Numerical Modeling in Biomedical Systems

Transcription

1 Numeial Modeling in Biomedial Systems BME 15:35 Letue 7 9/6/17

2 Nonlinea Systems Dunn Chapte 5 Nonlinea equations Root finding Baketing methods Open methods Gaphial Bisetion False Position Newton s method Fied-point iteation Newton- Raphson Bent s method

3 How do we solve this nonlinea equation to find? = Plot f () = vesus and see whee it osses zeo? 1 f() -1 - f() = Roots nea = -3,, Gaphial methods ae not vey auate but they an give us a lue about the egion whee the oot(s) lie - Guess a value fo, evaluate whethe f () =? - This will take a long time - Bette to develop systemati methods to seah fo f () = until ou equied level of auay is eahed

4 Baketing Methods We want to find the oots of f(), i.e. the values of whee f() = Whee to stat looking? Baketing methods use the fat that the value of f() hanges sign at eah oot f() f() = Stategy: Find two values of, ( 1, ) giving positive and negative values fo f() The oot must lie somewhee between these values of Naow the distane between ( 1, ) to lose in on the tue oot

5 Eample - Falling objet fom Letue m F up = dag foe mass, m F down = gavitational foe Using Newton s nd law leads to an analytial epession fo veloity v as a funtion of time t and system paametes mass (m) and dag oeffiient (): v gm / m t 1 e Given g,m,,t we an alulate v t But what if we ae given g,m,v,t and we want to alulate? v(t) If an objet has mass m = 68.1 kg, what dag oeffiient () is neessay in ode to eah a veloity of v = 4 m/s at time t = 1 seonds? 1 t Poblem: We an t eaange the epession in the fom = f (g,m,v,t)

6 Reap gm / m How to solve a nonlinea equation? vt 1 e Eg., find when g = 9.8 m/s, m = 68.1 kg, v = 4 m/s, t = 1 s t Solve this nonlinea equation fo : e / What ae the value(s) of whih satisfy: / e 4 Whee does the funtion f() equal zeo? f 68.1 / e These ae just diffeent ways of asking the same question

7 We need to find the oot(s) of f 68.1 / e I.e. the value(s) of whih lead to f () = We an stat to make some guesses fo, and see what we get fo f (): Guess fo f() f() Dag oeffiient, / m looks lose to zeo. v t gm 1 e t gives v = 4.59 m/s when = Gaphial methods ae not vey auate, but they an povide us with useful initial guesses to use with othe methods

8 Bisetion Method In geneal: If f () is eal and ontinuous in the ange between l and u, and f( l ) f( u ) <, then thee is at least one oot in the ange l to u f() Dag oeffiient, Step 1: Choose lowe ( l ) and uppe ( u ) guesses fo the loation of the oot. Step : If f( l ) f( u ) <, then the oot is estimated to lie at = ( l + u ) / Step 3: If f( l ) f( ) <, then the tue oot must lie between l and If f( l ) f( ) >, then the tue oot must lie between and u Retun to step with the new guesses fo the inteval, epeat until the inteval width is suffiiently small (i.e. the estimate fo the oot is within an aeptable toleane)

9 Repeat the pevious (falling mass) eample using the bisetion method: Eample - Bisetion Method f 68.1 / e Step 1: Make initial guesses fo l and u l = 1, u = 16 Chek: f (1) f (16) = (i.e. < ) a oot eists in this ange Step : Estimate the oot as lying at the midpoint of this ange: = ( l + u ) / = 14 f() Dag oeffiient, Step 3: Calulate f( l ) f( ) = f (1) f (14) = I.e. >, the tue oot does not lie between l and the tue oot must lie between and u (between 14 and 16) Retun to step with the new guesses fo the inteval: new estimate fo = (14+16)/ = 15 test whethe the oot lies between o 15-16

10 6 The tue value fo the oot is = We an evaluate the eo in the value geneated by the bisetion method at eah iteation: As peviously when evaluating the eo in iteative methods, we an use the uent and pevious values to geneate an appoimate elative eo, a : a new new old 1% f() Dag oeffiient, Iteation l u a [%] t [%] Afte 6 iteations, the appoimate elative eo a <.5% (and t is even lowe)

11 Compae the appoimate elative eo to the tue elative eo: Iteation a [%] t [%] With the bisetion method, the tue elative eo is always less than the appoimate elative eo: (This is a good thing fo engineeing design / poblem solving) If we teminate the alulation when a < s, we an be onfident that the tue value is within ou stopping iteion Relative Eo [%] Appoimate Eo Tue Eo Iteation #

12 With the bisetion method, the tue elative eo is always less than the appoimate elative eo. Why is this the ase? Choose lowe and uppe guesses fo the oot: Tue oot: l u The bisetion method gives a new estimate at l Bisetion method: Tue eo, t u This tue eo must always be less than /: Ou estimated (appoimate) eo a is always equal to /: new old u l So ou elative appoimate eo will always oveestimate the tue eo: a new new old 1%

