Applicable Mathematics 2A

Transcription

1 Applicable Mahemaics A Lecure Noes Revised: Augus 00 Please noe ha hese are my lecure noes: hey are no course noes. So I will be following hese noes very closely, supplemened by he homework eamples ec ha I se in each lecure. These lecure noes are provided: (a) o ensure ha everyone has an accurae se of noes, and, (b) o allow much more ime for running hrough eamples in class. I s probably no a good idea o refer o hese noes during he lecures (hey will disrac you from wha I am saying), alhough I do recommend ha you skim hrough hem before-hand so you have an idea of wha is coming. I is bes o use your ime in lecures o aim o undersand he maerial and relae i o wha you already know. Someimes, I like o give eamples for you o solve in lecures - o see if you have absorbed he maerial which has jus been delivered, and o reinforce i. I will be doing los of eamples in class and here are many more (60) in he uorials, as well as abou 5 homework eamples. I am aware ha here are an enormous differences in mahs abiliy (and ineres!) beween differen sudens. I do ry o pich hings for he average suden so for some sudens, I may be going oo fas and oo slow (simulaneously!). Whaever your mahs abiliy/ineres, I canno sress enough ha he key o success in his course lies in he amoun of ime you spend problem solving. In he end, ha is wha he eaminaions are abou. You can (and should?) es he ruh of his by siing some pas eaminaions and honesly assessing he resuls. Some pars of his course are essenial for oher engineering courses, while some ohers are ineresing in heir own righ. I hope ha you will learn somehing useful and worhwhile in hese lecures. Dr. T.G. Davies Deparmen of Civil Engineering Glasgow Universiy (Please le me know if you spo any errors or ambiguiies in hese noes. TGD)

2 Conens: Lecure. Funcions of wo (or more) variables. Parial differeniaion:. Chain rule: 4. Successive differeniaion: 5. Toal differenials: 6. Eac differenials: 7. Line inegrals: 8. Line inegrals: 9. Line inegrals: 0. Double inegrals:. Parameric represenaions:. Parameric represenaions:. Numerical inegraion: 4. Classificaion of PDEs 5. Separable, Eac: 6. Linear (s order): 7. Euler's mehod: 8. Review: Tuorial soluions and oher maerial (including some pas eams) can be found a: hp://people.civil.gla.ac.uk/~davies/mahs/inde.hm

3 Lecure : Funcions of wo (or more) variables Summary: - review of funcions of one variable (meaning of differeniaion / inegraion) - picorial represenaion of funcions of wo variables (by means of 'level curves', i.e., conours) - review of he principles of differeniaion (scalar producs rule, sum rule, produc rule, quoien rule, chain rule, parameric differenaion - some eamples. Theory Figure. Funcion of one variable Figure. Funcion of wo variables Figure. Level Curves (conours) Figure.4 Eample [ z 4 y ]. Review of differeniaion (a). Scalar (consan) producs: y f() c cons d d (cy) dy c d cf ' () (b). The Sum rule u f() v g() d d (u v) f '() g'() u' v'

4 (c). The Produc rule d d (u.v) u v' vu' (d). The Quoien rule d d u v [ vu' uv' ] v (e). The Chain rule y f(), g() dy dy d d (f). Parameric differeniaion y f(), g() dy d dy d dy. Eercises Skech level curves (conours) for: (a) f(, y) 4y (b) f(, y) ln ( y - ) (c) f(, y) an - [y/( )] 4

5 Lecure : Parial Differeniaion Summary: - parial differeniaion implies slope of a surface z F(, y) - slope a poin P varies depending on direcion considered - general epression for slope - epression for slope in direcion ("parial differeniaion wih respec o ") Parial differeniaion wih respec o is very similar o "normal" differeniaion, bu now y, and any oher variables, are reaed as consans.. Theory Figure. Funcion of wo variables Figure. Definiion of parial differenials P' has coordinaes (, y, 0), Q' has coordinaes (, y, 0) where: y y L cos α L sin α and: L P'Q' The slope of PQ is: f (, y) L f (, y) Now le m α slope a P in direcion PQ m α lim L 0 f (, y) L f (, y) Remember his is he slope a P, wih coordinaes (, y) 5

6 Now epress his in erms of slopes in and y direcions: m α lim L 0 f (, y) f (, L cos α y) cos α lim L 0 f (, y) f (, L sin α y) sin α m cos α m sin α y Now consider m : m lim L 0 f (, y) f (, L cos α y) where: L cos α Thus, we can wrie: m lim 0 f (, y) f (, y) where y is replaced by y, because m (a Q) m (a P) as 0 Finally, in he limi: m f ( d, y) f (, y) d This is known as he "parial derivaive wih respec o " We simply rea y as a consan. The noaion used is usually f or f (no m ) 6

7 Finally, he derivaive (slope) in he direcion PQ is: m a f f cos α sin α y. Eamples: (a). z f(, y) Sin ( - y) z Cos ( y) ( y) Cos ( - y). z y? (b). f(, y, ) yz y - z f yz y f y, f z 7

8 Lecure : Chain rule of parial differenaion Summary: - developmen of he chain rule - some eamples. Theory The chain rule of parial differeniaion is an eension of he familiar rule of Year. Consider: z f(, y) g(s, ) y h(s, ) Wha is: z s z? From firs principles, by subsiuion, we obain: z F(s, ) where F F(f,g,h) By definiion: z s lim s 0 F(s s, ) F(s, ) s Clearly: (s, ) (, y) (a one-o-one correspondence). Hence: ( s s, ) (, y y) - a change in s resuls in a change in boh and y. 8

