C. J. PAPACHRISTOU ASPECTS OF

Transcription

1 C. J. PAPACHRISTOU ASPECTS OF INTEGRABILITY OF DIFFERENTIAL SYSTEMS AND FIELDS HELLENIC NAVAL ACADEMY

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3 Aspects of INTEGRABILITY of Dfferental Systems and Felds Costas J. Papachrstou Department of Physcal Scences Hellenc Naval Academy Hellenc Naval Academy 15

4 Costas J. Papachrstou, 15

5 PREFACE Ths monograph, wrtten at an ntermedate level for educatonal purposes, serves as an ntroducton to the concept of ntegrablty as t apples to systems of dfferental equatons (both ordnary and partal) as well as to vector-valued felds. We stress from the outset that ths s not a treatse on the theory or the methods of soluton of dfferental equatons! Instead, we have chosen to focus on specfc aspects of ntegrablty that are often encountered n a varety of problems n Appled Mathematcs, Physcs and Engneerng. The followng general cases of ntegrablty are examned: (a) Path-ndependence of lne ntegrals of vector felds on the plane and n space; (b) ntegraton of a system of ordnary dfferental equatons (ODEs) by usng frst ntegrals; and (c) ntegrable systems of partal dfferental equatons (PDEs). Specal topcs nclude the ntegraton of analytc functons and some elements from the geometrc theory of dfferental systems. Certan more advanced subjects, such as Lax pars and Bäcklund transformatons, are also dscussed. The presentaton sacrfces mathematcal rgor n favor of smplcty, as dctated by pedagogcal logc. For a deeper study of the subject, the reader s referred to the lterature cted at the end. A vector feld s sad to be ntegrable n a regon of space f ts lne ntegral s ndependent of the path connectng any two ponts n ths regon. As wll be seen n Chapter 1, ths type of ntegrablty s related to the ntegrablty of an assocated system of PDEs. Smlar remarks apply to the case of analytc functons on the complex plane, examned n Chapter. In ths case the ntegrable system of PDEs s represented by the famlar Cauchy-Remann relatons. In Chapter 3 we ntroduce the concept of a frst ntegral of an ODE and we demonstrate how ths quantty can be used to fnd the soluton of ths equaton. As a characterstc example, the prncple of conservaton of mechancal energy s used to ntegrate the ODE expressng Newton s second law of moton n one dmenson. Ths dscusson s generalzed n Chapter 4 for systems of frst-order ODEs, where the soluton to the problem s agan sought by usng frst ntegrals. The method fnds an mportant applcaton n frst-order PDEs, the soluton process of whch s brefly descrbed. Fnally, we study the case of a lnear system of ODEs, the soluton of whch reduces to an egenvalue problem. Chapter 5 examnes systems of ODEs from the geometrc pont of vew. Concepts of Dfferental Geometry such as the ntegral curves and the phase curves of a system, the dfferentaloperator representaton of a vector feld, the Le dervatve of a functon, etc., are ntroduced n smple terms. The geometrc sgnfcance of frst-order PDEs s also studed, revealng a close connecton of these equatons wth systems of ODEs and vector felds. Two notons famlar from the theory of ntegrable nonlnear PDEs are Bäcklund transformatons and Lax pars. In both cases a PDE s expressed as an ntegrablty conon for soluton of an assocated system of PDEs. These deas are brefly dscussed n Chapter 6. In the fnal secton, a famlar system of PDEs n four dmensons, namely, the Maxwell equatons for the electromagnetc feld, s shown to consttute a Bäcklund transformaton connectng solutons of the wave equatons satsfed by the electrc and the magnetc feld. The soluton of the Maxwell system for the case of a monochromatc plane electromagnetc wave s derved n detal.

6 I would lke to thank my colleague and frend Arstds N. Magoulas for an excellent job n drawng a number of fgures, as well as for several frutful dscussons on Electromagnetsm! Praeus, Greece November 15 Costas J. Papachrstou

7 CONTENTS CHAPTER 1 INTEGRABILITY ON THE PLANE AND IN SPACE Smply and Multply Connected Domans 1 1. Exact Dfferentals and Integrablty 1.3 Lne Integrals and Path Independence Potental Vector Felds Conservatve Force Felds 1 CHAPTER INTEGRABILITY ON THE COMPLEX PLANE 13.1 Analytc Functons 13. Integrals of Complex Functons 15.3 Some Basc Theorems 16.4 Antdervatve and Indefnte Integral of an Analytc Functon 1 CHAPTER 3 ORDINARY DIFFERENTIAL EQUATIONS The Concept of the Frst Integral 3 3. Exact Equatons Integratng Factor Hgher-Order Dfferental Equatons Applcaton: Newton s Second Law n One Dmenson 6 CHAPTER 4 SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS Soluton by Seekng Frst Integrals 9 4. Applcaton to Frst-Order Partal Dfferental Equatons System of Lnear Equatons 37 CHAPTER 5 DIFFERENTIAL SYSTEMS: GEOMETRIC VIEWPOINT Dynamcal Systems Geometrc Sgnfcance of the Frst Integral Vector Felds Dfferental Operators and Le Dervatve Exponental Soluton of an Autonomous System Vector Felds as Generators of Transformatons Geometrc Sgnfcance of Frst-Order PDEs 56

8 CHAPTER 6 INTEGRABLE SYSTEMS OF PARTIAL DIFFERENTIAL EQUATIONS Notaton Bäcklund Transformatons Lax Par for a Nonlnear PDE The Maxwell Equatons as a Bäcklund Transformaton 63 APPENDIX: MATRIX DIFFERENTIAL RELATIONS 67 BIBLIOGRAPHY 7 INDEX 71 v

9 CHAPTER 1 INTEGRABILITY ON THE PLANE AND IN SPACE 1.1 Smply and Multply Connected Domans We begn wth a few basc concepts from Topology that wll be needed n the sequel: A doman D on the plane s sad to be smply connected f, for every closed curve C wthn ths doman, every pont of the plane n the nteror of C s also a pont of D. Alternatvely, the doman D s smply connected f every closed curve n D can be shrunk to a pont wthout ever leavng ths doman. If ths conon s not fulflled, the doman s called multply connected. ( α ) ( β ) ( γ ) In the above fgure, the regon (α) s smply connected, the regon (β) s doubly connected whle the regon (γ) s trply connected. Notce that there are two knds of closed curves n regon (β): those that do not encrcle the hole and those that encrcle t. (We note that the hole could even consst of a sngle pont subtracted from the plane.) By a smlar reasonng, the trple connectedness of regon (γ) s due to the fact that there are three knds of closed curves n ths regon: those that do not encrcle any hole, those that encrcle only one hole (no matter whch one!) and those that encrcle two holes. A doman Ω n space s smply connected f, for every closed curve C nsde Ω, there s always an open surface bounded by C and located entrely n Ω. Ths means that every closed curve n Ω can be shrunk to a pont wthout ever leavng the doman. If ths s not the case, the doman s multply connected. Examples: 1. The nteror, the exteror as well as the surface of a sphere are smply connected domans n space. The same s true for a sphercal shell (the space between two concentrc sphercal surfaces).. The space n the nteror of a torus s doubly connected (explan why!). 1

