FDTD - Simulation time and memory dependence on mesh size

fdtd

#1

Hi,

I’m wondering if anyone could help with this question: “why do the simulation time and memory requirements scale as λ/dx^4 & λ/dx^3 respectively for a 3D simulation?”

thanks,

Nigel


#2

Hi @n.phillipson2
In the FDTD method, the simulation time scales as 1/dx4 for 3D simulations and 1/dx3 for 2D simulations. dx is the mesh size. You can find many information in the following page :
https://kb.lumerical.com/en/index.html?ref_sim_obj_mesh_refinement.html


#3

Hi thanks for your help. The link was helpful, but I’m looking for some details for why there is a 1/dx^4 dependence on mesh size for the simulation time. Thanks, Nigel.


#4

The 1/dx^4 dependences is because the time step, dt, depends on the mesh size as well.


#5

Here is a link to a whitepaper about the conformal mesh technique:
https://www.lumerical.com/support/whitepaper/fdtd_conformal_mesh_whitepaper.html
We need to remember that the 1/dx^4 is only an approximation. It makes intuitive sense that there is at least 1/dx^3 dependence due to 3D structure. Then the differences could also be due to how the convergence works for the specific structure you are trying to simulate. Link to convergence process:
https://kb.lumerical.com/en/layout_analysis_test_convergence_fdtd.html
I hope that this helps!


#6

Yes thanks! that’s really helpful,

kind regards,

Nigel