I have bee stuck in this problem since more than a week. During my study I kinda understand how to find the adjoint operator for the linearization. But I have no Idea how to find the linearization to this problem. I have been studying nonlinear dynamical systems and I know how analyze it. But I am new to the diffusion reaction. Any hint or solutions will help me. Thanks in advance.
Let the equation
$$u_t = \left(\int_{-\infty}^{\infty} K(x-y) u(y) dy\right) \cdot u + u^3$$
have stationary solution $S(x)$, where $K(x)$ is sufficiently smooth function and decay exponentially as $|x| \to { \infty}$.
1- Find the linearized operator $L$ w.r.t $S(x)$.
2-Find $L^*$, where $L^*$ is the adjoint operator of $L$, that is,
$$\langle Lf,g \rangle_{L^2} = \langle f ,L^* g\rangle_{L^2}.$$
Updated
Let $u(x,t) = S(x) + \varepsilon v(x,t)$, substituting it in the equation, we get
$$\varepsilon v_t = S(x) \int_\mathbb{R} K(x-y) S(y) dy + \varepsilon S(x)\int_\mathbb{R} K(x-y) v(y) dy + \varepsilon v(x,t) \int_\mathbb{R} K(x-y) S(y) dy + \varepsilon^2 v(x,t) \int_\mathbb{R} K(x-y) v(y) dy + S^3 (x) + 3 \varepsilon S(x)^2 v(x,t)+ 3 \varepsilon^2S(x) v^2(x,t) + \epsilon^3 v^3(x,t).$$
The terms of order $\varepsilon^0,\varepsilon^2$, and $\varepsilon^3$ are vanishes.
\begin{equation} \label{p2eq1} v_t = S(x)\int_\mathbb{R} K(x-y) v(y) dy + \int_\mathbb{R} K(x-y) S(y) dy . v(x,t)+ 3 S(x)^2 v(x,t). \end{equation} Then the linearized operator;
$$L_{s} = S(x)\int_\mathbb{R} K(x-y) v(y) dy + \int_\mathbb{R} K(x-y) S(y) dy . v(x,t)+ 3 S(x)^2 v(x,t).$$
Now we find the adjoint operator.
\begin{align} \langle Lf,g \rangle_{L^2} & = \int_\mathbb{R} \left( \int_\mathbb{R} K(x-y) S(x) f(y) dy + \int_\mathbb{R} K(x-y) S(y) dy . f(x)+ 3 S(x)^2 f(x)\right) \overline{g(x)} dx.\\ &= \int_\mathbb{R} \left( \int_\mathbb{R} K(x-y) S(x) f(y) dy \right) \overline{g(x)} dx + \int_\mathbb{R} \left( \int_\mathbb{R} K(x-y) S(y) dy . f(x)\right) \overline{g(x)} dx \\ &+ 3\int_\mathbb{R} \left( S(x)^2 f(x)\right) \overline{g(x)} dx.\\ &= \int_\mathbb{R} f(y) \left( \int_\mathbb{R} K(x-y) S(x) \overline{g(x)} dx \right) dy + \int_\mathbb{R}f(x) \left( \int_\mathbb{R} K(x-y) S(y) dy. \overline{g(x)}\right) dx\\ &+ 3 \int_\mathbb{R} f(x) \left(S(x)^2 \overline{g(x)} \right) dx.\\ &= \int_\mathbb{R} f(y) \overline{\left( \int_\mathbb{R} \overline{K(x-y)} \overline{S(x)} g(x) dx \right)} dy + \int_\mathbb{R} f(x) \overline{\left( \int_\mathbb{R} \overline{K(x-y)} \overline{S(y)} dy . g(x)\right) } dx\\ &+3 \int_\mathbb{R} f(x) \overline{ \left( \overline{S(x)^2} g(x) \right)} dx.\\ \end{align}
Then for $v(x) \in L^2 (\mathbb{R})$ we have the adjoint operator $L^*$ of the operator $L$ as;
$$L^*(v(x))= \int_\mathbb{R} \overline{K(x-y)} \overline{S(x)} v(x) dx + \int_\mathbb{R} \overline{K(x-y)} \overline{S(y)} dy . v(x) + 3 \overline{S(x)^2} v(x).$$