q k11 tend k21

0 0 2100 0.1 2094.737 2097.3680.1 209.7368 2094.737 0.2 2089.447 2092.092

0.2 418.946 2089.447 0.3 2084.131 2086.789

0.3 627.625 2084.131 0.4 2078.788 2081.4590.4 835.7709 2078.788 0.5 2073.418 2076.103

0.5 1043.381 2073.418 0.6 2068.02 2070.719






34/43

4

nonlinear"

2

1'&

22'%'8 %

dt

d%=

linear"

d%

dt %= 8 12

&1''

'

A few steps for the solution (uler) are shown in the followin table, which also includes the

anal!tical solution fro# =a#ple 1'1' A plot of the result is also shown below' 0ote, the

nonlinear solution is the bolder line'

linear nonlinear analytical t v dvdt v dvdt v

0 0 9.8 0 9.8 00.1 0.98 9.620117 0.98 9.796827 0.971061

0.2 1.942012 9.443537 1.959683 9.787312 1.92446

0.3 2.886365 9.270197 2.938414 9.771473 2.860518

0.4 3.813385 9.100039 3.915561 9.749345 3.779552

0.5 4.723389 8.933005 4.890496 9.720979 4.681871

28.3$The differential e$uations to be sol/ed are

tN 1 s" d%

dt%= 8 12

&1''

'

t1 s" %dt

d%

1'&

'8 =

The first few steps for the solution (Heun) are shown in the followin table, alon with thesteps when the parachute opens' A plot of the result is also shown below'

t v k11 tend vend k21

0 0 9.8 0.1 0.98 9.620117 9.7100590.1 0.971006 9.621768 0.2 1.933183 9.445157 9.533463

0.2 1.924352 9.446778 0.3 2.86903 9.273379 9.360079





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0.3 2.86036 9.274971 0.4 3.787857 9.104725 9.189848

0.4 3.779345 9.106288 0.5 4.689974 8.939138 9.022713

0.5 4.681616 8.940673 0.6 5.575683 8.776563 8.858618

12.8 48.2953 0.935223 12.9 48.38883 0.918057 0.9266412.9 48.38797 0.918215 13.0 48.47979 -29.354 -14.2179

13 46.96618 -28.1316 13.1 44.15302 -25.8596 -26.9956

13.1 44.26662 -25.9513 13.2 41.67149 -23.8554 -24.9033

13.2 41.77629 -23.94 13.3 39.38228 -22.0065 -22.9733

13.3 39.47896 -22.0846 13.4 37.2705 -20.301 -21.1928

28.3%This proble# can be sol/ed with +AT6A7"

a?$ed "$ing ?a"" "%"e?

?a""* ?=1 :g

da?$ing, n!n'inea* a a"(dd) (dd), a=2 @(?")A2

"$ing, n!n'inea* A3, =5 @?A3

BCD 5 ve"i!n

nde$enden aia'e , "$an=["a "!$]

iniia' c!ndii!n" [(1)=ve'!ci%, (2)=di"$'ace?en];

0=0;

f=8;

"$an=[0 8]; ic=[1 0.5];

a) 'inea "!'ui!n

[,]=!de45(


36/43

&

'ae'(


37/43

a?$ed "$ing ?a"" "%"e? - ? d2d2 c dd : =0

?a""* ?=1 :g

da?$ing* n!n'inea- c dd = a "gn(dd) (dd)A2, a=2 @(?")A2

"$ing* n!n'inea- : = A3, =5 @?A3

(1)=ve'!ci%, (2)=di"$ace?en

funci!n d=n'c:(,);

d=[-2#a"((1)).#(1)-5#(2).#(2).#(2); (1)];





38/43

28.38 This proble# can be sol/ed with +AT6A7"

E!ced da?$ed "$ing-?a"" "%"e? / ?aeia' da?$ing

?a""* ?=2 :g

da?$ing, n!n'inea ai* a a"(dd) dd, a=5 @(?")A2

"$ing, 'inea* : = 6f!cing funci!n* E=E!("in(/)), E!=2.5 @, /=0.5 ad"

nde$enden aia'e , "$an=["a "!$]

iniia' c!ndii!n" [(1)=ve'!ci%, (2)=di"$'ace?en];

