Here I consider a system of reaction-diffusion equations of the form , . The functions are defined on where is a bounded domain in with smooth boundary. Let the be denoted collectively by . I assume that the diffusion coefficients are non-negative. If some of them are zero then the system is degenerate. In particular there is an ODE special case where all are zero. If this system really describes chemical reactions and the are concentrations then it is natural to assume that for all . It should then follow that in the presence of suitable boundary conditions for all . I assume that is a classical solution and that it extends to the boundary with enough smoothness that the boundary conditions are defined pointwise. It is necessary to implement the idea that the system is defined by chemical reactions. This can be done by requiring that whenever and it follows that . (This means that if a chemical species is not present it cannot be consumed in any reaction.) It turns out that this condition is enough to ensure positivity.
I will now explain a proof of positivity. The central ideas are taken from a paper of Maya Mincheva and David Siegel (J. Math. Chem. 42, 1135). Thanks to Maya for helpful comments on this subject. The argument is already interesting for ODE since important conceptual elements can already be seen. I will first discuss that case. Consider a solution of the equation on the interval where and are continuous functions with and suppose that . I claim that for all . Let . Since it follows by continuity that . Assume that . By continuity . If were greater than zero then by continuity it would also be positive for slightly larger than , contradicting the definition of . Thus . The evolution equation then implies that . This implies that for slightly less than , which also contradicts the definition of . Hence in reality and this completes the proof of the desired result.
Suppose now that we weaken the assumptions to and . We would like to conclude that for all . To do this we define a new quantity for positive constants and . Then . Hence and . Now and so . It follows that if is large enough satisfies the conditions satisfied by in the previous argument and it can be concluded that for all . Letting tend to zero shows that for all , the desired result.
This is different, and perhaps a bit more complicated than, the proof I know for this type of result. That proof involves considering the derivative of on . It also involves approximating non-negative data by positive data. A difference is that the proof just given does not use the continuous dependence of solutions of an ODE on initial data and in that sense it is more elementary. In Theorem 3 of the paper of Mincheva and Siegel the former proof is extended to a case involving a system of PDE.
Now I come to that PDE proof. The system of PDE concerned is the one introduced above. Actually the paper requires the be positive but that stronger condition is not necessary. This equation is solved with an initial datum and a boundary condition . Here is a diagonal matrix with non-negative entries and the function is non-negative. The derivative is that in the direction of the outward unit normal to the boundary. We assume that is a classical solution, i.e. all derivatives of appearing in the equation exist and are continuous. Moreover has a continuous extension to and a extension to . We now replace the differential equation by the differential inequality . We assume that the initial data are non-negative, . The assumption that is non-negative, together with the boundary condition, gives rise to the inequality . The aim is to show that all solutions of the resulting system of inequalities are non-negative. We assume the condition for a system of chemical reactions already mentioned.
The proof is a generalization of that already given in the ODE case. The first step is to treat the case where each of the inequalities is replaced by the corresponding strict inequality. In contrast to the proof in the paper we assume that that is strictly positive on . We define as in the ODE case so that is the longest interval where the solution is non-negative. We suppose that is finite and obtain a contradiction. Note first that, as in the ODE case, by continuity. Now . If were strictly positive on then by continuity would be strictly positive for slightly greater than , contradicting the definition of . Hence there is an index and a point with and for all . Suppose first that . Then and . This contradicts the strict inequality related to the evolution equation for and so cannot happen. Suppose next that is on the boundary of . Then and . This contradicts the strict inequality related to the boundary condition for . Thus in fact and is strictly positive for all time.
Now we do a perturbation argument by considering . Here is the vector all of whose components are and is a positive function. It is obvious that . We now choose which ensures its positivity and require that the outward normal derivative of on the boundary is equal to one. (Here I will take for granted that a function of this kind exists. A source for this statement is cited in the paper. For me the positivity statement is already very interesting in the case that is a ball and there the existence of the function is obvious.) Then the fact that satisfies the non-strict boundary inequality implies that satisfies the strict boundary inequality. It remains to derive an evolution equation for . A straightfoward calculation gives where is a Lipschitz constant for on the image of the compact region being considered. Choosing large enough ensures that the right hand side and hence the left hand side of this inequality is strictly positive. We conclude that and, letting , that .
If we want to prove an inequality for solutions of a PDE it is common to proceed as follows. We deform the problem continuously as a function of a small parameter so as to get a simpler problem. When that has been solved we let tend to zero to get a solution of the original problem. Often it is the equations which are deformed. Then we need a theorem on existence and continuous dependence to get a continuous deformation of the solution. The above proof is different. We perturb the solution in a way whose continuity is obvious and then derive a family of equations of which that is a family of solutions. This is easy and comfortable. The hard thing is to guess a good deformation of the soluion.