9 Thus: z s lim s 0 f(, y y) f(, y) s lim s 0 f (, y y) f (, y y) s lim s 0 f (, y y) f (, y) y y s f s f y y s Similarly: z z z y y The eension o funcions of hree or more variables is obvious: eg. z f(, y), g(r, s, ), y h(r, s, ), ec.. Eamples (a). Find T r and T θ when T(, y) - y y, and r cos θ and y r sin θ By chain rule T r T r T y y r T y, T y y 9

10 and r cos θ, y r sin θ so ha ( r T y ) cos θ ( y ) sinθ Subsiuing for and y in erms of r and θ gives r T r (cos θ sin θ) r cos θ sin θ (b). Righ circular cone Radius r increases a rae mm/s Heigh h increases a rae mm/s Find volume epansion rae when r 5 mm, h 5 mm V V r r V h h Now v r π r h, v h π r dr dh, so ha dv. π r h. π r 4 π r h π r For r 5 mm and h 5 mm dv 4 π.5.5 π.5 5 π mm /s. 0

11 Lecure 4: Successive Parial Differeniaion Summary: - definiion of successive differeniaion - equaliy of f y and f y - successive differeniaion of funcions of funcions - eamples 4. Theory Given z f(, y), and he parial derivaives y f, f find he second derivaives: f f y f y f f y ec... Noe ha in engineering, we may assume: y y f alhough his is no universally rue! 4. Chain rule: Le u f(s, ), s g(, y), h(, y) Find u. u s s u u All erms are funcions of (direcly or indirecly).

12 Using he produc rule: u s s u s s u u u (A) Now (chain rule): s u s u s s u s s u s s u Similarly, u. Subsiuing hese resuls ino (A): u s s u s s u u s s u u or, using he obvious condensed noaion,: s ss u s u s u u s u s u Similarly: u yy, u y (No o be memorised!) 4. Eamples Given: f(, y) y y, find f, f y, f ec.

13 Lecure 5: Toal Differenial Summary: - definiion - oal differenial in erms of parial differenials - applicaion o error analysis - relaive error - eamples 5. Theory Le u f(, y) Wha is u, given, y? By definiion: u f(, y y) - f(, y) Le, y y y Thus, u f (, y) f (, y) Rewrie his as: u f (, y) f (, y) f (, y) f (, y) y y f f y y Now, in he limi, ( 0, y 0) du f d f dy y The quaniy "du" is called he "oal differenial".

14 If we consider funcion of hree or more variables, u f(, y, z.. ), hen: f d f dy f dz du.. y z 5. Eamples (a) Given u y, find du. du u d u y dy y u y bu wha is u y Consider: u a y u (e ln a ) y e y ln a (raher han ) NB: a e ln a u y u y ln a. e y ln a ln a. a y ( Swiching from a o... y ln (b) Figure 5. Cylinder If, r 0 ± 0. mm, and, h 50 ± 0.05 mm, wha is maimum error in V? V πr h dv V dr V dh πrh.dr πr dh r h 4

15 Noing ha V dv, and dr 0. mm, dh 0.05 mm V 45 π mm (c) Relaive error Cylinder: r 0 mm, h 50 mm Given relaive errors r 0.%, and r h 0.%, h wha is he relaive errror in volume V V?? dv V dr V dh πrh.dr πr. dh r h dv V dr r dh h V V r r h h 0.76% 5

16 Lecure 6: Eac Differenials Summary: - comparison wih oal differenials - is a differenial epression "eac"? - wha is he "paren" funcion f? - he es for an eac differenial - he process (inegraion) o deermine f, if i eiss - eamples - applicaion o "fluid" flows 6. Theory Recall ha if z f(, y), he oal differenial is: dz f d f dy P d Q dy y where, P P(, y), Q Q (, y) Now, consider a general epression, P d Q dy This is an "eac differenial" if a paren funcion f can be found such ha: P f Q f y Two quesions arise:. Given P, Q can we find f? - is he epression an "eac differenial"?. If so, wha is f? 6

17 6. Eisence of paren funcion f: If f eiss: P f, Q f y Take he parial differenials: P y f y, Q f y For coninuous funcions, f y f y Hence, if f eiss: P y Q This is he es we employ. 6. Find he paren funcion f By definiion: f P, Q f y Inegrae: Pd F (, y) g(y) f where g(y) is an arbirary funcion of y (no jus a consan). Bu f is also given by he inegral: f Qdy F (, y) h() Now, deermine g(y) by inspecing nd epression. consan C o he resul. Check by deermining h() Add arbrirary 7

18 6. Eamples: (a) Is (6 9y) d (9 ) dy an eac differenial? If so, find he paren funcion f. Le: P 6 9y, Q 9 "Eac" if P y Q P y 9 Q 9 So his is an eac differenial. Now o find f: f 6 9y f ( 6 9y) d 9y g(y) Check by parial differeniaion w.r... consan. Also: g(y) can also include f y 9 f ( 9 ) dy 9y y h() Bu, f 9y g(y). So, comparing hese wo epressions: f 9y y C (b) Is (y y ) d (y ) dy an eac differenial? If so, find is paren funcion. 8

19 6.4 Some applicaions: Laer in his course o "pah inegrals", and differenial equaions In pracice, o fluid flows (incl. plasic flow) Figure 6. Fluid Flow (non-urbulen) By definiion, along a sreamline: V V y dy d V y d - V dy 0 (A) The "coninuiy equaion" demands ha: V y y V So equaion (A) conains an "eac differenial" and hence a simple soluion is possible. 9

