Ordinary Differential Equations

Transcription

1 Ordinary Differential Equations Arvind Saibaba Institute for Computational and Mathematical Engineering Stanford University September 16, 2010

2 Lorenz Attractor dx dt dy dt dz dt = σ(y x) = x(ρ z) y = xy βz σ is Prandtl number and ρ is Rayleigh number.

3 Quantum mechanics Time dependent Schrödinger s equation i t Ψ = ĤΨ ) = ( 2 2m 2 +V(r) Ψ Time independent Schrödinger s equation ĤΨ = ÊΨ

4 Vampire Population dv dt dh dt = av +dvh = nv dvh a is the death rate of vampires due to contact with sunlight, crucifixes, garlic and vampire hunters. a 1992 a J. Optimization Theory and Applications: Vol. 75, No.3,

5 What is an ODE? In general, an n th order ODE can be written as F(x,y,y,y,...,y (n) ) = 0 We shall assume that the differential equations can be solved explicitly for y (n) in terms of the remaining qunatities y (n) = f(x,y,y,...,y (n 1) ) A differential equation is said to be linear if it is linear in y and all its derivatives. Thus, an n th order ODE can be written as P n [y] = p 0 (x)y (n) + +p n (x)y = r(x) If r(x) = 0, it is called homogenous.

6 An Existence Theorem Theorem Let y 0 B, an open subset of R n, I R an interval containing t 0. Suppose F is continuous on I B and satisfies the following Lipschitz estimate in y: F(t,y 1 ) F(t,y 2 ) L y 1 y 2 for t I,y j B. Then the equation dy dt = F(t,y) y(t 0) = y 0 has a unique solution on some t interval containing t 0.

7 Separation of Variables If, the equation dy dx = f(x,y) can be written as dy dx = g(x)h(y) then, the ODE is separable. If h(y) 0, then 1 h(y) dy = g(x)dx is the desired solution, y is either an implicit or explicit function of x, upto an unknown constant of integration.

8 Example Solve Solution: dy dx = (1+y2 )e x 1 1+y 2dy = ex dx (1) 1 1+y 2dy = e x dx tan 1 (y) = e x +C As such, this is an implicit solution. Taking tan on both sides y = tan(e x +C)

9 Exact Equations Theorem If the functions M(x,y) and N(x,y) along with their partial derivatives M y (x,y) and N x (x,y) be continuous in the rectangle S : x x 0 < a, Y y 0 < b (0 < a,b < ). Then the ODE M(x,y)dx +N(x,y)dy = 0 or M(x,y)+N(x,y)y = 0 is exact iff M x = N y If the ODE is exact the implicit solution is M +Ny = u x +u y y = 0 u(x,y) = c then we must have u xy = M y and u yx = N x. Since M y and N x are continuous, we must have that u xy = u yx

10 Proof Start with the equation u X = M. Integrating both sides u(x,y) = x x 0 M(s,y)ds +g(y) We shall obtain g(y) from the other equation u y = N. We have Thus, g(y) = y x M(s,y)ds +g (y) = N(x,y) y x 0 y 0 N(x,t) x So that the solution is given by y N(x,t)+ y 0 x 0 M(s,y)ds + x x x 0 M(s,y 0 )ds = c x 0 M(s,y 0 )ds +g(y 0 )

11 Example Solve the ODE (y +2xe y )+x(1+xe y )y = 0 Solution Here, M = y +2xe y and N = x(1+xe y ). We have M y = N x = 1+2xe y, (x,y) S = R 2. Thus, the given ODE is exact in R 2. Taking (x 0,y 0 ) = (0,0) we have y x (x +x 2 e t )dt + 2sds = xy +x 2 e y = c 0 0

12 Integrating Factors Consider the following ODE dy +p(x)y = r(x) dx Set r(x) = 0, to obtain the homogenous solution y h (x) dy h dx +p(x)y h = 0 (2) dyh = p(x)dx y h ( ) y h = c 0 exp p(x)dx for a well defined constant c 0

