Uniform Permanence in Weakly Reversible Mass-Action Systems

Proof. We induct on m. Write \mu(u)=\min_i u_i and \one=(1,\ldots,1). Since the largest coordinate of X_h(u) is h^{\mu(u)}, we first construct a family for each fixed value t of the smallest exponent. We then extend its guarantee to a neighborhood of t and combine finitely many such families by shifting their offsets. The fixed-minimum construction is a singleton when m=1; the extension and combination therefore also prove the base case.

Throughout the proof, asymptotic comparisons along h_\nu\to0+ and p^\nu\to p use fixed gaps between limiting exponents. In particular, if p_i>q, then eventually p_i^\nu\geq q+\delta for some \delta>0, so h_\nu^{p_i^\nu}=o(h_\nu^q); if p_i<q, the ratio h_\nu^{p_i^\nu}/h_\nu^q tends to infinity. We shall use these comparisons only with strict limiting gaps.

A family for a fixed minimum. Fix t\in[a,b]. We construct a finite family \Lambda_t containing (t\one,0), with \mu(r)=t for every slope, such that the conclusion of the lemma holds for activity within \Lambda_t whenever \mu(p)=t.

If m=1 or t=b, the family \{(t\one,0)\} suffices: a point p\in[a,b]^m with \mu(p)=t then equals t\one. Otherwise m>1 and t<b. Say a label has type k if exactly k entries of its slope exceed t, the remaining entries being t. Start with the type 0 label (t\one,0), and construct types k=1,\ldots,m-1 successively. After type k-1 has been constructed, choose \beta_k satisfying

These conditions use only finitely many positive values of E, so such a choice is possible. The new type k labels will force every limiting exponent on a coordinate left at t by a type k-1 label that stays active to be at most \beta_k. The last inequality will then bound the error by that label’s tolerance.

For each k-element coordinate set U\subset\{1,\ldots,m\} and s\in[\beta_k,b]^U, let r(s) equal s on U and t off U. Apply the induction hypothesis in dimension k to this cube with tolerance s\mapsto E(r(s)), identifying U with its naturally ordered coordinates. For every resulting label (s,d), add to \Lambda_t the type k label

Here and below c denotes a local offset. The resulting family is finite, and all its offsets are o(h^t). The negative offset favors raising slope entries on coordinates where X_h(p) is small; the decreasing \beta_k make that preference stronger with each successive type.

To verify the guarantee, let p^\nu\to p\in[a,b]^m with \mu(p)=t, and suppose a type k label (r,c) stays active within \Lambda_t along h_\nu\to0+. Let U be its raised coordinate set, and abbreviate h=h_\nu. For k>0, comparison with (t\one,0) gives

Every coefficient on the left is at least \beta_k-t>0. A limiting exponent p_i<\beta_k would make the corresponding term divided by h^{\beta_k} tend to infinity. Therefore

Among labels made from this same U, all contributions off U and the term -h^{\beta_k} are common. Thus (s,d) is active in the family furnished by induction. Its limiting exponent vector lies in [\beta_k,b]^U, so

The approximating vectors p_U^\nu may approach a face of that cube from outside. This is why the induction statement allows them outside the cube, including when p_i=\beta_k or p_i=b.

It remains to bound the coordinates off U. This argument also includes type 0, for which U=\varnothing. If k<m-1, we claim that

Otherwise choose j\notin U with p_j>\beta_{k+1} and any type k+1 label raised on U\cup\{j\}. Every limiting exponent on this set exceeds \beta_{k+1}, by (2.2) and \beta_k>\beta_{k+1} when k>0. The competitor’s affine value minus that of (t\one,0) is therefore

For the supposed active label, the same difference is zero if k=0, and is bounded below by -h^{\beta_k}+o(h^{\beta_k}) if k>0. Since h^{\beta_k}=o(h^{\beta_{k+1}}), the competitor is strictly smaller for all sufficiently small h along the sequence, a contradiction. Consequently, off U,

If k=m-1, there is just one coordinate off U. Its limiting exponent is t, since \mu(p)=t and all coordinates on U have limiting exponents greater than t. Its error is zero. This proves the required strict bound for every active label at the fixed minimum.

