By Jean-Pierre Kernevez
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Extra resources for Enzyme Mathematics
We n e e d t h e f o l l o w i n g Lemma: Lemma 2 . 1 L2(n) a s t +. Let y s a t i s f y ( 2 . 2 7 ) -. and ( 2 . 2 8 ) . Then y ( . , t ) +.
MATHEMATICAL ANALYSIS 49 f is increasing - Suppose ( u l , v l ) 5 (u2,v2) and let w = s 2 - s l , z = p2 - p l , where (si,pi) = f(ui,vi), i = 1 ,2. ,O) = 0. The second member in ( 2 . 2 5 ) can be written: where the quan-tities within brackets are positive, the first one because of the Lipschitz continuity of F , the second one because F is a decreasing function of p. Thus, from the Maximum Principle for parabolic problems, w 0. 26) is positive, hence z 0. 21), may be considered as built up from three mappings.
T) 5 4u$(t), $(O) = 0 , which, together with $(t) 2 0 , imply $(t) = 0. ii/ Existence - Existence of a solution on any bounded interval [O,Tl, together with uniqueness, implies existence and uniqueness of a solution on LO,+-[. Thus, let us prove existence of a solution for ( 2 . 1 3 ) - ( 2 . 1 6 ) on the time interval [O,T], T < m . e. ) 5 9 results from the fact that 0 and p E 0. is an upper solution for pro- ~ blem ( 2 . 1 3 ) - ( 2 . ) + r(l) = - a + a - 1 + m -> 0. The boundary and initial conditions for (q,r) to be an upper solution are trivially satisfied.