I came across many problems in my course and I solved them but the forth one, it seems the hardest for me. I will show the problem I want to solve at first, after that I will show my solution for problem three. The main difficulty for now is the linearization of the system, since it has no explicit formulas for the equations. Could you please tell me how to attack this problem?
The problem I want to solve
Consider $\begin{cases} & u_t = u_{xx} + \partial_u W(u,v),\\ & v_t = d v_{xx} - \partial_v W(u,v). \end{cases}$
Where $W(u,v) \in \mathbb{R}$. Assume that this equation has a stationary solution $S(x)= (\overline{U}(x), \overline{V}(x))$.
-Find the linearized operator $L$ w.r.t $S(x)$ and its adjoint operator $L^*$. 2-Find $\Phi^*(x)$ such that $L^* \Phi^* =0.$ (\textit{Hint:} $L S_x=0$).
The solved problem
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,
$$\langle Lf,g \rangle_{L^2} = \langle f ,L^* g\rangle_{L^2}.$$
Solution
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;
$$L^*(v(x))= \int_\mathbb{R} \overline{K(x-y)} \overline{S(x)} v(x) dx + \int_\mathbb{R} \overline{K(x-y)} \overline{S(y)} dy . v(x) + 3 \overline{S(x)^2} v(x).$$
Let $u(x,t) = \overline{u}(x) + \varepsilon h(x,t)$, and $v(x,t) = \overline{v}(x) + \varepsilon g(x,t)$. Then it is easy to see that $u_{xx} = \overline{u} + \varepsilon h_{xx}$, and $v_{xx} = \overline{v} + \varepsilon g_{xx}$. Now we write $w(u,v) = w (\overline{u} + \varepsilon h, \overline{v}+ \varepsilon g)$ in terms of $h$ and $g$;
\begin{align} \tag{1} \frac{\partial w}{\partial h}&= \frac{\partial w}{\partial u}\frac{u}{\partial h}=\varepsilon \frac{\partial w}{\partial u} \Rightarrow \frac{\partial w}{\partial u} =\frac{1}{\varepsilon} \frac{\partial w}{\partial h}.\\ \tag{2} \frac{\partial w}{\partial g}&= \frac{\partial w}{\partial u}\frac{u}{\partial g}=\varepsilon \frac{\partial w}{\partial u} \Rightarrow \frac{\partial w}{\partial u} =\frac{1}{\varepsilon} \frac{\partial w}{\partial g}. \end{align}
Now we find the Taylor series for $w(\overline{u}+ \varepsilon h, \overline{v} + \varepsilon g)$.
\begin{align*} &w(\overline{u}+ \varepsilon h, \overline{v} + \varepsilon g) \\ &= w(\overline{u}, \overline{v}) + \varepsilon h w_h(\overline{u}, \overline{v}) + \varepsilon g w_g(\overline{u}, \overline{v}) + \varepsilon^2 h^2 w_{hh}(\overline{u}, \overline{v}) + \varepsilon^2 h g w_{hg}(\overline{u}, \overline{v}) + \varepsilon^2 g^2 w_{gg}(\overline{u}, \overline{v})+\dots\\ \end{align*} Now we find the derivatives $w_u$ and $w_v$
\begin{align*} w_h(\overline{u}+ \varepsilon h, \overline{v}+ \varepsilon g) &= \varepsilon w_h(s(x)) + 2 \varepsilon^2 h w_{hh}(s(x))+ \varepsilon^2 g w_{hg}(s(x))+ \dots,\\ w_g(\overline{u}+ \varepsilon h, \overline{v}+ \varepsilon g) &= \varepsilon w_g(s(x)) + \varepsilon^2 h w_{hg}(s(x))+ 2 \varepsilon^2 g w_{gg}(s(x))+ \dots,\\ \end{align*} using (1) and (2), and consider terms with order $\varepsilon^1$ we get
\begin{align*} w_{u}(u,v)&= 2 h w_{hh}(s(x)) + g w_{hg}(s(x)) \\ w_{v}(u,v)&= h w_{hg}(s(x))+ 2 g w_{gg}(s(x)) \end{align*}
Now we can express the right hand side of the main system as;
\begin{align*} u_{xx} + \partial_u w(u,v) =& h_{xx} + 2 h w_{hh}(s(x)) + g w_{hg}(s(x)),\\ d v_{xx} - \partial_v w(u,v)=& d g_{xx}- h w_{hg}(s(x))- 2g w_{gg}(s(x)). \end{align*}
Thus the the linearization operator;
$$L = \begin{bmatrix} \partial^2 x +2 \partial^2_u w(s(x)) && \partial_u \partial_v (s(x))\\ - \partial_u \partial_v (s(x)) && d\partial^2 - 2\partial^2_v w(s(x)) \end{bmatrix} $$