I have a 2D vector $\boldsymbol{u}$ and it's norm is $\lambda$. I have this function: $$\boldsymbol{T}=\dfrac{\boldsymbol{u}}{\lambda} e^\lambda$$ I need to compute $\boldsymbol{T}$ and it's Jacobean around zero as part of my Newton finite element problem.

The function is poorly defined around 0 and this is causing divergence issues.

This problem is closely associated to the fenics Cohisve zone model example except that I am looking to use the law above when fracture starts to happen and not prior. At that instant $\boldsymbol{u}$ and $\lambda$ are close to 0.

The literature uses laws like these but as far as I know, it is not explained how one could work around this issue.

  • 1
    $\begingroup$ Could you use the function $T_{\varepsilon} = \frac{u}{\sqrt{\lambda^2 + \varepsilon}} e^\lambda$? This sort of regularization is often done when dealing with total variation functionals, which divide by the norm of the gradient. $\endgroup$
    – whpowell96
    Apr 14, 2023 at 20:00
  • $\begingroup$ Since $T(u)$ is not differentiable at $u=0$, what do you need the derivative for? $\endgroup$ Apr 14, 2023 at 21:02
  • $\begingroup$ @whpowell96 I will try that @w $\endgroup$ Apr 15, 2023 at 14:11
  • $\begingroup$ @whpowell96 this regularization works!! Do you have a quick reference I can read about this a bit more? $\endgroup$ Apr 15, 2023 at 16:25
  • 1
    $\begingroup$ It comes from the fact that $|x| \approx \sqrt{x^2 + \varepsilon}$ but the approximation is differentiable. This citation uses it for TVD regularization and discusses the impact on optimization via Newton's method: Chan, Tony F.; Golub, Gene H.; Mulet, Pep_, A nonlinear primal-dual method for total variation-based image restoration. $\endgroup$
    – whpowell96
    Apr 15, 2023 at 21:23

1 Answer 1


@whpowel has the question answered in the comments. I will re-post it here for the sake of completeness:

Use the function: $$\boldsymbol{T}_{\varepsilon}=\dfrac{\boldsymbol{u}}{\sqrt{ \lambda^2+\varepsilon}} e^{\sqrt{\lambda^2+\varepsilon}}$$


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.