1. Introduction
In the sequel, we use the notations.
R+=0∞,R+∗=0∞.
N=01…the set of natural numbers and N∗=N\0.
For p∈R:pthe integer part of p.
For n∈N∗and x=x1…xn∈Rn:x2=∑j=1nxj2.
Z=⋯−1,0,1⋯the set of integers.
For x0∈Rnand ρ∈R+∗,:
B′x0ρ=x∈Rn:x−x0≤ρthe closed ball of center x0and radius ρ.
Sx0ρ=x∈Rn:x−x0=ρthe boundary of B′x0ρ.
Let Q⊂Rnn∈N∗a subset. ∂Qdenote the boundary of Q.
ln: the natural logaritm function.
ωnρ=2πn/2ρn−1Γn/2the surface area of S0ρ, where Γx=∫0∞e−tt−xdtx∈R+∗is the Gamma function.
BUCRnthe Banach space of bounded and uniformly continuous functions on Rnwith the supremum norm ∥u∥∞=supx∈Rnux.
X=BUCRn×BUCRnwhich is a Banach space endowed with the norm uvX=∥u∥∞+∥v∥∞.
For u∈LpRn(p∈1∞), we denote by ∥u∥pp=∫Rnupdx.
For u,v:Rn→Rtwo regular functions, ∇u=∂u∂x1…∂u∂xnand ∇u.∇v=∑j=1n∂u∂xj.∂v∂xj.
Reaction-Diffuison equations are nonlinear parabolic partial differential equations arises in many fields of sciences like chemistry, physics, biology, ecology and even medicine. It appears usually as coupled systems.
The somewhat general form of these systems of two equations is
uttx=aΔutx+f1txuv,vttx=dΔvtx+f2txuv,
where t>0,x∈Ωwith Ω⊂Rnn∈N∗is an open set, Δis the Laplacian operator, a, dare two real positive constants called the coefficients of the diffusion. For a chemical reaction where two substances S1and S2, uand vrepresent their concentrations at time tand position xrespectively, and f1and f2represent the rate of production of these substances in the given order. For more details see [1, 2].
In this chapter, we are concerned with the existence of global solutions to the reaction–diffusion system
uttx=aΔutx−guvm,tx∈R+∗×Rn,E1
vttx=dΔvtx+λtxguvm,tx∈R+∗×Rn,E2
with initial data
u0x=u0x,v0x=v0x,x∈Rn.E3
Whe assume that.
(H1)The constants a, dare such that a,d∈R+∗.
(H2)λ:R+×Rn→Ris a non-null, nonnegative and bounded function on R+×Rnsuch that λt.∈BUCRnfor all t∈R+. We denote λ∞=supt≥0λt∞.
(H3)nand mare positive integers, i.e. n,m∈N∗.
(H4)g:BUCRn→BUCRnis a function defined on BUCRnsuch that:
i. g0=0and gu≥0pour u≥0.
ii. gis of class C1and dguduis bounded on R.
(H5)The initial data u0, v0are nonnegative and are in BUCRn.
One of the essential questions for (1)–(3) is the existence of global solutions and possibly bounds uniform in time. Recently, Collet and Xin in their paper [3] published in 1996 have studied the system (1)–(3) but with a=λ=1, d>1and φu=u.In this particular case, this system describes the evolution of uthe mass fraction of reactant Aand that vof the product Bfor the autocatalytic chemical reaction of the form A+mB→m+1B.They proved the existence of global solutions and showed that the L∞norm of vcan not grow faster than Olnlntfor any space dimension.
If we replace guvmby uexp−E/vwhere E>0is a constant and take λ=1, there are many works on global solutions, see Avrin [4], Larrouturou [5] for results in one space dimension, among others.
It is worth mentioning here the result of S. Badraoui [6] who studied the system
ut=aΔu−uvm,
vt=bΔu+dΔv+uvm,
where a>0, d>0, b≠0, x∈Rn, n∈N∗,m∈2N∗is an even positive integer. He has proved that if u0, v0are nonnegative and are in BUCRnthat:
If a>d,b>0,v0≥ba−du0on Rn, then the solution is global and uniformly bounded.
