Advanced Topic in Numerical solution of Differential Equations? i) IF $\frac{dy}{dt} = - \frac{∂H}{∂z}, \frac{dz}{dt}= \frac{∂H}{∂y}$ 
 where H is a function of $y$ and $z$, show that $H(y,z)$ is constant in time. 
ii) Take a $H(y,z) =Ay^2 + 2Hyz + Bz^2$ where $A,B,H$ are constants and show that solutions of the system lie on ellipses. 
iii) Apply the explicit Euler and the symplectic Euler schemes to the system in (ii) and check whether the area is preserved. 
can anyone tell me how to start. totally confused.
=> I have try it for iii)
 when explicit euler is applied to part i) equations, it gives:
$ y_{n+1} = y_{n} - h z_{n}$
$ z_{n+1} = z_{n} + h y_{n}$
now squaring both sides and rearranging both equations gives
$ y^{2}_{n+1}+z^{2}_{n+1} = (1+h^{2}) (y^{2}_{n} + z^{2}_{n})$--------------(6)
after that I really don't know, I guess I need to find area of ellipses by comparing 
(6) with $H(y,z) =Ay^2 + 2Hyz + Bz^2$ but how?
and same for the symmetric euler applied to part i)equations gives:
$ y_{n+1} = y_{n} - h z_{n}$
$ z_{n+1} = z_{n} + h y_{n+1}$
now squaring both sides and rearranging both equations gives
$ y^{2}_{n}+z^{2}_{n} = (1-h^{2}+h^{4}) (y^{2}_{n+1}- 2h^{3} y_{n+1} z_{n+1}+(1+h^{2}) z^{2}_{n})$--------------(7)
again  to find area of ellipses by comparing (7) with $H(y,z) =Ay^2 + 2Hyz + Bz^2$ , but this time it gives
$A= 1-h^{2}+h^{4}, H= -h^{3}, B=1+h^{2}$
 so,
$AB-H^{2}= (1-h^{2}+h^{4})(1+h^{2})-(-h^{3})$
$= 1-h^{2}+h^{4}+h^{2}-h^{4}+h^{6}+h^{6}$
$= 1+2h^{6}$
where $AB-H^{2}=\lambda_{1} \lambda_{2}$
and $a= \frac{1}{\sqrt{\lambda_{1}}} , b=\frac{1}{\sqrt{\lambda_{2}}}$
SO, AREA = $\frac{\pi}{\sqrt{(1+2h^{6})}}$ ( area of $ellipse = \pi a b)$
I am sure with my answer for sympetic euler and not sure for explicit euler to find area.
 please someone help will be really grateful.
THANK YOU
 A: I can offer some information concerning (i) and (ii), but not much on (iii), for reasons which will be amplified in the following.
For item (i), observe that
$\dfrac{dH}{dt} = \dfrac{\partial H}{\partial y}\dfrac{dy}{dt} + \dfrac{\partial H}{\partial z}\dfrac{dz}{dt} \tag{1}$
by the chain rule.  Now we use the given equations
$\dfrac{dy}{dt} = - \dfrac{∂H}{∂z}, \tag{2}$
$\dfrac{dz}{dt}= \dfrac{∂H}{∂y}; \tag{3}$
if these are inserted into (1), we obtain
$\dfrac{dH}{dt} = -\dfrac{\partial H}{\partial y} \dfrac{\partial H}{\partial z} + \dfrac{\partial H}{\partial z} \dfrac{\partial H}{\partial y} = 0 \tag{4}$
thus resolving item (i).
As for item (ii), it is not exactly true as stated; hopefully the following will explain.  From (i), we see that $H(y, z)$ is constant along the trajectories of the system (2)-(3); thus these solutions lie in the level sets of $H(x, y)$.  They will in fact lie in ellipses if the sets
$H(y, z) = Ay^2 + 2Hyz + Bz^2 = \text{constant} \tag{5}$
are ellipses.  But whether (5) defines a family of ellipses or some other conic section depends on the values of $A, B, H$, to wit:  writing the function $H(y, z)$ defined in (5) in vector-matrix form we have
$H(y, z) = (y, z) \begin{bmatrix} A & H \\ H & B \end{bmatrix} \begin{pmatrix} y \\ z \end{pmatrix} = \begin{pmatrix} y \\ z \end{pmatrix}^T \begin{bmatrix} A & H \\ H & B \end{bmatrix} \begin{pmatrix} y \\ z \end{pmatrix}. \tag{6}$
The matrix
$M = \begin{bmatrix} A & H \\ H & B \end{bmatrix} \tag{7}$
occurring in (6), being symmetric, may be diagonalized by some orthogonal matrix $O$; it will then take the form
$O^T M O = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}; \tag{8}$
$\lambda_1$ and $\lambda_2$, being the eigenvalues of $M$, satisfy its characteristic equation
$\lambda^2 - (A + B)\lambda + (AB - H^2) = 0, \tag{9}$
and in the coordinate system $x_1$-$x_2$ which is the $y$-$z$ system transformed by $O$, (6) takes the form
$H(x_1, x_2) = (x_1, x_ 2) \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} =  \lambda_1 x_1^2 + \lambda_2 x_2^2; \tag{10}$
the equation
$H(x_1, x_2) = \lambda_1 x_1^2 + \lambda_2 x_2^2 = \text{constant} \tag{11}$
represents an elliptical curve precisely when $\lambda_1$, $\lambda_2$ are of the same sign, in which case their sign must be shared with $\text{constant}$ lest there be no curve whatsoever.  The signs of $\lambda_1$, $\lambda_2$, however, are governed by (9), and via the quadratic formula it is easy to see that these signs agree precisely when
$\det M = AB - H^2 = \lambda_1 \lambda_2 > 0. \tag{12}$
When (12) applies, the level sets of $H(y, z)$ are ellipses but with their axes tilted to align themselves with the $x_1$ and $x_2$ axes in the transformed coordinates; if on the other hand $\lambda_1 \lambda_2 < 0$, the level sets will be hyperbolas; the cases when one of the eigenvalues is zero may be similarly analyzed but I will not do so here; my point is to show that level sets of $H(y, z)$ are not necessarily ellipses without further qualification on the coefficients present in (5).  Nevertheless, the orbits of (2), (3) will always lie in a level set of $H(y, z)$, as we have seen.
Regarding item (iii), I obviously can't apply these methods within the scope of this answer, but there are undoubtedly code packages which implement such techniques; I'd recommend finding a good one and trying it out.  That's about all I can give on this, but this wikipedia page has a lot more than I to say about symplectic integrators.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
