Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question that I have to solve is found below. However, I do not know how to start the solution since I am unsure about the defintion of a Gradient/Hamiltonian System. What must I check first to know whether it is a Gradient or Hamiltonian System?

Which of the following is gradient/Hamiltonian( Conservative) system or other, or neither, and why? And if they are find the potential:

$\dot{x}=-2xe^{-x^2-y^2} \\ \dot{y}=-2ye^{-x^2-y^2}$

$\dot{x}= y+x^2y \\ \dot{y}=-x+2xy$

$\dot{x}=y \\ \dot{y}=y^3$

$\dot{x}=y\\ \dot{y}=-y-x^3$

share|cite|improve this question
up vote 2 down vote accepted

A system of ordinary differential equations of the form

$$ \dot{x}=f(x,y)\\ \dot{y}=g(x,y) $$

is called Hamiltonian if there exists a function $H(x,y)$ such that

$$ f(x,y)=\frac{\partial}{\partial y}H(x,y)\\ g(x,y)=-\frac{\partial}{\partial x}H(x,y). $$

To check if such a function exists, we need to check the following:

$$ \frac{\partial }{\partial x}f(x,y)+\frac{\partial}{\partial y}g(x,y)=0 $$

Why do we check this? Well, (assuming that $H$ has continuous second partials), we must have $$ \frac{\partial^2H}{\partial x\partial y}=\frac{\partial^2 H}{\partial y\partial x}\\ \Longrightarrow\frac{\partial}{\partial x}f(x,y)=\frac{\partial^2H}{\partial x\partial y}=\frac{\partial^2 H}{\partial y\partial x}=-\frac{\partial }{\partial y}g(x,y) $$ and hence $f_x+g_y=0$.

On the other hand, a system is called a Gradient system if there is a function $F$ such that

$$ \dot{X}=-\nabla F $$ where $X=[x,y]^\intercal$ and $F:\Bbb{R}^2\rightarrow\Bbb{R}$. Some books will use $\dot{X}=\nabla F$ instead (no negative sign) so watch out for that.

This would mean that we need a function such that $f(x,y)=-\frac{\partial}{\partial x}F$ and $g(x,y)=-\frac{\partial}{\partial y}F$.

One way to check if such a funciton exists is to check if the vector field $\vec{F}(x,y)=f(x,y)\hat{\imath}+g(x,y)\hat{\jmath}$ is path-independent (or conservative). This is done by checking the condition:

$$ \frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial x}g(x,y) $$

Then, if this holds for all $(x,y)\in\Bbb{R}^2$, you can start constructing a potential $H$; for simple systems, this can usually be done by "inspection". For more complicated systems, you'll need to go through an integration process to find $H$. For a video on this process, see here.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.