I have an ODE for a scalar function $u=u(t)$ of the form: $$ \frac{du}{dt}=L(u). $$ Here the function $L=L(u)$ satisfies: $$ L(0)=0, \quad L'(u)\le0. $$ Then it is easy to see that the solution $u=u(t)$ has the following property:

(i) If the initial value $u(0)\ge0$, then $u(t)\ge0$ for any $t>0$;

(ii) If the initial value $-M\le u(0)\le M$ with $M>0$, then $-M\le u(t)\le M$.

Question: I need to find a numerical method to solve this ODE and under arbitrary step size maintain the two properties or only one of two. I only know that the Euler backward method has these two properties. Does anyone of you know high order (e.g. third-order) methods to solve this problem?

Any links to literature or for further reading would be greatly appreciated.

  • $\begingroup$ @DavidKetcheson The solution of an ODE has the property of monotonicity-preserving, i.e., if $u(t)$ and $v(t)$ are solutions to the same ODE and $u(0)\le v(0)$, then $u(t)\le v(t)$ for any $t>0$. Since $u\equiv0$ is a solution to this ODE, property (i) holds for both positivity-preserving and negativity-preserving with the aid of monotonicity-preserving property. $\endgroup$
    – Michael
    Commented Jun 16, 2016 at 18:34

1 Answer 1

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    $\begingroup$ There are also a more recent review and a book (disclaimer: I'm an author on both). The review is here: brown.edu/research/projects/scientific-computing/sites/…. $\endgroup$ Commented Jun 16, 2016 at 17:03
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    $\begingroup$ The result regarding backward Euler goes back to a 1979 paper of Bolley and Crouzeix (in French). $\endgroup$ Commented Jun 16, 2016 at 17:04
  • $\begingroup$ Thanks for your wonderful answers! I just reviewed these literature briefly. In summary, it seems that there is NO ode solver which is greater than first-order and has the positivity-preserving and/or monotonicity-preserving property under arbitrary step size. $\endgroup$
    – Michael
    Commented Jun 16, 2016 at 18:38
  • $\begingroup$ @Michael See also this paper (reference [79] in the review David Ketcheson linked to): math.leidenuniv.nl/~spijker/PUBLICATIONS-GENERAL/… $\endgroup$
    – Kirill
    Commented Jun 16, 2016 at 18:45
  • $\begingroup$ @Kirill Thanks for the reference. I went through this paper and found in Section 1.4 Remark in the first paragraph: the authors explained that, the contractivity property concerns arbitrary pairs satisfying (1.2) but not for some pairs of specific form. I think the conclusion in that paper does not completely rule out the existence of solver for the ODE in my problem. $\endgroup$
    – Michael
    Commented Jun 16, 2016 at 18:52

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