I want to make sure I am doing this problem correctly, especially when it comes to drawing the potential function V(x).
Consider the system of differential equations:
$$\dot {x}=y$$ $$\dot {y}=-x+x^3$$
a. Find the fixed points
$\dot {x}=y$ $y=0$
$\dot {y}=-x+x^3$ $x=-1,1,0$
b. Find the potential function V(x) and draw its graph. Classify the type of each fixed point.
$V(x)=-\int_{x_0}^x \! F(s) \, \mathrm{d}s$ $V(x)=-\int_{0}^x \! -s+s^3 \, \mathrm{d}s$
$V(x)=\frac{x^2}{2}-\frac{x^4}{4}$
Jacobian:
$DF_{(x,y)}=\begin{pmatrix} 0 & 1 \\ -1+3x^2 & 0 \end{pmatrix}$
$DF_{(0,0)}=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$
$λ=±i$
(0,0) is a center.
$DF_{(±1,0)}=\begin{pmatrix} 0 & 1 \\ 2 & 0 \end{pmatrix}$
$λ=±\sqrt {2}$
(±1,0) are saddle points.
$V(0)=0$ (local minimum) and $V(±1)=1/4$ (local maxima)
c. Draw the phase portrait for the system of differential equations.
We are given:
$$x'= y \\ y'=-x+x^3$$
This system has three critical points at:
$$(x, y) = (-1, 0), (0,0), (1, 0)$$
The Jacobian is given by:
$$J(x, y) = \begin{pmatrix} 0 & 1 \\ -1+3x^2 & 0 \end{pmatrix}$$
Evaluating $J(x, y)$ at $(\pm~1, 0)$, we have eigenvalues:
$$\lambda_{1, 2} = \pm ~ \sqrt{2} \implies ~\mbox{saddle point}$$
Evaluating $J(x, y)$ at $(0, 0)$, we have eigenvalues:
$$\lambda_{1, 2} = \pm i \implies ~\mbox{center}$$
A phase portrait shows:
The potential function is given by:
$$V(x) = \displaystyle -\int_{x_0}^x F(s) ~ds = -\int_{0}^x (-s+s^3) ~ds = \dfrac{x^2}{2}-\dfrac{x^4}{4}$$
A plot of this shows:
I tried to line up the phase portrait and potential function. See what is going on there?