13 Funtions in Matlab Matlab has many built-in funtions eg. sin( ), ep( ), mean( ), binde( ), plot( ).., We an also define ou own funtions whih an be alled (by name) fom the ommand line, o fom within othe sipts / m-files

14 Anonymous Funtions in Matlab Suppose we want to find the minimum of the following funtion a when a = 1, b = -, = 1 f b f() f()=a +b+, a=1,b=-,=1 lea all; lose all; a = 1; b = -; = 1; f + b* + ); Ceate an anonymous funtion (@): = [-*pi:.1:*pi]; plot(,f()); % Plots funtion f() ove [-pi < < pi] title('f()=a^+b+, a=1,b=-,=1'); label(''); ylabel('f()'); hold on; % Find and plot the minimum minimum = fminbnd(f,-,); % We an pass ou funtion dietly to the minimization outine plot(minimum,f(minimum),'o'); % We an evaluate ou funtion at the value minimum Anonymous funtions an have moe than one vaiable eg: f

15 Eample: A evesible hemial eation involving eatants A and B foming podut C an be witten as: A + B C (Eg. H + CO CH 3 OH) and haateized by the equilibium elationship:, whee C i epesents the onentation of onstituent i. C K CaCb If the vaiable epesents the numbe of moles of podut C geneated by this eation, we an wite the onentation of podut C as (C, + ), whee C, is the initial onentation of C,. We an efomulate the equilibium elationship as K C, C C a, b,

16 K C, C C a, b, Question: If the equilibium onstant fo this eation is K =.16, and the initial mola onentations of onstituents a, b, and ae C a, = 4, C b, = 8, C, = 4, espetively, detemine the numbe of moles of podut C (epesented by vaiable ) geneated unde these onditions. We would like to wite = and plug in ou values fo K, C a,, C b,, C,, But we an t. Equivalently, ou poblem is to find the values of whih satisfy: f C, C C a, b, I.e. find the oot(s) of f() = K

17 Step 1: Plot the funtion and estimate whee the oot lies K =.16; Ca = 4; Cb = 8; C = 4; = linspae(,,5); % This is the funtion we ae inteested in (afte eaanging): f (C+)./(((Ca-*).^).*(Cb-)) - K; figue; plot(,f(),'b-'); label('amount of C podued, [Mol]'); ylabel('f()'); ylim([-.5.5]); It looks like the oot is aound 16

18 Step : Implement the baketing method to get an auate estimate fo the oot Xl = 1; Xu = ; i = 1; i_ma = 1; Es = 1; while (1) X = (Xl + Xu)/; % Initial lowe and uppe guesses fo the oot % Iteation ounte stats at one % Ma # iteations allowed % Stopping eo (when % elative appo eo < Es) % New oot estimate fom Baketing Method if i > 1 % Ensue that i > 1 (an t alulate Ea on 1st loop) Ea = abs((x - X_old)/X)*1; % Calulate appo. elative eo (%) Ea_stoe(i) = Ea; % Stoe the appo. elative eo end test = f(xl)*f(x); if test < Xu = X; else Xl = X; end X_old = X; oot_stoe(i) = X; if i > 1 if Ea < Es i > i_ma beak end end i = i + 1; end % Test whethe f() hanges sign between Xl and X % If test <, oot must be between Xl and X % The alulated oot beomes ou new uppe guess (Xu) % If test is not <, oot must be between X and Xu % The alulated oot beomes ou new lowe guess (Xl) % Assign the latest oot to X_old fo use in the net loop % Stoe the latest oot fo plotting late % If we ae on the,3,4 iteation, hek the appo. eo % If eithe ondition is tue, eit the loop % Inement the iteation ounte fo the net loop

19 Step 3: Plot out the esults disp('calulated oot'); X disp('# of Iteations'); i disp('appoimate (Relative) Eo'); Ea figue; subplot(1,3,1); plot([1:i],oot_stoe,'-b'); label('iteation #'); ylabel('estimated Root, X'); ais squae; title('bisetion Algoithm fo oot of f()'); lim([1 i]); subplot(1,3,); plot([:i],ea_stoe(:end),'b-'); label('iteation #'); ylabel('\epsilon_a [%]'); ais squae; title('appoimate Eo, \epsilon_a [%]'); lim([1 i]); legend(['\epsilon_s = ',numst(es),'%']); subplot(1,3,3); plot([:i],ea_stoe(:end),'b-'); label('iteation #'); ylabel('\epsilon_a [%]'); ais squae; title('appoimate Eo, \epsilon_a [%]'); ylim([ 5]); lim([3.5 7]); legend(['\epsilon_s = ',numst(es),'%']);

