- #1
ResiRadloff
- 2
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- Homework Statement
- Let us denote by ## \textbf{z} = (x,y) \in \mathbb{R}^2 ## the Cartesian coordinates of a point in the plane.
1. Given the Hamiltonian ## H(\textbf{z}) = \frac{|\textbf{z}|^2}{2} ##, write down the corresponding canonical Hamiltonian system for ##\textbf{z}(t)##.
2. Write down the analytical solution of this system.
In the following, assume that the system reads ##\frac{d}{dt}\textbf{z} = \textbf{F}((\textbf{z}(t)))## where ## \textbf{F} = (F_x,F_y) ##, and let ##t^n = n\Delta t## where ##\Delta t > 0## is the time step size. Also let ## \textbf{z}^n = \textbf{z}(t^n) ##.
3. Solve the Hamiltonian system numerically by using the following numerical methods. For each of them write down the scheme, plot your result vs. the analytical result, and plot energy (= Hamiltonian) and ##L^2 ##-error of the scheme. Start with a random point ##(x_0,y_0) \in [0,1]^2 ## at ## t = 0 ## and evolve ## \textbf{x}(t) ## forward in time. You may use ##\Omega = 2, \Delta t = 0.03 ##, and do 500 time steps.
(a) The explicit Euler method: ##\textbf{z}^{n+1} = \textbf{z}^n + \Delta t \textbf{F}(\textbf{z}^n) ; \textbf{z}^0 = \textbf{z}(t=0) ##
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- Relevant Equations
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Good evening,
unfortunately I can't get to the solution of my task
I wrote for the system:
## \frac{dz}{dt} = \nabla_p H ; \\
\frac{dp}{dt} = - \nabla_z H ##
Then the solution would be (as ## \nabla_p H = 0) ##:
## \frac{dz}{dt} = 0 \Rightarrow z = const. ## and ## p = zt + p_0 ##.
But that can't be as now finding a numerical solution doesn't make sense?.
I would be really happy if someone yould help me?
Thanks a lot
Resi
Danke schonmal und viele Grüße
unfortunately I can't get to the solution of my task
I wrote for the system:
## \frac{dz}{dt} = \nabla_p H ; \\
\frac{dp}{dt} = - \nabla_z H ##
Then the solution would be (as ## \nabla_p H = 0) ##:
## \frac{dz}{dt} = 0 \Rightarrow z = const. ## and ## p = zt + p_0 ##.
But that can't be as now finding a numerical solution doesn't make sense?.
I would be really happy if someone yould help me?
Thanks a lot
Resi
Danke schonmal und viele Grüße