💼 Case: consider a particle of mass $m$ moving along the $x$-axis subject to a nonlinear force $F(x)$

$$ m\ddot x = F(x) $$

where $F$ is independent of $\dot x$ and $t$ (no damping, friction or time dependent driving force)

Now this is path independent so it is obviously conserved but lets explicitly show this

• We denote $V(x)$ as the potential energy where $F(x)=-\frac{\text d V}{\text dx}$ allowing us to write

  $$ \begin{aligned} m\ddot x + \frac{\text dV}{\text dx}&=0 \\ m\dot x \ddot x + \frac{\text dV}{\text dx} \dot x&=0 \\ \frac{\text d}{\text dt}\left ( \frac 12 m\dot x^2 + V(x) \right )&=0 \end{aligned} $$

  Thus the total energy $E=\frac{1}{2} m\dot x^2+V(x)$ is constant over time so conserved

<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/792ab7ae-a73b-496d-8a7c-b5b8ee3fd4b2/Conservative.gif" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/792ab7ae-a73b-496d-8a7c-b5b8ee3fd4b2/Conservative.gif" width="40px" />

Conservative: If for equations

$$ \begin{aligned} \dot x &= f(x,y) \\ \dot y & = g(x,y) \end{aligned} $$

A function $E(x,y)$ exist which

• Is constant along trajectories

  $$ \frac{\text d}{\text dt} E(x(t),y(t))=0 $$

• Is not constant on any open set of points </aside>

💃 Example: Particle in a double potential well with $m=1$

$$ V(x)=-\frac 12 x^2+ \frac 14 x^4 $$

• Start by finding the force

  $$ F=m\ddot x = \ddot x =-\frac{\text dV}{\text dx}=x-x^3 $$

• We can rewrite this system $\ddot x=x-x^3$ as a vector field

  $$ \dot x=y \qquad \qquad \dot y=x-x^3 $$

• We find the equilibrium points where $(\dot x,\dot y)=(0,0)$

  $$ (0,0) \qquad (1,0) \qquad (-1,0) $$

• Now we can can define the Jacobian to find their stability

  $$ A=\begin{bmatrix} 0 & 1 \\ 1-3x^2 & 0 \end{bmatrix} $$

• For each point we find $\Delta = J_{11} J_{22} - J_{12} J_{21}$ and $\tau = J_{11} +J_{22}$

• $(0,0)$

• $\Delta = -1<0$

• $\tau =0$

💎 Saddle point

• $(1,0)$
• $\Delta =2>0$
• $\tau =0$

💎 Centre

• $(-1,0)$
• $\Delta =2>0$
• $\tau =0$

💎 Centre

• Trajectories are closed curves defined by the contours of constant energy, ie:

  $$ E=\frac 12 y^2 - \frac 12 x^2+ \frac 14 x^4 $$

• Solutions are typically periodic

• The solutions that go to the origin are homoclinic orbits because they approach the origin as $t \to \pm \infin$

Now we can make 2 plots A plot of the motion of the ball and of the energy

💫 Theorem: A conservative system cannot have attractive fixed points

🫀 Proof: Suppose the system is conservative and $(x^,y^)$ is an attractive fixed point

• All points in its basin of attraction would be at the same energy $E(x^,y^)$ because $E(x^,y^)$ must be constant on all trajectories in a conservative system
• However this contradicts with the definition of an attractive fixed point

💎 Conclusion: Conservative systems typically only have saddles and centres