20 Lecure 7: Line inegrals - I Summary: - inegrals as "area under he curve" - pah of inegraion beween limis A and B - line inegral relaed o "area of wall" under he surface - ypes of line inegral - normal ype (d, or dy) - differenial ype - inrinsic - parameric - eamples 7. Theory Figure 7. Inegraion in one dimension In one dimension (Year ), we deal wih funcions such as: y f() and inegrals such as: I b a f () d This definie inegral" was defined as he "area under he curve" Now consider eension o wo variables: and he inegral: z f(, y) I b a f (, y) d This makes sense only if we can subsiue for y (e.g., y g()). Noe ha a, b are limis along he ais 0

21 Subsiuing: b I f (, g() ) d a a b h () d Evidenly, he inegral depends on he "pah" of inegraion (ha is, he funcion, y g ()) Figure 7. Pah of inegraion Noe ha he limis on he -ais are (a, b) and he limis on he y-ais are (c, d). We are no inegraing under his curve! Our inegral is: I b a z d where z f (, y) Figure 7. Line inegral The physical meaning of his (line) inegral is he area of he wall under he curve: more precisely, i is he area of he shadow of he wall projeced on o o he z plane. 7. Types of line (conour) inegral (a). Ordinary f (, y) d c c a b B A

22 (b). Differenial B A [ f (, y) d g (, y) dy] (cf. "eac differenials") (c). Inrinsic B A f (, y) ds where s is he local ais direced along AB. (Yields he acual area of he wall) Figure 7.4 Inrinsic inegral (d). Parameric f (, y) where, g (), y h () Figure 7.5 Parameric inegral This ype of inegral is closely linked o he "inrinsic" form.

23 7. Eamples B (a). [ ( y ) y ] d I A where, A (½, 0), B (0, ). The pah of inegraion from A B is given by y - 4 Figure 7.6 Eample 7(a) Subsiue for y : limis in 0 I ( 4 ) ( 4 ) d d (b) I 0 y d given y 5 Ans:.4

24 Lecure 8: Line Inegrals - II Summary: - eample of differenial form - consideraion of differen pah of inegraion - special resul for "eac" differenials (line inegral independen of pah) - parameric line inegrals - eamples 8. Theory Consider he eample: B I [ ( y) d () dy ] A Figure 8. Pah of inegraion On he he pah AB (from A(0, ) B(, )), y A. Separae ino wo inegrals: I I I y where: I ( y) d I y c c dy By subsiuion: I ( ( ) ) d 4 0 Similarly, I y 6 I 0 4

25 B. Alernaively, reduce o a single variable: Since y, hen dy d Inegrae w.r.. : I {[ ( ) ] d d } 0 ( 7 ) d 0 0 Now we consider a differen pah of inegraion: Figure 8. Various pahs of inegraion We know ha I AB 0 (along line AB). Wha is I ADB? Evidenly: I ADB I AD I DB On pah AD: y, dy 0 D I AD ( ) A d 0 8 5

26 On pah DB:, d 0 I DB B D 6 dy 6y Hence: I 0. This is he same answer as before. Does his mean ha hese inegrals are no a funcion of he pah of inegraion? (No - no necessarily) 8. Pah independence of line inegrals of eac differenials Consider: B I [ Pd A Q dy ] For an eac differenial, P f, Q f y Thus, df P d Q dy Hence, we can wrie: I B A df B f fb fa A We see ha he inegral depends only on he funcion values a he endpoins A and B! And, in paricular, for a closed pah (AB), he inegral of an eac differenial mus be zero. Figure 8. Closed pah inegral 6

27 8. Parameric line inegrals These are of he form: I C f (, y) where, g (), y h () An eample: f (, y) y, 5, y Figure 8.4 Parameric line inegral Consider he case where he lower limi, and he upper limi 6. Subsiue: 6 I (5 ) 6 The pah of inegraion is along he curve defined in he -y plane by he parameer. This curve can be easily found in his case: y 5 y 5 Figure 8.5 Pah of inegraion 7

28 Lecure 9: Line inegrals - III Summary: - inrinsic line inegrals (ds) - eample - limis in "s" are undefined - relaionship beween ds and d - eample (solving in he domain) - an alernaive approach (and where, y are epressed in erms of parameer ) - relaionship beween ds and (he "Jacobian") - eample (solving in he domain) 9. Theory Figure 9. Inrinsic line inegrals I A B f (, y) ds where ds is he incremenal lengh along he pah of inegraion. In his case, he inegral is he area of he wall". 9. Eample Consider, f(,y) y, y, limis A(0,) o B(,5). Figure 9. Pah of inegraion We know how o deal wih : I f () d Bu how do we deal wih: I f () ds. Wha is he relaionship beween ds and d? How do we deal wih he limis in s? 8

29 To be eplici, we can deal wih: I f () d Bu, how do we deal wih: I Figure 9. Relaionship beween ds and d s s... ds Wha is s, s? From Pyhagaoras: (ds) (d) (dy) Hence: ds d dy d 9. Soluion o eample Given: I B A y ds and y We find: dy dy ds d d d 5 d Hence: I ( ) 0 5 d 5 ( ) d.9 0 9

30 9.4 An alernaive approach Epress and y in parameric form - if his is possible. y y Figure 9.4 Parameric represenaion I B A f (, y) ds f (, y) J (s, ) where J(s,) compensaes for he swich of variable from s, and is known as he "Jacobian of he ransformaion". Here, i is simply: ds J(s, ) Figure 9.5 Relaionship beween ds, d and dy As before, we have: (ds) (d) (dy) Hence: Thus: ds J ds d dy d dy Going back o he eample (now using he parameer as our variable of inergraion): d Since: dy and: y 0

31 ds d dy This gives: J 5 We see ha he lower (A) and upper (B) limis correspond o: - and, respecively. On subsiuion: B I y ds A ( ) ( ) Same answer as before - of course. Ofen i is beer o work in erms of a parameer raher han in erms of original variables, y.

