Let us considered the following ODEs \begin{align*} \dfrac{dX}{dt} = F(X), \tag{1.1} \end{align*} where the unknown $X\in \mathcal{D} \subset \mathbb{R}^l$ and it can be stiff. These ODEs can be solved numerically by classical Newton's method if $F$ is differentiable almost everywhere on the considered domain \begin{align*} \dfrac{X^{n+1}-X^n}{\Delta t} = F\left(X^{n+1}\right). \end{align*} In cases of $F$ is not differentiable everywhere, I have read in a lecture that there is a way to solve numerically by replacing the system (1.1) with the following system \begin{align*} X^{n+1} - X^n - (1-\nu)\Delta t F(X^{n+1}) & = 0 \in \mathbb{R}^l\tag{1.2a}\\ \det\left[I - (1-\nu)\Delta t \nabla F(X^{n+1}) \right] - w &= 0 \in \mathbb{R} \tag{1.2b} \\ \min(\nu,w) &= 0 \in \mathbb{R} \tag{1.2c} \end{align*} where $\nu$ and $w$ are two new unknowns to be sought for simultaneously to $X^{n+1}$. I wonder how they can do like this and what is technique they use?

  • 5
    $\begingroup$ Can you give a link to where you read this? $\endgroup$ Commented Feb 28, 2023 at 4:33
  • $\begingroup$ If $F$ is not differentiable, then $\nabla F$ in (1.2b) doesn't exist either. $\endgroup$ Commented Feb 28, 2023 at 12:49
  • $\begingroup$ Perhaps making $\nu=1$ at those points where $\nabla F$ does not exist is the strategy there. Looks like some kind of regularization, replacing the original problem by something close to it but with a differentiable RHS. $\endgroup$ Commented Feb 28, 2023 at 16:32
  • $\begingroup$ @MaximUmansky If you choose $\nu=1$ at these points, then (1.2b) boils down to $\text{det}[I]-w=0$, so $w=l$ ($l$=size of the vector $X$). But then (1.2c) isn't satisfied. This whole set of equations doesn't make any sense to me. $\endgroup$ Commented Feb 28, 2023 at 22:16
  • $\begingroup$ @WolfgangBangerth But 1.2c just says that the minimum for $(\nu,w)$ on the domain is zero, at a particular point $w$ can be $l$ (or any real number larger than zero) - am I wrong about this? $\endgroup$ Commented Feb 28, 2023 at 22:55


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.