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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.

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    $\begingroup$ 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
    Commented Apr 14, 2023 at 20:00
  • $\begingroup$ Since $T(u)$ is not differentiable at $u=0$, what do you need the derivative for? $\endgroup$
    – Wolfgang Bangerth
    Commented Apr 14, 2023 at 21:02
  • $\begingroup$ @whpowell96 I will try that @w $\endgroup$
    – CuteCompute
    Commented 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? $\endgroup$
    – CuteCompute
    Commented Apr 15, 2023 at 16:25
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    $\begingroup$ 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
    Commented Apr 15, 2023 at 21:23

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

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

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