n n. ( t) ( ) = = Ay ( ) a y

Transcription

1 Sstems of ODE Example of sstem with ODE: = a+ a = a + a In general for a sstem of n ODE = a + a + + a n = a + a + + a n = a + a + + a n n n nn n n n Differentiation of matrix: ( t) ( t) t t = = t t ( t) ( t) a a Usuall can take the form ( t) = = A = a a a a a n a a an Or in general A = = an an a nn n Example : mixing problem with tanks Each contain gal of water In Twe have pure water and in T we have 5 lb of fertilizer Circulation of water gal min (and stirring instantaneous mixing) Concentration of fertilizer with time t and ( t )

2 inflow outflow Model: ( t) = min min = This gives ODEs: =.. Equivalent to =.. We tr a solution of the form: = xe = λxe = Ax e This is an eigenvalues problem: Ax= λx Ax λx = Det A Iλ = With nontrivial solution if λt λt λt. λ. = (. λ) (.) = λ( λ +.4) =.. λ Two eigenvalues λ = and λ =.4 and two eigenvectors The solution constants x = and x λt λt.4t = c x e + c x e = c + c e where Initial conditions: = and = 5 c+ c = = c c c 75 and c 75 + = = = 5 c c = 5 = c and c are arbitrar At what time t concentration is ½ that of t? This happens when t contains /3 of total.4t.4t ln(3) amount 75 75e = = 5 e = t = = 7.5min 3.4

3 Example Electric network We want to know the currents I ( t ) and I we assume all currents and charges to be zero at t = The model is based on Kirchhoff s voltage law: In the left loop t in the two loops of the above circuit and the voltage drops over the inductor LI = I [ V] the voltage drops over the resistor R ( I I ) = 4( I I )[ V] + 4 = The sum of voltage must be equal to that of the batter I ( I I ) In the right loop the voltage drops over the first resistor R( I I) = 4( I I)[ V] the voltage drops over the resistor RI = 6I [ V] the voltage drops over the capacitor Idt = 4 Idt [ V] The sum must be equal to zero 6I + 4 I I + 4 Idt = or I 4I + 4 Idt = C Dividing b ten and deriving I.4I +.4I = Replacing b the value of I =.6I +.I In matrix form I.4 4 I I +.4I = I in left loop: I. = A + = +.6. I 4.8 J J g This is a nonhomogeneous sstem 3

4 We first look for a solution of the homogenous sstem tring J = xe J = λxe = Ax e λt λt λt I = A =.6. I J J The nontrivial solution ( A Iλ) ( λ ) det = =.6. λ λ +.8λ+.6 = with two eigenvalues And two eigenvectors x = and x =.8 Consistent with general homogenous solution Jh = cx e + cx e = c e + c e.8 t.8t t.8t λ = and λ =.8 To find a particular solution, we note that since g is a constant we can tr a constant J p = asuch that Aa g 4.a + 4.a +.=.6a +.a + 4.8= a. + = +.6. a 4.8 a = 3 and a = or 3 a = So the general solution 3 t.8t t.8t J = Jh + Jp = cx e + cx e + a = c e + c e +.8 t.8t t.8t I = ce + ce + 3 and I ce = +.8ce Appling the initial conditions I = c+ c+ 3= and I = c +.8c = c = 4 and c = 5 4

5 Solution 3 t.8t t.8t J = Jh + Jp = 4.x e + 5.x e + a = 4. e + 5. e +.8 t.8t t.8t I = 8.e + 5.e + 3 and I 4.e = + 4.e Solutions shown individuall look like two damp vibrations Phase plane Shows the solution using I and I as degrees of libert of sstem (parametric representation) where the arrow shows the sense of variation with time The current Irises almost linearl in time with I both reaching a maximum But then Idrops to zero while the current I reach a lower constant value below its maximum 5

6 Conversion of nth-order ODE to a sstem Permits stud and solution of single ODE b method of sstem Open wa of including theor of higher order ODE into that of first order sstem Nth-order ODE ( n) n = (,,, ) F t Transformations: =, =, 3 =,, n = n Sstem of the form = = 3 = n Such that = F( t,,, ) n n n And consequentl ( n) = n Example 3: mass on a spring c k m + c + k = or = m m Appling the transformations: =, =, 3 c k = and = 3 = = m m = Setting = A = = = k c m m The characteristic equation det λ A I k c λ m m λ = = λ + λ+ c m k m 6

