Stability of equilibrium of a reaction-diffusion equation

Question

I'm trying to find the equilibrium of system
$$\left\{ \begin{aligned} &u_t = (2x-1)uv + u_{xx}\\ &v_t =(1-2x)uv + v_{xx}\\ &u+v = 1, \quad \forall x\in[0,1],\forall t\in \mathbb{R}^+ \end{aligned} \right. $$ with $\left(x,t\right) \in \left[0,1\right]\times\mathbb{R}^+$, Neumann boundary condition, with initial values $u\left(x,0\right) = u_0\left(x\right)$ and $v\left(x,0\right) = v_0\left(x\right)$.
It's trivial that $\left(u,v\right) = (0,1)$ and $\left(u,v\right) = (1,0)$ are two equilibria. But I don't know that they are stable or not.
Alternatively, there is another equilibrium, which is the non-trivial solution of the right-hand side of the system. For example, the system for equilibrium for $u$ is $$u_{xx} + (2x-1)u(1-u) = 0.$$ Can we solve for it explicitly, because this equation is a second-order nonlinear differential equation?
For stability of $(0,1)$, I tried to linearize the system to $$\dfrac{\partial}{\partial t}\begin{pmatrix}u \\ v\end{pmatrix} = \begin{pmatrix}0 & -1 \\ 1 & 1\end{pmatrix}\begin{pmatrix}\Delta & 0 \\ 0 & (2x- 1) + \Delta\end{pmatrix}\begin{pmatrix}1 & 1 \\-1 & 0\end{pmatrix}\begin{pmatrix}u \\ v\end{pmatrix}$$ which leads to $$\dfrac{\partial}{\partial t}\begin{pmatrix}U \\ V\end{pmatrix} = \begin{pmatrix}\Delta & 0 \\ 0 & (2x- 1) + \Delta\end{pmatrix}\begin{pmatrix}U \\ V\end{pmatrix}$$ with $\begin{pmatrix}U\\V \end{pmatrix} = \begin{pmatrix}1 & 1 \\-1 & 0\end{pmatrix}\begin{pmatrix}u \\ v\end{pmatrix}$.
I don't know how to continue with this because the spectrum of $\begin{pmatrix}\Delta & 0 \\ 0 & (2x- 1) + \Delta\end{pmatrix}$ is not sure to be positive for all $x \in [0,1]$.
Thank you very much for your help!