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[SOLVED] damping coefficient

dwsmith

Well-known member
Feb 1, 2012
1,673
I don't understand how to start.

Consider the damped double-well potential model
$$
\ddot{x} - x + x^3 + \gamma\dot{x} = 0,
$$
where $\gamma$ is the damping coefficient.
This model has two fixed points at $(x,\dot{x}) = (1,0)$ and $(-1,0)$. In the phase plane $(x,\dot{x})$, determine the attraction basins of these fixed points. You can decide which $\gamma$ value to use (try to choose one which gives the best picture/graph). Note: the attraction basin of a fixed point is the set of initial points which go to the fixed as $t\to\infty$.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I don't understand how to start.

Consider the damped double-well potential model
$$
\ddot{x} - x + x^3 + \gamma\dot{x} = 0,
$$
where $\gamma$ is the damping coefficient.
This model has two fixed points at $(x,\dot{x}) = (1,0)$ and $(-1,0)$. In the phase plane $(x,\dot{x})$, determine the attraction basins of these fixed points. You can decide which $\gamma$ value to use (try to choose one which gives the best picture/graph). Note: the attraction basin of a fixed point is the set of initial points which go to the fixed as $t\to\infty$.
Change the ODE to a system of ODEs, we have
\begin{alignat}{3}
x_1' & = & x_2\\
x_2' & = & x_1 - x_1^3 + \gamma x_2
\end{alignat}
When $\gamma = 1$, how do I find the attraction basin?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Change the ODE to a system of ODEs, we have
\begin{alignat}{3}
x_1' & = & x_2\\
x_2' & = & x_1 - x_1^3 + \gamma x_2
\end{alignat}
When $\gamma = 1$, how do I find the attraction basin?
So with reference to >>this<< thread your system of equations should be,

\[x_1'=x_2\]

\[x_2'=x_1-x_1^3-\gamma x_2\]

I don't know if this is helpful, but I have drawn the phase portrait of this system. Note that I have replaced, \(x_1\) by \(x\) and \(x_2\) by \(y\). If you have Maxima you can animate the graph so that you can find the phase portraits for different values of \(\gamma\). Here I have included graphs for several values of \(\gamma\).

\[x'=y\mbox{ and }y'=x-x^3-\gamma y\]

Code:
load("plotdf");

plotdf([y,x-x^3-a*y],[x,-5,5],[y,-5,5],[sliders,"a=-10:10"]);

\(\mathbf{\underline{\mbox{When }\gamma=0}}\)




\(\mathbf{\underline{\mbox{When }\gamma=2}}\)




\(\mathbf{\underline{\mbox{When }\gamma=-2}}\)

 

dwsmith

Well-known member
Feb 1, 2012
1,673
So with reference to >>this<< thread your system of equations should be,

\[x_1'=x_2\]

\[x_2'=x_1-x_1^3-\gamma x_2\]

I don't know if this is helpful, but I have drawn the phase portrait of this system. Note that I have replaced, \(x_1\) by \(x\) and \(x_2\) by \(y\). If you have Maxima you can animate the graph so that you can find the phase portraits for different values of \(\gamma\). Here I have included graphs for several values of \(\gamma\).

\[x'=y\mbox{ and }y'=x-x^3-\gamma y\]

Code:
load("plotdf");

plotdf([y,x-x^3-a*y],[x,-5,5],[y,-5,5],[sliders,"a=-10:10"]);

\(\mathbf{\underline{\mbox{When }\gamma=0}}\)




\(\mathbf{\underline{\mbox{When }\gamma=2}}\)




\(\mathbf{\underline{\mbox{When }\gamma=-2}}\)

I have the phase portraits too but I don't know how to find the attractor basin still.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Change the ODE to a system of ODEs, we have
\begin{alignat}{3}
x_1' & = & x_2\\
x_2' & = & x_1 - x_1^3 + \gamma x_2
\end{alignat}
When $\gamma = 1$, how do I find the attraction basin?
Typo $-\gamma x_2$.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Typo $-\gamma x_2$.
Note that I have corrected that in post #3.

I have the phase portraits too but I don't know how to find the attractor basin still.
Here is a figure that roughly depicts the attraction basins for the fixed points, \((1,0)\) and \((-1,0)\). The green region is the attractor basin of \((-1,0)\) and the red region is the attractor basin of \((1,0)\).

 

dwsmith

Well-known member
Feb 1, 2012
1,673
How do you find the attractor basin?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621