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Consider the two-dimensional autonomous system \begin{equation*}\vec{\frac{dx}{dt}}(t)=\vec{f}(\vec{x}(t)):\mathbb{R}^{2}\to\mathbb{R}^{2}.\end{equation*}

Suppose the system has a conserved quantity of motion associated with it (i.e. an energy conservation law): \begin{equation*}E(\vec{x}(t))=E_{0}\;\forall\;t\Longrightarrow\frac{dE}{dt}=0.\end{equation*}

Then here are some questions.

(1) Is it possible to have a second conserved quantity of motion which is different from $E$?

(2) Is it necessarily the case that $E(x_{1},x_{2})=E_{0}$ corresponds to the integral family of curves which satisfes \begin{equation*}\frac{dx_{2}}{dx_{1}}=\frac{f_{2}(x_{1},x_{2})}{f_{1}(x_{1},x_{2})}\mapsto F(x_{1},x_{2})=E_{0}\end{equation*} (i.e. the solution curves in the phase-plane?). For conservative mechanical systems such as $x''=-f(x)$ this is evidently the case (if one uses the standard physical conservation of energy). But what about for general systems. And why or why not is this the case (i.e. a proof)?

Actually, I think I can prove (2): If $E$ represents the conserved quantity, then along trajectories, $E(\vec{x})=E_{0}\;\forall\;t$, so that the trajectories must correspond to the level sets of the function $E(\vec{x}):\mathbb{R}^{2}\to\mathbb{R}$. From another point of view, the initial conditions determine (and are determined by) the initial energy $E_{0}$. Then...

(3) Assuming (2) is indeed true, then if the system has a conservative quantity, presumably it can always be obtained by integrating the differential equation obtained by eliminating $t$ from the system? Of course there are other ways to obtain energy functions (for example, in the mechanical system one can multiply both sides of the equation by $\frac{dx}{dt}$ and integrate). But what (2) tells me, is that any method which is used to obtain $E$ is necessarily equivalent to integrating the trajectories.

(4) Assuming (3), once one integrates the t-eliminated ODE, then all one must do is check that $\frac{dE}{dt}=0$. If this holds, then the system is conservative and you have your energy function (I am tacitly assuming (1) is NOT true; i.e. that the conserved quantity is unique). If it does not, then you have your trajectories yes, but there is no conserved quantity along such trajectories.

I might add some more questions to this later, but I'll leave it at that for now.



(4) Was kind of a dumb question/observation. Of course, when you get the integral family of curves you get something like $F(x_{1},x_{2})=C$. Differentiating this with respect to $t$ of course gives you $0$, so $F$ is obviously constant along trajectories. But doesn't this effectively say every system is conservative then? ... Now I'm confused. ><

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This hinges on what you mean by an 'energy conservation law'. For physical systems, this is meaningful, otherwise you need to be more explicit. Clearly $E$ needs to place some constraints on $x(t)$. – copper.hat Jul 28 '12 at 8:16
up vote 3 down vote accepted

(1) No, you can't have a second conserved quality because the system is two-dimensional. Each solution curve is contained in a level set of each conserved quantity: this is essentially a tautology. In two dimensions, a generic level set of a function is a curve, and intersection of two such level sets gives isolated points - no room for trajectory.

(2) See the bold above. Level sets of E can have complicated structure even when E is smooth. In particular they can have branching points which solutions don't (when uniqueness holds). To take a mild example, $E=x_1x_2$ is conserved by $\dot x=(x_1,-x_2)$. The level set $E=0$ is the union of two lines (of two separate solution curves).

(3). In a neighborhood of each point where the field does not vanish, the solution curves indeed coincide with level sets of some E. Look up "flow box theorem". But you don't always have such E globally, or in a neighborhood of equilibrium. Simple example: $\dot x=x$. There is no E that describes trajectories near the equilibrium.

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Ad (1): If some $E$ is conserved then any $f(E)$ is conserved also. The truth is expressed in (3). – Christian Blatter Jul 28 '12 at 14:52
Thank you guys! This was helpful in clarifying my thoughts on this. – Taylor Martin Jul 29 '12 at 6:59

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