If a<d,b<0,v0≥ba−du0on Rn,then the solution is global.
Our work here is a continuation of the work of Collet and Xin [3]. We treat the same question in a slightly general case. Inspired by the same ideas in [3] we prove that the system (1)–(3) under the assumptions (H1) to (H5) has a unique global nonnegative classical solution.
The chapter is organized as follows: In section 2, we treat the existence of local solution and reveal its positivity using the maximum principle.
In section 3, firstly, we prove by a simple comparison argument that if a≥d, the solution is uniformly bounded and we give an upper bound of it. Afterwards, we attack the hard case in which a<dwhere we used the Lyapunov functional Luv=α+2u−ln1+ueεv(α,ε>0)and the cut-off function φx=1+x2−n.We show that for αsufficiently large and εsmall enough we can control the Lp-norms of vp>max1n/2on every unit spacial cub in Rnfrom which we deduce the L∞-norm of vat any time t>0.
We emphazise here that I have engaged to calculate the constants encountered in all equations and inequalities exactly.
3. Existence of a global solution
For this task we will use the fact that the solution is nonnegative.
Theorem 3.1.Let uvbe the solution of the problem (1)–(3) under the assumptions Hjj=15and such that
a≥d.E12
Then, the solution is global and uniformly bounded on R+×Rn. More precisely, we have the estimates
ut∞≤u0∞,forallt∈R+,E13
vt∞≤v0+λ∞adn/2u0∞,forallt∈R+.E14
Proof.By the comparison principle we get (13).
The solution uvsatisfies the integral equations
utx=S1tu0−∫0tS1t−τguτvmτdτ,E15
vtx=S2tv0+∫0tS2t−τλτguτvmτdτ.E16
Here S1tand S2tare the semigroups generated by the operators aΔand dΔin the space BUCRnrespectively. As uis nonnegative, then from (15) we get
∫0tS1t−τguτvmτdτ≤S1tu0.E17
Since a≥d, using the explicit expression of S1t−τguτvmτand S2t−τguτvmτ, one can observe that (see [10])
∫0tS2t−τλτguτvmτdτ≤adn/2∫0tS1t−τλτguτvmτdτ≤λ∞adn/2∫0tS1t−τguτvmτdτ.E18
From (17) and (18) into (16) we get
vt≤S2tv0+λ∞adn/2S1tu0.E19
This last inequality leads to the veracity of (14).
Thus, from (13) and (14), we deduce that the solution uvis global and uniformly bounded on R+×Rn. ♦
In the case where d>a, it seems that the idea of comparison cannot be applied. Nevertheless, we can prove the existence of global classical solutions; but it appears that their boundedness is not assured.
Theorem 3.2.Let uvbe the solution of the problem (1)–(3) with the assumptions Hjj=15. If
a<d,E20
the solution uvis global. More precisely we have the estimates (13) and (83).
Proof.In this case, it is not easy to prove global existence. But can derive estimates of solutions independent of t1by using the same method used in [3] and the same form of the functional used in [6] but with different coefficients.
We need some lemmas.
Lemma 3.3.Let uvbe the solution of the problem (1)–(3) under the assumptions Hjj=15on the local interval time 0t1. Define the functional
Luv=α+2u−ln1+ueεvwithα,ε∈R+∗.E21
Then for any φ=φxx∈Rna smooth nonnegative function with exponential spacial decay at infinity, we have
ddt∫RnφLdx=d∫RnΔφLdx+d−a∫RnL1∇φ.∇udx−∫RnφaL11∇u2+a+dL12∇u∇v+dL22∇v2dx+∫RnφλL2−L1guvmdx,E22
where
L1≡∂L∂u=2−11+ueεv,L2≡∂L∂v=εα+2u−ln1+ueεv,L11≡∂2L∂u2=11+u2eεv,L12≡∂2L∂u∂v=ε2−11+ueεv,L22≡∂2L∂v2=ε2α+2u−ln1+ueεv.E23
Proof.Note that L>0, L1>0, L2>0, L11>0, L12>0and L22>0. We can differentiate under the integral symbol
ddt∫RnφLdx=a∫RnφL1udx+d∫RnφL2Δvdx+∫RnφλL2−L1guvmdx.E24
Using integration by parts, we get
∫RnφL1Δudx=∫RnφL1Δudx=−∫Rn∇φL1∇udx=−∫RnL1∇φ∇udx−∫RnφL11∇u2dx−∫RnφL12∇u∇vdx,E25
In fact, let ρ∈R+∗,then we have by the Geen theorem
∫B′0ρφL1Δudx=∫B′0ρφL1Δudx=−∫B′0ρ∇φL1.∇udx+∫S0ρφL1∂u∂νdx,E26
where ∂u∂νis the derivative of uwith respect to the unit outer normal νto the boundary S0ρ.
