Let me start with a disclaimer. This is the first time ever dealing with Gateaux derivative, hence my knowledge around this topic is poor.
As I was going through this paper, I got stuck in the following part:
Here the function $f$ is given by: $f(u_1)=-u_1(u_1-\alpha)(u_1-1)$. In my understanding, $A_\rho$ is simply the linear part of $(2.2)$. How come and in addition represents the Gateaux derivative of $F_\rho$? In any case, I see no differentiation with respect to anything here.
I'm having a hard time getting my head around this one. Could you please share more details on why $A_\rho=dF_\rho(0)$? Any help is much appreciated. Thanks in advance!
EDIT: After the comment I received, I realized that I certainly did not phrase my question correctly. I do not see the second equality in $(2.3)$, not the first one. I get that $A_\rho$ is simply defined. But in my understanding, the matrix in $(2.3)$ is only the linear part of $(2.2)$, without any differentiation involving somehow. But then again, in $(2.3)$, it is claimed that it is the Gateuax derivative...
The definition of $F_\rho$ is $$ F_\rho(u_1,u_2) = \pmatrix{ \Delta u_1 + −u_1(u_1−\alpha)(u_1−1)- u_2 \\ \epsilon u_1 + \epsilon\gamma u_2}, $$ so that its Gateaux derivative at $(u_1,u_2)=0$ is the operator in block form $$ F'_\rho(0,0) = \pmatrix{ \Delta + \alpha\ id & -id \\ \epsilon id & \epsilon\gamma\ id}. $$ here, $\alpha \ id = f'_\rho(0)$.