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 $\ce{Cu+ <-> Cu^{2+}}$ there but I do not know if having an irreversible reaction “$\ce{Cu^+ + Cu^{2+} ->}$ some product” in this configuration is possible.

Can you think of any “real” chemical system that would satisfy such kinetic equations?

  • $\begingroup$ This one seems to be a homework problem. Please show your efforts before you expect an answer here. $\endgroup$ May 28, 2020 at 4:56
  • $\begingroup$ You need to have rate constants that are common between the two reactions. $\endgroup$
    – porphyrin
    May 28, 2020 at 6:27
  • $\begingroup$ @MitradipDas: I tried but I have to admit that I am no chemist. I think that the $k_1$ and $k_6$ part corresponds to a reversible monomolecular reaction. The $k_4$ and $k_9$ part could produce a third quantity, let’s say $U+V\to W$, and similarly we could have $2U\to X$ and $2V\to Y$. Is that more convincing? $\endgroup$
    – Elvith
    May 28, 2020 at 6:41
  • $\begingroup$ @porphyrin: yes, some rates can be equal if needed, it’s fine by me. You can also have $k_2-k_3=k_7-k_8=0$ and forget about this part of the equations, if needed. $\endgroup$
    – Elvith
    May 28, 2020 at 6:44
  • $\begingroup$ @MitradipDas: I added more details in the opening post. Hope this is satisfying. $\endgroup$
    – Elvith
    May 28, 2020 at 13:21

1 Answer 1


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?


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