## The deficiency one theorem

Here I continue with the discussion of chemical reaction network theory begun in the previous post. After having presented a proof of the Deficiency Zero Theorem in my course I proceeded to the Deficiency One Theorem following the paper of Feinberg in Arch. Rat. Mech. Anal. 132, 311. If we have a reaction network we can consider each linkage class as a network in its own right and thus we can define its deficiency. If we denote the deficiency of the full network by $\delta$ and the deficiency of the $i$th linkage class by $\delta_i$ then in general $\delta\ge\sum_i \delta_i$. The first hypothesis of the Deficiency One Theorem is that the deficiencies of the linkage classes are no greater than one. The second is that equality holds in the inequality relating the deficiency of the network to those of its linkage classes. The stoichiometric subspace $S$ of the full network is the sum of the corresponding spaces $S_i$ for the linkage classes. The sum is direct precisely when equality holds in the inequality for the deficiencies. The third condition is that there is precisely one terminal strong linkage class in each linkage class ( $t=l$). The first conclusion is that if there exists a positive stationary solution there is precisely one stationary solution in each stoichiometric compatibility class. The second is that if the network is weakly reversible there exists a positive stationary solution. Thus two of the conclusions of the Deficiency Zero Theorem have direct analogues in this case. Others do not. There is no statement about the stability of the stationary solutions. Networks which are not weakly reversible but satisfy the three conditions of the Deficiency One Theorem may have positive stationary solutions. In the paper of Feinberg the proof of the Deficiency One Theorem is intertwined with that of another result. It says that the linearization about a stationary solution of the restriction of the system to a stoichiometric compatibility class has trivial kernel. A related result proved in the same paper says that for a weakly reversible network of deficiency zero each stationary solution is a hyperbolic sink within its stoichiometric class. Statements of this type ensure that these stationary solutions possess a certain structural stability.

In the proof of the Deficiency One Theorem the first step (Step 1) is to show that when the assumptions of the theorem are satisfied and there is positive stationary solution $c^*$ then the set of stationary solutions is equal to the set of points for which $\log c-\log c^*$ lies in the orthogonal complement of the stoichiometric subspace. From this the conclusion that there is exactly one stationary solution in each stoichiometric compatibility class follows just as in the proof of the Deficiency Zero Theorem (Step 2). To complete the proof it then suffices to prove the existence of a positive stationary solution in the weakly reversible case (Step 3). In Step 1 the proof is reduced to the case where the network has only one linkage class by regarding the linkage classes of the original network as networks in their own right. In this context the concept of a partition of a network $(\cal S,\cal C,\cal R)$ is introduced. This is a set of subnetworks $({\cal S},{\cal C}^i,{\cal R}^i)$. The set of species is unchanged. The set of reactions $\cal R$ is a disjoint union of the ${\cal R}^i)$ and ${\cal C}^i$ is the set of complexes occurring in the reactions contained in ${\cal R}^i)$. The partition is called direct if the stoichiometric subspace of the full network is the direct sum of those of the subnetworks. The most important example of a partition of a network in the present context is that given by the linkage classes of any network. That it is direct is the second condition of the Deficiency One Theorem. The other part of Step 1 of the proof is to show that the statement holds in the case that there is only one linkage class. The deficiency is then either zero or one. Since the case of deficiency zero is already taken care of by the Deficiency Zero Theorem we can concentrate on the case where $\delta=1$. Then the dimension of ${\rm ker} A_k$ is one and that of ${\rm ker} (YA_k)$ is two. The rest of Step 1 consists of a somewhat intricate algebraic calculation in this two-dimensional space. It remains to discuss Step 3. In this step the partition given by the linkage classes is again used to reduce the problem to the case where there is only one linkage class. The weak reversibility is preserved by this reduction. Again we can assume without loss of generality that $\delta=1$. The subspace $U={\rm im} Y^T+{\rm span}\omega_{\cal C}$ is a hyperplane in $F({\cal C})$. We define $\Gamma$ to be the set of functions of the form $\log a$ with $a$ a positive element of ${\rm ker} (YA_k)$. The desired stationary solution is obtained as a point of the intersection of $U$ with $\Gamma$. To show that this intersection is non-empty it is proved that there are points of $\Gamma$ on both sides of $U$. This is done by a careful examination of the cone of positive elements of ${\rm ker} (YA_k)$.

To my knowledge the number of uses of the Deficiency Zero Theorem and the Deficiency One Theorem in problems coming from applications is small. If anyone reading this has other information I would like to hear it. I will now list the relevant examples I know. The Deficiency Zero Theorem was applied by Sontag to prove asymptotic stability for McKeithan’s kinetic proofreading model of T cell activation. I applied it to the model of Salazar and Höfer for the dynamics of the transcription factor NFAT. Another potential application would be the multiple futile cycle. This network is not weakly reversible. The simple futile cycle has deficiency one and the dual futile cycle deficiency two. The linkage classes have deficiency zero in both cases. Thus while the first and third conditions of the Deficiency One Theorem are satisfied the second is not. Replacing the distributive phosphorylation in the multiple futile cycle by processive phosphorylation may simplify things. This has been discussed by Conradi et. al., IEE Proc. Systems Biol. 152, 243. In the case of two phosphorylation steps the system obtained is of deficiency one but the linkage classes have deficiency zero so that condition (ii) is still not satisfied. It seems that the Deficiency One Algorithm may be more helpful and that will be the next subject I cover in my course.