Harmonic oscillator and symplectic Euler method

[ra_forever8]

Given the equations for the harmonic oscillator
$\frac{dy}{dz}=z, \frac{dz}{dt}= -y$if the system is approximated by the symplectic Euler method, then it gives$z_{n+1}= z_{n}-hy_{n}, \\ y_{n+1}= y_{n}+hz_{n+1}$which shows that the circle $y^2_{n} + z^2_{n} = 1$ is mapped into an ellipse.
Deduce that to order $h^2$ the ellipse has the same enclosed area as the circle.=> my attempt so far
in this case, area of ellipse = area of circle.Truncation error from the symplectic euler method gives$hT_{z}= h(z'+y) +O(h^2)$$hT_{z}= O(h^2)$ since, $h(z'+y)$ goes to zero$T_{z}= O(h)$Similarly,$hT_{y}= h(y'-z) +O(h^2)$$hT_{y}= O(h^2)$ since, $h(y'-z)$ goes to zero$T_{y}= O(h)$Both have first order
did my answer satisfy the question?

 

[jvicens]

Your attempt is on the right track, but there are a few things that need to be clarified. First, the truncation error is not equal to $hT_z$, it is equal to $z_{n+1}-z(t_{n+1})$. Similarly, the truncation error for $y$ would be $y_{n+1}-y(t_{n+1})$. Also, when you say $h(z'+y)$ goes to zero, it should be $h(z_n + y_n)$ since this is the actual value of $z$ and $y$ at the next time step. Additionally, it is not correct to say that the truncation errors are $O(h^2)$, they are actually $O(h)$ as you correctly deduced later. Finally, you need to show that the area of the ellipse is equal to the area of the circle to order $h^2$, not just that it is equal. This means you need to show that the difference in areas is of order $h^2$. Here is a possible solution:

We know that the area of a circle is given by $A = \pi r^2$, where $r$ is the radius. The area of an ellipse is given by $A = \pi ab$, where $a$ and $b$ are the semi-major and semi-minor axes, respectively. We want to show that $ab = r^2$ to order $h^2$, which would mean that the areas are equal.

Using the equations for the harmonic oscillator, we can write $y_{n+1} = y_n + hz_n + O(h^2)$ and $z_{n+1} = z_n - hy_n + O(h^2)$. Substituting these into the equation for the ellipse, we get:

$ab = \sqrt{(y_{n+1})^2 + (z_{n+1})^2} = \sqrt{(y_n + hz_n + O(h^2))^2 + (z_n - hy_n + O(h^2))^2} = \sqrt{(y_n^2 + z_n^2 + h^2(z_n^2 + y_n^2) + O(h^3)) + (z_n^2 + y_n^2 - h^2(z_n^2 + y_n^2) + O(h^3