13 For r(x) 0, try solution of form Integrating Factors contd. y p = u(x)exp ( ) p(x)dx By chain rule, { } ( du dyp = dx dx +p(x)y p exp ( ) = r(x)exp p(x)dx ( u(x) = r(x)exp p(x)dx ) p(x)dx ) dx +C (3) So that, y p (x) = exp ( ){ p(x)dx ( r(x)exp ) } p(x)dx dx +C

14 Example dy dx 4 x y = x5 e x I.F. exp ( ) p(x)dx (4) = e 4log(x) = x 4 Now try y p = u(x)x 4 du = xe x (5) dx u = xe x e x To complete the solution y = x 4 [(x 1)e x +C]

15 Consider the first order linear ODE Duhamel s Principle dy dt = a(t)y +b(t) y(0) = y 0 where, a(t) and b(t) are continuous real valued functions. Define A(t) = t 0 a(s)ds The above ODE can be written as e A(t) d ( ) e A(t) y = b(t) dt which yields t y(t) = e A(t) y 0 +e A(t) e A(s) b(s)ds 0

16 Linearity of Solutions Consider the homogenous second-order ODE p 0 (x)y +p 1 (x)y +p 2 (x)y = 0 (6) where p 0 (x)(> 0), p 1 (x) and p 2 (x) are continuous in [a,b]. We have Theorem There exist two solutions y 1 (x) and y 2 (x) of eqn. 6 which are linearly independent in [a,b]. Because of linearity, for arbitrary constants c 1 and c 2, y(x) = c 1 y 1 (x)+c 2 y 2 (x) is also a solution to eqn. 6.

17 Linear Independence Define the Wronskian W(x) as W(x) = y 1(x) y 2 (x) y 1(x) y 2(x) = y 1(x)y 2(x) y 2 (x)y 1(x) Theorem Two solutions y 1 (x) and y 2 (x) of eqn. 6 are linearly independent if the Wronskian, as defined above, is non-zero for some x 0 [a,b].

18 Variation of Parameters Suppose, one solution y 1 (x) of eqn. 6 is known. Substitute y(x) = u(x)y 1 (x), p 0 (uy 1 ) +p 1 (uy 1 ) +p 2 (uy 1 ) = 0 (7) p 0 y 1 u +(2p 0 y 1 +p 1 y 1 )u +(p 0 y 1 +p 1 y 1 +p 2 y 1 ) = 0 }{{} =0. Why? Let v = u and let y 1 0 p 0 y 1 v +(2p 0 y 1 +p 1 y 1 )v = 0 It can be shown that v(x) = 1 ( x ) p 1 (t) y1 2 exp (x) p 0 (t) dt So that, x ( 1 t ) p 1 (s) y 2 (x) = y 1 (x) y1 2 exp (t) p 0 (s) ds dt

19 VOP Example It is easy to verify that y 1 (x) = x 2 is a solution of the differential equation x 2 y 2xy +2y = 0 x 0 For the second solution, x 1 y 2 (x) = y 1 (x) y1 2 exp (t) ( = x 2 x 1 t 4 exp = x 2 x 1 t 4t2 dt = x ( t p 1 (s) t ( 2s s 2 p 0 (s) ds ) ) ds dt ) dt (8)

20 Inhomogenous equations Theorem Let y 1 (x) and y 2 (x) be two linearly independent solutions to the homogenous equation 6. Let y p be a particular solution to the ODE. Then, the general solution to the ODE is given by Further, y p can be computed as y p (x) = y 1 (x) y(x) = c 1 y 1 (x)+c 2 y 2 (x)+y p (x) x r(t)y 2 (t) x r(t)y 1 (t) dt +y 2 (x) dt W(t) W(t) where, W(x) is the Wronskian defined before.

