Numerical Methods for Ordinary Dierential Equations Episode 1 docx

[...]... J(x) = −k1 − k2 y3 − 2k3 y2 2k3 y2 −k1 − k2 y2 0 , and the characteristic polynomial is λ2 + (k1 + k2 y3 + 2k3 y2 )λ + 2k3 y2 (k1 + k2 y2 ) (10 4i) 16 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS 1. 0 0.5 y −0.0 10 4 −0.2 10 4 10 1 y3 10 −2 10 −3 10 −4 10 −5 0 −0.6 10 4 λ y2 0 0.2 0.5 1 2 −0.4 10 4 y1 5 10 20 50 10 2 10 3 λ −0.8 10 4 10 4 1. 0 10 4 x Figure 10 4(i) Solution and most negative eigenvalue for the... exist Theorem 10 1A The quantities 1 2 2 2 2 y + y4 − (y1 + y2 ) 1/ 2 , H= 2 3 A = y1 y4 − y2 y3 are constant Proof We verify that the values of dH/dx and dA/dx are zero if y satisfies (10 1a)– (10 1d) We have dH dy3 dy4 dy1 2 dy2 2 2 2 = y3 + y4 + y1 (y + y2 )−3/2 + y2 (y + y2 )−3/2 dx dx dx dx 1 dx 1 y1 y3 y2 y4 y1 y3 y2 y4 =− 2 − 2 + 2 + 2 2 2 2 2 (y1 + y2 )3/2 (y1 + y2 )3/2 (y1 + y2 )3/2 (y1 + y2 )3/2... ) i =1 N (yj − zj ) (yj 1 − 2yj + yj +1 − zj 1 + 2zj − zj +1 ) = (N + 1) 2 j =1 N 1 N (yj − zj )(yj +1 − zj +1 ) − 2(N + 1) 2 = 2(N + 1) 2 j =1 (yj − zj )2 j =1 N = −(N + 1) 2 (yj +1 − yj − zj +1 + zj )2 j=0 ≤ 0 Another aspect of the discretization that might be explored is the spectrum of the matrix A, in comparison with the spectrum of the linear operator 2 u → d u on the space of C 2 functions on [0, 1] for. .. Period of simple pendulum for various amplitudes Θ 0◦ 3◦ 6◦ 9◦ 12 ◦ 15 ◦ 18 ◦ 21 24◦ 27◦ 30◦ T 6.28 318 53072 6.28426208 31 6.28749444 21 6.2928884880 6.3004544 311 6. 310 20664 31 6.32 216 37356 6.3363486630 6.35278885 01 6.3 715 163462 6.3925680085 where q = θ and p = dθ/dx The second order equation (10 3n) is now equivalent to the first order system p q = 0 1 1 0 ∂H ∂p ∂H ∂q Differential index and index reduction Carry... an ordinary differential equation system, which mimics the behaviour of the parabolic equation, let y1 (t), y2 (t), , yN (t), denote the values of u( N1 , t), u( N2 , t), , u( NN , t), respectively That is, +1 +1 +1 yj (t) = u j ,t , N +1 j = 0, 1, 2, , N + 1, 8 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS where we have included y0 (t) = u(0, t), yN +1 (t) = u (1, t) for convenience For. .. dr dθ dy1 = cos(θ) − r sin(θ), dx dx dx dr dθ dy2 = sin(θ) + r cos(θ), y4 = dx dx dx √ so that, because y1 y4 − y2 y3 = 1 − e2 , we find that y3 = r2 dθ = dx 1 − e2 (10 1f) From (10 1e) and (10 1f) we find a differential equation for the path traced out by the orbit dr 2 1 = r 2 e2 − (1 − r)2 , dθ 1 − e2 DIFFERENTIAL AND DIFFERENCE EQUATIONS 7 and we can verify that this is satisfied by 1 − e2 = 1 + e cos(θ)... + 2y3 + 2y4 = y4 + 2y3 (−y1 y5 ) + 2y4 (−y2 y5 + 1) , dx dx dx dx which simplifies to dy5 = 3y4 dx (10 3q) 14 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS Given that consistent initial values are used, it seems that the equations (10 3f)– (10 3i) together with any of (10 3j), (10 3o), (10 3p) or (10 3q) give identical solutions Which of the possible formulations should be used? From the point of view... vN +1 = 0, we find that it is possible to write (10 2c) in the form vj 1 − qvj + vj +1 = 0, j = 1, 2, , N, (10 2d) where q = 2 + λ/(N + 1) 2 The difference equation (10 2d) has solution of the form vi = C(µi − µ−i ), where µ + µ 1 = q, unless q = ±2 (which is easily seen to be impossible) Because vN +1 = 0, it follows that λ2N +2 = 2 Because µ = 1, it follows that µ = exp with i = √ kπi N +1 , k = 1, ... = y4 , (10 3g) = −y1 y5 , (10 3h) = −y2 y5 + 1, (10 3i) = 1 (10 3j) It will be convenient to choose initial values defined in terms of θ = Θ, with 12 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS the velocity equal to zero That is, y1 (0) = sin(Θ), y2 (0) = cos(Θ), y3 (0) = y4 (0) = 0, y5 (0) = cos(Θ) The five variables are governed by four differential equations (10 3f)– (10 3i), together with the single... convenience For j = 1, 2, , N , ∂ 2 u/∂x2 , evaluated at x = j/(N + 1) , is approximately equal to (N + 1) 2 (yj 1 − 2yj + yj +1 ) Hence, the vector of derivatives of y1 , y2 , , yN is given by dy1 (t) = (N + 1) 2 dt dy2 (t) = (N + 1) 2 dt dy3 (t) = (N + 1) 2 dt α(t) − 2y1 (t) + y2 (t) , y1 (t) − 2y2 (t) + y3 (t) , y2 (t) − 2y3 (t) + y4 (t) , dyN 1 (t) = (N + 1) 2 yN −2 (t) − 2yN 1 (t) + yN (t) , . device 11 1 246 Starting methods 11 2 247 Numerical examples 11 3 25 Taylor Series Methods 11 4 250 Introduction to Taylor series methods 11 4 2 51 Manipulation of power series 11 5 252 An example of a. Explicit Methods 17 0 320 Methods of orders less than 4 17 0 3 21 Simplifying assumptions 17 1 322 Methods of order 4 17 5 323 New methods from old 18 1 324 Order barriers 18 7 325 Methods of order 5 19 0 326. solution 11 6 253 Other methods using higher derivatives 11 9 254 The use of f derivatives 12 0 255 Further numerical examples 12 1 26 Hybrid Methods 12 2 260 Historical introduction 12 2 2 61 Pseudo