7 Numerical example: m =, c = and k =.75 c k λ + λ + = λ + λ+ = λ + λ+ m m Two eigenvalues λ =.5 and λ =.5 And two eigenvectors x = and x =.5 Yields general solution = c e + c e.5.5t.5t Therefore = ce + ce and.5t.5t = = ce.5ce.5t.5t Basic theor of sstem ODEs General sstem = f t,,,, ( n ) (,,,, ) = f t (,,,, ) = f t n n n We can write the sstem as = f ( t ) n A solution on some interval a< t< bis a set of differentiable functions = h ( t),, n = hn( t) In vector form h = h n h such that = h ( t ) Initial value problem n-initial conditions ( t) = K, ( t) = K, ( ) In vector form ( t ) = K, n t = Kn 7

8 Theorem : existence and uniqueness of solution Let f, f = f t be continuous functions having continuous partial derivatives in, n δ f δ fn δ fn,,, in some domain R of t n -space containing the point δ δ δ ( t K K ), then = ( t ) n,,, n f has a solution on some interval t α < t < t + α, satisfing the initial value problem and this solution is unique Linear sstem Has the form = A+ g which is nonhomogeneous or homogenous when g = n For a linear sstem we have = a ( t),, = a ( t) f Theorem : existence and uniqueness of solution Let a jk and g j be continuous functions on open interval t = t, then = A + g has a solution ( t) f nn in theorem n which is unique α < t < β containing the point Superposition principle If and are solutions of the homogeneous linear sstem = A then the combination = c + c is also a solution PROOF: We simpl take the derivative = c + c = ca + ca = Ac + c = A 8

9 Basis, General solution and Wronskian A basis or fundamental sstem of solutions of homogeneous ODEs interval J is a linearl independent set of n-solutions and the general solution is a combination arbitrar constants ( n),, ( n) c c n = A in some of homogeneous ODEs = + + where c,, cn are We can write the n-solutions ( n),, as column of n nmatrix ( n) Y = And the determinant of this matrix is Wronskian W ( n) (,, ) ( n) ( n) = ( n) n n n The solution form a basis on J if and onl if W at an time t in this interval Then ( n) Y = is a fundamental matrix Then = Ycwhere c c = c n Particular case: if and z are solutions of -order homogeneous ODE then z W(, z) = z 9

10 Constant coefficient sstems Assume = A With constant coefficients n nmatrix A= a jk all independent of t Solution = K using = ce Eigenvalues problem Ax λx = kt = = = λt λt xe λxe A If A has linearl independent set of eigenvalues, example A smmetric ( a a kj jk ) skew smmetric ( a = a kj jk ) or n -different eigenvalues λ,, λn then we have ( n) x,, x The Wronskian W (,, ) eigenvectors and a related set of solutions λt λ t ( n) λt λ t ( n) λt λ t ( n) = x = x = x e λ t e e λt n n λnt = or ( n) x e x e x e x x x λnt ( n) x e x e x e x x x = = e λnt n λt+ λt+ + λnt ( n) x e x e x e x x x λnt n n n n n n The first term is an exponential which is never zero The second term is not zero because columns are independent eigenvectors General solution If constant matrix in ( n),, = A has linearl independent set of eigenvalues then form a basis and the general solution is λ t λ t ( n) = cx e + c x e + + c x e λ n nt

11 Phase plane Qualitative method of obtaining general qualitative information on solutions without actuall solving ODE or sstem Created b Henri Poincaré (854-9) French mathematician who worked on complex analsis, divergent series, topolog and astronom In phsics phase plane is rmr - = rmv - = rp - Consider two ODEs then a a = a a = A takes the form With solution ( t) ( t) t = which can be graphed as two simple curves, or together in the phase plane The parametric curve in this phase plane is the trajector Example 4 3 = A = 3 The characteristic equation ( A λi) λ λ ( x )( x ) Two eigenvalues λ = and λ = 4 corresponding to two eigenvectors x = and x = and the general solution t = c e + c e 4t Phase portrait two straight trajectories c = and c = Other values of constants ields other trajectories det = = =

12 Critical points In example 4, the point is a critical point = common point for all the trajectories From d dt a + a = A we get that = = = d dt a + a This associates with ever point (, ) passing through P, except at, d P a unique tangent direction of trajector d d P = where d = There are 5 different tpes of critical points: which is a critical point Improper node All the trajectories except two have the same limiting direction of the tangent The two exceptional trajectories also have limiting direction of tangent at P but the are different The common limiting direction at zero is that of x = T 4t because e goes faster to [ ] zero than e t The two exceptional directions are x = T and x = T [ ] [ ] Proper node Ever trajector as limiting direction and for an given direction dat P there is a trajector having das its limiting direction Unit matrix = Characteristic equation ( λ ) = t t Solution = c e + c e Trajectories c = c (linear)

