Ordinary Differential Equations

Transcription

1 Ordinary Differenial Equaions 5. Examples of linear differenial equaions and heir applicaions We consider some examples of sysems of linear differenial equaions wih consan coefficiens y = a y a y n y = a y a y 1 1 n = + + yn an 1y1... anny n n n (5.1) and he linear n-h-order differenial equaion (n > 1) wih consan coefficiens x + a x a x + a x= ( n) ( n 1) n 1 1 (5.) We discuss suiable mehods o compue soluions and he fundamenal marix. In paricular, we compue he fundamenal marix for all planar sysems (n = ) and classify hese sysems according o heir properies in phase space. We also inroduce he noion of sabiliy for linear differenial equaions Planar sysems 1. Equaion of small oscillaions. We sudy he linear second-order differenial equaion x+ a x + a x= (5.3) 1 for real coefficiens a, a 1. This equaion arises from problems in mechanics and elecriciy. Oscillaions of a spring. A mass m coupled o a spring is submied o a linear force. Newon s equaion of moion reads mx = Dx where D is a posiive consan. I is of he form (5.3) wih a = D/m and a 1 =. If fricion is aken ino accoun hen a 1 is a posiive consan. Small oscillaions of he mahemaical pendulum. We consider a weighless rod of lengh l aached a one end and bearing a poin mass m a he oher end. If x denoes he angle by which he rod deviaes from he verical hen ml x = mgl sin x. For small deviaions sin(x) x and we ge an equaion of he form (5.3) wih a = g/l and a 1 =. 1

2 Again, if we ake ino accoun a force due o fricion ha is proporional o he velociy hen a = a 1 is a posiive consan. Invered pendulum. If we urn he pendulum upside down, hen he sign of he force due o graviaion changes sign and herefore a = - g/l. Elecrical circuis. The charge y of a condenser of capaciy C in a closed elecrical circui wih induciviy L and resisance R saisfies he second order differenial equaion 1 Ly + Ry + C y = which is of he form (5.3) wih a = 1/LC and a 1 = R/L Transformaion o a firs-order sysem. The second-order differenial equaion (5.3) is equivalen o a wo-componen linear firs-order sysem wih marix A given by 1 A = a a1 (5.4) I is convenien o rewrie A using he race and he deerminan or is eigenvalues. Then where 1 1 A = = de + ( A) r( A) λ1λ λ1 λ (5.5) 4de r A r A A λ1 = + 4de r A r A A λ = (5.6) are he (possibly complex) eigenvalues of A. If here is no fricion (no resisance), hen r(a) = which means ha he sysem is area preserving in phase space. 1.. Equaion in phase space. The equaion (5.3) describes a moion in he wodimensional phase space wih coordinaes x, y = x. The marix A defines he direcion field, which is a composiion of a roaion and a dilaion of he plane. The orbi of a x (), y (), or if we consider y as a funcion of x he soluion is he se of poins { } se of poins {x, y(x)} where y is (a leas formally) a soluion of he non-linear firs-order differenial equaion dy ( x ) = a x a1 = de A x + r A dx yx yx

