DifferentialEquations: enforcing realistic solution?

Specific Domains Modelling & Simulations
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I have a (large) ODE where each state x must lie in the range [0,1]. Due to numeric inaccuracies, the solution does drift outside of this range. I have tried to map x into the allowed range at the start of the function defining the ODE, but this does (of course) not solve the problem.

Question: what is the right procedure to constrain the solution to the range [0,1]? Use some callback structure and map x to the allowed range before x is stored? Other tricks?

NOTE: this mapping into [0,1] needs to be done after each update of the ODE solver, and not only at certain events.

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There is isoutofdomain, maybe that can handle it.

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3

Well if isoutofdomain “doesn’t handle it”, that means the ODE doesn’t actually have a solution that lies in that range. So isoutofdomain is a rather safe way to do it, but sometimes overkill and can hurt performance because of its safety. A quicker way are the domain callbacks which will use interpolations at the cost of some safety.

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4

“isoutofdomain : Specifies a function isoutofdomain(u,p,t) where, when it returns true, it will reject the timestep. Disabled by default.”

That is not really what I need — I don’t want the timestep to be rejected. Would there be any help in that? Won’t the next timestep just do the same thing? I could understand that a timestep perhaps should be rejected if there is some stochasticity involved, but my equation is deterministic, and I’d guess the solver would do the same the next time??

So a little bit more concrete, in my case, state u (or x or whatever) is a matrix, and I have a model with a first attempt similar to:

function model(du,u,p,t)
    #... current attempt
    u_ = constrain.(u)
    du = f(u_)
end

where function constrain does:

function constrain(u)
    if u < 0
        return 0
    elseif u > 1
        return 1
    else
        return u
    end
end

But this doesn’t hinder the solution to drift out of the range [0,1].

So, what I look for is something like:

function model(du,u,p,t)
    #... new attempt
    du = f(u)
end

where I instead fix the problem via a solver modification.

Question: would the correct be to use a DiscreteCallback with condition and affect! functions:

condition(u,t,integrator) = minimum(u) < 0 || maximum(u) > 1
affect!(integrator) = integrator.u = constrain.(integrator.u)

?

5

Hm… with this DiscreteCallback, I get a solution:

image
image949Ă—256 64.8 KB

while if I don’t do this, I get a solution:

image
image714Ă—350 75.9 KB

So the problem is not completely solved, but it is better. Perhaps if I fine-tune some simulation conditions, I can avoid the problem…

6

A DiscreteCallback will enforce the condition only after every time step and not for internal stages of a Runge-Kutta discretization (assuming that you are using a Runge-Kutta method).

If satisfying the bounds is a hard requirement for you that you want to enforce for every internal stage, you should look for methods allowing a stage_limiter! such as strong stability preserving (SSP) methods, see ODE Solvers · DifferentialEquations.jl. They act similar to callbacks but are activated after every stage of these Runge-Kutta methods. That should do what you ask for but you should still think about your ODE per se - does it really satisfy the bounds property you ask for and if not - why? Working on this theoretical side may be more beneficial than this kind of band aid based on limiters.

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7

If your equation does not mathematically allow the solution to “drift” outside of the domain, then if it does, it means that the solver took too large a step and overstepped the domain boundary. Thus rejecting the step and taking a smaller one which does not overstep is a good thing to do. It’s easy to try, so I would go for it; if it impacts performance too much you can still try other approaches mentioned.

However, as @ranocha said, if your equation is not mathematically constrained to the domain but only by your if-elseif-else statement, then that means that the solution is non-smooth at the boundary which is something many solvers may struggle with (and I think isoutofdomain will not help as it assumes smoothness). Then you probably need a more fancy solution. But, again as @ranocha said, it would be best if you can re-formulate your equation such that it is smooth at the boundary while still respecting the boundary.

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Ah, I found a bug in the code… I applied a fitted function to argument x, and didn’t constrain the fitted function when x was outside of the range…

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