Advanced Numerical solution of differential equations

[ra_forever8]

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.
=>

for i)
by using the chain rule, we observe that
$\frac{dH}{dt} = - \frac{∂H}{∂y} \frac{dy}{dt} + \frac{∂H}{∂z} \frac{dz}{dt}$ --------(1)
now we use the given equations $\frac{dy}{dt} = - \frac{∂H}{∂z}, \frac{dz}{dt}= \frac{∂H}{∂y}$
into (1), to obtain
$\frac{dH}{dt} = - \frac{∂H}{∂y} \frac{∂H}{∂z} + \frac{∂H}{∂z} \frac{∂H}{∂y}$
$\frac{dH}{dt}=0$
$H(t)= constant$
Therefore, $H(y,z)$ is constant in time.

ii)
From part i) $H(y,z)$ is constant. So,
$H(y,z) =Ay^{2 }+ 2Hyz + Bz^{2}= constant -------(2)$
are ellipses. but whether $(2)$ defines a family of ellipses or some other conic section depends on the values of $A,B,H$
Writing the function $H(y,z)$ defined in $(2)$ in vector-matrix form we have
\begin{equation}
H(y,z)= (y,z) \begin{bmatrix}
A & H\\
H & B\\
\end{bmatrix} \begin{pmatrix}
y\\
z\\
\end{pmatrix} -------------------(3)
\end{equation} Let the matrix \begin{equation}
M= \begin{bmatrix}
A & H\\
H & B\\
\end{bmatrix}
\end{equation}
occurring in (3), being symmetric may be diagonalised by some orthogonal matrix $O$, it will then take the form
\begin{equation}
O^{T} M O= \begin{bmatrix}
\lambda_{1} & 0\\
0 &\lambda_{2}\\
\end{bmatrix}
\end{equation}
$\lambda_{1}$ and $\lambda_{2}$ being the eigenvalues of $M$, satisfy its characteristics equation.
\begin{equation} \lambda^{2} -(A+B) \lambda + (AB-H^{2})=0 -------------------(4)
\end{equation}
and in the coordinate system $x_{1} - x_{2}$ which is the $y-z$ system transformed by $O$,
So $(3)$ takes the form
\begin{equation}
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}
\end{equation}
\begin{equation}
= \lambda_{1} x^2_{1} + \lambda_{2} x^2_{2}
\end{equation}
The equation is
\begin{equation}
H(x_{1},x_{2})= \lambda_{1} x^2_{1} + \lambda_{2} x^2_{2}=constant
\end{equation}
represents an elliptical curve precisely when $\lambda_{1},\lambda_{2}$ are the same sign, in which case their sign must be shared with constant lest there be no curve whatsoever
The signs of $\lambda_{1},\lambda_{2}$ are governed by $(4)$ and via the quadratic formula, these signs are precisely when
$ det M = AB-H^{2}= \lambda_{1},\lambda_{2} >0$ ----------------(5)
when $(5)$ applies, the level of 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.

(IS THIS ENOUGH TO ANSWER THE PART ii), did I missed anythings, ? do I need to calculate $\lambda_{1},\lambda_{2}$?)
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

 

[jvicens]

for reading my answer.
For part ii), you have correctly shown that the solutions of the system lie on ellipses. To determine the specific shape and orientation of these ellipses, you need to find the values of $A$, $B$, and $H$ that satisfy the condition in equation (5).
To do this, you can use the characteristic equation (4) and the fact that the determinant of a symmetric matrix is equal to the product of its eigenvalues. This means that $AB-H^2 = \lambda_1 \lambda_2$, where $\lambda_1$ and $\lambda_2$ are the eigenvalues of the matrix $M$.
Solving the characteristic equation (4) for $\lambda_1$ and $\lambda_2$, we get:
$$\lambda_1 = \frac{1}{2}(A+B + \sqrt{(A+B)^2 - 4(AB-H^2)})$$
$$\lambda_2 = \frac{1}{2}(A+B - \sqrt{(A+B)^2 - 4(AB-H^2)})$$
Now, we need to consider the different cases based on the sign of the expression inside the square root. This will determine the shape and orientation of the ellipses.
If $(A+B)^2 - 4(AB-H^2) > 0$, then both eigenvalues will be positive, and the level sets of $H(y,z)$ will be ellipses with axes aligned with the $x_1$ and $x_2$ axes, as you have correctly stated.
If $(A+B)^2 - 4(AB-H^2) < 0$, then both eigenvalues will be negative, and the level sets of $H(y,z)$ will be hyperbolas.
If $(A+B)^2 - 4(AB-H^2) = 0$, then one eigenvalue will be positive and one will be negative, and the level sets of $H(y,z)$ will be parabolas.
Therefore, to ensure that the solutions lie on ellipses, we need to have $(A+B)^2 - 4(AB-H^2) > 0$. This condition can be rewritten as:
$$\frac{(A+B)^2}{4} > AB-H^2$$
$$\frac{A^2 + 2AB + B^2}{4} >