3 which is of homogeneous ype and can be solved (a leas in principle) by elemenary inegraion mehods (see ch..3). Anoher imporan quaniy defined on he phase space is he energy of he sysem. The energy funcional associaed o (5.3) is defined by (, ) 1 1 E x x = x + a x (5.7) If x() is a soluion of (5.3) hen d E ( x, x ) = x ( x + a x ) = a x r 1 = ( A ) x (5.8) d If here is no fricion ( a 1 = -r(a) = ) he energy of a soluion remains consan and he soluion follows he lines of consan energy in phase space. In his case he sysem is called conservaive. Since he equaion is linear he energy defines a quadraic form and he orbis are eiher ellipses (de(a) = a > ), hyperbolas (de(a) = a < ), or sraigh lines (de(a) = a = ). In he presence of fricion in a mechanical sysem or elecrical resisance in an elecrical circui ( a 1 = -r(a) > ) he energy is decreasing Explici soluions of he second order equaion. According o he resuls of he previous chaper (heorem 4.7) he general complex valued soluion is of he form () x = cexp λ+ cexp λ, λ λ x = cexp λ + cexp λ, λ = λ λ 1 1 (5.9) The corresponding real soluions are lised in he following able. λ 1, λ real, r(a) 4de(A) λ 1 λ r(a) > 4de(A) λ 1 = λ = λ, r(a) = 4de(A) () = ( λ ) + ( λ ) = ( λ ) + ( x c1exp 1 cexp λ 1, λ complex conjugae, r(a) < 4de(A) x c exp c exp λ) 1 λ 1 = µ iω, λ = µ + iω λ 1 = iω, λ = iω, r(a) = () = exp( µ ) 1sin ( ω ) + cos( ω ) or 1 ( ω ) ( () = exp( µ ) sin ( ω + δ) x = Csin ( ω+ δ ) x c c x C x = c sin + c cos ω) or The invered pendulum belongs o he firs class since de(a) = a <. In all oher models de(a) = a > and damping -a 1 = r(a) For weak damping i.e. small r(a) here are wo complex conjugae, criical damping corresponds o a double real eigenvalue If boh eigenvalues are real and differen he sysem is called overdamped. 3

4 1.4. Compuing exp(a). A complex valued fundamenal marix of he associaed firs order sysem is easily compued if he eigenvalues are differen. The marix A defined in (5.4) resp. (5.5) has eigenvecors 1 1 v1 =, v = λ1 λ Therefore a fundamenal marix Y() is given by Since we ge ( λ1) exp( λ) exp exp Y() = ( v1exp( λ1) vexp( λ) ) = λ1exp λ1 λ λ 1 λv1 λ1v ( λ λ1) =, v v1= ( λ λ1) 1 (5.1) exp ( A) If λ 1 = µ iω, λ = µ + iω (5.11) reads λ1 λ λ λ1 1 λe λ1e e e = λ λ λλ( e e ) λ e λe λ λ λ λ (5.11) exp ( A) µ e ωcos ω µ sin ω sin ω = ω sin ( ω )( µ + ω ) ωcos( ω ) + µ sin ( ω ) (5.1) To compue exp(a) in case λ 1 = λ = λ we le λ 1 λ = λ. We obain exp 1 λ = λ 1 + λ (5.13) λ ( A) e The direc compuaion goes as follows: By heorem 4.7. he funcions exp(λ) and exp(λ) are soluions of he second order equaion. The corresponding soluion vecors are λ 1 λ 1 = e, v = e v λ 1+ λ from which we easily consruc a fundamenal marix. 4

5 . General firs order sysems and heir classificaion. Le A be an arbirary real marix. Then A is eiher similar o he marix given in (5.5) or A is a muliple of he uni marix E. In he firs case all resuls are similar o he resuls for he second order equaion. To classify he sysems i is sufficien o consider heir normal forms, which are one of he following real marices: λ1 λ 1 µ ω,, λ λ ω µ (5.14) where we used he noaions inroduced above. Diagonal case λ 1 < λ < λ 1 < < λ < λ 1 < λ Sable node Saddle poin Unsable node λ 1 = λ < λ 1 < = λ < λ 1 = λ Sable sar Marginally sable Unsable sar 5

6 Jordan case λ 1 = λ = λ < λ1 = λ = λ = λ 1 = λ = λ = Sable node Marginally unsable Unsable node Complex case µ <, ω > µ =, ω > µ >, ω > Sable focus Cener Unsable focus.1. Sabiliy. We precise he meaning of sabiliy and insabiliy of linear equaions. We noe ha y() = is always a saionary soluion of a linear homogeneous sysem. We define sabiliy and insabiliy as a propery of soluions wih iniial condiions close o he zero soluion. We say ha y() = is sable if all soluions remain bounded, asympoically sable if all soluions end o zero as ends o infiniy, and unsable if here is an unbounded soluion of he linear equaion. We have seen ha y() = is sable if here is no eigenvalue wih posiive real par ( wih excepion of he marginally unsable Jordan block), asympoically sable if all eigenvalues have negaive real par, and unsable if here is an eigenvalue wih posiive real par or a marginally unsable Jordan block. A more appropriae formulaion of sabiliy of a saionary soluion, which also exends o nonlinear sysems is he concep of Lyapunov sabiliy. Definiion: The saionary soluion y() = is called sable or Lyapunov sable if for every ε > here is a δ > such ha for every y wih y < δ he soluion y() wih 6