We have
∫S0ρφL1t∂ut∂νdx≤2eεvt∞∂ut∂ν∞∫S0ρφdx≤2eεvt∞∂ut∂ν∞11+ρ2n2πn/2ρn−1Γn/2.E27
From (27) we obtain
limρ→∞∫Sx0ρφL1t∂ut∂νdx=0.E28
We pass to the limit for ρ→∞in (26) taking into account (28) we obtain (25).
By the same way we get
∫RnφL2Δvdx=−∫RnL2∇φ.∇vdx−∫RnφL22∇v2dx−∫RnφL12∇u∇vdx,E29
∫RnLΔφdx=−∫RnL1∇φ.∇udx−∫RnL2∇φ.∇vdx.E30
From (30) we find that
∫RnL2∇φ.∇vdx=−∫RnL1∇φ.∇udx−∫RnLΔφdx.E31
From (25), (29) and (31) into (24) we get our basic identity (22). ♦
Lemma 3.4.There exist two positive real constants α=αadγ1u0∞and ε=εadγ1γ2λ∞u0∞such that
ddt∫RnφLdx≤d∫RnLΔφdx+d−a∫RnL1∇φ.∇udx−γ1∫RnφaL11∇u2+dL22∇v2dx−γ2∫RnφL1guvmdx,E32
where γ1,γ2∈01are two arbitrary constants.
Proof.We seek Lsuch that
aL11∇u2+a+dL12∇u∇v+dL22∇v2≥γ1aL11∇u2+dL22∇v2E33
and
λL2−L1≤−γ2L1E34
for γ1,γ2∈01.
The inequality (33) is satisfied if
a+d2L1224ad1−γ12L11L12≤1.E35
From (23); (35), then (33) is satisfied if
α≥a+d21+2u0∞24ad1−γ12.E36
Also, (34) is satisfied if ελ∞α+2u0∞1−γ2≤1, i.e. ε≤1−γ2λ∞α+2u0∞,and from (36) we get
0<ε≤1−γ2λ∞4ad1−γ12a+d21+2u0∞2+8ad1−γ12u0∞.E37
Whence, if αsatisfies (36) and εsatisfies (37), we obtain (32). ♦
As a consequence of (33) we have
ddt∫RnφLdx≤d∫RnLΔφdx+d−a∫RnL1∇φ.∇udx−γ1a∫RnφL11∇u2dx.E38
Lemma 3.5.With the functional Ldefined in (21) and α, εdefined in (36) and (37) respectively and with the truncation function φ:Rn→Rdefined by
φx=11+x−x02n.E39
We have
ddt∫RnφLdx≤dk1n∫RnφLdx+14γ1ad−a2k22n∫RnφL12L11dx,E40
where
k1n=2n3n+2,k2n=2n.E41
Proof.Calulate Δφand estimate it
Δφ=−2n21+x−x02n+1−4nn+1x−x021+x−x02n+2;
whence
Δφ≤2n3n+2φ.E42
Calulate ∇φand estimate it
∇φ2=4n2x−x021+x−x022n+2;
whence
∇φ≤2nφ.E43
Using the Cauchy-Schwarz inequality ∇φ.∇u≤∇φ∇uand the inequalities (42) and (43) into (38) we get
ddt∫RnφLdx≤dk1n∫RnφLdx+d−ak2n∫RnφL1∇φdx−γ1a∫RnφL11∇u2dx.E44
We pove that
d−ak2nφL1∇φ−γ1aφL11∇u2≤14γ1d−a2ak22nφL12L11.E45
To do this, it sufficies to compute the discriminant of the trinoma in ∇φ
Δ=−γ1aφL11∇u2+d−ak2nφL1∇φ−14γ1d−a2ak22nφL12L11.
