# How does the error work for the Strang Splitting?

We know in Strang splitting that the splitting error in the steady state solution is proportional to $$h^2$$. I want ask 2 things: If this error in the steady state solution is the global error? If we are in in a previous step before the steady state solution for example $$t=0.3$$ sec (with $$t_{total}=1$$ sec) will be the splitting error proportional to $$h^2$$ or it changes ? Thanks!

• If you are considering a stable steady state, then the error will likely oscillate and remain bounded, as your numerical solution will oscillate around the true steady state. Therefore the notion of global error ist probably not so relevant. If you want to have a better steady state preservation, search for "rebalanced splitting", which corrects the basic'splitting to ensure accurate steady states. – Laurent90 Apr 13 at 20:25
• Thank you for your response. I also worked rebalanced splitting. What i want to examine is how the global error changes when i change the timestep. So for example i have an analytical solution of the problem with timestep dt=10^(-6) and dx=0.001. Now i implement the strang splitting or rebalanced splitting for different values of dt ( for example dt=[10^(-5) - 10^(-3)] ) and keep dx close to the value of 0.001. Now for example i want to examine the global error in t=0.3 which is not the steady state solution. So i estimate a global error with dt=10^(-4) and a global error with dt=10^(-3). – Giannis Kavroulakis Apr 13 at 21:12
• So i ask if we expect in this case the ratio of the global errors be close to 100 (because splitt error in steady state solution is proportional to h^2 and i ask if this happens and when we are not in the steady state solution) or we have to expect something different for example the global error when we are not in steady state solution be linear an in this example the ratio of the global errors for dt=10^(-4) and dt=10^(-3) be close to 10. I hope my question is clear. – Giannis Kavroulakis Apr 13 at 21:19