Tải bản đầy đủ - 0trang
348
10 Pattern Formation and the Gierer–Meinhardt Model in Molecular Biology
⎧
up
⎪
⎪
Δ
u
−
α
u
+
+ ρ (x) = 0, u > 0
⎪
⎪
vq
⎨
r
u
Δ v − β v + s = 0, v > 0
⎪
⎪
v
⎪
⎪
⎩
u = ε , v = Γs,r (ε )
in Ωε ,
(10.41)
in Ωε ,
on ∂ Ωε .
The existence of a classical solution to (10.41) is obtained by using the
Schauder’s fixed point theorem. For 0 < ε < ε0 and m1 , m2 < 1 < M1 , M2 consider
A =
⎧
⎪
⎪
⎨
⎪
⎪
⎩
m1 ϕ1 ≤ u ≤ M1 ϕ1
(u, v) ∈ C(Ω ε ) × C(Ω ε ) : m2Γs,r (ϕ1 ) ≤ v ≤ M2Γs,r (ϕ1 )
u = ε , v = Γs,r (ε )
⎫
in Ωε , ⎪
⎪
⎬
in Ωε , .
⎪
⎪
⎭
on ∂ Ωε
Next, we define the mapping T : A → C(Ω ε ) × C(Ω ε ) as follows. For (u, v) ∈
A we set
T (u, v) = (Tu, T v),
(10.42)
where Tu and T v satisfy
⎧
(Tu) p
⎪
⎪
⎪
⎨ Δ (Tu) − α (Tu) + vq + ρ (x) = 0, Tu > 0 in Ωε ,
ur
Δ (T v) − β (T v) +
= 0, T v > 0
in Ωε ,
⎪
⎪
(T v)s
⎪
⎩
Tu = ε , T v = Γs,r (ε )
on ∂ Ωε .
(10.43)
Using the definition of A , by the sub and supersolution method combined with
Theorem 1.2 and Corollary 1.3, the above system has a unique solution (Tu, T v)
with Tu, T v ∈ C2 (Ω ε ). Basic to our approach are the following two results which
allows us to apply Schauder’s fixed point theorem.
Lemma 10.5 There exist m1 < 1 < M1 and m2 < 1 < M2 which are independent of
ε such that T (A ) ⊆ A , for all 0 < ε < ε0 .
Proof. Let w ∈ C2 (Ω ) be the unique solution of problem (10.11). In view of
(10.7) and (10.12) we have
w(x) ≤ c2 ϕ1 ≤
c2
c2
d(x) = ε
C
C
on ∂ Ωε .
Hence, if δ1 = min{1, cC2 } then δ1 w ≤ ε on ∂ Ωε . Furthermore,
Δ (Tu) − α (Tu) + ρ (x) ≤ 0 ≤ Δ (δ1 w) − α (δ1 w) + ρ (x)
in Ωε ,
10.3 Case 0 ≤ p < 1
349
Tu = ε ≥ δ1 w
on ∂ Ωε .
By the maximum principle, we obtain Tu ≥ δ1 w in Ωε . In view of (10.12), let us
choose m1 = δ1 c1 in the definition of A (where c1 is the constant in (10.12)). Then,
(10.12) combined with the last estimates yields
Tu ≥ m1 ϕ1
in Ωε .
(10.44)
From the second equation in (10.43) and the fact that u ≥ m1 ϕ1 in Ωε we have
Δ (T v) − β (T v) +
mr1 ϕ1r
≤ 0 in Ωε .
(T v)s
Let us consider the problem
⎧
⎨ Δ ξ − β ξ + ϕ1r ξ −s = 0 in Ω ,
ξ >0
in Ω ,
⎩
ξ =0
on ∂ Ω .
(10.45)
(10.46)
Using Proposition 10.3 (ii), there exists ξ ∈ C2 (Ω ) ∩ C(Ω ) a unique solution of
(10.46) with the additional property
c3Γs,r (ϕ1 ) ≤ ξ ≤ c4Γs,r (ϕ1 )
in Ω ,
(10.47)
for some c3 , c4 > 0. Moreover, by (10.47), (10.7) and the property (10.26) of Γs,r we
can find c5 , c6 > 0 such that
c5Γs,r (d(x)) ≤ ξ ≤ c6Γs,r (d(x))
r/(1+s) 1
, c }.
6
Let δ2 = min{1, m1
in Ω .