From (45) into (42) we find the desired result (40). ♦
Lemma 3.6.For αand εdefined in (36) and (37) respectively and for all real constant γ
γ≥max1a8u0∞+4,E46
we have
∫RnφLdx≤βeσt,forallt∈R+;E47
where
β=2nα+2u0∞ωneεv0∞,E48
and
σ=dk1n+γ4γ1ad−a2k22n.E49
Proof.We seek a constant γ∈R+∗such that
L12L11≤γL,forallu∈0u0∞.E50
The inequality (50) is equivalent to 2u+12eεv≤γα+2u−ln1+u.We prove that if γsatisfies (46) then (50) follows.
Whence, from (50) into (40) we obtain
ddt∫RnφLdx≤dk1n+γ4γ1ad−a2k22n∫RnφLdx,forallt∈R+.E51
As
∫RnφLt=0dx=∫Rnφα+2u0−ln1+u0eεv0dx;E52
then, from (51) and (52) we get
∫RnφLdx≤α+2u0∞φ1expεv0∞eσt,forallt∈R+,E53
where σis defined by (49).
Now, let us estimate φ1.We have ([11] on page 485)
φ1=∫Rnφdx=∫Rn11+x2ndx=ωn∫0∞rn−111+r2ndr.
As
∫0∞rn−11+r2ndr=∫01rn−11+r2ndr+∫1∞rn−11+r2ndr≤∫01rn−1dr+∫1∞1rn+1dr≤2n,
then
φ1≤2nωn.E54
Thus, from (54) in (53) we get the estimate (47) with βand σgiven by (48) and (49). ♦
In the following step we trie to control the second component vof the solution on any unit spacial cube in the Lp−norms with p∈1∞.
Let x0=x10…xn0∈Rnbe an arbitrary fixed point and
Q=x=x1…xn∈Rn:xk−xk0≤12forallk=1…n.E55
Lemma 3.7.Let uvbe the solution of the problem in consideration. For αand εsatisfying (36) above and (63) below respectively, then for any unit cube Qof Rnof the form (55) we have
∫Qvpdx≤βp+1p+1αεp+14+n4neσt,forallpt∈1∞×R+.E56
Proof.It’s obvious that
φx≥44+nn,forallx∈Rn,E57
and
eεv≥εkk!vk,forallk∈N∗.E58
Then
∫RnφLdx≥αεkk!44+nn∫Qvkdx.E59
Let us combine (47) and (59)
∫Qvkdx≤βk!αεk4+n4neσt,forallkt∈N∗×R+.E60
By induction we prove that
k!≤pp,forallk∈N∗andp≥k.E61
Let p≥1and k=p+1,then we have by the imbedding theorem for Lp−spaces
∫Qvpdx≤∫Qvkdxp/k.E62
Taking εenough small such that βk!αεk4+n4n≥1. Combining this with (37)
0<ε≤min1−γ2λ∞4ad1−γ12a+d21+2u0∞2+8ad1−γ12u0∞,1αβk!4+n4n1/k.E63
From (60), (61) and (63) into (62) we get (56). ♦
Lemma 3.8.Let Qiet Qjbe two different unit cubes of center xi=x1i…xniand xj=x1j…xnjrespectively of the form
Qi=x=x1…xn∈Rn:xk−xki≤1/2,forallk=1,…,n,Qj=x=x1…xn∈Rn:xk−xkj≤1/2,forallk=1,…,n,E64
with xj=xi+l,where l=l1…ln∈Zn\0Zn. Then, there exists a positive constant
δn=2+n2,E65
such that
distxiQj2≤xi−y2≤δndistxiQj2,forally∈Qj.E66
Proof.By Pythagorean theorem we have
xj−y≤n2.E67
As xi−xj≥1, then from (67)
xj−y≤n2xi−xj.E68
Also, it’s clear that distxiQj=distxi∂Qj, but every point z=z1…zn∈∂Qjis of the form
z=xj+s,E69
where s=s1…sn≠0and sk∈−1212,for all k=1,…,nwith at least one of the sk∈−1212.