Extension to nearby minima. For a fixed label \lambda=(r,c)\in\Lambda_t, let A_\lambda\subset[a,b]^m consist of the limits of all sequences p^\nu\in\R^m for which h_\nu\to0+ and \lambda is active within \Lambda_t at X_{h_\nu}(p^\nu). This set is closed. Indeed, if z_n\in A_\lambda and z_n\to z, choose from a sequence realizing z_n a parameter 0<h_n<1/n and a vector u_n with \norm{u_n-z_n}_\infty<1/n such that \lambda is active at X_{h_n}(u_n). Then u_n\to z, so this selected sequence realizes z\in A_\lambda. This argument uses actual active points and requires no continuity of the offsets in h.

The set of limits violating the desired estimate,

is compact: the union is finite and each E(r) here is a fixed number. The preceding argument shows that B_t contains no point with \mu(p)=t. Continuity of \mu and compactness therefore give \rho_t>0 such that B_t contains no point with

If B_t is empty any positive radius initially suffices; otherwise take \rho_t<\min_{p\in B_t}\abs{\mu(p)-t}. Thus \Lambda_t has the asserted limiting activity property whenever \mu(p)\in I_t.

Shrink \rho_t further, if necessary, so that every local offset obeys

For a nonsingleton family, choose t+\rho_t<\min_k\beta_k and use (2.1); for a singleton the offset is zero. We have obtained a family valid near each possible minimum exponent. The remaining task is to ensure that a globally active label comes from a family valid at its own limiting point.

Combining the families. Take a finite subcover of [a,b] from the intervals I_t. Discard intervals contained in others, keeping just one if two coincide, and order the remaining centers as t_1<\cdots<t_N. Write I_{t_j}=(L_j,R_j). Noncontainment forces both endpoints to have the same strict order; since t_j=(L_j+R_j)/2, it is the order of the centers. Hence

Consecutive intervals overlap. If R_j\leq L_{j+1}, a point of [R_j,L_{j+1}]\subset(t_j,t_{j+1})\subset[a,b] would be uncovered. The first interval contains a and the last contains b.

For 1\leq j<N, choose a cut

The interval displayed is nonempty because consecutive intervals overlap and each contains its own center. With q_0=a and q_N=b, we have

For N=1, the sole interval contains [a,b] and no interior cuts are needed. The closed intervals in (2.4) are the ranges of minimum exponents assigned to the respective families; see Figure 1.

Combine the selected families, replacing every local offset c(h) in \Lambda_{t_j} by

These offsets are o(h^a): (2.3) has t_j+\rho_{t_j}>a, and each shift exponent satisfies q_\ell>t_\ell\geq a. A common shift preserves activity within each family. Between consecutive families, the new term -h^{q_j} will force any limiting minimum exponent to the correct side of the cut.

Suppose a fixed label from \Lambda_{t_j} is active in the combined family along h_\nu\to0+ and p^\nu\to p\in[a,b]^m. Write c(h) for its original local offset and h=h_\nu. We claim

If j<N and \mu(p)>q_j, each coordinate h^{p_i^\nu} is o(h^{q_j}). Also c(h)=o(h^{q_j}), by (2.3) and q_j<R_j=t_j+\rho_{t_j}. Subtracting the value of the shifted label (t_{j+1}\one,0) from the active label’s value gives

a contradiction.

If j>1 and \mu(p)<q_{j-1}, compare instead with the shifted label (t_{j-1}\one,0). The difference is

Every coefficient in the scalar product is at least t_j-t_{j-1}>0. Some coordinate has p_i<q_{j-1}, so its exponential divided by h^{q_{j-1}} tends to infinity. Meanwhile, c(h)=o(h^{q_{j-1}}) because q_{j-1}<t_j<t_j+\rho_{t_j}. Thus the displayed difference divided by h^{q_{j-1}} tends to +\infty, again contradicting activity. The outer bounds for j=1 or j=N follow from \mu(p)\in[a,b].