32 Lecure 0: Double (and higher order) Inegrals Summary: - definiion of double inegrals - inegral as "volume under he surface" - domain of inegraion (limis in and y) - dealing wih non-consan limis (inegrae w.r.. variable wih non-consan limis firs) - eample (consan limis) - eample (non-consan limis) 0. Theory A ypical double inegral: I y y f (, y) d dy where limis in, y are no consans. (NB: da d dy) Figure 0. Definiion of double inegral The inegral can be undersood as he, "volume under he surface". Figure 0. Inegraion process Here we inegrae wr. y firs: I y () y () f (, y) dy d

33 0. A special (simple) case Figure 0. Inegraing over a recangle I ( y R ) d dy I 0 ( y )d dy - inegraing wr. firs y dy 6 y 0 0 dy 6y y 8/ 0. A more ypical eample 0 Figure 0.4 Inegraing over a riangle I ( y R ) da where R is known as he "domain of inegraion". y y ( y ) d dy Here i is bes o inegrae wr. firs, because he limis are more easily epressed as a funcion of y.

34 Lower limi: y, Upper limi: - y y Le g(y) y ( y ) d y y y 8 y y 8y I 0 g (y) dy 4 8y 6y 4y y 4/ 0 4

35 Lecure : Linear Inerpolaion Summary: - eample (from numerical modelling in engineering) - linear inerpolaion: f(t) - linear inerpolaion: f() (where is defined o be ± a he wo known poins) - he inerpolaion ("shape") funcions N() - general properies of N() - eample - an alernaive approach for deriving he shape funcions. Theory Figure. Figure. Figure. Figure.4 Machine componen Typical elemen pach Deformed mesh Linear inerpolaion The problem is o deermine a any arbirary poin T, given wo known values. The equaion of a sraigh line is: T T T T This reduces o: (T (T T) T ) (T (T T ) T ) f (T) f (T) (A) where, f (T), f (T) are linear funcions of T. Noe ha: f (T) f (T) (for all values of T) I is convenien o work in erms of a new parameer (insead of T), where varies from - o over he region of inerpolaion. We ransform T by a linear ransformaion. 5

36 Le: T a b and deermine consans a, b by observing ha T T when -, and, T T when. Thus: T a ( ) b T a ( ) b which yields: a (T - T )/ ( half-range ) b (T T )/ ( mid-poin ) Subsiue back in o (A): N N (B) where: N ( - )/, N ( )/ These funcions (N, N ) are called "shape funcions" Like he funcions f, f, hey are linear in and heir sum is uniy. One oher imporan propery: N (a ) N (a ) 0 and conversely for N Figure.5 Shape funcions N, N. Eamples Figure.6 Linear inerpolaion A T 4, 0; a T 8, 40. Wha is (5)? Transform T T a b b midpoin (T T )/ 6 a half-range (T - T )/ Hence: T 6 (check) For T 5 -½ 6

37 Now deermine N, N N ( - )/ ¾ N ( )/ ¼ So: N N ¾ (0) ¼ (40) 7.5. Alernaive approach for finding shape funcions Begin wih: N N (B) We know ha N N are linear in, where [ ] Also: N a node (zero a oher) N a node (zero a oher) We can herefore derive N and N direcly. We know ha: N c d and: N when -. N 0 when Subsiuing ino (C): c (-) d 0 c() d which yields: c -½, d ½ Hence: N ( - )/ (Deermine N, similarly, as an eercise) 7

38 Lecure : Quadraic Inerpolaion Summary: - definiion of he problem - review of linear inerpolaion - direc derivaion of shape funcions N for quadraic inerpolaion - inerpolaion equaions viewed as parameric equaions - eample: replacing a complicaed y f(), by wo simple equaions ( f(), y g()). Theory Figure. Quadraic inerpolaion Eample. Given: (), () 4, (5) 0, find (4). Recall linear inerpolaion: N N X (B) where: N, N are linear, and we make use of ransformed ais, which yields: N ( - )/, N ( )/ For quadraic inerpolaion, we epec o obain: N N N (C) where: N N N are quadraic funcions of. Also: N a node and zero a all oher nodes N a node and zero a all oher nodes N a node and zero a all oher nodes Taking N as an eample: Clearly: N a b c 8

39 We obain he consans a, b, c by subsiuing he condiions above. Thus: - a b (-) c 0 0 a 0 a b c Solving: Hence: a 0, b -½. c ½ N ( - )/ Similarly: N -, N ( )/ (Show his) Figure. Quadraic shape funcions. Eamples Given: (), () 4, (5) 0, find (4). Transform T (linear ransformaion only in his course) T a b [ ] b mid-range (T T )/ a half-range (T - T )/ Hence: T (check) So for T 4, 0.5 Compue shape funcions: N ( - )/ -⅛ N N ( )/ ⅜ (Check ha N ) Finally: N N N This mehod can be easily eended o cubic inerpolaion (and above). 9

