Lecture 20: ODE V - Examples in Physics
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1 Lecture 20: ODE V - Examples in Physics Helmholtz oscillator The system. A particle of mass is moving in a potential field. Set up the equation of motion. (1.1) (1.2) (1.4) (1.5)
2 Fixed points Linear stability analysis about,. (1.6) (1.7) (1.8) (1.9) Hence, this fixed point is a center node. Linear stability analysis about,. (1.10) (1.11) Hence, this fixed point is a saddle point. Phase Portrait.
3 Quantum free falling A particle of mass is released. The quantum state of the particle is determined by a Schrödinger equation Assuming that initial kinetic and potential energy are both zero, thus the total energy is, and introducing a new variable, the Schrödinger equation is simplified to This is the Airy differential equation and its general solution is give by
4 where and are Airy functions. To determine the coefficients we need to find an appropriate boundary condition. Noting that is probability density, it must decreasing function in the positive direction of. (Otherwise, the particle is moving upward against the gravity.) Therefore, we demand a boundary condition. Since,. The remaining coefficient is determined by a certain normalization, which we do not discuss here. Expressing in the original coordinate,. This probability density suggests that there is a small chance that the particle is found slightly above the initial height. This is due to quantum uncertainty principle. Exercise Solve the free falling problem with non-zero initial energy. Quantum harmonic oscillator Consider a simple harmonic oscillator of mass for the oscillator is and spring constant. The Schrödinger equation where is the energy of the oscillator. The boundary condition is vanishes as. Introducing the frequency of the oscillator, we introduce normailzed energy and position
5 and. Then, the Schrödinger equation is simplified to Furthermore, we introduce a new function. Inverting this, the boundary condition is satsified if is a polynomial of. Substituting this into the Schrödinger equation, we obtain The general solution to this equation is where and are 1st and 2nd kind of Kummer functions. (3.1) (3.2) (3.3)
6 If (n=non-negative integer) then odd order terms in both series vanish above a certain order. However, the even order terms in the second series diverges. To satisfiy the boundary condition, must be null. On the other hand, if then, the even order terms in the second series vanish above a certain order. A problem is that the odd order terms in both serises diverge. However, if, the odd order terms are all cancelled out. (3.4) As you can see, wll odd terms disappeared. In conclusion, only when, polynomial solutions exist. (3.5)
7 (3.6) (3.7) As you can see, non polynomial terms can be elliminated by seeting either or. The solutions are known as Hermite polynomials. Going back to the original expression, the solution exists only when and the eigenfunction is where the normalization constant is simplify = simplify e = = 1 4 simplify = simplify = = 1 4 simplify = = simplify =
8 Lorenz attractor Lorentz Equation Parameter Values Initial Conditions
9 Numerically solve the equation Plot the result Lorenz attractor (chaotic attractor)
10 Henon-Heiles Model 0. The system and Hamiltonian While at Princeton in 1964, Michel Hénon and Carl Heiles published a paper that describes the nonlinear motion of a star around a galactic center where the motion is restricted to a plane. In their model a star of mass traveling in a two-dimensional potential field and thus the Hamiltonian of the system is given by
11 (5.1) 1. Derive the equation of motion Hamilton's equations of motion These equations are much simpler than those of the double pendulum. However, these simple looking equations essentially share the same properties as the double pendulum. (5.2) 2, Parameter values For simplicty, we assume 1 (5.4)
12 3. Initial conditions Since the degrees of freedom is 4, we need to specify 4 initial conditions. Since the energy conserves, energy is a convenient initial condition which remains constant. We then specify,, and at the initial time. Note that the initial value of is determined by the conservation of energy. We investigate a trajectory for a given initial condition: 4. Solving the equation (5.5) We can solve the ODEs analytically. However, for learning purpose, we solve it numerically and
13 discuss how to visualize the trajectories. 5. Plot coordinates as a function of time First, we look at the coordinates, and as a function of time. This is the traditional way to look at the solution. Can you tell from these plots if the trajectory is periodic or not? 6. Projection of the trajectory onto a plane Since we cannot plot four dimensional phase trajectory, we display its projection on to twodimensional planes.
14 q1-p1 plane q2-p2 plane q1-q2 plane p1-p2 plane 7. Torus The trajectories are not confined on a torus, allowing chaotic trajectories.
15 8. Poincare Map
16
17 Driven Duffing Oscillator (Bistable Potential) See also this web site Antonomous Duffing Oscillator 0. The system
18 A particle of mass is moving in a potential field (6.1.1). and are positive constants. Th Hamiltonian of system (no dissipation) is (6.1.2)
19 The equations of motion with friction are given by,. where, and is a frictional coefficient. This problem can be solved analytically. However, the trajectory is only given in an implicit form. Our analysis is baswed on numerical solutions. 1. Trajecyoties
20 It looks like a dumped harmonic oscillator.
21 Linear stability analisys Find nullclines = =
22 Find fixed points = 1st fixed point - saddle node Linear expansion around = 0. and = 0.. ( ) Eigenvalues ( ) This is a saddle point. Corresponding eigenvector ( ) 2nd fixed point - center node
23 Linear expansion around = 1. and = 0.. ( ) ( ) Eigenvalues ( ) This is a center node. 3rd fixed point - center node Linear expansion around = and = 0.. ( ) Eigenvalues ( )
24 This is another center node. Driven Duffing Oscillator
25
26 Stroboscopic Map and Strange Attractor
27
28 Bifurcation Diagram
29 Homework Due 11/13, 11am 20.1 In quamtum free falling, (a) Drive from the original Schrödinger equation. (b) In example, we assumed. Find a solution for A Schrodinger equation for the radial variable in the spherical coordinates is given by
30 (a) Show that the general solution to this equation is where and are spherical Bessel and Nuemann functions, and. (b) When the boundary condition is, show that.
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