By Jean-Pierre Kernevez
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Additional resources for Enzyme Mathematics
It is very important to take into account this fact when explaining the behavior of biological membranes: interacting diffusion and reaction may cause substrate (resp. product) concentrations to be much lower (resp. higher) within membranes than at boundaries. And in any case, one cannot speak of a single internal concentration for each chemical species involved, but rather of a concentration profile. A modification of the environment causes the concentration profiles to undergo a fast transient evolution, during a period of the order of L2/DS, the diffusion characteristic time, and to approach exponentially a stable steady state.
51). 50) is a function O(x,t) such that: whereas a lower solution satisfies the reversed inequalities. 53) O(x) = a(1-x) + Bx. 50), while y(x,t) = 0 is a lower solution. 54) 0 5 s(x,t) 5 $ J ( x ) . 51) are functions O(x) and v(x) satisfying respectively: and the reversed inequalities. 3. To begin with, let us state four propositions, which will be proven later in this chapter. Proposition 1 . 1 - Let 7 and 7 be a pair of lower and upper solutions such that 7 5 9 for the evolution problem: st - s xx + U F ( S ) = 0, s(0,t) = a, S(X,O) = s(1,t) 0 < x < 1, t > 0, = B, so(x) (So' L2(0,l)) where F is Lipschitz-continuous.
In fact c < < 1 since the membrane volume vM is small compared to the receptor compartment volume vR. We may distinguish two epochs in the evolution of the system. 10) -P"(x) - F(x,s(x),~(x) 0 , p'(0) This fast transition obeys ( 2 . 2 ) , ( 2 . 6), = 0. ( 2 . 7 ) and B. Fig. 4 shows the S and P concentration profiles during this evolution, and the numerical simulations show that these profiles are unchanging after about t = 10 (remember that the unit of time, 8 = L 2 / D S , corresponds to 18 seconds if, as is the case, L = 5 10-3cm and DS = 5 10-3cm2h-').