What does Arg {Inf I(d)} means? I am currently studying the phase-field method for fracture modeling. In an article by Miehe -"Thermodynamically consistent phase-field models of fracture: Variational principles and multi-field FE implementation", I came across on an identity that I don't understand basically it says:
$$
d = \mbox{Arg}\left\{\inf_{d \in W} I(d)\right\}.
$$
This expression came from the Euler type differential equation
$$
d(x) - l^2d''(x) = 0, 
$$
which has the solution as an exponential function
$$
d(x)\ = \ \exp(-|x|/l)
$$
Variational principle of this differential equation is ,
$$
d = \mbox{Arg}\left\{\inf_{d \in W} I(d)\right\}.
$$
with $$ W = \{d\ |\ d(0)=1,\ d(\pm \infty)=0\}$$
where I(d) is a functional defined as an integral
$$
I(d) = \frac{1}{2}\int [d^2 + l^2d'^2] dV,
$$
and $l$ is a parameter.

Can anyone tell me what does that mean?
It returns the argument that makes the functional $I\equiv I(d)$ infimum such that the function $d$ belongs to the space of kinematically acceptable functions $W$, i.e., it satisfies the prescribed boundary conditions. It just explicitly emphasises that we are after the function (or argument) that makes the functional infimum and not the infimum value itself.
It returns the argument that makes the functional $I\equiv I(d)$ infimum such that the function $d$ belongs to the space of kinematically acceptable functions $W$, i.e., it satisfies the prescribed boundary conditions. It just explicitly emphasises that we are after the function (or argument) that makes the functional infimum and not the infimum value itself.