40 . Applicaion o parameric equaions Thus: Similarly: N N N g () y g () (where are consans) (i.e. a pair of parameric equaions) Figure. Parameric represenaion of geomery Noe ha he pair of equaions: g () and y g () is accurae and simple, while he alernaive, y g (), may be very difficul o deermine and is likely o be very complicaed..4 Eample Figure.4 Parameric represenaion of a quadran Noe ha a Cos 45º. Inerpolaion: N N N From he figure: 0, a, Recalling: N ( - )/ ec. we obain: (.5 - a).5 a Similarly: y N y N y N y (.5 - a) -.5 a Thus, he quadran is now represened by his pair of parameric equaions. How accurae are hey? Consider ½ (say) By subsiuion:.76, y.6 Disance o origin is: R y. 976 So he error <% [Theoreical ] Advanages of he parameric equaions: highly accurae, independen of aes (,y), simple (universal) equaions. 40

41 Lecure : Numerical Inegraion (of complicaed inegrals) Summary: - inegraion as "area under he curve" - numerical approimaion of area: mid-poin rule rapezium rule Simpson's rule use of sub-areas - eample. Theory Figure. Inegral as area under curve I f () d area under curve Figure. Mid-poin rule Le h - I f ( ). h (by inspecion, recangle) Figure. I Trapezium rule f () f ( ).h - area of rapezium Figure.4 Simpson s rule I h 6 f ( ) 4f ( ) f ( ) -area under quadraic curve 4

42 In fac, Simpson's rule inegraes polynomials up o order (cubic) eacly, as will now be proved: Consider: I (a b c d) d a 4 4 b c d e b d Simpson s rule saes ha: I h 6 f ( ) 4(0) f ( ) Here: h f (-) -a b - c d f (0) d f () a b c d Hence: b I d QED. Use of sub-areas o increase accuracy Figure.5 Subdivision of domain of inegraion I f () d I A I B I C AB BC AB BC 4

43 . Eamples B I A f (, y) ds where he pah of inegraion is he quadran of a circle (radius ). Figure.6 Numerical inegraion Here we consider, for simpliciy, he case, f (, y) We also use he (approimae) parameric eqauions: (.5 - a).5 a y (.5 - a) -.5 a So: B I ds A ds ds [ J ] J ds d dy d Clearly: (.5 a). 5 Hence: I dy.088 (.5 a) No immediaely obvious how o inegrae his, so we use Simpson's rule: 4

44 The inegral akes he form: I f () h f (-).75 f (0). f ().75 I h 6 f ( ) 4 f (0 ) f ( ) 4.67 If we had inegraed analyically, we would have obained 4.68 (so he error in Simpson s rule in his case is very small ( 0.%)) Quie separaely, in his eample we chose o use a parameric approimaion for he geomery of he quadran. If we had used he eac epression for he quadran, hen we would ge: B I ds A lengh of quadran π r/ 4.7 This approimaion herefore produces a small error. The poin is hough ha we ofen have o use a parameric represenaion: he eac epression of he curve is eiher no known or oo complicaed. 44

45 Lecure 4: Differenial Equaions - I Summary: - soluion, by inegraion - dependen and independen variables - classificaion of differenial equaions ordinary/parial order linear/non-linear If linear: homogenous/non-homogenous - soluion "by inspecion" - soluion characerisics (number of consans) - firs-order differenial equaions - he "direcion field" - eamples 4. Inroducion An eample of a differenial equaion: u y µ u u 5 Sin We find: u f(, y, ) by inegraion. The unknown (arbirary) consans are deermined from he boundary condiions: (e.g. u 7, when, y, ) 4. Classificaion (i). Parial - Ordinary (ii) Order (iii) Linear - Nonlinear (iv) Homogenous / nonhomogenous 45

46 (i) Parial/ Ordinary f f y 4 y f f (, y), dependen variable f, independen variables, y d f d 4 df d Cos f f() (ii) Order: (highes derivaive) f f y 0 f y f f y 0 d 4 d 0 (iii) Linear/Nonlinear: Nonlinear: dependen variable (or derivaive) occurs as: producs, raised o powers, nonlinear funcion f y d f 4 df d d f 4 y Cos d 4 d 0 d d 4 d 4 Sin sin 0 46

47 If linear, hen we classify as eiher homogenous or nonhomogenous Homogenous: all erms involve eiher he dependen variable or is derivaives. f y f 4 y d Sin 0 d d Sin 0 d f d 4 df d Cos 0 4. Mehods of soluion "by inspecion" d 4 Soluion: f(). Try: Ae -4, where A is an arbirary consan Figure 4. Soluions for -4 If a boundary condiion is specified (e.g., (0).5), hen we can deermine he arbirary consan (here, A.5, by subsiuion) 4.4 Definiions The general soluion is he mos general funcion which saisfies he differenial equaion. (The number of arbirary consans will be he same as he "order" of he equaion.) 47

48 The paricular soluion is he specific soluion obained when he general equaion has o saisfy cerain consrains (e.g. boundary condiions), which hen yields numerical values for he arbirary consans. 4.5 Firs-order differenial equaions In his course, we only consider firs-order equaions: d f (, ) d For specific values of (, ), we can deermine (and plo). This plo is called a "direcion field". Figure 4. Direcion field Consider: d ( ) Firs draw a direcion field, by ploing he gradiens (d/) a arbirary poins (bu ypically on a grid paern). Saring a any poin, draw a smooh curve "parallel" o he direcion field. The equaion of his curve: g() mus be a soluion of he differenial equaion. The family of such soluions is he general soluion, and helps o deermine he analyical soluion. Figure 4. Some paricular soluions 48