10 CHAPTER 1 1. Exact Dfferentals and Integrablty We consder the plane R wth coordnates (x,y). Let D R be a doman on the plane and let P(x,y) and Q(x,y) be functons dfferentable at every pont of D. The expresson P(x, y) dx + Q(x, y) dy s an exact (or total, or perfect) dfferental f there exsts a functon u(x,y), dfferentable everywhere n D, such that Necessary conon for the exstence of u(x,y): In general, du = P(x, y) dx + Q(x, y) dy (1) u u du= dx+ dy x y () By comparng (1) and () and by takng nto account that the dfferentals dx and dy are ndependent of each other, we fnd the followng system of partal dfferental equatons (PDEs): u x u = P( x, y), = Q( x, y) y (3) In order for the system (3) to have a soluton for u (that s, to be ntegrable), ts two equatons must be compatble wth each other. The compatblty conon or ntegrablty conon of the system s found as follows: We dfferentate the frst equaton wth respect to y and the second one wth respect to x. By equatng the mxed dervatves of u wth respect to x and y, we fnd the PDE P y = Q x (4) If conon (4) s not satsfed, the system (3) [or, equvalently, the dfferental relaton (1)] does not have a soluton for u and the expresson Pdx+Qdy s not an exact dfferental. Example: ydx xdy du, snce P=y, Q= x, and P/ y=1 whle Q/ x= 1. Note: Conon (4) s necessary for the exstence of a soluton u of the system (3) or, equvalently, of the dfferental relaton (1). Ths conon wll also be suffcent f the doman D R s smply connected (by assumpton, ths s a doman where the functons P and Q are dfferentable). Examples: 1. We consder the dfferental relaton du=ydx+xdy. We have P=y, Q=x, so that P/ y= Q/ x=1. Moreover, the functons P and Q are dfferentable everywhere on the plane R, whch s a smply connected space. Thus, the conons for exstence of u are fulflled.

11 INTEGRABILITY ON THE PLANE AND IN SPACE Relatons (3) are wrtten { u/ x=y, u/ y=x}. The frst one yelds u=xy+c(y), where C s an arbtrary functon of y. Substtutng ths nto the second relaton, we fnd C (y)= C=constant. Thus, fnally, u(x,y)=xy+c.. We consder the relaton du=(x+e y )dx+(xe y y)dy. The functons P=x+e y and Q=xe y y are dfferentable on the entre plane R, whch s a smply connected space. Furthermore, P/ y= Q/ x=e y. Relatons (3) are wrtten { u/ x=x+e y, u/ y= xe y y}. By the frst one we get u=(x /)+xe y +φ(y) (wth arbtrary φ). Then, the second relaton yelds φ (y)= y φ(y)= y +C. Thus, fnally, u(x,y)=(x /)+xe y y +C. Consder, now, a doman Ω R 3 n a space wth coordnates (x,y,z). Also, consder the functons P(x,y,z), Q(x,y,z) and R(x,y,z), dfferentable at each pont (x,y,z) of Ω. The expresson P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz s an exact dfferental f there exsts a functon u(x,y,z), dfferentable n Ω, such that du = P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz (5) Equvalently, snce du=( u/ x)dx+( u/ y)dy+( u/ z)dz, the functon u wll be a soluton of the system of PDEs, u u u = P( x, y, z), = Q( x, y, z), = R( x, y, z) x y z (6) The ntegrablty (compatblty) conons of the system (necessary conons for exstence of soluton for u) are P Q P R Q R =, =, = y x z x z y (7) Conons (7) are also suffcent for soluton f the doman Ω, wthn whch the functons P, Q, R are dfferentable, s smply connected. Example: Consder the dfferental relaton du=(x+y+z)(dx+dy+dz), wth P=Q=R=x+y+z. We notce that relatons (7) are satsfed, as well as that the functons P, Q, R are dfferentable n the entre R 3, whch s a smply connected space. Thus, the gven dfferental relaton admts a soluton for u. Relatons (6) are wrtten: { u/ x=x+y+z, u/ y=x+y+z, u/ z=x+y+z}. The frst one yelds u=(x /)+xy+xz+φ(y,z) (wth arbtrary φ). Substtutng ths nto the second relaton, we fnd φ/ y=y+z φ(y,z)=(y /)+yz+ψ(z) (arbtrary ψ). Makng the necessary replacements nto the thrd relaton, we have: ψ (z)=z ψ(z)=(z /)+C. Fnally, u=(x +y +z )/ +xy+xz+yz+c. 3

12 CHAPTER Lne Integrals and Path Independence Consder the plane R wth coordnates (x,y). Let L be an orented curve (path) on the plane, wth ntal pont A and fnal pont B. y L B A x The curve L may be descrbed by parametrc equatons of the form { x=x(t), y=y(t) } (1) Elmnatng t between these equatons, we get a relaton of the form F(x,y)= whch, n certan cases, may be wrtten n the form of a functon y=y(x). Example: Consder the parametrc curve { x = R cost, y = R snt }, t π. The orentaton of the curve depends on whether t ncreases ( counterclockwse ) or decreases ( clockwse ) between and π. y R t R x By elmnatng t, we get x +y R = y = (R x ) 1/. Gven a plane curve L from A to B, we now consder a lne ntegral of the form IL = P( x, y) dx+ Q( x, y) dy () L In the parametrc form (1) of L, we have: dx=(dx/)=x (t), dy=y (t), so that t { [ ( ), ( )] ( ) [ ( ), ( )] ( )} B IL = P x t y t x t + Q x t y t y t ta In the form y=y(x) of L, we wrte dy=y (x)dx and (3) 4

13 INTEGRABILITY ON THE PLANE AND IN SPACE x { [, ( )] [, ( )] ( )} (4) B IL = P x y x + Q x y x y x dx xa In general, the value of the ntegral I L depends on the path L connectng A and B. For every path L: A B, we can defne the path L: B A wth opposte orentaton. By (3) t tb ta follows that, f IL = ( ), then I L = ( ). Thus, t A t B I L = I L (5) If the end ponts A and B of a path concde, then we have a closed curve C and, correspondngly, a closed lne ntegral I C, for whch we use the symbol. We then have: C ( ) = ( ) (6) C C where the orentaton of C s opposte to that of C (e.g., f C s counterclockwse on the plane, then C s clockwse). Example: The parametrc curve represents a crcle on the plane. { x = R cost, y = R snt }, t π y R t x If the counterclockwse orentaton of the crcle (where t ncreases from to π) corresponds to the curve C, then the clockwse orentaton (wth t decreasng from π to ) corresponds to the curve C. Proposton: If Pdx+ Qdy= for every closed curve C, then the lne ntegral C Pdx+ Qdy L s ndependent of the path L connectng any two ponts A and B. The converse s also true. Proof: We consder any two ponts A and B on the plane, as well as two dfferent paths L 1 and L connectng these ponts (there s an nfnte number of such paths). 5

