Determine if the autonomous system is Hamiltonian Consider the autonomous system $\dot{x}=-y-\alpha ^2xy^2$ and $\dot{y}=x^3$, where $\alpha$ is a real parameter.

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*(a) For which values of $\alpha$ is this system Hamiltonian? For each case, find the Hamiltonian.

*(b) For each value of $\alpha$, find all equilibrium solutions of the above system. Can the principle of linearized stability be used to determine their stability?

*(c) Show that for all $\alpha \in R$ the origin is a stable equilibrium. (Hint: Can you use Hamiltonian functions from (a)?).

So the way to solve this is to use the fact that $\dot{x}=\partial H/\partial y$ and $\dot{y}=-\partial H/\partial x$, where $H(x,y)$ is the Hamiltonian for this system. But then $H(x,y)=-\frac{x^4}{4}+V(y)$ and $H(x,y)=-\frac{y^2}{2}+V(x)$ but I don't know how to solve this further.  The only way to have Hamiltonian is if $\alpha =0$, ie. $H(x,y)=-\frac{x^4}{4}-\frac{y^2}{2}$. Then we will have system $\dot{x}=-y$ and $\dot{y}=x^3$. Is this right?
So only $\alpha$ that would work is if $\alpha =0$ and then only equilibrium will be the origin $(0,0)$. Then the Jacobian would be $Df=\begin{pmatrix} 0 & -1 \\ 3x^2 & 0 \end{pmatrix}=\begin{pmatrix} 0 & -1 \\ 0 &0\end{pmatrix}$. So then what is equilibrium solution of this system and can principle of linearized stability be used to determine its stability?
How do we show for $\alpha \in R$, the origin is a stable equilibrium?
I am not sure if I got the Hamiltonian correct, since I am assuming only $\alpha =0$ will give a Hamiltonian. But the questions assume there are other $\alpha$ values that will work just as well.
Please help.
 A: A general Hamiltonion system in the two configuration variables $x$ and $y$ takes the form
$\dot x = \dfrac{\partial H(x, y)}{\partial y}, \tag 1$
$\dot y = -\dfrac{\partial H(x, y)}{\partial x}, \tag 2$
where $H(x, y)$ is a scalar function of $x$ and $y$.  If we assume $H(x, y)$ is of class $C^2$, then we may form the divergence of the vector field
$(\dot x, \dot y) = \left (\dfrac{\partial H(x, y)}{\partial y}, -\dfrac{\partial H(x, y)}{\partial x} \right ) \tag 3$
and find
$\nabla \cdot (\dot x, \dot y) = \nabla \cdot \left (\dfrac{\partial H(x, y)}{\partial y}, -\dfrac{\partial H(x, y)}{\partial x} \right )$
$= \dfrac{\partial^2 H(x, y)}{\partial x \partial y} - \dfrac{\partial^2 H(x, y)}{\partial y \partial x} = 0; \tag 4$
this is a necessary condition for $(\dot x, \dot y)$ to be Hamiltonion; applying this criterion to the vector field given by
$\dot x = -y - \alpha^2 xy^2, \tag 5$
$\dot y = x^3 \tag 6$
$\nabla \cdot (\dot x, \dot y)  = \dfrac{\partial (-y - \alpha^2xy^2)}{\partial x} + \dfrac{\partial x^3}{\partial y} = -\alpha^2 y^2 = 0 \tag 7$
if and only if
$\alpha = 0, \tag 8$
in which case the system becomes
$\dot x = -y, \tag 9$
$\dot y = x^3, \tag{10}$
and it is easy to see that taking
$H(x, y) = -\dfrac{y^2}{2} - \dfrac{x^4}{4} \tag{10.5}$
yields the equations (9)-(10).  Thus is part (a) resolved.
As for part (b), for any $\alpha$ the equilibrium points satisfy
$0 = \dot x = -y - \alpha^2 xy^2 \tag{11}$
and
$0 = \dot y = x^3; \tag{12}$
now (12) forces
$x = 0, \tag{13}$
and substituting this into (11) in turn forces
$y = 0 \tag{14}$
as well.  Thus the only critical point occurs at
$(x, y) = (0, 0) \tag{15}$
no matter what value
$\alpha \in \Bbb R \tag{16}$
may take.  We may attempt to investigate its stability by forming the Jacobian matrix
$J(x, y) = \begin{bmatrix} \dfrac{\partial{\dot x}}{\partial x} & \dfrac{\partial{\dot x}}{\partial y} \\ \dfrac{\partial{\dot y}}{\partial x} & \dfrac{\partial{\dot y}}{\partial y} \end{bmatrix}; \tag{17}$
using (5)-(6) and (15) we find
$J(0, 0) =  \begin{bmatrix} 0 & -1 \\ 0  & 0 \end{bmatrix},\tag{18}$
the characteristic polynomial of which is
$\det(J(0, 0) - \lambda I) = \det \left ( \begin{bmatrix} -\lambda & -1 \\ 0  & -\lambda \end{bmatrix} \right ) = \lambda^2, \tag{19}$
the (repeated) root of which is
$\lambda = 0; \tag{20}$
$\lambda$ is of course the eigenvalue of $J(0, 0)$, and since it has $0$ real part, linearization cannot be used to determine the stability of $(0, 0)$.
We turn to part (c).  We consider the function $H(x, y)$ as in (10.5); we observe that this function takes its maximum value $0$ uniquely at $(0, 0)$, and that the other level sets of $H(x, y)$ are of the form
$-\dfrac{y^2}{2} - \dfrac{x^4}{4} = \text{constant} < 0, \tag{21}$
that is, they are closed "ellipsoidal" curves symmetrically surrounding the origin.  We compute $\dot H(x, y)$ along the trajectories of (5)-(6):
$\dot H(x, y) = -y\dot y - x^3 \dot x; \tag{22}$
in light of (5), (6) this becomes
$\dot H(x, y) = -yx^3 - x^3 (-y - \alpha^2 xy^2)$
$= -yx^3 + x^3y + \alpha^2 x^4y^2 = \alpha^2x^4y^2 \ge 0; \tag{23}$
in accord with (23), we see that $H(x, y)$ is nondecreasing along the trajectories of (5)-(6); thus any orbit of (5)-(6) which passes through a point interior to an "ellipsoidal" level set of $H(x, y)$ forever remains in this interior region of the plane.  Since such regions may be taken arbitrarily
small by choosing $H(x, y)$ sufficiently close to yet less than $0$, $(0, 0)$ is a stable point of the system (5)-(6).