20 Nonlinea Systems Dunn Chapte 5 Net time: False-position, Newton s method, Fied point iteation Nonlinea equations Root finding Baketing methods Open methods Gaphial Bisetion False Position Newton s method Fied-point iteation Newton- Raphson Bent s method

21 Open Methods fo Root Finding: Newton-Raphson The most widely used method fo oot finding 1) Stat with one initial guess fo the oot ( i ) f () Slope = f () f() ) Daw a line tangent to the point at [ i, f( i )] f ( i ) y = f ( i ) 3) Ou new estimate fo the oot oesponds to the -value whee the tangent fom [ i, f( i )] osses the -ais i+1 i = ( i i+1 ) The slope of the tangent line f ( i ) equals f i y i f i i1 Reaanging this in tems of the new oot estimate ( i+1 ): i1 i f i f i

22 Eample: Use the Newton-Raphson method to find the oot of f() = e - Given a funtion f(), its deivative f (), and an initial guess fo the oot ( i, i = ), the Newton-Raphson equation gives us a new oot estimate ( i+1 ): i1 i f f i i Funtion: f () = e - Deivative: f () = e - 1 Initial guess: = Use these epessions in the N-R equation: i1 i f f i i i1 i i e e i i 1

23 Oiginal poblem: Find the oot of f() = e -.5 f() = e f() = e (Tue oot = ) y.5 y Newton-Raphson Iteation i t [%] < 1-8

24 Tanspot Eample: Kogh Tissue Cylinde Model fo studying tanspot of metabolites fom apillaies to suounding tissue Tissue z Capillay The itial adius it is the distane at whih the metabolite onentation is edued to zeo: C m C it vaies with distane along the apillay aoding to: it R ln R R 1 R 4D C T t m D 4 V T DT R 1z R 1 K whee R it t m.

25 Question (i) : What is the itial tissue adius ( it ) whee gluose is no longe supplied to ells, at a distane z =.5 m unde the onditions listed in the table? ln R K D z R V D t R C D R R R T T m T. m it t R whee We annot eaange this equation in the fom it = We need to find the oot of f ( it ) =. (The value of it whih satisfies f ( it ) = ). 1 ln R Dz B A R R 4 m T t R C D A 1 K D B T 4 T V D D Let s simplify this nonlinea epession: whee (A, B, and D ae just onstants whih an be found fom values given in the table) and z is given in the question.

26 Let s make this epession even simple: Let = R R ln R A B DzR 1 A B Dz ln 1 If we solve this nonlinea equation fo (find the oots of f ( ) ln( ) A E( 1) ) then we an easily onvet the answe fom R it : E B Dz Wite Matlab ode to implement the Newton-Raphson method to solve this poblem: Fist steps: 1) Define the elevant onstants ) Define the funtion we want to solve (find the oot(s) of) 3) Plot out the funtion to obtain a ough idea of whee the oot(s) lie

27 Net steps: 4) Define additional onstants (initial guess fo the oot ( 1 ), stopping eo ( s ), ) 5) Set up a loop to keep geneating and stoing new estimates fo the oot, using the Newton-Raphson method i1 6) Estimate and stoe the eo ( a ) at eah iteation of the loop (if i >1). Eit the loop if this eo is small enough ( a < s ) 7) Plot out the oot estimates and eo estimates as a funtion of iteation numbe i f f i i

28 Question (pat ii): You just alulated it fo a single aial loation of z =.5 m. Tissue z How does it vay as z anges fom 1 m to.15 m? Capillay Modify you ode fom pat (i) to alulate it at lots of z values between 1 m and.15 m Stoe the values alulated fo it at eah z value. Plot it vesus distane z.

29 Poblems with the Newton-Raphson method: (1) Slow Convegene Eample: f () = 1 1 Find the oot fom an initial guess of = y 4 y y The new guess is the value whee the tangent fom f( ) osses the -ais Iteation i The algoithm is slowly onveging on the tue oot ( = 1) Due to (1) the natue of the funtion f () () ou initial guess fo the oot,

30 Poblems with the Newton-Raphson method: () No Convegene f () f () 1 1 Point of infletion nea the oot Tapped at loal ma / min f () f () 1 1 Slope of f() nea oot Slope of f() = at estimated oot

31 Poblems with the Newton-Raphson method: No Convegene Thee is no geneal onvegene ule fo the Newton-Raphson method Convegene depends on the natue of f() and the initial guess Computational methods using N-R should inlude some safety featues: 1. Substitute the final estimate fo the oot bak into f (). Chek whethe the esult is lose to zeo. Limit the numbe of iteations to some maimum value 3. Chek whethe the slope f () = at any iteation 4. Make the pogam plot the estimated oot at eah iteation fo inspetion

32 Nonlinea Systems Dunn Chapte 5 Nonlinea equations Root finding Baketing methods Open methods Gaphial Bisetion False Position Newton s method Fied-point iteation Newton- Raphson Bent s method