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?