(10.48)
Then
Δ (δ2 ξ ) − β (δ2ξ ) + mr1ϕ1r (δ2 ξ )−s ≥ δ2 Δ ξ − β ξ + ϕ1r ξ −s = 0 in Ω ,
(10.49)
and by (10.48) we have
δ2 ξ ≤ δ2 c6Γs,r (d(x)) ≤ Γs,r (ε )
on ∂ Ωε .
(10.50)
Therefore, from (10.45), (10.49) and (10.50) we have obtained
Δ (T v) − β (T v) + mr1ϕ1r (T v)−s ≤ 0 ≤ Δ (δ2 ξ ) − β (δ2ξ ) + mr1ϕ1r (δ2 ξ )−s
T v = Γs,r (ε ) ≥ δ2 ξ
on ∂ Ωε .
in Ω ,
350
10 Pattern Formation and the Gierer–Meinhardt Model in Molecular Biology
By Corollary 1.3 it follows that T v ≥ δ2 ξ in Ωε . In view of (10.47), the last inequality leads us to T v ≥ δ2 c3Γs,r (ϕ1 ) in Ωε . Thus, we consider
m2 = min{1, δ2 c3 } > 0
in the definition of the set A . Note that m2 is independent of ε and T v ≥ m2Γs,r (ϕ1 )
in Ωε .
The definition of A and (10.25) yield
v ≥ m2Γs,r (ϕ1 ) ≥ m2 ϕ1σ
in Ωε .
Using the estimate v ≥ m2 ϕ1σ in the first equation of (10.43) we get
−qσ
Δ (Tu) − α (Tu) + m−q
(Tu) p + ρ (x) ≥ 0 in Ωε .
2 ϕ1
As above, we next consider the problem
⎧
−q −qσ
⎨ Δ ζ − αζ + m2 ϕ1 ζ p + ρ (x) = 0 in Ω ,
ζ >0
in Ω ,
⎩
ζ =0
on ∂ Ω .
(10.51)
(10.52)
Since qσ < p + 1, by Proposition 10.2 there exists ζ ∈ C2 (Ω ) ∩ C(Ω ) a unique
solution of (10.52) such that
c7 ϕ1 ≤ ζ ≤ c8 ϕ1
in Ω ,
(10.53)
for some c7 , c8 > 0. Note that qσ < p + 1, (10.53) and Lemma 10.1 imply Δ ζ ∈
L1 (Ω ). Let A1 = max{1, Cc1 7 }. Then
−qσ
Δ (A1 ζ ) − α (A1 ζ ) + m−q
(A1 ζ ) p + ρ (x) ≤ 0 in Ωε .
2 ϕ1
Also by (10.7) and (10.53) we have
A1 ζ ≥ A1 c7 ϕ1 ≥ A1Cc7 d(x) ≥ ε
on ∂ Ωε .
Define
−qσ
Ψ (x,t) = −α t + m−q
(x)(A1t) p + ρ (x),
2 ϕ1
Then Ψ satisfies the hypotheses in Theorem 1.2 and
(x,t) ∈ Ωε × (0, ∞).
10.3 Case 0 ≤ p < 1
351
Δ (A1 ζ ) + Ψ (x, A1 ζ ) ≤ 0 ≤ Δ (Tu) + Ψ (x, Tu)
in Ωε ,
Tu, A1 ζ > 0 in Ωε , Tu = ε ≤ A1 ζ on Ωε ,
Δ (A1 ζ ) ∈ L1 (Ωε ).
By Theorem 1.2 it follows that Tu ≤ A1 ζ in Ωε . In view of (10.53), let us take
M1 := max{1, A1 c8 } in the definition of the set A . Then M1 does not depend on ε
and by (10.53) we have
Tu ≤ M1 ϕ1
in Ωε .
The definition of A yields u ≤ M1 ϕ1 in Ωε . Then, the second equation of system
(10.43) produces
Δ (T v) − β (T v) +
M1r ϕ1r
≥ 0 in Ωε .
(T v)s
(10.54)
Let A2 = max{1, M1r , c1 }. If ξ is the unique solution of (10.46), then
5
Δ (A2 ξ ) − β (A2ξ ) + M1r ϕ1r (A2 ξ )−s ≤ 0 in Ωε ,
and, by (10.48) we also have
A2 ξ ≥ A2 c5Γs,r (d(x)) ≥ Γs,r (ε )
on ∂ Ωε .