14 CHAPTER 1 B L 1 L A We form the closed path C=L 1 +( L ) from A to B through L 1 and back agan to A through L. We then have: Pdx+ Qdy= Pdx+ Qdy + Pdx+ Qdy= C L L 1 Pdx+ Qdy Pdx+ Qdy= Pdx+ Qdy = Pdx+ Qdy. L L L L 1 1 Theorem 1: Consder two functons P(x,y) and Q(x,y), dfferentable n a smply connected doman D of the plane. Then, the followng 4 conons are equvalent to one another (f any one s true, then the rest are true as well): (a) Pdx+ Qdy=, for any closed curve C wthn D. C (b) The ntegral A and B of D. Pdx+ Qdy s ndependent of the curved path L connectng two fxed ponts L (c) The expresson Pdx+Qdy s an exact dfferental. That s, there exsts a functon u(x,y) such that du = Pdx + Qdy u/ x = P, u/ y = Q. (d) At every pont of D, P y = Q x. Comment: In the case where the doman D s not smply connected, conon (d) does not guarantee the valy of the remanng three conons. However, conons (a), (b), (c) are stll equvalent to one another and each of them separately guarantees (d). Note that (d) s the ntegrablty conon for the valy of (c). (Remember that the former conon s necessary but not suffcent n the case where the doman D, n whch the functons P and Q are dfferentable, s not smply connected.) Example: Consder the dfferental expresson ydx xdy ω= + x + y. Here, P= y/(x +y ), Q= x/(x +y ), and conon (d) s satsfed (show ths!). We notce that the functons P and Q are dfferentable everywhere on the plane except at the orgn O of our 6

15 INTEGRABILITY ON THE PLANE AND IN SPACE coordnate system, at whch pont (x,y) (,). Now, every doman D of the plane not contanng pont Ο (left dagram n the fgure below) s smply connected (explan why!). A closed curve C wthn D wll then not contan pont Ο n ts nteror. For such a curve, ω= and ω=du, C wth u(x, y)=arctan(y/x). A curve C, however, contanng Ο (rght dagram) cannot belong to a smply connected doman (why?). For such a curve, ω. C y D y C O x O x Let us now consder lne ntegrals n space. Let L: {x=x(t), y=y(t), z=z(t)} be a parametrc curve from pont A of R 3 to pont B. Let P(x,y,z), Q(x,y,z), R(x,y,z) be functons dfferentable n the doman Ω R 3 n whch the curve L s located. We consder the lne ntegral or, n parametrc form, t IL = P( x, y, z) dx+ Q( x, y, z) dy+ R( x, y, z) dz (7) L { [ ( ), ( ), ( )] ( ) [ ( ), ( ), ( )] ( ) [ ( ), ( ), ( )] ( )} (8) B IL = P x t y t z t x t + Q x t y t z t y t + R x t y t z t z t ta Theorem : If the doman Ω s smply connected, then the followng 4 conons are equvalent to one another: (a) Pdx + Qdy + Rdz =, for any closed path C wthn Ω. C (b) The ntegral Pdx + Qdy + Rdz s ndependent of the curved path L connectng two fxed L ponts A and B of Ω. (c) The expresson Pdx+Qdy+Rdz s an exact dfferental. That s, there exsts a functon u(x,y,z) such that (d) At every pont of Ω, du = Pdx + Qdy + Rdz u/ x = P, u/ y = Q, u/ z = R. P Q P R Q R =, =, = y x z x z y. Comment: If the doman Ω s not smply connected, conon (d) does not guarantee the valy of the remanng three conons. However, conons (a), (b), (c) are stll equvalent to one another and each of them separately guarantees (d). Note that (d) s the ntegrablty 7

16 CHAPTER 1 conon for the valy of (c). (As we have sad, the former conon s necessary but not suffcent n the case where the doman Ω, n whch the functons P, Q and R are dfferentable, s not smply connected.) From (c) t follows that, for any open curve L lmted by two fxed ponts A and B, B B Pdx+ Qdy+ Rdz = du = u( B) u( A) u( xb, yb, zb ) u( xa, ya, za) (9) A Note, n partcular, that ths automatcally verfes (b). A 1.4 Potental Vector Felds Consder a vector feld n a doman Ω R 3 : A( r ) = P( x, y, z) uˆ + Q( x, y, z) uˆ + R( x, y, z) uˆ x y z (1) where r s the poston vector of a pont (x,y,z) of the doman, and where u ˆx, u ˆy, u ˆz are the unt vectors on the axes x, y, z, respectvely. The functons P, Q, R are assumed to be dfferentable n the doman Ω. We wrte: A ( P, Q, R), r ( x, y, z). We say that the feld (1) s potental f there exsts a functon u(x, y, z) such that A= u () or, n components, u u u P=, Q=, R= x y z (3) The functon u s called potental functon or smply potental of the feld A. If () s vald, then A = u = (4) That s, a potental vector feld s necessarly rrotatonal. In component form, Eq. (4) s wrtten P Q P R Q R =, =, = y x z x z y (5) Relatons (5) are the ntegrablty conons for exstence of a soluton u of the system (3), thus also of the vector relaton (). 8

17 INTEGRABILITY ON THE PLANE AND IN SPACE Conon (4) s necessary n order for the feld A to be potental. Is t suffcent also? That s, s an rrotatonal feld always potental? Proposton: An rrotatonal feld A ( P, Q, R) n a smply connected doman Ω s potental. Proof: By assumpton, the system of PDEs (5) s satsfed at every pont of a smply connected doman. Hence, accordng to the dscusson n prevous sectons, the expresson Pdx+Qdy+Rdz s an exact dfferental. That s, there exsts a functon u(x, y, z) such that Pdx + Qdy + Rdz = du (6) Takng nto account the ndependence of the dfferentals dx, dy, dz, we are thus led to the system (3), thus to the vector equaton (). Theorem of Sec. 1.3 can be re-expressed n the language of vector felds as follows: Theorem: Consder a vector feld A ( P, Q, R), where the functons P, Q, R are dfferentable n a smply connected doman Ω R 3. Let L be an open curve and let C be a closed curve, both lyng n Ω. Then, the followng 4 conons are equvalent to one another: (a) A d r Pdx+ Qdy+ Rdz=. C (b) The ntegral any two fxed ponts n Ω. C L A d r Pdx+ Qdy+ Rdz L s ndependent of the curved path L connectng (c) There exsts a functon u(x, y, z) such that, at every pont of Ω, A= u. (d) At every pont of Ω, A= Comments: (.e., the feld A s rrotatonal). 1. From (c) we have: A d r = u d r = du. Thus, f L s a curved path wth lmt ponts b and c, L c A d r = du = u ( c) u ( b) b (ndependent of the path b c).. Assume that the doman Ω n whch conon (d) s vald s smply connected. Then, for every closed curve C n Ω, there s an open surface S bounded by C. da d r C S da 9