21 Constant coefficient case Consider, ay +by +cy = 0 (9) We try solutions of the form y = e rx. This results in the algebraic quadratic equation known as the characteristic equation. ar 2 +br +c = 0 Real, distinct roots r 1,r 2. e r1x and e r2x are two solutions of eqn. 9 and the general solution is y(x) = c 1 e r1x +c 2 e r2x

22 Constant coefficient case contd. Repeated, real roots r 1 = r 2 = r. One solution is, of course, e rx. To find the other solution, we use the method of Variation of Parameters. It is easy to show that y 2 = xe rx, so that y(x) = e rx (c 1 +c 2 x) Complex, conjugate roots r 1 = µ+iν,r 2 = µ iν Using Euler s formula e iθ = cosθ +sinθ, we can write the general solution as y(x) = c 1 e µx cosνx +c 2 e µx sinνx

23 Euler-Cauchy equations x 2 y +axy +by = 0 x > 0 Also, known as equidimensional ODE s. Plugging y = x m, we have x 2 m(m 1)x m 2 +axmx m 1 +bx m = 0 or, m(m 1)+am+b = 0 Depending on the roots, we have three cases 1. Real, distinct roots, m 1,m 2 : y(x) = c 1 x m1 +c 2 x m2 2. Real, repeated roots, m 1 = m 2 = m: y(x) = x m1 (c 1 +c 2 logx) 3. Complex conjugate roots, m 1 = µ+iν,m 2 = µ iν: y(x) = x µ (c 1 cos(νlogx)+x 2 sin(νlogx))

24 Higher Order Homogenous Constant Coefficient ODE s Consider the ODE Define, d n y a 0 dt n +a d n 1 y 1 dt n 1 + +a ny = 0 z 1 = y (10) z 2 = z 1 = y. z n. = z n 1 = y (n 1) z n = a 1 a 0 z n a 2 a 0 z n 1 a n a 0 z 1

25 System of ODEs z 1 z 2.. = z n 1 z n Can be written as which has the solution z(t) = e At z an a } 0 an 1 a 0... a2 a {{ 0 a1 a 0 } A dz dt = Az z(t = 0) = z 0 z 1 z 2.. z n 1 z n }{{} z

26 Exponential of a Matrix e At = I+At + (At)2 + (At) ! 3! This series always converges and ( e As )( e At) = e A(s+t) ( e At )( e At) = I If the matrix A is diagonalizable, i.e. A = SΛS 1 e At = I +SΛS 1 t +SΛ 2 S 1t2 2! +SΛ3 S 1t (11) ) 3! = S (I +Λt + (Λt)2 + (Λt) S 1 2! 3! = Se Λt S 1 Otherwise, appeal to Jordan Canonical form.

27 Euler s method We want to approximate the solution of the equation Define the points y (t) = f(t,y(t)) y(t 0 ) = y 0 t 0 = a, t i = a+ih, i = 1,2,...,N h = b a N Starting with the differential equation, we replace the derivative with a finite difference approximation y (t) = y(t i+1) y(t i ) h +O(h) So, using this approximation we have the forward Euler method which defines the approximate solution {y i } N i=0 y i+1 = y i +hf(t i,y i )

28 Properties Define the error, e i = y(t i ) y i,i = 1,...,n. For the forward Euler method we expect e i = O(h) and assuming the function is Lipschitz continuous, one can also derive the following error bound where, e i+1 Mh 2L [el(t i+1 a) 1] M = maxa t b y (t)

29 Boundary Value problems Consider the Poisson s equation u (x) = f(x) x [0,1] u(0) = u(1) = 0 We start with a central difference approximation to the second derivative. Discretizing the domain into equal intervals with N + 1 points, x i = ih,h = 1/N,i = 0,...,N u (x i ) = f(x i ) = u i+1 2u i +u i 1 h 2 +O(h 2 )

30 This results in a tridiagonal system of equations 2 1 u 1 f(x 1 ) h u 2... = f(x 2 ) u n f(x N ) The solution to the problem using Finite Differences κu +βu = f