Example of chemical reaction with a given form for the kinetic equations I’m an applied mathematician looking for examples from various applications fields (right now, chemistry, obviously) to illustrate the following reaction-diffusion system
$$
\begin{cases} 
\frac{\partial u}{\partial t} - d_u \Delta u = k_1 v + (k_2-k_3) u - k_4 u^{p_1} v^{p_2} - k_5 u^{p_3+p_4} \\
\frac{\partial v}{\partial t} - d_v \Delta v = k_6 u + (k_7-k_8) v - k_9 u^{q_1} v^{q_2} - k_{10} v^{q_3+q_4} 
\end{cases}
$$

Above, all constants $d_u$, $d_v$, $k_i$, $p_i$, $q_i$ are positive, and $p_i,q_i\geq 1$. $k_2-k_3$ and $k_7-k_8$ might be of any sign. The system need not be self-contained — if you need to add a third equation in order to make these two exist, feel free to do so.

If the diffusion part is bothering you, then I’m also fine with 
$$
\begin{cases} 
\frac{\text{d} u}{\text{d} t} = k_1 v + (k_2-k_3) u - k_4 u^{p_1} v^{p_2} - k_5 u^{p_3+p_4} \\
\frac{\text{d} v}{\text{d} t} = k_6 u + (k_7-k_8) v - k_9 u^{q_1} v^{q_2} - k_{10} v^{q_3+q_4} 
\end{cases}
$$

And if the $p_i$ and $q_i$ are bothering you, then one example where they are all equal to $1$ is 
$$
\begin{cases} 
\frac{\text{d} u}{\text{d} t} = k_1 v + (k_2-k_3) u - k_4 u v - k_5 u^2 \\
\frac{\text{d} v}{\text{d} t} = k_6 u + (k_7-k_8) v - k_9 u v - k_{10} v^2 
\end{cases}
$$

One more precision: if $k_2-k_3\leq 0$, then $k_5=0$ is allowed, and similarly, if $k_7-k_8\leq 0$, then $k_{10}=0$ is allowed. Therefore the following system would be a satisfying answer:
$$
\begin{cases} 
\frac{\partial u}{\partial t} - d_u \Delta u = \tilde{k}_1 v - \tilde{k}_2 u - \tilde{k}_3 uv \\
\frac{\partial v}{\partial t} - d_v \Delta v = \tilde{k}_2 u - \tilde{k}_1 v - \tilde{k}_4 uv
\end{cases}
$$
(where the constants $\tilde{k}_i$ differ a priori from the previous constants $k_i$). 

I looked for actual examples of first-order reversible reactions. I found $\ce{Cu+ &lt;-&gt; Cu^{2+}}$ there but I do not know if having an irreversible reaction “$\ce{Cu^+ + Cu^{2+} -&gt;}$ some product” in this configuration is possible.

Can you think of any “real” chemical system that would satisfy such kinetic equations?
A chemist friend found an answer for the simplest case:
$$
\begin{cases} 
\frac{\partial u}{\partial t} - d_u \Delta u = \tilde{k}_1 v - \tilde{k}_{-1} u - \tilde{k}_2 uv \\
\frac{\partial v}{\partial t} - d_v \Delta v = \tilde{k}_{-1} u - \tilde{k}_1 v - \tilde{k}_2 uv
\end{cases}
$$


$u$ is the concentration of ethenol;
$v$ is the concentration of ethanal;
the tautomerization of ethenol and ethanal gives the first-order part of the reaction;
the aldol reaction of ethanal with ethenol, giving the 3-hydroxybutanal, gives the second order part.


This is definitely a satisfying answer.

Nevertheless I’m still interested in other answers, especially with autocatalysis of either $u$ or $v$ (that would decrease the first-order consumption of it) or with autoreactions (that would give the term $-k_i u^2$ or $-k_i v^2$). As a matter of fact, the French Wikipedia claims that an aldol reaction of ethanal with itself in presence of a base can also produce 3-hydroxybutanal, so that the same example would also give a term $-k v^2$? But maybe these two aldol reactions correspond in fact to the same elementary mechanism?