"$an=[0 15]; ic=[0 1];

[,]=!de45(


39/43

8

funci!n d=n'E(,);

d=[-2.5#a"((1)).#(1)-3#(2)1.25#"in(0.5#); (1)];

28.3&

E++ata: In the #oos ,i+st -+inting the )i,,e+ential equation as e++oneousl5 -+inte) as

%22

2

=+ p&

dxd&

xdx&d

The (o++e(t equation hi(h a--ea+s in late+ -+intings is

%2

2

2

=

+ p&

dx

d&

xdx

&d

This proble# can be sol/ed with +AT6A7"

FG D!unda% a'ue !'e?

a$eed c!nica' c!!'ing fin

u[](2)(u[]-$u)=0 D. u(=0)=0 u(=1)=1

i="$aia' inde, f!? 1 ! H

nu?eing f! $!in" i" i=1 ! i=H f! (H-1) d "$ace"

u(i=1)=0 and u(i=H)=1

H=21;

!n"an"

d=1(H-1);





40/43

4%

d2=d#d;

aa?ee"

$(1)=10; $(2)=20; $(3)=50; $(4)=100;

"i+ing ?aice"

u=+e!"(1,H); =+e!"(1,H);

a=+e!"(1,H); =+e!"(1,H); c=+e!"(1,H); d=+e!"(1,H);

a=+e!"(1,H); ga=+e!"(1,H);

nde$enden aia'e

=0*d*1;

D!unda% !ndii!n"

u(1)=0; u(H)=1;

f! :=1*4;

!efficien"

(2)=-2-2#$(:)#d2d;

c(2)=2;

f! i=3*H-2,

a(i)=1-d(d#(i-1));

(i)=-2-2#$(:)#d2(d#(i-1));

c(i)=11(i-1);

end

a(H-1)=1-d(d#(H-2));

(H-1)=-2-2#$(:)#d2(d#(H-2));

d(H-1)=-(11(H-2));

!'ui!n % !?a" 'g!i?

a(2)=(2);

ga(2)=d(2)(2);

f! i=3*H-1,

a(i)=(i)-a(i)#c(i-1)a(i-1);

ga(i)=(d(i)-a(i)#ga(i-1))a(i);

end

ac: "u"iui!n "e$

u(H-1)=ga(H-1);

f! i=H-2*-1*2,

u(i)=ga(i)-c(i)#u(i1)a(i);

end

'!

$'!(,u) i'e(


41/43

41

28.4'The second-order e$uation can be e=pressed as a pair of first-order e$uations,

)cos(1

tm

Px

m

kx

m

k%

m

c

dt

d%

%dt

dx

+=

=

3ubstitutin the para#eter /alues"

)'%cos('%4'% 1 txkxk%dt

d%

%dt

dx

+=

=

!a"6inear (k1: 1I k: %)"

)'%cos('%4'% tx%dt

d%

%dt

dx

+=

=

Here are the results of sol/in these e$uations with the initial condition, x: %: %, usin the

Heun #ethod without corrector and a step-si>e h: %'12'





42/43

42

!#"0onlinear (k1: 1I k: %')"

)'%cos('%'%4'% txx%dt

d%

%dtdx

+=

=

Here are the results of sol/in these e$uations with the initial condition, x: %: %, usin the

Heun #ethod without corrector and a step-si>e h: %'12'





43/43

4

28.41The followin +AT6A7 +-file can be set up to co#pute the riht-hand-sides of the @s,

funci!n d%d = ungee(,%,C,cd,?,:,ga??a)g = 9.81;

c!d = 0;

if %(1) > C dee?ine if e c!d ee" a f!ce

c!d = :?#(%(1)-C)ga??a?#%(2);

end

d%d = [%(2); g - "ign(%(2))#cd?#%(2)A2 - c!d];

?e can use !de45to obtain the solutions and displa! the# on a plot,

>> [,%]=!de45(&ungee,[0 50],[0 0],[],30,0.25,68.1,40,8);

>> $'!(,-%(*,1),

Chapter 28 metodos numericos para ingenieros

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Transcript of Chapter 28 metodos numericos para ingenieros