7 iniial condiion y() = y exiss for all > and saisfies y() < ε for all >. The saionary soluion is called asympoically sable if in addiion all soluions saisfy lim y =. As an applicaion of he linear resul we presen he following resul for nonlinear planar auonomous sysems. Theorem 1: Le A M (, ) and g:, g( y) = O( y ) a differeniable funcion. Consider he differenial equaion y = Ay+ g( y) (5.15) If all eigenvalues of A have negaive real pars hen he saionary soluion y = of (5.15) is asympoically sable. Proof: The proof of he sabiliy resul will be a beauiful applicaion of Gronwall-s inequaliy. If y() is a soluion of (5.15) hen i saisfies he inegral equaion () = exp + exp ( ( )) ( ) y A y A s g y s ds (5.16) Le. be a norm on R. Since all eigenvalues of A have negaive real pars here exis posiive consans a, C such ha a exp( A ) x Ce x for any x R. By he assumpio n on g here is a posiive consan K such ha g( y) Therefore, if y r, hen K y if y r. g( y) Kr y Laer we shall choose r sufficienly small such ha he consan Kr will be small. Taking he norm on boh sides of (5.16) and using hese esimaes we obain he following inegral inequaliy which holds for all such ha y rin [,]: () a + ( ) a s y Ce y CKr e y() s ds 7

8 We se z () = y () e a. Then z() saisfies he inegral inequaliy + CKr By Gronwall s inequaliy z () Cy e. z() C y CKr z( s) ds We choose r sufficienly small such ha CKr = a/. We ge If we choose y such ha y y () rfor all > concluding he proof. y () Cy e a r C, hen he above inequaliy shows in paricular ha Remark: I is clear ha heorem 1 also holds for higher dimensional auonomous sysems. For he second order differenial equaion (5.3) he rivial soluion x = is asympoically sable if de(a) = a > and -a 1 = r(a) <. In his case he energy funcional defined in (5.7) is posiive definie and sricly decreasing for nonrivial soluions. Funcionals on phase space having his propery will be called Lyapunov funcionals. The exisence of an appropriae Lyapunov funcional will imply he sabiliy resul. 3. Applicaions An example from pharmacokineics. The reacion equaion k1 k 1 A k1 A describes a basic model for sudying he dynamics of a drug in he body where A 1 and A represen he drug concenraion in issue and blood, respecively. The rae k is due o removal from he body by he kidneys. The corresponding sysem of differenial equaions is hen y = k y k y = y y k y k y k (5.17) Hence A k k 1 1 = k1 k1 k (5.18) 8

9 and r(a) <, de(a) = k 1 k >. The rivial soluion is asympoically sable: he drug will compleely disappear as one expecs. 3.. The mahemaical pendulum wih fricion. The equaion for he mahemaical pendulum wih fricion is given by 1 x + γ x = gl sin ( x) (5.19) By heorem 1 he saionary soluion x = is asympoically sable. 5.. How o solve linear equaions wih consan coefficiens We consider he differenial equaion y = Ay for a real n n marix A. While i is sraighforward o find all soluions if all eigenvalues o f A have algebraic mulipliciy one (see heorem 4.5) he compuaion of soluions in he case of muliple eigenvalues is a lile bi more involved. 1. A soluion algorihm. We describe a rouine o compue a fundamenal sysem of any linear differenial equaions wih consan coefficiens. ( sep 1) Compue he characerisic polynomial P A(σ) and compue is zeros which are he eigenvalues of A. (sep ) For simple eigenvalues λ find he corresponding eigenvecors w. consruc he soluion as in heorem 4.5, ha is se y = e λ wand ake real and imaginary par in case o f complex conjugae eigenvalues λ, λ. (sep 3)Le λ be an eigenvalue wih m ulipliciy k > 1. Firs of all, solve for corresponding eigenvecors, i.e. solve ( A λe) w= The number of linear independen soluions n 1 is equal o he dimension of he nullspace (or kernel) of A λe. The soluions of he differenial equaion are consruced as in (sep ) yielding n 1 linearly independen soluions for he eigenvalue λ. (sep 4) If n 1 = k, hen you have found all soluions for his eigenvalue. If n 1 < k, compue he nullspace of (A λe), i.e. solve 9

