Recent content by ResiRadloff

Okay, thanks, yeah that makes more sense. Does anyone know how to calculate the Energy-error and what is meant by ##L^2## error?

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...