# About the adaptive time steps of Tsit5()

**URL:** https://discourse.julialang.org/t/about-the-adaptive-time-steps-of-tsit5/133036
**Category:** General Usage
**Tags:** question
**Created:** [October 10, 2025, 7:24am UTC](https://discourse.julialang.org/t/about-the-adaptive-time-steps-of-tsit5/133036 "2025-10-10T07:24:14Z")
**Posts on this page:** 1
**Page:** 1

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### Author: ![k\_zhou](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/k_zhou/32/216695_2.png) [@k\_zhou](https://discourse.julialang.org/u/k_zhou)
#### Post date: [October 10, 2025, 7:24am UTC](https://discourse.julialang.org/t/about-the-adaptive-time-steps-of-tsit5/133036/1 "2025-10-10T07:24:14Z")

</div>

Hello！

I have a short problem about the time steps of Tsit5(). Would the maximum-allowed time step of Tsit5() be larger than the limit of CFL condition?

> N = Int(ceil(40 \* Lc / lamda))  
> x\_grid = range(0, stop = Lc, length = N)  
> dx = step(x\_grid)  
> dtCFL = n \* dx / c  
> sol1 = solve(prob, Tsit5(); dt=dtCFL, save\_everystep=false, dense=false,abstol=1e-6, reltol=1e-9)  
> sol2 = solve(prob, Tsit5(); dt=dtCFL, dtmax = dtCFL, save\_everystep=false, dense=false,abstol=1e-6, reltol=1e-9)

The set of prob is completely same, but the results of `sol2` is accuracy than the `sol1`, and the time of `sol2` is more 2 times than the `sol1`. Does it mean the time step of `sol1` larger than the step of CFL?

Well, I just curious it that If the maximum time step is smaller than the adaptive time step of Tsit5() without the set of `dtmax`, would a fixed time step solver such as ROCK4 be more efficient?