18 CHAPTER 1 By Stokes theorem we then have A d r = ( A) da =. C S 3. If the doman Ω s not smply connected, conon (d) does not guarantee the valy of the remanng three conons. However, the frst three conons are stll equvalent to one another and each of them separately guarantees (d). 1.5 Conservatve Force Felds In Physcs, a statc (tme-ndependent) force feld F( r) s called conservatve f ts work W AB on a test partcle movng from pont Α to pont Β s ndependent of the path connectng these ponts. Equvalently, the work on the partcle along a closed path C s zero: W AB B = F d r A s ndependent of the path A B F d r= (1) C Let S be an open surface nsde the feld, bounded by the closed curve C (cf. fgure n Sec. 1.4). By Stokes theorem, relaton (1) yelds F d r= ( F) da=. C S In order for ths to be vald for every open surface bounded by C, we must have F = () That s, a conservatve force feld s rrotatonal. (The valy of the converse requres that the doman of space n whch the feld s defned be smply connected.) From (1) t also follows that, accordng to the Theorem of Sec. 1.4, there exsts a functon such that F( r) s the grad of ths functon. We wrte F = U The functon U ( r ) = U( x, y, z) s called the potental energy of the test partcle at the pont r ( x, y, z) of the feld. [The negatve sgn n (3) s only a matter of conventon and has no specal physcal meanng. One may elmnate t by puttng U n place of U. Note also that U s arbtrary to wthn an adve constant, gven that U and (U+c) correspond to the same force F n (3).] The work W AB s wrtten (3) B B B W = F d r = ( U ) d r = du AB A A A W = U ( r ) U ( r ) U U AB A B A B (4) 1

19 INTEGRABILITY ON THE PLANE AND IN SPACE Now, by the work-energy theorem, W AB = E k,b E k,a (5) where E k =mv / s the knetc energy of the partcle (m and v are the partcle s mass and speed, respectvely). By combnng (4) wth (5), we have: E k,a +U A = E k,b +U B (6) The sum E=E k +U represents the total mechancal energy of the partcle. Relaton (6), then, expresses the prncple of conservaton of mechancal energy; namely, the total mechancal energy of a partcle movng nsde a conservatve force feld assumes a constant value durng the moton of the partcle. Example: We consder the electrostatc Coulomb feld due to a pont charge Q located at the orgn O of our coordnate system. Let q be a test charge at a feld pont wth poston vector r = r rˆ, where r s the dstance of q from O and where ˆr s the unt vector n the drecton of r [n ths problem t s convenent to use sphercal coordnates (r, θ, φ)]. a r a q r dr ˆr O r b b The force exerted on q by the feld, at a momentary poston r of the charge, s F = kqq rˆ r (where k s a constant that depends on the system of unts). As can be shown, F =. That s, the force feld F s rrotatonal. Ths feld s defned n a smply connected doman of space (the whole space wth the excepton of the sngle pont O where the charge Q producng the electrostatc feld s located). Hence, the consdered rrotatonal force feld wll also be conservatve. Indeed, puttng F(r)=kqQ/r, we wrte F( r) F( r) F = F( r) rˆ = r F d r = r d r r r. But, 1 1 r d r = d( r r ) = d( r ) = r d r, 11

20 CHAPTER 1 so that F d r = F( r) d r. Thus, the work produced durng the moton of q from a feld pont a to a feld pont b s b b 1 1 Wab F d r F( r) d r kqq = = a = a r a r b. Ths expresson allows us to defne the potental energy U(r) of q at a gven feld pont by usng (4): W ab =U a U b. As s easy to see, We notce that kqq U ( r) = ( + const.). r U kqq U = rˆ = rˆ = F. r r The total mechancal energy of the charge q remans fxed durng the moton of the charge nsde the feld and s equal to E = E k +U(r) = mv / + kqq/r = const. 1

21 CHAPTER INTEGRABILITY ON THE COMPLEX PLANE.1 Analytc Functons We consder complex functons of the form w = f (z) = u (x, y) + v (x, y) (1) where z=x+y (x, y) s a pont on the complex plane. Let z be a change of z and let w=f (z+ z) f (z) be the correspondng change of the value of f (z). We say that the functon (1) s dfferentable at pont z f we can wrte w = f ( z ) + ε ( z, z ) wth lm ε ( z, z ) = z z () Then, the functon w f ( z + z) f ( z) f ( z) = lm = lm z z z z (3) s the dervatve of f (z) at pont z. Evdently, n order for f (z) to be dfferentable at z, ths functon must be defned at that pont. We also note that a functon dfferentable at a pont z s necessarly contnuous at z (the converse s not always true). That s, lm f ( z) = f ( z ) (assumng that the lmt exsts). A functon f (z) dfferentable at every pont of a doman G of the complex plane s sad to be analytc (or holomorphc) n the doman G. The crteron for analytcty s the valy of a par of partal dfferental equatons (PDEs) called the Cauchy-Remann relatons. Theorem: Consder a complex functon f (z) of the form (1), contnuous at every pont z (x, y) of a doman G of the complex plane. The real functons u(x,y) and v(x,y) are dfferentable at every pont of G and, moreover, ther partal dervatves wth respect to x and y are contnuous functons n G. Then, the functon f (z) s analytc n the doman G f and only f the followng system of PDEs s satsfed: z z u v u v =, = x y y x (4) It s convenent to use the followng notaton for partal dervatves: φ φ φ φ φ φ, φ, φ, φ, φ x y x y x y x y xx yy xy, etc. 13

22 CHAPTER Relatons (4) then read u x = v y, u y = v x (4 ) The dervatve of the functon (1) may now be expressed n the followng alternate forms: f (z) = u x +v x = v y u y = u x u y = v y +v x (5) Comments: 1. Relatons (4) allow us to fnd v when we know u, and vce versa. Let us put u x =P, u y =Q, so that {v x = Q, v y =P}. The ntegrablty (compatblty) conon of ths system for soluton for v, for a gven u, s P/ x = Q/ y u xx +u yy =. Smlarly, the ntegrablty conon of system (4) for soluton for u, for a gven v, s v xx +v yy =. We notce that both the real and the magnary part of an analytc functon are harmonc functons,.e., they satsfy the Laplace equaton Φ xx + Φ yy = (6) Harmonc functons related to each other by means of the Cauchy-Remann relatons (4) are called conjugate harmonc.. Let z*=x y be the complex conjugate of z=x+y. Then, x = (z+z*) /, y = (z z*) / (7) By usng relatons (7) we can express u(x,y) and v(x,y), thus also the sum w=u+v, as functons of z and z*. The real Cauchy-Remann relatons (4), then, are rewrtten n the form of a sngle complex equaton, w / z* = (8) One way to nterpret ths result s the followng: The analytc functon (1) s lterally a functon of the complex varable z=x+y, not just some complex functon of two real varables x and y! Examples: 1. We seek an analytc functon of the form (1), wth v(x,y)=xy. Note frst that v satsfes the PDE (6): v xx +v yy = (harmonc functon). Thus, the ntegrablty conon of the system (4) for soluton for u s satsfed. The system s wrtten { u/ x=x, u/ y= y}. The frst relaton yelds u=x /+φ(y). From the second one we then get φ (y)= y φ(y)= y /+C, so that u=(x y )/+C. Puttng C=, we fnally have w=u+v= (x y )/+xy. [Exercse: Usng relatons (7), show that w=f (z)= z /, thus verfyng conon (8).]. Consder the functon w=f (z)= z defned on the entre complex plane. Here, u(x,y)=x +y, v(x,y)=. As s easy to verfy, the Cauchy-Remann relatons (4) are not satsfed anywhere on the plane, except at the sngle pont z= where (x,y) (,). Alternatvely, we may wrte w=zz*, 14