By (2.4), the claimed bounds place \mu(p) inside I_{t_j}, including when it equals a cut. The label is active within its original local family, so the guarantee on I_{t_j} gives \norm{p-r}_\infty<E(r). This completes the induction.□

Proof. Otherwise there are h_\nu\to0+, points p^\nu\in[a,b]^m, and active labels \lambda_\nu such that \norm{p^\nu-r_{\lambda_\nu}}_\infty\geq E(r_{\lambda_\nu}). By finiteness and compactness, pass to a subsequence on which \lambda_\nu=\lambda is fixed and p^\nu\to p\in[a,b]^m. The inequality passes to the limit and contradicts Lemma 2.1. No continuity of E is needed because its argument r_\lambda is fixed on this subsequence.□

Proof. Write p_c=\log c/\log h. If a label active at c had a nonzero coordinate r_{\lambda,i}, (3.5) with v=e_i would give \abs{p_{c,i}}\geq H, contrary to (3.6). Thus every label active at c has zero slope. At least one label is active there, so one of the affine functions is the constant M_h. It follows that F_h\leq M_h everywhere, and that M_h is the least value of any constant label. Hence

is closed and convex, and contains c.

If x=X_h(p)\in K_h\cap B_h, the constant label with value M_h is active at x. Applying (3.3) to its zero slope gives

In particular, K_h misses the boundary of B_h. If some point of K_h lay outside B_h, the segment joining it to c\in\operatorname{int}B_h would meet that boundary at a point of K_h, by convexity. This is impossible. Thus K_h\subset B_h, and the preceding argument proves (3.7) for every point of K_h. Closedness and boundedness give compactness.□

Proof. First, let x\in P\setminus K_h. If r_\lambda\in S^\perp, then x-c\in S implies

But F_h(x)<M_h by Proposition 3.1. Such a label cannot be active at x.

If R is empty, then f=0 and S=\{0\}, so (3.8) holds and (3.9) is vacuous. For the flux comparison, assume R\ne\varnothing and fix x=X_h(p)\in B_h and an active slope r. Whenever y,z\in C satisfy r\cdot y<r\cdot z, (3.5) applied to z-y gives

List the distinct values of r\cdot y, for y\in C, as a_1<\cdots<a_m. For 1\leq j<m, define the net upward flux across the cut at a_j by

An edge crossing the cut in the first sum is called upward, and an edge in the second sum is called downward.

If there is a downward edge u\to u', weak reversibility supplies a directed path from u' back to u. This path starts below the cut and ends above it, so it contains an upward edge, say w\to w'. Every source u of a downward edge has r\cdot u>r\cdot w. By (3.10) and (3.6),

Thus one upward edge already contributes more than the entire downward sum, and Q_j(x)>0. If there is no downward edge, then Q_j(x)\geq0, with strict inequality whenever there is an upward edge. In all cases Q_j(x) is nonnegative, and it is positive if any reaction crosses the cut.

The contribution of an edge to r\cdot f(x) is its flux multiplied by the difference between the levels of its target and source. Writing this difference as the signed sum of the intervening gaps gives

Every term is nonnegative. If r\notin S^\perp, some reaction y\to y' satisfies r\cdot(y'-y)\ne0 and therefore crosses at least one cut; the corresponding term is positive. This proves (3.8). When m=1, the sum is empty and every reaction has zero scalar product with r, as required. Self-loops never cross a cut and contribute zero to the vector field.

For the uniform bound, the active set of a fixed label,

is compact. If r_\lambda\notin S^\perp and T_{\lambda,h} is nonempty, the continuous function r_\lambda\cdot f is strictly positive on this set and has a positive minimum there. Take the minimum of these positive numbers over the finitely many eligible labels. If there are none, choose any \delta_h>0. This proves (3.9), including the case R=\varnothing.□

Proof. For a solution x(t) and each fixed time t before its maximal existence time, the right derivative of its affine minimum is

Indeed, all inactive labels have a positive gap above the minimum at time t. There are finitely many of them, so continuity excludes them from the minimum for a sufficiently short time to the right. For each active label, differentiability of the solution gives

Taking the minimum of these finitely many expansions proves (3.13).

