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FIGURE 226
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The solution o second-order of CDE with o conveciive boundorycondltionot one end ond o fixed lemoerolure the other ot
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Linear interpolationcan then be employedto computethe correctinitial temperature 1 5 0- 3 0 0
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: - 417544 @00 6835088) 2413643 K - 683s088'
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which corresponds a gradientof Zu - 02068Using theseinitial conditions,ode45can to be employedto generate conect solution,as depictedin Ftg 226 the Note that we can verify that our boundaryconditionhasbeensatisfied substituting by the initial conditions into Eq (2212)to give | JJK ir x(02m)'x(200K-2413643K):-200 --- rr xt02m)'x02068* mrKs mKs m
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which can be evaluated yield *51980J/s: -51980J/s Thus, conduction conto and vectionareequalandtranSferheatoutoftheleftendoftherodatarateof
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2222 The Shooting Merhod for Nonlineor ODEs
For nonlinearboundary-value problems,linear interpolationor extrapolation through two solutionpoints will not necessarily result in an accurate estimateof the requiredboundary condition to attain an exact solutionAn alternative to perform threeapplications is ofthe shooting method and use a quadratic interpolating polynomial to estimatethe proper boundarycondition However,it is unlikely that such an approachwould yield theexact answer,and additionaliterationswould be necessary home in on the solution to Another approach a nonlinearprobleminvolvesrecasting as a rootsproblem for it Recall that the generalgoal of a rootsproblemis to find the value of r that makesthefunction how the shooting method f (x) :0 Now, let us usethe heatedrod problemto understand can be recastin this form First, recognizethat the solution of the pair of differentialequations alsoa "funcis tion" in the sense that we guessa conditionat the left-handend of the rod zn, andtheintegration yields a predictionof the temperature the right-hand end 76Thus, we canthink at of the integrationas
Tu: f k,)
METHOD 222 THE SHOOTING
That is, it represents processwherebya guessof 2,, yields a predictionof 16 Viewed in a this way, we can seethat what we desireis the value of ;,, that yields a specificvalue of 71, I1',as in the example,we desireTo : 400 the problem can be posedas 400 : l k) a By bringing the goal of 400 over to the right-handside of the equation,we generate new function res(zn)that represents difference,or residual,betweenwhat we have, l (2,,), the and what we want 400 r e s ( 7 , , )J ' k a ) - 4 0 0 : If we drive this new function to zero, we will obtain the solutionThe next exampleillustratesthe approach EXAMPLE 224 The Shooting Method for Nonlineor ODEs Problem Stotement Although it served our pu{posesfor illustratingthe shooting rnethod,F,q(226) was not a completelyrealisticmodel for a heatedrod For one thing, suchas radiationthat are nonlinear such a rod would lose heatby mechanisms of Suppose that the following nonlinearODE is usedto simulatethe temperature the heatedrod:
_ 0 : = + h,(T& Tt + o,'q! _ ra1 dx'
' r ' , parameter reflectingthe relative impactsof radiation and where o' : a bulk heat-transfer how the shooting x l0 e K 3 m-2 This equation can serveto illustrate conduction:2J problem The remaining two-point nonlinear boundary-value method is used to solve a 2, problem conditions are as specified in Example 222: L : 10 r7t,h':005 m 7 - : 2 0 0 K , f ( 0 ) : 3 0 0 K , a n d7 ( 1 0 ) : 4 0 0 K equationis first exSolution Just as with the linear ODE, the nonlinear second-order oressed two first-orderODEs: as tlT
d: ' : - 0 0 5 t 2 0 0 T ) - 2 1x l 0 u ( 1 6x l O q- 1 4 ) tlx
(1:T
to An M-file can be developed computethe right-handsidesof theseequations:
dy=ty dy=flydxn (x, Y) function (200-y(1,) (2) ;-005* ) 21e-9* (1 6e9 y(1)^ ) I ;
Next, we can build a function to hold the residual that we will try to drive to zero as r=res (za) function fx,yl=ode45(@dydxn, t0 101, l3Aa zal); r=y (length (r), 1) -400 ;
BOU DARY N VALU PROBLEMS E
lK 400 300
10 r rn
FIGURE 22,7 The to result using shooting of fie method solve non ineor em prob a
Notice how we use the ode45 function to solvethe two ODEs to generate temperature the at the rod's end:y (length (x) , 1) We can then find the root with the f zero function:
>> fzero (@res, -50)
41,1
T h u s , w e s e e t h a t iw e s e t t h e i n i t i atlr a j e c t o r y z ( 0:) - 4 1 1 4 3 1 t h e r e s i d u a l f u n c t i o n w i l l f boundarycondition Z(10) :400 at the endof the be driven to zero and the temperature rod should be satisfied This can be verified by generating entire solutionandplotting the the temperatures versusx: > > l x , y l = o d e 4 5 ( @ d y d x n ,t 0 1 0 1 , 1 3 0 0 f z e r o ( @ r e s , - 5 t l ) l ) ;
>> plot(x,t'(:,1))
The resultis showninFig227 alongwith the originallinearcase 222 fionr Example As expected,the noniinearcaseis depressed linear model due to theaddilower than the tional heatlost to the surroundinggas by radiation
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