Lecture 15: Ordinary Differential Equations: Second Order

Transcription

1 Lecture 15: Ordinary Differential Equations: Second Order 1. Key points Simutaneous 1st order ODEs and linear stability analysis. 2nd order linear ODEs (homogeneous and inhomogeneous. Maple DEplot Eigenvectors 2. General Remarks Second order ODEs are much harder to solve than first order ODEs. First of all, a second order ODE has two linearly independent solutions and a general solution is a linear combination of these two solutions. In adition, many popular second order ODEs have singular points. Except for a few cases, the solutions have no simple methamtical expression. However, there is a systematic way to find solutions in a form of series (Frobenius method). Although simple mathematical expressions are not available, the properties of solutions to poular second ODEs are well investigated and the solutions are often knwon as special functions such as Bessel functions. 3. Simultaneous 1st order ODEs A set of 1st order ODEs coupled to each others are ubiqutous in many fields of sciece beyond physics. Consider autonomous ODEs (1) There is no general solution to this type of ODEs. However, there is a nice way to get a flow diagram (parametric plot of a solution) using linear stability analysis. (2) 1. Fixed points Find fixed points and defined by and. We need to solve simulataneous equations (3) and 2. Linearization Introducing new variables and the ODEs near the fixed point are (4) (5) and

2 where higher order terms are ignored. The matrix is defined by (6) (7) 3. Geneal solutions Equations (5) and (6) are linear and can be solved when two eigenvalues/eigenvectors exist. The general solution is given by (8) where the eigenvalue and eigenvector are defined by 4. Characterization of solutions (9) Six cases Unstable node ( and ) (10) (11)

3 (12) (13) v 0 x

4 Stable node ( and ) (14) (15) (16) (17)

5 v 0 x Saddle node ( and ) (18) (19)

6 (20) (21) v 0 x Center node ( ) (22) (23)

7 (24) (25) v 0 x Unstable spiral ( ) (26) (27)

8 (28) (29) v 0 x Stable spiral ( ) (30) (31)

9 (32) (33) v 0 x Example 1: Harmonic Oscillator

10 The Newton equations for a simple harmonic oscillator is given by (34) This 2nd order ODE can be written as a coupled 1st order ODEs: (35) where and. Equations (35) and (36) have only one fixed point at and and are already linear. We have (36) (37) (38) If, the eigenvalues are real and both negative. Hence, the fixed point is a stable node. This is the case of overdumped oscillator. On the other hand, if, we have a stable spiral.

11 v v 0 x 0 x Example 2: Pendulum (39) (40) The fixed points are and. (n=integer) For,, the linearized eqution has a matrix (41) This case is the same as the previous example (simple harmonic oscillator). For,,

12 (43) One of the eigenvalue is always positive and the other is always negative. Therefore, this is a saddle node.. Without friction With friction v x v x Exercise Consider the system (44) Find fixed points and describe qualitatively the beviour of and. (45) (46) (47)

13 Hence, there is only one fixed point at and. = 1 = at Y=0 = 0 = (48) (49) (50). That means the flows move away from the fixed point parallel to the x axis. For the negative eigenvalue, the eigenvector is. The flows move toward the fixed point along the eigenvector.

14 2 y x 4. 2nd-order linear homogeneous ODEs with constant coefficients First we consider a homogeneous equation: (51) A popular way to solve this equation is to assume. Substituting it to (51), we obtain (52) which is called a characteristic equation. Solving this equation, we find two solutions

15 (53) Then, the general solution to (51) is given by where and are constants of integration. When, there is only one solution. In that case, the general solution is given by (54) (55) where. Proof Using the solution to the quadratic equation (52), Eq. (51) can be written as (56) where. If we have immediately a first solution. Solving the first order ODE, we obtain (57) For the second solution,. Let. Then,. Solving the 1st order ODE, we have. Now, we have another 1st order ODE for, The particluar solution to this ODE (the homogeneous solution is nothing but ) is (58) (59) which leads to

16 (60) Example Consider a simple harmonic oscillator of mass m and spring constant k subject to a frag force -bx. Its equation of motion is The corresponding characteristic equation is (61) and its solutions are (62) (63) Damped oscillation: (64) where and. The general solution is then given by which indicates damped oscillation. (65) Critial damping: The general solution is (66)

17 (67) Overdamping: (68) where. The general solution is which shows no oscillatary behaviour. (69) Using Maple The Maple solution is valid for but it misses the critical damping case. In Eq. (70), if the quantity inside the squre root is zero, obviously the second term coincides with the first term. Hence, Eq. (70) cannot be a general solution under that condition. However, if you force that condition, (70) Maple gives a correct answer. (71) 5. 2nd-order linear inhomogeneous ODEs with constant coefficients We consider a simple infomogeneouos ODE (72) Even such a simple looking ODE is hard to solve in general. There is a systematic method using Green's functions. However, we study only one case where solutions can be obtained without using the Green's function method. In general, a general solution consists of a complementary function and a particular solution. The complementry function is nothing but a general solution to the corresponding homogeneous equation ( and constants of integration appears only in the complementary function as

18 Since we already learned how to find the homogeneous solution, we focus on the particular solution. If the inhomogeneous term is exponential where is a complex number, we have a simple solution. Assuming, Eq. (72) becomes (73) (74) Hence, and we obtain a general solution (75) Example: Forced oscillator Consider a harmonic oscillator driven by an external force. The equation of motion is The right hand side of Eq. (76) is not exponential. Noting that solve the following ODE:, we first (76) (77) The real part of z(t) is the solution to Eq. (76). The particular solution to Eq. (77) is (78) Hence, the particular solution to (76) is given by

19 (79) simplify = Simplifying the expression, we have (80) where and. Since the homogeneous solution decays to zero as, the general solution approaches to the particular solution. Using Maple (81) which agrees with our hand calculation after simplifying the expression. 6. General 2nd-order linear homogeneous ODEs: Frobenius' method We consider a general form of linear homogeneous equation: (82) where and are given functions. Unlike 1st-oder ODEs, there is no general expression for the solutions of 2nd-order ODEs. However, we can construct at least one solution in a series. We assume a solution in the following form: where the lowest order term is not zero, i.e.,. The basic idea is to convert the ODE to a recursive relation for the expansion coefficients. This method is called Frobenius' method. (83)

20 The Frobenius' method provides at least one solution but the second solution is not guaranteed depending on singularities in and. For Eq. (82), one can find the second solution if the first one is known. Without proof, here is the second solution (84) Example As an example, we solve the simplest 2nd order ODE using the Frobenius' method. (85) By substituting Eq. (83) into Eq. (85), we have (86) Based on the uniquness of the power series, the coefficients of each power of x must vanish. (87) (88) (89) (90) (91)

21 Equation (91) is the recursive equation and the lowest two orders (87) and (88) determine. Since by definition, we have or from Eq. (87). Similarly from Equation (88) we have or. Our objective is to find one or two series that satisfes the oroginal ODE. Therfore, we can chose any value for at our disposal as long as it satisfies the recursive relation. For convenience, we chose Then, the recursive relation (91) immediately tells that all odd order terms is zero (. Now, we determine the even oder term using Eq. (91) For, (92) (93) (94) Now we have the first solution is (95) Similarly, for k = 1, (96) (97)

22 (98) (99) The second solution is (100) Homework A particle of mass is traveling in a potential field. The equation of motion is given by (101) where is friction constant. This second order equation can be written as a set of simultaneous ODE's. (102) 1. Find all fixed points 2. Characterize the types of the fixed points. 3. Plot a phase portrait. (103)