ENGI 9420 Engineering Analysis Assignment 3 Solutions

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1 ENGI 9 Egieerig Aalysis Assigmet Solutios Fall [Series solutio of ODEs, matri algebra; umerical methods; Chapters, ad ]. Fid a power series solutio about =, as far as the term i 7, to the ordiary differetial equatio d y π + y = sec, < d [This ODE is also i Assigmet Questio.] You may quote the Maclauri series epasio for sec : sec = ( < π ) 7 86 This ODE is aalytic for all i ( π + π ),. Therefore = is ot a sigular poit. Attempt a simple power series solutio: = = ( ) y= a y = a = = y = + + a = a Substitute the series for y, y" ad sec ito the ODE: ( a+ ( + )( + ) + a ) = = Matchig coefficiets: : a + a = a = a a a a a : + = = 6 : + = = = ( ) = + a a a a a a : = 5 = = 6 = a a a a a a

2 ENGI 9 Assigmet Solutios Page of 8. (cotiued) : 7 6a 7 + a5 = a7 = a5 = a = 5 a 6 : 8 7a a6 = 7 : 6 5a + a = a = a = a = 5 a 6 6 ( 5 ) ( 56 ) a = a = a = + a The leadig coefficiets a ad a are arbitrary costats, which are also the iitial values y ad The geeral solutio is y respectively. Rewrite them as A ad B respectively. 6 = ( ) B( ) ( ) y A 7 Upo recogizig the presece of factorial fuctios, we ca write the geeral solutio as y = 6 A + +!! 6! B + +!! 5! 7! ! 6! 8! Oe ca recover the complemetary fuctio quicly, upo recogizig the Maclauri series for the sie ad cosie fuctios: y = Acos + Bsi ! 6! 8! The particular solutio turs out to be the Maclauri series epasio for si + ( cos ) l cos A brief Maple worsheet provides this Maclauri series epasio.

3 ENGI 9 Assigmet Solutios Page of 8. Use the method of Frobeius to fid the geeral solutio of the ordiary differetial equatio d y dy + ( + ) + ( ) y = d d as a series about =, as far as the fourth o-zero term. d y dy + ( + ) + ( ) y = P d d F Q R Q R F = ( + ), = ad = are all aalytic. P P P Therefore = is a regular sigular poit of this ODE. + r + r = = y a ( r)( r + r ) = y= a y = a + r = + + Substitute the series ito the ODE: + r + + r + a( + r)( + r ) + a + r = = + r + + r+ + r a( r) a a = = = = Adjust the ide of summatio, such that the epoet of is +r i all cases. The ODE becomes + r a ( + r)( + r ) + ( + r) = + a ( ( + r ) + r + ) = = + r + r ( r )( r ) a ( ( r) ) a = = r ( r )( r+ ) a + + r ( + r )( + r+ ) a + ( ( + r) ) a = = This equatio must be true for all values of. Therefore the coefficiet of every power of must be zero. But a caot be idetically zero. r r+ = r = or

4 ENGI 9 Assigmet Solutios Page of 8. (cotiued) Case r = : This value of r is a positive iteger, so that the series becomes a ordiary power series. ( + )( + + ) a + ( ( + ) ) a = > ( + ) a = a ( + ) Employig this recurrece relatio to fid a to a, ( + ) 6 a = a = a + 5 ( + ) ( + ) ( + ) a = a = a = + a a = a = a = a a = a = a = + a y = a = a = Case r = /: ( )( + ) a + ( ( ) ) a = > a ( ) = a ( ) > ( ) a = a ( ) = a = > This series is therefore fiite, with oly oe term: y = a / The fuctio y ( ) is clearly idepedet of the fuctio Therefore the geeral solutio is y. A 5 = + ( ) y B (where A ad B are arbitrary costats).

5 ENGI 9 Assigmet Solutios Page 5 of 8. Fid the etire series solutio (usig the method of Frobeius, adapted to a first order ODE) about = of the ordiary differetial equatio dy y + = d P=, Q=, F = Q F = ad = are both aalytic P P Therefore = is a regular sigular poit of this ODE. + r + r = = y= a y = a + r Substitute the series ito the ODE: r y + + y = a + r+ = = r a r The leadig term ( = ) is ( + ) 5 If r = the a + = a = The ODE the becomes 5 ( 5 + a + ) = a = > 5 = + this is a particular solutio. y = a = 5 = If r = the a ( r+ ) = r = ( a caot be zero) = 5 a = a = ecept for =, where a is arbitrary, (relabel it as A ) ad for = 5, where 5a 5 = a5 = 5 y = a = A = 5 which icludes the case r =. Therefore the geeral solutio is 5 A = + y [Note that this ODE is liear. It is easy to verify that the solutio above is correct.]