49 Lecure 5: Differenial Equaions - II Summary: - separable ODEs - direcly - by subsiuion - "eac" ODEs - eamples 5. Separable equaions: If: d h () g () hen, crudely speaking, we can "cross-muliply" and obain: g() d h () The soluion is: g () d h () 5. Eample d 4 4 ( ) d 4 log e C C e De 49

50 5. Separable by subsiuion (quasi-separable) Consider: d g This is no separable (e.g., g Sin Use he subsiuion: y ( y) From he produc rule: d dy d y Bu g ( y ) Hence: dy g (y) y Now his is separable. 5.4 Eample d Divide by : d g Le, y. So, d y y g(y) Recall (or calculae from firs principles): dy g (y) y y This is separable... dy gives: log C e y y Recalling: y /, gives: log e C 50

51 5.5 "Eac" equaions Consider a funcion f(, ): df f d f h d g in which: h h (, ), g f (, ) If: df 0, hen he soluion is f consan. Now, suppose ha we have equaion: h d g 0 ( or equivalenly, d g ) h Then, is soluion is: f consan, provided ha: h f, g f (i.e., equaion is of he "eac" form). We can es for his "eac" form in he usual way, by checking wheher: h g If "eac", hen, we can obain he paren funcion f as usual by inegraion: [f hd, and, f g ] 5

52 5.6 An eample: d ( ) 0 If eac f h [ ] f, g [ ] Tes for eac ype: h g, So his is eac. Inegrae o ge paren funcion: f d f ( ) p () - r() Hence: f - C The soluion is: f consan. So: ± C 5

53 Lecure 6: Differenial Equaions - III Summary: - for linear differenial equaions - soluion for homogenous ODEs - eample - soluion for non-homogenous ODEs - eample 6. Inroducion Linear firs-order differenial equaions in general can be wrien as: d p () r () 6. Homogenous form d p () 0 Muliply hroughou by h h (). (The soluion remains he same) h d hp 0 Recall he "eac" form: h d g 0 which is eac if h g In our equaion: g hp and so i will be "eac" if: h (hp) Noing ha: h h (), p p (), hen, dh hp 5

54 This now separable, wih soluion: dh h p Solving, we ge: log e h p which can be wrien as: h e k, where k p () Thus, choosing h e k (called he "inegraing facor") makes he he original equaion "eac" (i.e., when i is muliplied hroughou by h.) In oher words: k e d k p e 0 is eac. We now find he paren funcion f, as usual: f k k e d e g() Hence: f k k p e e h () f e k C (Noe ha p dk/) and he soluion is herefore: Ce -k (since he soluion of an eac differenial equaion is, f consan). 54

55 6. Eample d 0 (here, p ) h e k where k p In general, he paren funcion f is: Therefore: f e k C e f C And he soluion is: C e To summarize, in general, he differenial equaion is simplified o he form: k d ( e ) 0 and herefore he soluion is: e k C 55

56 6.4 Non-homogenous form d p r where r r () Muliply hroughou by he inegraing facor: k e d Therefore: p e k r e k d ( e k ) r e k Inegrae wih respec o : k k e r e C Soluion: k k e r e C 6.5 Eample: d (here p ) Inegraing facor: h e k where, k p d Thus: d (e k ) e k Inegraing wih respec o gives: e k e C e C Finally, e e C C e 56

57 Lecure 7: Differenial Equaions - IV Summary: - Euler's mehod: - he direcion field - approimaing he slope over a finie disance - he incremenal process (over large disance) - divergence and convergence - error in Euler's mehod - eample 7. Inroducion Figure 7. Direcion field Given a saring poin ( o, o ) and ODE d f (, ) our ask is o deermine () for any. 7. Euler s mehod Figure 7. Numerical soluion (one sep) d f (, ) h Le be an approimaion o he rue answer a he end of he firs incremen. (Noe: does no mean he derivaive of ) o o h f (, ) o If h 0 57

58 Thus, given a sar poin ( o, o ) we can esimae (a o h) provided ha h is small. o h f (, ) o where f(, ) o is he gradien a ( o, o ) If h is large, hen i is necessary o proceed in series of small seps (incremens). Figure 7. Incremenal soluion We normally choose (h) o be he same for all incremens (his is convenien, bu no necessary). For a series of incremens, he mehod can be described by hese equaions: Le: o h, h,..., n n h A he sar: Thereafer: o o ( o, o ) (, ) o h f h f h f, n n ( ) This mehod of soluion is called Euler's mehod. n n 58

59 7. Convergence of Euler s mehod This numerical soluion ends o drif (diverge) from he rue soluion. Figure 7.4 Convergence of Euler s mehod In general, beer soluions are obained as h decreases, bu his means more incremens are needed. The error in Euler s mehod can be esimaed from he Taylor series: h ( h ) ( ) h ( ) " ( )! where means d/, ec. The Euler mehod neglecs all erms of 0(h ) and higher. Bu he error is no proporional o h, because reducing h increases he number of seps. In fac, he error is error is proporional o h, which means ha reducing he sep size by half (say) reduces error by half. 7.4 Eample Given: d find (), given (). Take h 0.5 Euler s mehod: h f (, ) n n n n n n n n n n If h 0.5 (4 incremens), we obain.9 (check!). We epec ha he error will be halved (approimaely) in his second soluion. Thus (calling he rue soluion X, and he error ε): X.7 ε X.9 ε/ Hence: X. 59

