Potential in $2$ dimensional systems Given a $1$ dimensional dynamical system represented by $\dot{x}=f(x)$ we define the potential $V(x)$ to be the function that satisfies $\dot{x}=f(x)= -\frac{\partial V(x)}{\partial x}$. How to we extend this concept to $2$ dimensional systems? 
I also know that in $1$ dimensional systems we can't have oscillations but in $2$ dimensional systems we might have so what happens with the potential in that case? Does anyone have any suggested literature where I can read more about this?
 A: Probably the most straightforward generalization of the system $\dot{x}=f(x)= -\frac{\partial V(x)}{\partial x}$ to two or more dimensions is to consider gradient dynamical systems, that is, systems on $\Bbb R^n$ of the form
$\dot{\mathbf r} = -\nabla \phi({\mathbf r}), \tag{1}$
where $\phi(\mathbf r)$ is a differentiable function of $\mathbf r \in \Bbb R^n$.  A system such as (1) cannot in fact exhibit oscillatory behavior, since, as in the one-dimensional case, $\phi (\mathbf r)$ must decrease along the trajectories:  if $\mathbf r(t)$ satisfies (1), then we have
$\dfrac{d\phi(\mathbf r(t))}{dt} = \nabla \phi(\mathbf r) \cdot \dot{\mathbf r} = -(\nabla \phi(\mathbf r))^2 \le 0 \tag{2}$
by an easy application of the chain rule, again using (1).  (2) shows that along any non-trivial integral curve, that is, one where $\dot{\mathbf r} = -\nabla \phi({\mathbf r}) \ne 0$, we in fact have strict inequality.  Thus if there were a closed orbit, following the system around one cycle would bring us back to the starting point but with a lesser value of $\phi(\mathbf r)$, clearly an impossibility.  So (1) has no closed, periodic integral curves.
If one allows more general systems of the form
$\dot{\mathbf r} = \mathbf V(\mathbf r), \tag{3}$
where $\mathbf V(\mathbf r)$ is not necessarily a gradient vector field, i.e. then closed orbits are of course possible.  To see this, just take
$\mathbf V(\mathbf r) = \begin{pmatrix} y \\ -x \end{pmatrix} \tag{4}$
for $\mathbf r = (x, y)^T \in \Bbb R^2$.  (4) in fact has circular trajectories; it should be noted however that $\mathbf V(\mathbf r)$ is not a gradient, since 
$\nabla \times \mathbf V = -2 \mathbf k \ne 0 \tag{5}$ 
in the standard $\mathbf i-\mathbf j-\mathbf k$ basis of $\Bbb R^3$, considering $\Bbb R^2$ a subspace of $\Bbb R^3$ in the usual way.  It is well known that a vector field $\mathbf V(\mathbf r)$ on $\Bbb R^3$ (and here we include $\Bbb R^2$ as a subspace) is the gradient of some scalar field $\phi(\mathbf r)$ if and only if $\nabla \times V = 0$; this is a standard result of vector calculus.  This Widipedia page is a good place to start learning more; one might also check out the book Differential Equations, Dynamical Systems, and an Introduction to Chaos by Smale, Hirsch, and Devaney, published by Academic Press; see especially section 9.3 and any references cited therein.
Hope this helps.  Cheerio,
and a always,
Fiat Lux!!!