6 ENGI 9 Assigmet Solutios Page 6 of 8. I Chapter of the lecture otes, dimesioal aalysis is used to derive the fuctioal form of the Plac legth L i terms of the uiversal costats G, h ad c. P I the study of the steady flow of icompressible fluid through a pipe, it is ow that the volume of liquid issuig per secod from a pipe, Q, depeds o the coefficiet of viscosity η (measured i the uits g m - s - ), the radius a of the pipe (measured i metres) ad the pressure gradiet p/l set up alog the pipe (measured i the uits N m - m - or, equivaletly, g m - s - ). Use a similar dimesioal aalysis to derive the fuctioal form for Q i terms of η, a ad p/l. z y p Q = η a, where is a costat of proportioality that caot be determied l by dimesioal aalysis aloe. Q is a volume per uit time (m s - ). Coductig the dimesioal aalysis, y z y z z z LT ML T [ L] ML T + M L + = = T y z M R + R L R+ R T R ( ) R R R + R =, y =, z = Therefore Q a p = η l which is Poiseuille s formula. By methods beyod the scope of this course, it ca be show that = π. 8 Plai tet output from the liear system reductio program is available at this li.

7 ENGI 9 Assigmet Solutios Page 7 of 8 5. Fid the itersectio of the three plaes + y + z = + 5y + z + = + y + z = Where the three plaes all meet, all three equatios must be true simultaeously. Form ad reduce the augmeted matri that is equivalet to this triple of simultaeous liear equatios. A b R R = 5 R R + Upo reachig this triagular form, oe ca see that ra A = ra [A b] = so that there is a uique solutio. R R 9 R R R + R Geometrically, the three plaes itersect each other i a sigle poit. From this reduced echelo form, we ca read the uique solutio (, y, z ) = ( 9,, ) 6. By reducig the appropriate liear system to echelo form, show that the itersectio of the three plaes + y + z + = + y + z + = + y + z + = y ( ) z is the lie = =. The liear system for (, y, z) correspodig to the simultaeous equatios of the three plaes is

8 ENGI 9 Assigmet Solutios Page 8 of 8 6 (cotiued) y z Plae Plae Plae R R R R At this poit, oe ca see that (row ) = (row ), so that row will cacel dow to a row of all zeroes ad ra A = ra [A b] = <. There are therefore ifiitely may solutios. More precisely, oe ca detect a oe-parameter family of solutios. Completig the row reductio, R R R+ R This is the reduced row echelo form for [A b]. There are leadig oe s for ad y oly. z is therefore a free parameter (call it t ). Readig the solutios from this echelo form, z = t; y + t = y = t ; t = = t+ (, y, z) = ( t+, t, t) = (,, ) + (,, ) t This oe vector equatio is also a set of three simultaeous scalar equatios for, y, z i terms of t. Mae t the subject of all three equatios, the t ( ) y z = = = which is the equatio of a lie. Geometrically, a oe parameter family of solutios should geerate a oe-dimesioal object (the lie).

9 ENGI 9 Assigmet Solutios Page 9 of 8 7. Fid the value of the determiat ad, if it eists, the iverse matri, for the matri A = Form the augmeted matri [ ] A I = Trasform the matri to reduced echelo form: R R R + R R R The left matri is ow i triagular form. The oly row operatio so far that has chaged the value of the determiat is the row iterchage. The value of the determiat of the origial matri A is therefore ( ½) =. No-zero determiat A - eists. Resumig the row reductio: R

10 ENGI 9 Assigmet Solutios Page of 8 7. (cotiued) R R R+ R R R R R + R = I A R R The oly row operatios that chaged the value of the determiat were oe row iterchage ad two cacellig row multiplicatios (the first by a factor of ½ ad the secod by a factor of ). All other row operatios were of the type (from oe row subtract or add a multiple of aother row), which do ot affect the value of the determiat. Therefore the determiat of the origial matri A is the determiat of the echelo form I. But det I =. Therefore det A =. Also, the fact that the reduced echelo form of A is the idetity matri allows us to coclude that A is ivertible ad that the iverse is the right-had matri i the reduced echelo form of the augmeted matri. Therefore det A ad A = = [Oe may chec the aswer, usig A A = A A = I.]

11 ENGI 9 Assigmet Solutios Page of 8 7. (cotiued) Additioal Note: It is possible to use a cofactor epasio to evaluate the determiat ad iverse of the matri A = However, this is a much slower method tha row reductio (Gaussia elimiatio) to echelo form. The details of this alterative method are o a separate web page. 8. I Assigmet, Questio 5, fid, correct to the earest millimetre, the iitial head H of water eeded i the coical ta i order for the ta to drai completely i eactly oe miute, usig (a) the method of bisectio (or a graphical zoom-i, i which case provide setches or screeshots) ad eplai your choice of iitial values. The time T required to drai the coical ta of head H completely was foud i assigmet to be ( ) H T = H + H + g A ta of head cm was foud to drai i approimately 6 s. The fuctioal form for T (H) ivolves a higher power tha the square. It is therefore certai that a ta of head 6 cm will require more tha 6 secods to drai. I the bisectio method, tae as startig values H = ad H = 6. [et page]

12 ENGI 9 Assigmet Solutios Page of 8 8 (a) (cotiued) H T At this poit oe ca deduce that, correct to the earest millimetre, a draiage time of eactly oe miute will occur whe H =. cm A alterative to tabular bisectio is a sequece of graphical zooms, as illustrated here.