60 Lecure 8: A Review of he Course REVIEW:. Funcions of variables: Conour plos. Parial differeniaion: f. Chain rule: f f f y y 4. Successive differeniaion: f, y f 5. Toal differenials: f d f dy df y 6. Eac differenials: p d q dy 7. Line inegrals: I B A f (, y) d 8. Line inegrals: I B [ f d A g dy ] I B A f (, y) 9. Line inegrals: I B A f (, y) ds 0. Double inegrals: I f (, y) dy d. Parameric represenaions: N N 60

61 . Parameric represenaions: N N N. Numerical inegraion: f () w i Differenial Equaions: 4. Classificaion: Direcion fields 5. Separable: d f () g () "Eac": h d g 0 6. Linear (s order): "inegraing facor" d p () r () h e p 7. Euler's mehod: d f (, ) n n h f (, ) n 6

62 Applicable Mahemaics A Tuorials Quesions & Answers Revised, Sepember 009 6

63 Applicable Mahemaics A Eamples. Level Curves. Skech he level curves of he following funcions: a) f(,y) 4y b) f (, y) n( y ) c) f (, y) an y Answers: skeches no available 6

64 Applicable Mahemaics A Eamples. Parial differenials Q. Find f / and f / y when: a) y f (, y) e cos b) y f (, y) y 6 Q. Find z / and z / y when z(, y) saisfies: a) y z 0 b) yz y z Q. Find f, fy and fz when: z f (, y, z) e cosy Q4. Show ha: saisfies: Q5. Show ha: saisfies: / f (, y, z) ( y z ) f yfy zfz f (, y, z) f (, y y, z ) y z f fy fz Answers (parial): f Q. (a) e y f y y 6 cos (b) y ( y 6) z Q. (a) z z (b) y (y ) z yz y Q. f e z sin y y f z e cos y z Q4 & 5. Proof. 64

65 Applicable Mahemaics A Eamples. Parial differenials chain rule Find z / when: z y, and y. Find f / s and f / when: f (, y) e cosy, s, y s n. Show ha if z f ( u) where u y /, hen: z y z nz y 4 4 Verify his resul for: z y y 4. Show ha, if f is a funcion of he independen variables and y, and he laer are changed o independen variables u and v, where u e y / and v y, hen: a) f y f v f (b) y v f y f uv f y u 5. Show ha he oal surface area, S, of a cone of base radius, r, and perpendicular heigh, h, is given by: ( r h ) S πr πr If r and h are each increasing a he rae of 0.5 cm/s, find he rae a which S is increasing a he insan when r and h 4. Answers:. z ( ) / 4 f. s e (scos y sin y) f e ( cos y s sin y ). Proof. 4. Proof 5. S/. 8π 65

66 Applicable Mahemaics A Eamples 4. Parial differenials successive differeniaion Verify ha: f (, y) saisfies he equaion: y f f y 0 Verify ha: f f for he cases: y y a) f (, y) cos y b) f (, y) sinh cosy ( y ) Show ha: V(, y, z) ep z 4z saisfies he differenial equaion: V V V y z Answers: Q-. Proofs are required in all cases. 66

67 Applicable Mahemaics A Eamples 5. Parial differenials oal differenials. The funcion z is defined by, z(, y) y y Find z and dz when 4, y, 0. 0and y A bo has inernal dimensions m.5m 0.75m. I is made of shee meal 4mm hick. Find he acual volume of he meal used and compare i wih he approimae volume found using he differenial of he capaciy of he bo. r. The acceleraion, f, of a pison is given by: f rω cos θ cos θ l When θ π/6 radians and r/l 0.5, calculae he approimae percenage error in he calculaed value of he acceleraion if measured values of boh r and ω are % oo small. 4. In a coal-processing plan he flow, V, of slurry along a pipe is given by, pr V π 4 8 ηl If r and l boh increase by 5% and p and η decrease by 0% and 0% respecively, find he approimae percenage change in V. Answers:. z , dz 0.0. V m, dv m. df / f V / V 5% 67

68 Applicable Mahemaics A Eamples 6. Parial differenials eac differenials. Deermine which of he following are eac differenials of a funcion, and find, where appropriae, he corresponding funcion: a) ( y y ) d ( y ) dy b) (y y cos)d ( y sin ) dy c) ( z y) d ( y ) dy ( z ) dz Hin. In par (c), he differenial is hree-dimensional: Pd Qdy Rdz Generalising he resul for wo dimensions, i will be an eac differenial if Py Q AND Pz R AND Qz Ry (where Py means P/y ec.) The oher sligh complicaion is ha when we wan o find he funcion f, hen, when we inegrae wr o (say), hen he consan erm mus now be a funcion of y and z (i.e., g(y,z)), ec. Show ha he differenial: Answers: g(, y) is no eac. (0 6y 6y )d (9 4y 5y However, if we muliply g(,y) by ( y), show ha he new differenial is an eac differenial. Find he corresponding funcion. Q. (a) Eac. f (, y) y y C (b) Eac. f (, y) y y sin C (c) Eac. f z y 4y C )dy Q. f 8 6 y 6 y 6y 54y 7y C 68

69 Applicable Mahemaics A Eamples 7. Line inegrals (d, ds). Evaluae he line inegral: y d c along he parabola (y 4 ) from A(, 4) o B(, 6).. Evaluae he line inegral: ( y) dy c along he line beween A(,) o B (4,7).. Evaluae he line inegral: ds along he curve (y ) from A(0, 0) o B(, ). c Hin: I a d a 4 [ ( a ) a a Sinh ] 8a a 4. Evaluae he line inegral: ds along he line beween A(,) o B (4,7). c Now calculae he disance (using Pyhagoras) beween A & B. Eplain he resul. Answers: Q. 60 Q. Q. 4. Q

