# Homework Help: Differential Equation- Pendulum System

1. May 9, 2012

### Kamekui

1. The problem statement, all variables and given/known data

(a). Find the critical points
(b). Linearize the system and determine local stability
(c). Find a Liapunov Function (Hint - Use total energy)
(d) Use the Liapunov Function to conclude that the critical points are stable.
(e.) Use the LaSalle theorem to argue that every trajectory has to tend toward one of the critical points.

2. Relevant equations

x'=y
y'=-y-sin(x)

3. The attempt at a solution

(a). Let x',y'=0
x'=0 → x=0

If x=0, then:

y'=0=0-sin(x)→ x=0

So, this implies there are infinite critical points occurring at (nπ,0) where n is an integer.

(b). Let x'=f(x,y), y'=g(x,y), the the Jacobian Matrix is

\left[
\begin{array}{ccc}
fx & fy \\
gx & gy
\end{array}
\right]= \left[
\begin{array}{ccc}
0 & 1 \\
-cos(x) & -1
\end{array}
\right]

If n is even, then we get:

\left[
\begin{array}{ccc}
0 & 1 \\
-1 & -1
\end{array}
\right]

\left[
\begin{array}{ccc}
x' \\
y'
\end{array}
\right]= \left[
\begin{array}{ccc}
0 & 1 \\
-1 & -1
\end{array}
\right]*\left[
\begin{array}{ccc}
x \\
y
\end{array}
\right]= \left[
\begin{array}{ccc}
y \\
-x-y
\end{array}
\right]

The determinant yields: λ2+λ+1

→λ=-1/2-+ i(√3)/2

Since the eigenvalues are complex conjugates with negative real part, this implies the system is asymptotically stable locally at (nπ,0) (where n is even).

On the other hand, if n is odd we get: (I didn't do the Jacobian Matrix again because its a simple calc)

λ= -1/2+- (√5)/2

Since the eigenvalues are real and of opposite sign, this implies the critical points (nπ
,0) (where n is odd) are saddle points which means the critical point is locally unstable.

(c). This is where I get confused, finding Liapunov Functions is new to me. I think I'm supposed to assume V is a Liapunov Function of the form V = (1/2)ax2 + (1/2)by2.

So then ∇V= (2ax,2by)

v= ∇V ° (f,g)= axy-by2-by*sin(x)

So now I need to show that v≤kV, for some k<0