I restated the linear approximation method as follows. We have an autonomous system x' = f(x,y), y' = g(x,y) , and a critical point (a,b). Introduce new coordinates u = x-a, v = y-b. Then near (a,b) the system (usually) behaves like the linear sytem u' = J(a,b)u , were J(a,b) is the Jacobian evaluated at (a,b);
J = [df/dx df/dy ; dg/dx dg/dy]
where here d denotes partial differentiation, and we are using Matlab notation so that the two rows of the matrix are separated by a semicolon. (I did not talk about the issue of structural stability, described in the handout on phase portraits.)
I applied this to the nonlinear damped pendulum
theta'' + p theta' + q sin(theta) = 0.
I remarked that in deriving this I assumed that the pendulum would resist forces exerted towards the center. When the bob is above the center this says we should think of a rod rather than a rope.
One converts this into a first order system
theta' = omega
omega' = - q sin(theta) - p omega
The phase plane will thus descibe both position and velocity. The critical points occur where omega = 0, theta = k pi, k an integer. The whole system is unchanged if we increase theta by 2 pi, so there are two types of critical points: (0,0), (pi,0). J = [0,1 ; -q cos(theta),-p].
For definiteness take p = 1, q = 2.
J(0,0) = [0 1 ; -2 -1], whose characteristic polynomial lambda^2 + lambda + 2 has roots -1/2 +- sqrt((1/4)-2) with negative real part and nonzero imaginary part: so we get a stable spiral. The -2 in the lower left corner of J(0,0) implies a clockeise direction. I inserted little spirals at the critical points theta = 2 k pi.
J(pi,0) = [0 1 ; 2 -1], whose characteristic polynomial lambda^2 + lambda -2 has roots 1, -2. We get a saddle. The eigenvalue 1 has eigenvector [1;1], and -2 has eigenvector [1:-2]. I sketched the saddle and inserted it at the critical points theta = (2k+1) pi.
Then I joined up the little pictures and dramatized the result. I remarked that all solutions continue for all time. Some slowly approach a straight up position.
Then I sketched the undamped pendulum, and noticed that if we regard omega as a function of theta then by the chain rule
d omega/d theta = (d omega/dt)/(d theta/dt) so
omega d omega/dx = - q sin(theta)
This is a separable ODE, leading to the "first integral"
(1/2)omega^2 - q cos(theta) = constant.
Physicists call the constant of integration here the "energy" (up to choice of units; the first term is kenetic energy, the second is potential energy). These are the equations of the trajectories in the (theta,omega) plane.
New topic: IVPs.
Take A = [1 0 ; 2 -1] as above. Its normal modes are
a e^{-2t} [1;-2] and b e^t [1;1],
and the general solution is
x(t) = a e^{-2t} [1;-2] + b e^t [1;1] = [e^{-2t} , e^t ; -2e^{-2t} , e^t][a;b].
The matrix here is a "fundamental matrix" for the ODE: this means that the columns are independent solutions. It's a matrix-valued function. The notes use F(t) but I'll follow the book in writing Phi(t).
In an IVP x(t0) is specified, for some t0, and we wish to find a,b. Subsituting and solving,
[a;b] = Phi(t0)^{-1}x(t0).
Recollection on the inverse of a matrix A = [a b;c d]:
A^{-1} = (1/det A) [d -b;-c a].
For example t0 = 0, x(0) = [1;2], leads to Phi(0)=[1 1;-2 1], Phi(0)^{-1} = (1/3)[1 -1;2 1] and so [a;b] = (1/3)[-1;4], and
x(t) = (1/3)[-e^{-2t}+4e^t ; 2e^{-2t}+4e^t].