70 Applicable Mahemaics A Eamples 8. Line inegrals (ddy,, ds()). Evaluae he line inegral: (y d dy) along he arc C of he circle c ( y ) in he firs quadran, from A(,0) o B(0,). Eplain your resul. (,). Show ha he line inegral: ( y d y dy) is independen of he pah (0,) from (0,) o (,). Evaluae he inegral using wo differen pahs.. Evaluae he line inegral: ( y d y dy) around he closed conour formed by a recangle cenred on he origin and wih side lenghs of si (6) and four (4) along he & y aes, respecively. Eplain your resul. 4. Evaluae he line inegral: y along he pah defined by he pair of c parameric equaions: ( ; y ) beween A(, 0) and B(, ). 5. Evaluae he line inegral: y ds along he pah defined by he pair of c parameric equaions: ( ; y ) beween A(, 0) and B(, ). Answers: Q. Zero Q. 4 Q. Zero Q Q

71 Applicable Mahemaics A Eamples 9. Double inegrals. Evaluae he double inegral: I y ( 0 y)dyd. Evaluae he double inegral: I ( / y)ddy over he recangle bounded by he lines 0,, y and y. Check your answer by inegraing in he reverse order.. Skech he domain of inegraion and evaluae: I dy d ( y ) 4. Calculae he volume of an ocan (an eighh par) of a sphere of radius hree (). Check ha your resul agrees wih he heoreical soluion. (The volume of a sphere is 4 π r /) Hin: The equaion of a sphere cenred a he origin is, z y r. Inegrae over he y plane (in he posiive quadran) from he origin o he ouer limi - he circle, y r. Answers: Q. 4 Q. (8/)log e Q. 0. Q4. Proof required 7

72 Applicable Mahemaics A Eamples 0. Inerpolaion. Using linear shape funcions, inerpolae beween A(,0) and B(5, 70), o find D(, y). (i.e., find f(), given f() 0, f(5) 70).. The linear shape funcions N() are defined such ha he firs poin A corresponds o -, while he second poin B corresponds o. (i.e., (A) -, (B). Derive he linear shape funcions N(r) which we would obain if we choose o define a new parameer r where, r(a) 0, r(b).. Using quadraic shape funcions, inerpolae beween A(, 0), B(.5, 60) and C(5, 70) o find D(, y). (i.e., find f(), given f() 0, f(.5) 60, f(5) 70). 4. The quadraic shape funcions N() are defined such ha he firs poin A corresponds o -, while he hird poin C corresponds o. (i.e., (A) -, (C). Derive he quadraic shape funcions N(r) which we would obain if we choose o define a new parameer r where, r(a) 0, r(c). Answers: Q. 0. Q. N r, N r Q Q4. N ( r)( r), N 4r( r ), N r(r ) 7

73 Applicable Mahemaics A Eamples. Parameric Represenaions Q. The (, y) coordinaes of hree poins on a curved surface are measured o be: (7, 8), (, ) and (7, 8). Plo hese poins. (a) Using he quadraic shape funcions, [N (-)/, N -, N ()/) ], deermine he parameric forms, f() and yg() for his curve. (b) esimae wha he (, y) coordinaes are a he poin (on he curve) mid-way beween he las wo poins. (c) If he curve is, in fac, a circle, find is cenre and calculae is radius using he measured coordinaes - and your knowledge of geomery. (d) Using he same cenre, wha is he disance from he cenre o he (, y) coordinaes you calculaed in par (a)? (e) Hence, wha is he percenage error in radius (a his poin) using he parameric represenaion of he geomery? Does his surprise you? Q. Evaluae he line inegral: y ds along he conour defined by he parameric c equaions: 5, y 5, from A(7, 8) o B(7, 8), where hese limis are (, y) coordinaes. Hin: This inegral is difficul o inegrae analyically, so he final resul migh be obained graphically, or by using Simpson s rule, or some oher mehod. Answers: Q. (a) 5, y 5, (b) 4.5,.75, (c), 8, r5, (d) 4.5, (e) 0%, No - oo large an arc. Q. 94, appro. 7

74 Applicable Mahemaics A Eamples. Differenial equaions. Classify hese equaions (order, lineariy, homogeneiy (if linear), parial/ordinary) and name he dependen and independen variables: f f f h (a) y sin y 0 (b) h 0 y y (c) d s ds (sin ) ( cos ) s e (d) v uv u v (e) dr z 0 dz d (f) f () (g) d d y f () g() (h) dy. Plo he direcion field for he ODE [(d/) - ] a a grid spacing of / for values of in he range 0 4 and in he range 0 4. Find he general soluion of his equaion. If (0), plo he paricular soluion (equaion) passing hrough his poin. Check ha his curve is consisen wih he direcion field.. By inspecion (rial and error), find he general soluion of he ODEs: d 4 d d 4 4. Find he general soluion of he differenial equaions: (a) d d ( sin ) co (b) d (d) d 4 0 (c) ( ) 0 5. Calculae he value () using Euler s mehod (using sep sizes of 0., 0. & 0.05) for he following iniial-value problem. Hence, esimae he eac resul. d 4 given ha (0) Answers: Q. (a), p, nl (b), p, l, h (c), o, l, nh (d), p, l, nh (e), o, l, h (f), o, l, h (g), o, l, nh (h), o, nl Q.. Q. C 4 /, Ce 4, 5 /0 / C C e ) Q4. cos - (e ( - cos ) ) (log C { ± 4( C)}/ {( )}/ 8 Ce Q5. (.87,.7,.66,.60 ) 74