13 ENGI 9 Assigmet Solutios Page of 8 8 (a) (cotiued) This process of graphical zooms ca be cotiued, i order to fid the solutio to greater precisio: from which H =. cm, correct to four decimal places. T (.) = 6. s (b) Newto s method ad eplai your choice of iitial value. As i part (a), we ca deduce that the required value of H is somewhere i (, 6). The o-liear fuctio suggests that the value is closer to tha it is to 6. Therefore select H = as a iitial guess. + = f f ( ) ( ), where f = ( ) g ( 5 5 ) f = + + g f ( ) f ' ( ) f / f ' Therefore, correct to oe decimal place, H =. cm [A Ecel spreadsheet file for this solutio is available here. Oe may eperimet with differet startig values.]

14 ENGI 9 Assigmet Solutios Page of 8 9. Why should Newto s Method ot be used to fid a root of e = ta, (ecept whe the iitial guess is very ear the true value)? Demostrate the problem by usig Newto s Method to try to fid the first positive root, with iitial guesses of.99 ad.. The fuctio ta has a ifiite umber of ifiite discotiuities. It is quite possible for the taget lie from a value of to cross oe or more such discotiuities before reachig its ais itercept. Newto s Method ca thus become very ustable, jumpig from ear oe root to ear some other root. Graph of y = e ta Also ote the proimity to the first positive root of two turig poits (where f ' () = ). Usig Newto s Method ayway, with the startig values.99 ad.: f ( ) + =, where f = e ta f = e sec f ( )

15 ENGI 9 Assigmet Solutios Page 5 of 8 9 (cotiued) f( ) f '( ) f / f ' Note that the method has coverged o the wrog root! f( ) f '( ) f / f ' ad =.6... is the first positive root. Note how two startig values so close together have coverged to very differet roots. This is a illustratio of how ustable Newto s Method is i this case. [A Ecel spreadsheet file for this solutio is available here. Oe may eperimet with differet startig values.]

16 ENGI 9 Assigmet Solutios Page 6 of 8. Use the stadard fourth order Ruge-Kutta method, with a step size of h =., to fid the value at =.6 of the solutio of the iitial value problem dy ( y y), y. d = = Show your calculatios for the costats,,, ad for y i the first iteratio. You may submit output from a spreadsheet program (such as Ecel) for the other steps. y For this ODE, the RK algorithm becomes = f, y = y y = y y ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (, ) ( ) ( )(( ) ) = f + h, y + h = + h y + h y + h = f + h, y + h = + h y + h y + h = f + h y + h = + h y + h y + h h y+ = y + ( ) 6 with startig values = ad y =.. I the first iteratio, =... = = =.6 = =.6 = =.8. y(.) = y = + ( ) = Secod iteratio: = (.) (.9996) ((.9996) ) =.8 ( ) ( ) = (. +.) ( )( ) (.) = y = = =.7 = = = =.98978

ENGI 9 Assigmet Solutios Page 7 of 8 (cotiued) y Fial iteratio: = (.) (.98978) ((.98978) ) =. ( ) ( ) = (. +.) (.98978 +..9 7 )( ) (.6) = y =.98978 +. 6 =. +..98978 +...98978 +.. =.986 =. +..98978 +..986.98978 +..986 =.9 77 7.

17 ENGI 9 Assigmet Solutios Page 7 of 8 (cotiued) y Fial iteratio: = (.) (.98978) ((.98978) ) =. ( ) ( ) = (. +.) ( )( ) (.6) = y = = =.986 = = = =.9866 Therefore, correct to four decimal places, y(.6) =.99 A directio field plot from Maple cofirms the behaviour of the solutio to this iitial value problem: [A Ecel spreadsheet file for this solutio is available here.] [A Maple worsheet, from which the plot above was derived, is available here.]

18 ENGI 9 Assigmet Solutios Page 8 of 8 (cotiued) Note that the ODE is Beroulli: dy + y = y d P R ( ) h / h = P d = d = e = e y h / / e R d = e d = e h h + / / w = = e e R d + C = e e + C y The geeral solutio is y = / + Ae + The iitial coditio y () =. leads to A =. The eact value of y (.6) is The RK value is correct to seve decimal places! Retur to the ide of assigmets

ENGI 9420 Lecture Notes 3 - Numerical Methods Page 3.01

ENGI 9420 Lecture Notes 3 - Numerical Methods Page 3.01 ENGI 940 Lecture Notes 3 - Numerical Methods Page 3.0 3. Numerical Methods The majority of equatios of iterest i actual practice do ot admit ay aalytic solutio. Eve equatios as simple as = e ad I = e d