# [SOLVED]damping coefficient

#### dwsmith

##### Well-known member
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
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
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
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
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
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
How do you find the attractor basin?

#### Sudharaka

##### Well-known member
MHB Math Helper
How do you find the attractor basin?
I am not sure whether I understand your question here. The basins of attraction for $$\gamma=1$$ are depicted in the figure in post #6.