I am trying to solve a set of diffusion-advection equations with coupled decay (see unanswered coupled diffusion-advection-decay equations). The equations are as follows: \begin{align} u_t - \alpha u_{xx} + c u_x +\lambda (u-v) = 0 \tag{1} \end{align} \begin{align} v_t - \alpha v_{xx} - c v_x +\lambda (v-u) = 0 \tag{2} \end{align} on $\mathbb R \times \mathbb R_+$ with $\lambda \geq0$, $\alpha > 0$, $c>0$. The functions at $t = 0$ are $~u(x,0) = \delta(x), ~x \in \mathbb R$ and $v(x,0) = -\delta(x), ~x \in \mathbb R$.
I'm interested in the steady state ($v_t = u_t = 0$). This can be solved analytically if $\lambda = 0$ (just two uncoupled diffusion-advection solutions).
Am I wrong to conclude that in steady state at large $x$ , $u = v$? The analytic solutions without $\lambda$ do not have $u = v$. What surprises me is that turning on $\lambda$ by even the smallest amount forces $u = v$. I've tried to tackle these numerically and I don't get $u=v$ but I think it's because of numerical accuracy.
I'm also interested in long-time dependence of the solutions. Numerically I find the power law behavior.