It’s easy to prove that
xkj−xki≤2xkj−xki+sk,forallk=1,…,n.E70
Then
xj−xi≤2distxiQj.E71
As xi−y≤xi−xj+xj−ywe get from (68) and (71) the estimate
xi−y≤2distxiQj+n2xi−xj≤2+ndistxiQj.E72
We have obviously
xi−y≥distxiQj.E73
From (71) and (73) we get (66). ♦
Proof of theorem 3.2.
Let x∈Rnan arbitrary point and Qjj∈Nbe the family of pairwise disjoint measurable cubes of the form (64) covering Rnsuch that the center of Q0is x0=x.
Firstly, using the fact that Rn=∪j=0∞Qjand applying the left-hand inequality in (66)
∫Rne−x−y24dt−sλguvmdy=∑j=0∞∫Qje−x−y28dt−se−x−y28dt−sλguvmdy≤∑j=0∞e−distxQj28dt−s∫Qj−x−y28dt−sλguvmdy.E74
By Hölder ineguality with p>max1n2and q=1−1p
∫Qj−x−y28dt−sλguvmdy≤∫Qj−qx−y28dt−sdy1/q∫Qjλpgpuvpmdy1/p.E75
As
∫Qje−qx−y28dt−sdy≤∫Rn−qx−y28dt−sdy=8πdqn/2t−sn/2E76
and by (56) we have
∫Qjλpgpuvpmdy≤λ∞pg∞pβpm+1pm+1αεpm+14+n4neσt,E77
where
g∞=supu∈0u0∞gu.E78
Then, from (76) and (77) into (75)
∫Qj−x−y28dt−sλguvmdy≤K8πdqn21−1pt−sn21−1pλ∞g∞eσ/pt,E79
where
K=Kpmnαε=βpm+1pm+1αεpm+14+n4n1/p.E80
On the other hand, we deduce from the right-hand inequality in (66) that
∫Qje−x−y28dδnt−sdy≥e−distxQj28dt−s,forallj∈N∗.E81
Then
∑j=0∞e−distxQj28dt−s≤1+∑j=1∞e−distxQj28dt−s≤1+∑j=1∞∫Qje−x−y28dδnt−sdy≤1+∫Rne−x−y28dδnt−sdy≤1+8πdδnn/2t−sn/2.E82
We have from (79) and (82) into (74)
14πdt−sn/2∫Rne−x−y24dt−sλguvmdy≤2n/21−1pn21−1pt−s−n2pKλ∞g∞eσ/pt1+8πdδnn/2t−sn/2≤2n/21−1pn21−1pKλ∞g∞eσ/ptt−s−n2p+8πdδnn/2t−sn21−1p.
Whence
∫0tS2t−sλguvmds≤2n/21−1pn21−1pKλ∞g∞eσ/pt2p2p−nt1−n2p+8πdδnn/22ppn+2+2ptn21−1p+1
and finally we have for all t∈R+
vt∞≤v0∞+2n/21−1pn21−1pKλ∞g∞eσ/pt2p2p−nt1−n2p+8πdδnn/22ppn+2−ntn21−1p+1E83
As p>max1n2,the function in ton the right-hand side of the estimate (83) is continuous on R+. As ut∞≤u0∞on 0tmaxand vsatisfied (83), we conclude from (11) that tmax=+∞. Whence, the solution is global. ♦
Remark.We can extend the system to the case where instead of vmwe put vhvprovided that.
i. h:BUCR→BUCRis a locally continuous Lipschitz function, namely: for all constant ρ∈R+,there exists a constant cρ∈R+∗such that for all u,v∈BUCRnwith u∞≤ρand v∞≤ρwe have
hu−gv∞≤cρu−v∞.
ii.There exist two constants M∈R+∗and r∈Nsuch that:
0≤hv≤Mvr,forallv∈R+.
In this more general case, by examining the proof of the theorem 3.2; we see that under the same assumptions above, the system has also a global nonnegative classical solution. ♦