10 ( A λe) w= The number of linear independen soluions n is equal o he dimension of he nullspace (or kernel) of (A λe). Take he n n 1 linear independen soluions, which do no belong o he nullspace of A λe, i.e. A λe w. The corresponding soluion y() of he differenial equaion is given by y() = exp(a)w. I can be explicily compued as follows: T exp( A) w = exp( ( λe + A λe) ) w = exp( λ ) exp ( λ ) ( E A E w e λ = E+ ( λ ) ( λ ) = + A E + 1 A E +... w ( λ ) λ λ e w e A E w he las equaliy is a consequence of ( A λe) w= imaginary par in case of comp lex conjugae eigenvalues λ, λ. n ) for all n > 1. Again ake real and (sep 5) If n = k, hen you have found all soluions for his eigenvalue. If n < k, hen compue he nullspace o f (A λe) 3 and proceed as before. (sep 6) Repea he procedure for he following powers of A λe unil he dimension of he nullspace and hence he number of independen soluion of he differenial equaion equals k.. Examples. We presen a few explici compuaions of soluion in he case of muliple eigenvalues..1. Solve he Cauchy problem y = 1 1 y, y() = 1 3 (5.) If A denoes he marix in (5.) hen P A (λ) = (λ 1) 3 and herefore λ = 1 is a riple eigenvalue of A. Le {e 1, e, e 3 } denoe he sandard orhonormal basis of R 3. We compue 1

11 1 A E = 1 We see ha he nullspace of A E is spanned by e 1. Hence y 1 () = exp() e 1 solves (5.). We compue (A E). ( A E) = The nullspace of (A E) is spanned by e 1 and e. Only e yields a new soluion of he differenial equaion. According o he general procedure described above i is given by y () = ee + e A E e = ee + e e We noe ha since (A E) 3 = is nullspace is R 3. Take e 3 as he hird vecor spanning R 3. The hird linear independen soluion is y3() = ee3 + e ( A E) e3+ e ( A E) e3 = e + e 1 + e 1 We have compued he fundamenal marix 1 Y() ( y (), y (), y () ) e 1 1 = 1 3 = Since Y() = E we have Y() = exp(a). The soluion of he Cauchy problem is herefore () y = e 1 = e + 3 (5.1) Solve he Cauchy problem 11

12 y = y, y() = 1 (5.) If A denoes he marix in (5.) hen P A (λ) = λ 3 and herefore λ = is a riple eigenvalue of A. The nullspace of B is spanned by he vecors 1 1 w1 = 1, w = 1 Hence y 1 () = w 1 and y () = w solve (5.). We noe ha since A = is nullspace is R 3. Take w 3 = w = (,1,) as he hird vecor spanning R 3. The hird linear independen soluion is 1 y3() = e + Ae = = 1+ We have compued he fundamenal marix 1 1 Y() = ( y1(), y(), y3() ) = We easily check ha exp A ( 1+ = y3 y1, y3, y y1 + y3 ) = 1+ 1 The soluion of he Cauchy problem is herefore y () = 1+ = (5.3) 1

13 .3. Consider he equaions of moion for wo coupled pendulums x = ax k x y y = ay k( y x) (5.4) where a and k are posiive consans. The marix of he corresponding firs-order sysem T for he vecor x, xyy,, is given by 1 a k A = k 1 k a k i has degenerae purely imaginary eigenvalues ± i a, ± i a+ k and can be solved by he procedure described above. In his case, however, i is more convenien o decouple he equaions of moion by considering he equaions for he sum and he difference of x and y, which are given by ( y) x+ y = a x+ x y = a+ k ( x y) (5.5) which are easily solved. 13