I have this system of nonlinear reaction diffusion equations
$\begin{cases} & u_t = u_{xx} + \partial_u W(u,v),\\ & v_t = d v_{xx} - \partial_v W(u,v). \end{cases}$
I found the linearization to it as
$$L = \begin{bmatrix} \partial^2_x +2 \partial^2_u w(s(x)) && \partial_u \partial_v (s(x))\\ - \partial_u \partial_v (s(x)) && d\partial_{x}^2 - 2\partial^2_v w(s(x)) \end{bmatrix} $$
I believe that my linearization is right. My question is about the adjoint I found, is it correct? This is the adjoint I found:
updated 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} \label{p4eq1} \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}.\\ \label{p4eq2} \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 ($\ref{p4eq1}$) and ($\ref{p4eq2}$), 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 (\ref{p4maineq}) 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;
$$L = \begin{bmatrix} \partial^2_x +2 \partial^2_u w(s(x)) && \partial_u \partial_v (s(x))\\ - \partial_u \partial_v (s(x)) && d\partial_{x}^2 - 2\partial^2_v w(s(x)) \end{bmatrix} $$
or
$$L = \begin{bmatrix} \partial^2_x &&0\\ 0&& d\partial_{x}^2 \end{bmatrix} + \begin{bmatrix} 2 \partial^2_u w(s(x)) && \partial_u \partial_v (s(x))\\ - \partial_u \partial_v (s(x)) && - 2\partial^2_v w(s(x)) \end{bmatrix} $$
Let $U= \begin{bmatrix} u_1\\ u_2 \end{bmatrix} \in D(L):L^2(\mathbb{R}^2) \to L^2 (\mathbb{R}^2),$ and let $V= \begin{bmatrix} v_1\\ v_2 \end{bmatrix}$. Since the diffusion operator is self-adjoint in its domain, we are interested in the second part of $L$;
\begin{align*} \langle \begin{bmatrix} 2 \partial^2_{u_1} w(s(x)) && \partial_{u_1} \partial_{u_2} (s(x))\\ - \partial_{u_{1}} \partial_{u_{2}} (s(x)) && - 2\partial^2_{u_{2}} w(s(x)) \end{bmatrix}, \begin{bmatrix} v_1\\ v_2 \end{bmatrix} \rangle &=\\ \langle \begin{bmatrix} u_1\\ u_2 \end{bmatrix}, \begin{bmatrix} 2 \overline{\partial^2_{u_1} w(s(x))} && \overline{\partial_{u_1} \partial_{u_2} (s(x))}\\ - \overline{\partial_{u_{1}} \partial_{u_{2}} (s(x))} && - 2\overline{\partial^2_{u_{2}} w(s(x)) } \end{bmatrix}\rangle\\ \end{align*}
Then,
$$L^* = \begin{bmatrix} \partial^2_x &&0\\ 0&& d\partial_{x}^2 \end{bmatrix} + \begin{bmatrix} 2 \overline{\partial^2_{u_1} w(s(x))} && \overline{\partial_{u_1} \partial_{u_2} (s(x))}\\ - \overline{\partial_{u_{1}} \partial_{u_{2}} (s(x))} && - 2\overline{\partial^2_{u_{2}} w(s(x)) } \end{bmatrix}\\ $$
If my answers is right How can I find a function $\phi$ such that $L^* \phi=0$ given that $Ls(x)=0$ where $s(x)$ is the stationary solution. Thank you very much.