How should I show that the transformation makes the system Hamiltonian?

[Math100]

Homework Statement

Consider the system of equations $\frac{dx}{dt}=(a-by)x(1-x^2), 0<x<1, \frac{dy}{dt}=-(c-dx)y(1-y^2), 0<y<1$, where $a, b, c$ and $d$ are positive constants.

Show that the transformation $x=\frac{1}{\sqrt{1+e^{-2q}}}, y=\frac{1}{\sqrt{1+e^{-2p}}}$ makes the system Hamiltonian, with Hamiltonian function $H(q, p)=ap-b\cdot ln(e^{p}+\sqrt{1+e^{2p}})+cq-d\cdot ln(e^{q}+\sqrt{1+e^{2q}})$. (Hint: You may find the following indefinite integral useful: $\int \frac{dx}{\sqrt{1+e^{-2x}}}=ln(e^{x}+\sqrt{1+e^{2x}}).$)

Relevant Equations

By definition, a system $\dot{x}=X(x, y), \dot{y}=Y(x, y)$ is called a Hamiltonian system if there exists a function $H(x, y)$ such that $X=\frac{\partial H}{\partial y}$ and $Y=\frac{\partial H}{\partial x}$. Then $H$ is called the Hamiltonian function for the system. A necessary and sufficient condition for $\dot{x}=X(x, y)$ and $\dot{y}=Y(x, y)$ to be Hamiltonian is that $\frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}=0$.

Proof:

Consider the transformation $x=\frac{1}{\sqrt{1+e^{-2q}}}$ and $y=\frac{1}{\sqrt{1+e^{-2p}}}$ with the Hamiltonian function $H(q, p)=ap-b\cdot ln(e^{p}+\sqrt{1+e^{2p}})+cq-d\cdot ln(e^{q}+\sqrt{1+e^{2q}})$.
Let $\dot{x}=\frac{dx}{dt}=(a-by)x(1-x^2)=(a-by)(x-x^3)$ and $\dot{y}=\frac{dy}{dt}=-(c-dx)y(1-y^2)=-(c-dx)(y-y^3)$.
By using direct substitution of the transformation $x=\frac{1}{\sqrt{1+e^{-2q}}}$ and $y=\frac{1}{\sqrt{1+e^{-2p}}}$,
we have that $\dot{x}=\frac{dx}{dt}=(a-\frac{b}{\sqrt{1+e^{-2p}}})(\frac{1}{\sqrt{1+e^{-2q}}}-(\frac{1}{\sqrt{1+e^{-2q}}})^3)=(a-\frac{b}{\sqrt{1+e^{-2p}}})(\frac{e^{q}}{(e^{2q}+1)^{3/2}})$ and $\dot{y}=\frac{dy}{dt}=-(c-\frac{d}{\sqrt{1+e^{-2q}}})(\frac{1}{\sqrt{1+e^{-2p}}}-(\frac{1}{\sqrt{1+e^{-2p}}})^3)=-(c-\frac{d}{\sqrt{1+e^{-2q}}})(\frac{e^{p}}{(e^{2p}+1)^{3/2}})$.
Note that $X=\frac{\partial H}{\partial p}=a-b(\frac{1}{e^{p}+\sqrt{1+e^{2p}}})(e^{p}+\frac{e^{2p}}{\sqrt{1+e^{2p}}})$ and $Y=\frac{\partial H}{\partial q}=c-d(\frac{1}{e^{q}+\sqrt{1+e^{2q}}})(e^{q}+\frac{e^{2q}}{\sqrt{1+e^{2q}}})$.

Above is my proof for this problem. Is everything correct up to here? How should I finish this proof? What needs to be done?

 

[pasmith]

You need to show that \dot q = \frac{\partial H}{\partial p} nad \dot p = - \frac{\partial H}{\partial q} for some H(p,q) (and the question suggests one). Therefore you need to calculate \dot q and \dot p; the easiest way to do that is <br /> \dot q = \frac{dq}{dx} \dot x \qquad \dot p = \frac{dp}{dy}\dot y. Note, however, that x = f(q) and y = f(p) where f(z) = (1 + e^{-2z})^{-1/2}. Rearranging this we have e^{2q} = \frac{x^2}{1 - x^2} so that 2q = 2\ln x - \ln(1 - x^2) and hence \begin{split}<br /> \frac{dq}{dx} &amp;= \frac1x + \frac{x}{1 - x^2} \\<br /> &amp;= \frac{1}{x(1 - x^2)}. \end{split} The denominator neatly cancels the factor of x(1-x^2) in \dot x leaving <br /> \dot q = a - by = a - bf(p). In the same way you can find \dot p.

Now \dot q is a function of p alone and \dot p is a function of q alone, so setting <br /> H = \int \dot q \,dp - \int \dot p \,dq <br /> will give <br /> \dot q = \frac{\partial H}{\partial p} \qquad \dot p = -\frac{\partial H}{\partial q} as required.

 

[Math100]

You need to show that \dot q = \frac{\partial H}{\partial p} nad \dot p = - \frac{\partial H}{\partial q} for some H(p,q) (and the question suggests one). Therefore you need to calculate \dot q and \dot p; the easiest way to do that is <br /> \dot q = \frac{dq}{dx} \dot x \qquad \dot p = \frac{dp}{dy}\dot y. Note, however, that x = f(q) and y = f(p) where f(z) = (1 + e^{-2z})^{-1/2}. Rearranging this we have e^{2q} = \frac{x^2}{1 - x^2} so that 2q = 2\ln x - \ln(1 - x^2) and hence \begin{split}<br /> \frac{dq}{dx} &amp;= \frac1x + \frac{x}{1 - x^2} \\<br /> &amp;= \frac{1}{x(1 - x^2)}. \end{split} The denominator neatly cancels the factor of x(1-x^2) in \dot x leaving <br /> \dot q = a - by = a - bf(p). In the same way you can find \dot p.

Now \dot q is a function of p alone and \dot p is a function of q alone, so setting <br /> H = \int \dot q \,dp - \int \dot p \,dq<br /> will give <br /> \dot q = \frac{\partial H}{\partial p} \qquad \dot p = -\frac{\partial H}{\partial q} as required.

That was very informative and thorough! I have another question under this same problem.

Show that if $b>a$ and $d>c$, then the transformed system has a centre at $x=c/d$ and $y=a/b$, and prove that in this case, all orbits of the original system are periodic.

My question is, how should I construct a proof for this question? Does the given point $x=c/d, y=a/b$ indicate a fixed point for this system?

 

[pasmith]

That was very informative and thorough! I have another question under this same problem.

Show that if $b>a$ and $d>c$, then the transformed system has a centre at $x=c/d$ and $y=a/b$, and prove that in this case, all orbits of the original system are periodic.

My question is, how should I construct a proof for this question? Does the given point $x=c/d, y=a/b$ indicate a fixed point for this system?

Note that the transformation in part (a) has domain (q,p) \in [0,\infty)^2 and range [1,\infty)^2.

How you determine if a point is or is not a fixed point?

If a point is a fixed point, how do you classify it as a centre or otherwise? What are the possible types of trajectory which can occur in a Hamiltonian system?

 

[Math100]

The fixed points occur when $X(x, y)=Y(x, y)=0$.