23 INTEGRABILITY ON THE COMPLEX PLANE so that w/ z*=z (except at z=). We conclude that the gven functon s not analytc on the complex plane.. Integrals of Complex Functons Let L be an orented curve on the complex plane, the ponts of whch plane are represented as z=x+y (x, y). y L B A x The ponts z of L are determned by some parametrc relaton of the form z = λ(t) = x(t) + y(t), α t β (1) As t ncreases from α to β, the curve L s traced from A to B, whle the opposte curve L s traced from B to A wth t decreasng from β to α. We now consder ntegrals of the form f ( z ) dz, where f (z) s a complex functon. We wrte L dz=λ (t), so that Also, f ( z) dz = ( ) = ( ) L α β f ( z) dz = f [ λ( t)] λ ( t) () L β α β α f ( z) dz = f ( z) dz (3) L A closed curve C wll be conventonally regarded as postvely orented f t s traced counterclockwse. Then, the opposte curve C wll be negatvely orented and wll be traced clockwse. Moreover, Examples: L f ( z) dz = f ( z) dz (4) C C 1. We want to evaluate the ntegral counterclockwse, (b) clockwse. I = dz, where the crcle z a =ρ s traced (a) z a z a =ρ 15

24 CHAPTER (a) The crcle z a =ρ s descrbed parametrcally by the relaton z=a+ρe t, t π. Then, dz = (a+ρe t ) = ρe t. Integratng from to π (for counterclockwse tracng) we have: t π π ρe I = = t ρe dz = π. z a z a = ρ (b) For clockwse tracng of the crcle z a =ρ, we wrte, agan, z=a+ρe t ( t π). Ths tme, however, we ntegrate from π to. Then, I = = π. Alternatvely, we wrte z=a+ρe t ( t π) and ntegrate from to π, arrvng at the same result. z a =ρ dz. Consder the ntegral I =, where the crcle z a =ρ s traced counter- ( z a) clockwse. We wrte z=a+ρe t ( t π), so that π t π π ρe t I = = e = t ρ e ρ. In general, for k =, ±1, ±, and for a postvely (counterclockwse) orented crcle z a =ρ, one can show that dz π, f k = 1 = ( z a) k (5), f k 1 z a = ρ.3 Some Basc Theorems We now state some mportant theorems concernng analytc functons: Theorem 1 (Cauchy-Goursat): Assume that the functon f (z)=u (x, y)+ v (x, y) s analytc n a smply connected doman G of the complex plane. Then, for any closed curve C n G, Proof: Wrte dz=dx+ dy, so that f ( z ) dz = (1) C f ( z) dz = ( udx vdy) + ( vdx + udy) f ( z) d z = ( udx vdy) + ( vdx + udy). C C C Now, gven that f (z) s analytc n G, the Cauchy-Remann relatons wll be vald n ths doman. Moreover, snce G s smply connected, the conons of valy of Theorem 1 of Sec. 1.3 are satsfed. Hence, 16

25 INTEGRABILITY ON THE COMPLEX PLANE by whch relatons Eq. (1) follows. u = ( v) udx + ( v) dy =, y x C v = u vdx + udy =, y x C Corollary: In a smply connected doman G, the lne ntegral of an analytc functon f (z) s ndependent of the path connectng any two ponts A and B. B L 1 L A Proof: As n Sec. 1.3, we consder the closed path C=L 1 +( L ). By (1), then, we have: f ( z) dz = f ( z) dz + f ( z) dz = C L L 1 f ( z) dz f ( z) dz = f ( z) dz = f ( z) dz. L L L L 1 1 Let us assume, now, that the functon f (z) s analytc n a doman G that s not smply connected. G (For example, the doman G n the above fgure s doubly connected.) Let C be a closed curve n G. Two possbltes exst: (a) The curve C does not enclose any ponts not belongng to G. Then, C may be consdered as the boundary of a smply connected subdoman of G where the conons of valy of the Cauchy-Goursat Theorem are fulflled. Therefore, f ( z ) dz =. C (b) The curve C encloses ponts not belongng to G. Then, C may not belong to some smply connected subdoman of G and the conons of Theorem 1 are not fulflled. In such a case, relaton (1) may or may not be satsfed. 17

26 CHAPTER Example: Let G consst of the complex plane wthout the orgn O of ts axes (.e., wthout the pont z=). The functon f (z)=1/z s analytc n ths doman. Let C be the crcle z =ρ centered dz dz at O. Then, as we saw n Sec.., = π ( ). On the contrary, C k z = for k 1. C z Theorem (composte contour theorem): Consder a multply (e.g., doubly, trply,...) connected doman G of the complex plane and let Γ be a closed curve n G. Let γ 1, γ,..., γ n be closed curves n the nteror of Γ (but n the exteror of one another) such that the doman D between the γ k and Γ belongs entrely to G. Then, for every functon f (z) analytc n G, n f ( z) dz = f ( z) dz = f ( z) dz + f ( z) dz + + f ( z) dz () Γ = 1 γ γ γ γ 1 n where all curves Γ, γ 1,..., γ n, are traced n the same drecton (e.g., counterclockwse). G γ n γ 1 D γ Γ Problem: Show that Γ dz z = π where Γ s any postvely orented, closed curve enclosng the orgn O (z=) of the complex plane. (Hnt: Consder a crcle γ : z =ρ centered at O and lyng n the nteror of Γ.) Theorem 3 (Cauchy ntegral formula): Consder a functon f (z) analytc n a doman G. Let C be a closed curve n G, such that the nteror D of C belongs entrely to G. Consder a pont z D. Then, f ( z ) = 1 f ( z) dz π (3) z z C where C s traced n the postve drecton (.e., counterclockwse). 18

27 INTEGRABILITY ON THE COMPLEX PLANE C D z G Comments: 1. The value of the ntegral n (3) s ndependent of the choce of the curve C whch encloses z and satsfes the conons of the theorem. (Ths follows from the composte contour theorem for n=1.). More generally, we can wrte 1 f ( z) dz f ( z ), f z D π =. C z z, f z ( G D) Indeed: If z (G D) (that s, z D), then the functon f (z)/(z z ) s analytc everywhere nsde the smply connected doman D and thus satsfes the Cauchy-Goursat theorem. Applcaton: Puttng f (z)=1 and consderng a postvely orented path C around a pont z, we fnd dz = π. C z z Note: More generally, for k =, ±1, ±,, where the pont z s located n the nteror of C. dz π, f k = 1 = C ( z z ) k (4), f k 1 Theorem 4 (Laurent seres): Consder a functon f (z), analytc n an annular doman D: r < z z < R centered at z. Let C be a postvely orented, closed path around z and nsde the annulus D. 19