13 Saddle point Two incoming and two outgoing trajectories through P and all the other trajectories in neighborhood of P bpass P Sstem = λ λ = Characteristic equation t Solution = c e + c e t famil of hperbolas + coordinate axes Center + = constant famil of ellipses Spiral r = ce t famil of spirals for each real c Trajectories = constant (hperbolas) Critical point is enclosed b infinitel man trajectories Sstem = 4 Characteristic equation λ + 4 = it it with solution = c e + c e i i Next step = transformation to real form b Euler formula Shortcut: multipl equations = and = 4 4 = and integrate Critical point P about which the trajectories spiral approaching it as t Sstem = Characteristic equation λ + 4 = ( + it ) ( it ) Solution = c e + c e i i + = + = r in Shortcut: ( ) polar coordinates differentiating dr rr = r = dt ln r = + t c r * 3

14 Degenerated node no basis Example 4 = The matrix A is not skew smmetric ( A T A ) Characteristic equation ( λ 3) = double root λ = 3 With corresponding one eigenvector In fact x λt λt = x te + µ e with µ = [ µ µ ] = and some non zero multiple of another T Taking derivative = xe + x λte + λµ e = A = Ate x + Aµ e λ t λ t λ t λ t λ t Since Ax= λx the term xλte λt cancel then dividing b e λt ( λ ) x + λµ = Aµ A I µ = x Where λ = 3and = x so that ( A 3I) Thus we have µ + µ = and µ µ 4 3 µ µ = = 3 µ = which suggest µ = T 3t 3t And the solution = c + c = c e + c t+ e Critical point at origin is degenerate node c gives heav straight line lower part c > and upper part c < right part of heav curve from through second, first and fourth quadrant gives the other part from through fourth, third and second quadrant 4

15 Critical points vs. stabilit of solutions Consider an sstem of homogeneous ODEs with constant coefficients a a = = A = a a On the phase plane, the solution ( t) = ( t) ( t) portrait d dt a + a With critical point such that = = = d dt a + a T ields trajectories = phase The relation between critical point and eigenvalues pass b characteristic equation det a λ a A λi = = λ ( a+ a ) λ + det( A) = a a λ This is a quadratic equation λ pλ + q = with solutions λ = p + and λ = p Where p ( a a ) q= det( A) = ( a a a a ) = + is the trace of A = p 4q If we express the solution in terms of product λ pλ+ q= ( λ λ )( λ λ ) Then we have p = λ + λ q = λλ = ( λ λ ) Eigenvalue criteria for critical points p = λ+ λ q λλ = = λ λ eigenvalues Node q > Real same sign Saddle point q < Real opposite sign Center p = q > Pure imaginar Spiral point p < Complex, not pure imaginar 5

16 Stabilit of critical points Stable all the trajectories close to P remain close to P in future For ever disk D ε of radius ε > with center P, there is a disk D δ with radius δ > P such that ever trajector that has point P( t ) center points t t in D ε If condition not verified = unstable in D δ has all its subsequent Stable and attractive - P is stable and all trajectories with point within D ε approaches P as t (asmptoticall stable) and For example, q > implies λλ have same sign or complex conjugate In this case, p < means eigenvalues are negative or have negative real part If < λ = α+ iβ and λ = α iβ thus p = α < implies spiral attractive If p > unstable spiral point If p = λ = λ q = λ, but if q > λ = q< which implies that the two eigenvalues are pure imaginar periodic solutions with trajectories closed curves around critical point 6

17 3 Example 5 = A = 3 p = 6, q = 8, = 4 node whish is stable and attractive Example 6 mass on spring c k m + c + k = or = m m = A = k c m m λ A I k c λ m m det λ = = λ + λ+ = c k p =, q = and m m c k = 4 m m c m No damping c = which ields p = and q > the critical point is a center k m Underdamping c < 4mk which ields p <, q >, < stable attractive spiral Critical damping c = 4mk which ields p <, q >, = stable attractive node Overdamping c > 4mk which ields p <, q >, > stable attractive node 7

18 Numerical example: m =, c = and k =.75 c k λ + λ + = λ + λ+ = λ + λ+ m m Two eigenvalues λ =.5 and λ =.5 And two eigenvectors x = and x =.5 Yields general solution = c e + c e.5.5t.5t Therefore = ce + ce and.5t.5t = = ce.5ce.5t.5t 8

