# Nonlinear Oscillations

• Dec 29th 2009, 06:09 AM
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Nonlinear Oscillations
Hi guys! I think I posted a question similar to this before. I think i've made some progress in solving questions of this type, but i'm still getting stuck on bits of it.

Quote:

Find the general solution to $\displaystyle \ddot{x}-2\dot{x}-3x=10 \sin t$.
The method for this is long so i'll just put the answer.

This comes out to be $\displaystyle x(t)=Ae^{3t}+Be^{-t}-2 \sin t + \cos t$ where $\displaystyle A,B$ are constants.

Quote:

Write this equation in the three variables $\displaystyle x,y=\dot{x}$ and $\displaystyle \dot{s}=1$ so that the equations are autonomous and first order.
$\displaystyle \dot{x}=y$
$\displaystyle \dot{y}=10 \sin s +3x+2y$
$\displaystyle \dot{s}=1$

We know that $\displaystyle \frac{ds}{dt}=1$. This gives $\displaystyle \frac{dx}{dt}=\frac{ds}{dt}. \frac{dx}{ds}=\frac{dx}{ds}$.

Applying a similar method gives:

$\displaystyle \frac{dy}{dt}=\frac{ds}{dt}.\frac{dy}{ds}=\frac{dy }{ds}$

We also have:

$\displaystyle \frac{d^2x}{dt^2} = \frac{d}{dt} \left(\frac{dx}{dt}\right) = \frac{d}{ds} \left( \frac{dx}{ds}\right) \cdot \frac{ds}{dt} = \frac{d^2x}{ds^2} = \frac{dy}{ds}$ since $\displaystyle y=\frac{dx}{ds}$.

This gives my 3 equations as:

$\displaystyle \frac{dt}{ds}=1$
$\displaystyle \frac{dx}{ds}=y$
$\displaystyle \frac{dy}{ds}-2y-3x=10 \sin t$

(Kudos to Danny for this!)

Quote:

Write down the corresponding solution $\displaystyle (x(t),y(t),s(t))$.
$\displaystyle x(p) \in \mathbb{R}^3$
$\displaystyle x(p)=(\dot{x}, \dot{y},\dot{s})=(y,10 \sin s+3x+2y,1)$

But we know the general solution.

$\displaystyle (x(t),y(t),s(t))=(Ae^{3t}-2 \sin t+ \cos t, 3Ae^{3t}-2 \cos t - \sin t, t+C)$

Quote:

The vector field in $\displaystyle \mathbb{R}^3$ is now periodic in $\displaystyle s$ with period $\displaystyle 2 \pi$.

Write down the Poincare map in the plane $\displaystyle s= 2 \pi$ for a set of initial conditions $\displaystyle (x_0,y_0)$ lying in the plane $\displaystyle s=0$.
So we're going from $\displaystyle s=0$ to $\displaystyle s= 2 \pi$

$\displaystyle (x(0),y(0),0)=(x_0,y_0,0)$

This gives:

$\displaystyle A+1=x_0$
$\displaystyle 3A-2=y_0$
$\displaystyle C=0$

$\displaystyle \Rightarrow 4A-1=x_0+y_0$
$\displaystyle \Rightarrow A=\frac{1}{4}(x_0+y_0+1)$

So this gives:

$\displaystyle (x(t),y(t),s(t))=$$\displaystyle \left( \frac{1}{4}(x_0+y_0+1)e^{3t}-2 \sin t+ \cos t, \frac{3}{4}(x_0+y_0+1)e^{3t}-2 \cos t- \sin t, t \right)$

Quote:

Find it's fixed point and discuss it's stability in terms of eigenvalues.
This is where I get stuck. I only have one example in my notes that does not have an $\displaystyle e^{3t}$ in it. As for the part on eigenvalues, I have no idea how to even get a matrix from this!!

I'd appreciate some help on this, there are still more bits of the question that I have to try and do!!
• Jan 9th 2010, 04:21 PM
Are you solving the equation $\displaystyle \ddot{x}-\ddot{x}-3x=10\sin t$ or $\displaystyle \ddot{x}-2\ddot{x}-3x=10\sin t$? I found complex eigenvalues $\displaystyle \lambda=\frac{1}{2}\pm\frac{1}{2}\sqrt{13}$ instead of the $\displaystyle \lambda=3,-1$ that you did.
Also, for getting matrices, assume that $\displaystyle x=x$ and $\displaystyle y=x'$. Differentiating these gives $\displaystyle x'=y$ and $\displaystyle y'=x''$, which you can then use to turn any second order equation into a system of first order equations.
I'm so sorry! I did mean to write $\displaystyle \ddot{x}-2\ddot{x}-3x=10\sin t$.
I also think i've tried $\displaystyle x=x$ and $\displaystyle y=\dot{x}$. Haven't I just written it as $\displaystyle \dot{y}=10 \sin s +3x+2y$?