We use the elementary fact that a continuous function g on [a,b] with right derivative at least q at every point of [a,b) satisfies g(b)-g(a)\geq q(b-a). For completeness, for any \eta>0 the function g(u)-qu+\eta u has positive right derivative before b. Its maximum on the compact interval therefore cannot occur before b. Comparing its values at a and b and then letting \eta\downarrow0 proves the assertion. Formula (3.13) and Proposition 3.2 now give monotonicity in B_h, and give (3.12) when the segment lies in P\setminus K_h.

Suppose next that x(0)\in K_h, and consider the maximal solution of the polynomial vector field f. Since K_h\subset\operatorname{int}B_h, the solution initially stays in B_h. On any interval on which it does so, monotonicity and the global upper bound F_h\leq M_h force F_h(x(t))=M_h. Hence it remains in K_h. It cannot have a first exit from \operatorname{int}B_h within its interval of existence: continuity would place the exit point in both K_h and the boundary of B_h, contradicting Proposition 3.1. Thus the solution stays in the compact set K_h throughout its maximal interval. A solution of a smooth vector field that remains in a compact subset of its domain extends past any finite endpoint of its interval of existence. To see the relevant point here, f is bounded on a neighborhood of K_h, so a finite endpoint would give a limit of x(t) in K_h, at which local existence continues the solution. The solution is therefore global. It remains positive by (3.7); moreover, f(x)\in S implies x(t)-x(0)\in S, so a solution starting in P stays in P.

Finally, let a global solution in P remain in B_h for all t\geq t_0. If it never enters K_h after t_0, then (3.12) gives

contradicting F_h\leq M_h for sufficiently large t. It must enter K_h at a finite time, and the forward invariance just proved keeps it there.□

Proof of Theorem 1.1. Let x^0\in P. Since x^0 is positive, we may choose 0<h_0\leq h_P so small that

where the logarithm is componentwise and H>0 is the fixed separation constant from Section 3. In particular, x^0\in B_{h_0}. The coordinate sign comparisons force every label active at x^0 to have slope zero, just as they did at c. An active constant label must have the least value among all constant labels, namely M_{h_0}=F_{h_0}(c). Hence

Lemma 3.3 implies that the solution through x^0 is global, stays positive, and remains in K_{h_0}. Since f(x)\in S, it also remains in P.

Starting at h_0, define

We claim that for every j there is a finite time T_j such that x(t)\in K_{h_j} for all t\geq T_j. The case j=0 holds with T_0=0. Suppose it holds at j. Because h_{j+1}\leq\sqrt{h_j}, the boxes satisfy B_{\sqrt{h_j}}\subset B_{h_{j+1}}. Thus (4.1) yields

The finite-entry assertion of Lemma 3.3, followed by its invariance assertion, supplies a finite T_{j+1}\geq T_j with x(t)\in K_{h_{j+1}} for all t\geq T_{j+1}. This proves the claim. The argument uses containment in successive boxes; no nesting of the sets K_h is assumed.

Repeated square roots of h_0\in(0,1) tend to 1. Since h_P<1, the capped sequence (4.2) therefore reaches h_P after finitely many steps. We conclude that every solution through a point of P enters K_{h_P} in finite time and stays there.

Finally set

This is nonempty because it contains c, and it is compact and convex as the intersection of the compact convex set K_{h_P} with a closed affine subspace. Since K_{h_P} lies in the positive orthant, K_P\subset P. Both K_{h_P} and the affine subspace c+S are forward invariant, so K_P is forward invariant. The preceding argument proves its absorbing property. Moreover, K_P\subset B_{h_P}, and therefore \eps_P=h_P\in(0,1) gives

for all sufficiently large t, with the entry time allowed to depend on x^0. The scale h_P, and hence these bounds, was fixed before choosing the initial point.□