28 CHAPTER D z r z R C Then, at every pont z D, the functon f (z) s represented by a convergent seres of the form + n (5) n= f ( z) = a ( z z ) n where the coeffcents a n are gven by a n = 1 f ( z) dz (6) ( ) C n+ 1 π z z and where the value of the ntegral n (6) s ndependent of the choce of the curve C. Proof of the coeffcent formula: Assumng the valy of (5), we have: But, by (4), C f ( z) dz ( z z ) k+ 1 f ( z) ( z z ) k + 1 n k 1 = a ( z z ) n n n k 1 n C n nk n = a ( z z ) dz a I. n I nk π, f n = k = = π δ, f n k nk where δ nk s the Kronecker delta, assumng the values 1 and for n=k and n k, respectvely. Hence, C f ( z) dz 1 f ( z) dz = π a δ = π a a = ( z z ) ( z z ). k+ 1 n nk k k k 1 C + n π

29 INTEGRABILITY ON THE COMPLEX PLANE Comment: The annulus D: r < z z < R may be the regon between two concentrc crcles ( < r < R) ; a crcle wth ts center z deleted (r=, R >) ; the exteror of a crcle (r >, R= ) ; or the entre complex plane wth pont z deleted (r=, R= )..4 Antdervatve and Indefnte Integral of an Analytc Functon Let z and z be two ponts n a smply connected doman G of the complex plane. We regard z as constant whle z s assumed to be varable. Accordng to the Cauchy-Goursat theorem, the lne ntegral from z to z, of a functon f (z) analytc n G, depends only on the two lmt ponts and s ndependent of the curved path connectng them. Hence, such an ntegral may be denoted by z z f ( z ) dz or, for smplcty, z f ( z) dz. For varable upper lmt z, ths ntegral s a functon of z ts upper lmt. We wrte z f ( z) dz = I ( z) (1) z As can be shown, I(z) s an analytc functon. Moreover, t s an antdervatve of f (z); that s, I (z)= f (z). Analytcally, ( ) d z I z = f ( z ) dz f ( z dz = ) () z Any antdervatve F(z) of f (z) [F (z)= f (z)] s equal to F(z)=I(z)+C, where C=F(z ) s a constant [note that I(z )=]. We observe that I(z)=F(z) F(z ) z f ( z) dz = F( z) F( z) (3) z In general, for gven z 1, z and for an arbtrary antdervatve F(z) of f (z), we may wrte z f ( z) dz = F( z) F( z1) (4) z1 Now, f we also allow the lower lmt z of the ntegral n Eq. (1) to vary, then ths relaton yelds an nfnte set of antdervatves of f (z), whch set represents the ndefnte ntegral of f (z) and s denoted by f ( z) dz. If F(z) s any antdervatve of f (z), then, by relaton (3) and by puttng F(z )= C, { } f ( z) d z = F( z) + C / F ( z) = f ( z), C = const.. 1

30 CHAPTER To smplfy our notaton, we wrte f ( z ) dz = F ( z ) + C (5) where the rght-hand sde represents an nfnte set of functons, not just any specfc antdervatve of f (z)! Examples: 1. The functon f (z)=z s analytc on the entre complex plane and one of ts antdervatves s F(z)=z 3 /3. Thus, 3 z z dz = + C and 3 1 (1 ) 3 z dz =. 1. The functon f (z)=1/z s dfferentable everywhere except at the orgn O of the complex plane, where z=. An antdervatve, for z, s F(z)= 1/z. Hence, dz 1 z = + C and z z z 1 dz z 1 1 = z z 1 (where the path connectng the ponts z 1 and z does not pass through O).

31 CHAPTER 3 ORDINARY DIFFERENTIAL EQUATIONS 3.1 The Concept of the Frst Integral An ordnary dfferental equaton (ODE) s often easer to solve f we can fnd one or more frst ntegrals. In smple terms, a frst ntegral s a relaton (algebrac or dfferental) that gves us the nformaton that some mathematcal quantty retans a constant value as a consequence of the gven ODE. Ths quantty may contan the dependent varable y, the ndependent varable x, as well as dervatves y (k) (x) = d k y/dx k. When dervatves are contaned, a frst ntegral leads to an ODE of lower order than the ntal ODE. Thus, by usng a frst ntegral one may reduce the order of a gven ODE. If the ODE s of the frst order, a frst ntegral s an algebrac relaton expressng the soluton drectly. In general, an ODE of order n s completely solved f one manages to fnd n ndependent frst ntegrals. In Classcal Mechancs one often needs to fnd the soluton of a system of ODEs expressng Newton s second law of moton. Wth the excepton of some smple cases, ths system s hard to ntegrate; for ths reason one seeks as many frst ntegrals as possble. These quanttes are called constants of the moton and they express correspondng conservaton laws, such as, e.g., conservaton of total mechancal energy, of momentum or of angular momentum. 3. Exact Equatons Consder the frst-order ODE dy M ( x, y) = (N ), whch s wrtten, more symmetrcally, dx N( x, y) M (x, y) dx + N (x, y) dy = (1) Equaton (1) s sad to be exact f there exsts a functon u (x,y) such that M (x, y) dx + N (x, y) dy = du () (that s, f the expresson Mdx+Ndy s an exact dfferental). Then, by (1) and (), du= u (x, y) = C (3) Equaton (3) s an algebrac relaton connectng x and y and contanng an arbtrary constant. Thus t can be regarded as the general soluton of (1). Relaton (3) s a frst ntegral of the ODE (1) and drectly determnes the general soluton of ths equaton. Accordng to (), the functon u(x,y) satsfes the followng system of frst-order partal dfferental equatons (PDEs): 3

32 CHAPTER 3 u x u = M ( x, y), = N( x, y) y (4) The ntegrablty conon of ths system for exstence of a soluton for u, s M y N = x (5) If conon (5) s vald at all ponts of a smply connected doman D of the x-y plane, then t guarantees the exstence of a soluton for the system (4) or, equvalently, for the dfferental relaton (). The constant C n the soluton (3) s determned by the ntal conon of the problem: f the specfc value x=x corresponds to the value y=y, then C=C =u (x, y ). We thus get the partcular soluton u (x, y)=c. Example: We consder the ODE (x+y+1) dx + (x y +3) dy =, wth ntal conon y=1 for x=. Here, M=x+y+1, N=x y +3 and M/ y= N/ x (=1), at all ponts of the x-y plane. The system (4) s wrtten { u/ x=x+y+1, u/ y=x y +3}. The frst equaton yelds u=x /+xy+x+φ(y), whle by the second one we get φ (y)= y +3 φ(y)= y 3 /3+3y+C 1. Thus, u=x / y 3 /3+xy+x+3y+C 1. The general soluton (3) s u(x,y)=c. Puttng C C 1 C, we have x / y 3 /3+xy+x+3y=C (general soluton). Makng the substtutons x= and y=1 (as requred by the ntal conon), we fnd C=8/3 and x / y 3 /3+xy+x+3y=8/3 (partcular soluton). 3.3 Integratng Factor Assume that the ODE M (x, y) dx + N (x, y) dy = (1) s not exact [.e., the left-hand sde s not a total dfferental of some functon u(x,y)]. We say that ths equaton admts an ntegratng factor µ(x,y) f there exsts a functon µ(x,y) such that the ODE µ(mdx+ndy)= s exact; that s, such that the expresson µ(mdx+ndy) s a total dfferental of a functon u(x,y): µ (x, y) [M (x, y) dx + N (x, y) dy] = du () Then the ntal ODE (1) reduces to the dfferental relaton du= u (x, y) = C (3) on the conon that the functon µ(x,y) does not vansh dentcally when x and y are related by (3). Relaton (3) s a frst ntegral of the ODE (1) and expresses the general soluton of ths equaton. 4