19 Non linear sstems Solution b analtic method is difficult or impossible First order non linear sstems: = f and thus (, ) (, ) = f = f If sstem is autonomous (does not depend on t ) extended phase plane methods give characterization of various general properties of solutions Advantage over numerical methods, which give onl one (approximate) solution at a time (but with higher precision) If first order nonlinear solution has several critical points discuss one after the other each time moving point P :( ab, ) to be discussed at origin appling translation = a and = b P is isolated onl critical point within sufficientl small circular disk with center on origin Linearization of nonlinear sstems Assume fand f are continuous and have continuous partial derivatives in neighborhood of P A linear sstem approximating = two exceptions) Since P is critical point f f f have same kind of critical points (with, =, = Transformation = A+ h and thus (, ) (, ) = a + a + h = a + a + h If Det A then the tpe and stabilit of P is the same as that of critical point of linear sstem = a+ a = A and thus = a + a The above assumption on derivatives implies that h and h are small near P Two exceptions : eigenvalues are equal (degenerated) or pure imaginar, then in addition to the tpe of critical points of linear sstem the nonlinear sstem ma have a spiral point 9

20 Example free undamped pendulum Mathematical model mlθ + mg sinθ = g Dividing b mland writingk = : θ + k sinθ = L This is a nonlinear equation When small value of θ sinθ θ and approximate form θ + kθ = has solution Acos kt + Bsin kt Critical points b linearization: Set θ = and θ = then nonlinear sstem The right side both zero when = and ( nπ,) where n =, ±, ±, (, ) = f = = f, = ksin = nπ infinit of critical points Consider (,) the McLaurin series sin = + 6 = Linearized sstem = A = k and thus = k We need A = k > and = p 4q= 4k p =, q = det 3 From this we conclude that (,) is a center alwas stable sinθ = sin is periodic with period π all critical points (,) Since n =±, ± 4, are centers nπ with

21 Consider now ( π,) setting θ π = and ( θ π ) = θ = therefore 3 sinθ = sin( π) = sin = = And linearized sstem = A = k and thus = k Which ields that p =, q = det A = k ( < ) and = = p 4q 4k From which we conclude that (,) alwas unstable nπ with n =±, ± 3, are saddle points which are

22 Example damped pendulum Introduce damping term proportional to velocit θ + cθ + ksinθ = Linearized sstem = = ksin c Critical points at same locations ( π ) ( π ) Consider (, ) = = A = k c and = k c,, ±,, ±,, For small damping spiral point For critical point (,) = = + p 4q c 4k For c > a saddle point π we have p a a c = + =, q = det A = k ( < ) Damping = lost of energ instead of closed trajectories we spiraling ones no more trajectories connecting critical points and

23 Transformation to a first-order equation in phase plane,, " = Consider differential equation of second order F( ) Transform it to first order taking = and = and transforming b chain rule d d d d = = = dt d dt d d Then F,, = d Example for free undamped pendulum We get d = k sin d Separating the variables and integrating d = ksin d + C cos = k + C and multipling b ml m L ml kcos mlc = These terms are energies is the angular velocit so L is the velocit and first term is kinetic velocit Second term is the potential energ mlcis the total energ, which is constant (undamped sstem) 3

24 Tpe of motion depends on C Smallest possible C = k then = and cos =, the pendulum is at rest Change of direction of motion = θ = then kcos + C = If = π then cos = and C = k, hence if k< C< kthe pendulum oscillate (closed trajectories) If C > k then = is impossible and pendulum makes a whirl motion appearing a wav trajector If C = k correspond to two separating trajectories connecting saddle point 4

25 Example 3 van der Pol equation There are phsical sstems such that for small oscillations energ is fed into the sstem (negative damping) while for large oscillation energ is taken from the sstem We expect such sstem to approach a periodic behavior = closed trajector = limit ccle Van der Pol equation µ + = with µ > a constant First occurred in stud of electrical circuits containing vacuum tubes The damping term µ ( ) < for small oscillations > Setting = and = d µ ( ) + = d d Isoclines in plane K d = d = µ ( ) = K d Solving = µ ( ) K isoclines for µ =. Limit ccle almost a circle because + = and two trajectories approaching it from interior and exterior 5

26 For µ = limit ccle no longer a circle and approach on trajectories more rapid 6

27 Non homogeneous linear sstems 7