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I have been asked to consider a set of vector fields and determine whether these vectorfields are Hamilton or gradient and if possible to determine a Hamilton or potential function. I am a bit confused with the definitions of these concepts and can't find them clearly explained anywhere. For a system $x' = f(x,y)$ and $y' = g(x,y)$ I believe it is hamiltonian if $\partial_y f= -\partial_x g $, which is just some kind of exactness of the d.e.? But consider for example: $$ x' = y \\ y' =-x, $$ of course the solutions are cosines/sines but bare with me. A Hamiltonian $H = x^2 + y^2 = C$ satisfies $\partial H/\partial y = y $and $-\partial H/\partial x = -x $ but it also satisfies $x'' = -x = - \partial V /\partial x$ and $y'' = -y =- \partial V/\partial y$ for V the same function as the Hamilton above. I should know this rather easy concept, but I am utterly confused.

Funzies
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1 Answers1

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The system is a gradient system if there's a function $U(x,y)$ such that $$ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \partial{U}/\partial{x} \\ \partial{U}/\partial{y} \end{pmatrix} $$ and it's a Hamiltonian system if there's a function $H(x,y)$ such that $$ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} \partial{H}/\partial{x} \\ \partial{H}/\partial{y} \end{pmatrix} . $$ I think the reason for your confusion is that if $H(x,y) = \frac12 y^2 + V(x)$, then the Hamiltonian system is equivalent to the second-order ODE $x'' = -V'(x)$, where $V(x)$ is usually called the potential. (In this situation, $y$ is the momentum variable conjugate to the position variable $x$.) But when one speaks about gradient systems, the word "potential" just refers to what I called $U(x,y)$ above, with no mechanical interpretation of the variables $x$ and $y$.

(The example you give is a Hamiltonian system, not a gradient system.)

Hans Lundmark
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