33 ORDINARY DIFFERENTIAL EQUATIONS Example: The ODE ydx xdy= s not exact, snce M=y, N= x and M/ y=1, N/ x= 1. However, the equaton 1 ( ) ydx xdy = s exact, gven that the left-hand sde s equal to y d (x/y). Thus, d (x/y)= y= Cx. The soluton s acceptable snce the ntegratng factor µ=1/y does not vansh dentcally for y= Cx. 3.4 Hgher-Order Dfferental Equatons In the case of an ODE of second order or hgher, a frst ntegral leads to a reducton of order of the ODE. Consder an ODE of order n: F [ x, y, y, y,..., y (n) ] = (1) (where y (n) d n y/dx n ). We assume that the left-hand sde of (1) can be wrtten as the dervatve of an expresson Φ of order (n 1): Then (1) reduces to dφ/dx = ( n) d ( n 1) F [ x, y, y, y,, y ] = Φ [ x, y, y,, y ] () dx Relaton (3) s a frst ntegral of (1); t s an ODE of order (n 1). Φ [ x, y, y,..., y (n 1) ] = C (3) Example: Consder the second-order ODE yy +(y ) =. We notce that the left-hand sde s equal to d (yy )/dx. Hence, the gven equaton s wrtten yy = C (whch s a frst-order ODE), from whch we have y = C 1 x+c. Sometmes the left-hand sde of (1) s not n tself a total dervatve but can transform nto one upon multplcaton by a sutable ntegratng factor µ [x, y, y,..., y (n 1) ]: ( n 1) ( n) d ( n 1) µ [ x, y, y,, y ] F [ x, y, y, y,, y ] = Φ [ x, y, y,, y ] (4) dx Then dφ/dx =, so we are led agan to a frst ntegral of the form (3). Example: Consder the ODE yy (y ) =. Multpled by µ=1/y, the left-hand sde becomes (y /y). The gven equaton s then wrtten (y /y) = y /y= C (a frst-order ODE), by whch we get y= C 1 e Cx. 5

34 CHAPTER Applcaton: Newton s Second Law n One Dmenson In ths secton, one or more dots wll be used to denote dervatves of varous orders wth respect to the varable t : xɺ = dx /, ɺɺ x= d x /, etc. We consder the second-order ODE mɺɺ x= F( x) (1) wth ntal conons x(t )=x and v(t )=v, where v=dx/. Physcally, relaton (1) expresses Newton s Second Law for a partcle of mass m movng wth nstantaneous velocty v(t) along the x axs under the acton of a force (or, more correctly, a force feld) F(x). By solvng (1) we fnd the poston x=x(t) of the partcle as a functon of tme t. We defne an auxlary functon U(x) (potental energy of the partcle) by x d U ( x) = F( x ) dx F( x) = U ( x) () dx (wth arbtrary lower lmt of ntegraton). The functon U may always be defned n a onedmensonal problem, whch s not the case n hgher dmensons snce the ntegral correspondng to the one n () wll generally depend on the path of ntegraton and thus wll not be unquely defned (except n the case of conservatve felds; see Sec. 1.5). We also note that the functon U depends on tme t through x only;.e., U s not an explct functon of t ( U/ t=). Ths means that the value of U changes exclusvely due to the moton of the partcle along the x axs whle, at any fxed pont x, U s constant n tme. Equaton (1) s now wrtten du mɺɺ x+ =. dx The left-hand sde s not a perfect dervatve wth respect to t. Let us try the ntegratng factor µ=ɺ x : But, du du xɺ mɺɺ x+ = m xɺɺɺ x+ xɺ =. dx dx 1 d xɺɺɺ x= ( xɺ ) and du du dx du xɺ = =. dx dx We thus have d 1 m xɺ + U ( x) = 6

35 ORDINARY DIFFERENTIAL EQUATIONS 1 1 E m x + U x = mv + U x = ɺ ( ) ( ) constant (3) Equaton (3) expresses the conservaton of mechancal energy of the partcle. Notce that ths result s an mmedate consequence of Newton s law. (In hgher dmensons, ths prncple s vald only n the case where the force feld F( r ) s conservatve; cf. Sec. 1.5.) Relaton (3), whch consttutes a frst ntegral of the ODE (1), s a frst-order ODE that s easy to ntegrate. We have: dx = [ E U ( x )] m. Takng the case where v=dx/ >, we wrte 1/ x dx t [ ( )] x 1/ m { } t dx = E U x = (where we have taken nto account that x=x for t=t ). Fnally, x = t t x 1/ (4) dx [ E U ( x )] m Note: As derved, relaton (4) s vald for v>. For v< one must put dx n place of dx nsde the ntegral. In general, n cases where the velocty v s postve n a part of the moton and negatve n another part, t may be necessary to perform the ntegraton separately for each part of the moton. Relaton (4) represents a partcular soluton of (1) for the gven ntal conons. By puttng x=x and v=v n (3) and by takng nto account that E s constant, we can determne the value of the parameter E that appears n the soluton (4): 1 E = m v + U ( x) (5) Comment: It s evdent from () that the functon U(x) s arbtrary to wthn an adve constant whose value wll depend on the choce of the lower lmt n the ntegral defnng U. Through (3), the same arbtrarness s passed on to the value of the constant E; t dsappears, however, upon takng the dfference E U(x). Thus, ths arbtrarness does not affect the result of the ntegraton n (4). Example: Rectlnear moton under the acton of a constant force F. We take t =, x =x()=, v()=v, and we assume that v=dx/ > (n partcular, v >) for the part of the moton that nterests us. From () we have (makng the arbtrary assumpton that U= for x=): 7

36 CHAPTER 3 du dx U = F du = F dx U ( x) = Fx x. Equaton (4) then yelds x 1/ dx = t 1/ ( E+ Fx) m 1/ 1/ F 1/ ( E+ Fx) = t + E m. Squarng ths, we fnd 1/ F E x= t + t m m. We set F/m=a=const. (acceleraton of the partcle). Also, from (5) we have that E=mv / [snce U()=]. Thus, fnally (takng nto account that v >), 1 x= a t + v t, whch s the famlar formula for unformly accelerated rectlnear moton. Problem: Show that a result of the same form wll ensue n the case where v<. [Hnt: Use (4) wth dx n place of dx ; put v()=v wth v <.] 8

37 CHAPTER 4 SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS 4.1 Soluton by Seekng Frst Integrals We consder a system of n ordnary dfferental equatons (ODEs) of the frst order, for n unknown functons x 1 (t), x (t),..., x n (t): dx = f ( x, x,, x, t) ( = 1,,, n) (1) 1 n If the functons f are not explctly dependent on t (.e., f f / t = for =1,,...,n) the system (1) s called autonomous: dx = f ( x, x,, x ) ( = 1,,, n) () 1 n A conservaton law for system (1) s an ODE of the form d x1 x xn t Φ = (,,,, ) (3) whch s vald as a consequence of the system (.e., s not satsfed dentcally). Equaton (3) s mmedately ntegrable: Φ ( x, x,, x, t) = C (4) 1 The functon Φ s a frst ntegral of the system (1). It retans a constant value when the x 1, x,..., x n satsfy the system (that s, t s not dentcally constant but reduces to a constant on solutons of the system). If one or more frst ntegrals of the system are known, one can trvally produce a whole nfnty of frst ntegrals by takng sums, multples, products, powers, etc., of them. We are only nterested, however, n frst ntegrals that are ndependent of one another, snce t s n ths case that we obtan the most useful nformaton for the soluton of the problem. Let us now assume that we manage to fnd k ndependent frst ntegrals of the system (1) (where k n): n Φ ( x, x,, x, t) = C 1 1 n 1 Φ ( x, x,, x, t) = C 1 n Φ ( x, x,, x, t) = C k 1 n k (5) 9

38 CHAPTER 4 Relatons (5) allow us to express k of the varables x 1,..., x n n terms of the remanng (n k) varables and t. We thus elmnate k unknown functons from the problem, so that the system (1) reduces to one wth fewer unknowns, that s, (n k). If k=n, then all unknown functons x 1,..., x n can be determned algebracally from system (5) wthout the necessty of ntegratng the dfferental system (1) tself. The autonomous system () s wrtten dx f ( x,, x ) 1 n = ( = 1,,, n) (6) Snce the f do not contan t drectly, ths varable can be elmnated from the system. Indeed, snce all left-hand sdes n (6) are equal to, they wll be equal to one another as well. Hence, dx1 dx dxn f ( x,, x ) = f ( x,, x ) = = f ( x,, x ) 1 1 n 1 n n 1 n (7) Relaton (7) represents a system of (n 1) equatons n n varables x 1, x,..., x n. To solve t we seek (n 1) ndependent frst ntegrals of the form Φ ( x, x,, x ) = C ( j= 1,,, n 1) (8) j 1 n j We also seek a frst ntegral Φ n of the complete system (6): Φ ( x, x,, x, t) = C (9) n 1 n n Relatons (8) and (9) consttute a system of n algebrac equatons n (n+1) varables. By solvng ths system for the x 1, x,..., x n, we can express these varables as functons of t. Analytcally, one way of ntegratng the system (6) s the followng: Wth the ad of relatons (8), we express (n 1) of the n varables x 1,..., x n as functons of the remanng varable. Assume, for example, that the x 1, x,..., x n 1 are expressed as functons of x n. Takng (6) wth =n, we have: dxn dxn = F( xn) + c = t+ c f ( x,, x ) f ( x,, x ) n 1 n n 1 n Φ ( x, t) F( x ) t = C (1) n n n n Equaton (1) allows us to express the varable x n as a functon of t. Gven that the x 1, x,..., x n 1 are known functons of x n, the above (n 1) varables can, n turn, also be expressed as functons of t. 3

39 SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS Examples: 1. Consder the system dx d y = y ( a) = x ( b) (Here, x 1 x, x y.) We seek frst ntegrals of ths system. Two are suffcent for a complete soluton of the problem. Addng (a) and (b), we have d (x+y)/=x+y. Puttng x+y=u, we wrte du/=u, whch yelds u=c 1 e t (x+y) e t =C 1. Smlarly, subtractng (b) from (a) and puttng x y=u, we fnd du/= u u=c e t (x y) e t =C. We have thus found two ndependent frst ntegrals of the system: Φ 1 (x, y, t) (x+y) e t = C 1, Φ (x, y, t) (x y) e t = C. [Exercse: Verfy that d Φ 1 /= and d Φ /= when x and y are solutons of the system of (a) and (b). Note that Φ 1 and Φ are not dentcally constant!] By usng the frst ntegrals Φ 1 and Φ we can now express x and y as functons of t. Puttng C 1 and C n place of C 1 / and C /, respectvely, we fnd x = C 1 e t + C e t, y = C 1 e t C e t. Comment: One can easly fnd more frst ntegrals of the system (a), (b). For example, by elmnatng we have dx/y=dy/x xdx=ydy d (x y )=, so that Φ 3 (x,y) x y =C 3. Let us note, however, that Φ 3 =Φ 1 Φ. Thus, the relaton Φ 3 =const. s a trval consequence of Φ 1 =const. and Φ =const. In other words, Φ 3 s not an ndependent, new frst ntegral of the system; therefore, t does not furnsh any useful nformaton for the soluton of the problem.. Consder the system dx d y = y ( a) = x ( b) We seek two frst ntegrals. In ths case, we get no useful nformaton by addng or subtractng the two equatons of the system. However, snce ths system s autonomous, we can elmnate : dx/y = dy/x xdx+ydy= d (x +y )= Φ 1 (x, y) x +y = C 1. To solve the problem completely, we need another frst ntegral of the system; ths tme, one that contans t explctly. From (a) and (b) we have x (dy/) y (dx/) = (x +y ) d (y/x) / = [1 + (y/x) ]. Puttng y/x=u, we wrte du / (1+u ) = d (t + arctan u) =, from whch we fnd 31

40 CHAPTER 4 Φ (x, y, t) t + arctan (y/x) = C. We now use the frst ntegrals Φ 1 and Φ to solve the system (a), (b) algebracally. The relaton Φ =C yelds y= x tan (t C ). Then, by the relaton Φ 1 =C 1 we get x [1+tan (t C )]=C 1. By usng the dentty cos a=1/1+tan a, t s not hard to show that x = C 1 cos (t C ) so that y = C 1 sn (t C ). Comment: An alternatve way to solve the problem s by transformaton of coordnates from Cartesan (x,y) to polar (r,θ), where r and θ < π. The transformaton equatons are The system (a), (b) s wrtten x = r cosθ, y = r snθ r = (x +y ) 1/, θ = arctan (y/x). (dr/) cosθ r (dθ/) snθ = r snθ, (dr/) snθ + r (dθ/) cosθ = r cosθ. Solvng for the dervatves, we can separate the varables r and t, fndng a separate equaton for each varable: dr/ =, dθ/ = 1, wth correspondng solutons r=c 1, θ = t+c. Substtutng nto the transformaton equatons, we have: x = C 1 cos (t C ), y = C 1 sn (t C ), as before. The frst ntegrals of the system are easly found by solvng the above equatons for the constants C 1 and C. 3. Consder the system dx = y z ( a) d y = z x ( b) dz = x y ( c) Rather than tryng to solve t analytcally, we wll express the soluton mplctly wth the ad of three ndependent frst ntegrals (as requred for an algebrac soluton of the problem), at least one of whch wll contan the varable t explctly. Takng the sum (a)+(b)+(c), we have d (x+y+z) / = Φ 1 (x, y, z) x+y+z = C 1. On the other hand, the combnaton x.(a) + y.(b) + z.(c) yelds d (x +y +z ) / = Φ (x, y, z) x +y +z = C. Now, by usng the equatons Φ 1 =C 1 and Φ =C we can express two of the dependent varables, say x and y, n terms of the thrd varable z. Then, relaton (c) n the system s wrtten n the form of an equaton for a sngle varable z: 3