Therefore, by Corollary 1.3 it follows that T v ≤ A2 ξ in Ωε . Now, we take M2 :=
max{1, A2 c4 } in the definition of the set A . It follows that M2 is independent of ε
and, by virtue of (10.47), we obtain T v ≤ M2Γs,r (ϕ1 ) in Ωε . This finishes the proof
of our Lemma 10.5.
Lemma 10.6 The mapping T : A → A defined in (10.42)–(10.43) is compact and
continuous.
Proof. Let us fix (u, v) ∈ A . Then u, v, Tu and T v are bounded away from zero
in Ω ε which yields
(Tu) p
vq
L∞ (Ωε )
,
ur
(T v)s
L∞ (Ωε )
≤ cε = c(ε , m1 , m2 , M1 .M2 , p, q, r, s).
Hence, by Hăolder estimates, for all > N we obtain
Tu
W 2,τ (Ωε ) ,
Tv
W 2,τ (Ωε )
≤ c1,ε ,
352
10 Pattern Formation and the Gierer–Meinhardt Model in Molecular Biology
for some c1,ε > 0 independent of u and v. Since the embedding W 2,τ (Ωε ) →
C1,γ (Ω ε ) , 0 < γ < 1 − N/τ is compact, we derive that the mapping T : A →
A ⊂ C(Ω ε ) × C(Ω ε ) is also compact.
It remains to prove that T is continuous. To this aim, let {(un , vn )}n≥1 ⊂ A be
such that un → u and vn → v in C(Ω ε ) as n → ∞. Since T is compact, there exists
(U,V ) ∈ A such that up to a subsequence we have
T (un , vn ) → (U,V )
in A as n → ∞.
r
(Tun ) p
and (Tuvnn )s
, it follows that (Tun )n≥1
q
vn
n≥1
n≥1
2,
τ
and (T vn )n≥1 are bounded in W (Ωε ) for all τ > N. As before, this implies that
(Tun )n≥1 and (T vn )n≥1 are bounded in C1,γ (Ω ε ) (0 < γ < 1 − N/τ ). Next, by
Schauder estimates, it follows that (Tun )n≥1 and (T vn )n≥1 are bounded in C2,γ (Ω ε ).
Since C2,γ (Ω ε ) is compactly embedded in C2 (Ω ε ), we deduce that up to a subse-
Using the L∞ (Ωε ) bounds of
quence, we have that
Tun → U
and T vn → V
in C2 (Ω ε ) as n → ∞.
Passing to the limit in (10.43) we get that (U,V ) satisfies
⎧
Up
⎪
⎪
Δ
U
−
α
U
+
+ ρ (x) = 0, U > 0
in Ωε ,
⎪
⎪
vq
⎨
r
u
Δ V − β V + s = 0, V > 0
in Ωε ,
⎪
⎪
V
⎪
⎪
⎩
U = ε , V = Γs,r (ε )
on ∂ Ωε .
Using the uniqueness of (10.43), it follows that Tu = U and T v = V . Thus, we have
obtained that any subsequence of {T (un , vn )}n≥1 has a subsequence converging to
T (u, v) in A . But this implies that the entire sequence {T (un , vn )}n≥1 converges
to T (u, v) in A , whence the continuity of T . The proof of Lemma 10.6 is now
complete.
We now come back to the proof of Theorem 10.4. According to Lemmas 10.5
and 10.6 we are now in position to apply Schauder’s fixed point theorem. Thus,
for all 0 < ε < ε0 , there exists (uε , vε ) ∈ A such that T (uε , vε ) = (uε , vε ). By
standard elliptic regularity arguments, we deduce uε , vε ∈ C2 (Ω ε ). Therefore, for all
0 < ε < ε0 we have proved the existence of a solution (uε , vε ) ∈ C2 (Ω ε ) × C2 (Ω ε )
of system (10.41). Next, we extend uε = ε , vε = Γs,r (ε ) in Ω \ Ω ε . Furthermore, by
10.3 Case 0 ≤ p < 1
353
the definition of A we have
m1 ϕ1 ≤ uε ≤ M1 ϕ1 + ε ≤ M1 ϕ1 + ε0
in Ω ,
m2Γs,r (ϕ1 ) ≤ vε ≤ M2Γs,r (ϕ1 ) + Γs,r (ε ) ≤ M1 ϕ1 + cε0
in Ω .
(10.55)
(10.56)
As above, L bounds together with Hăolder estimates yield (u )0<ε <ε0 , (vε )0<ε <ε0
2,τ
are bounded in Wloc
(Ω ), for all τ > N. With similar arguments, there exist u, v ∈
C2 (Ω ) such that for all ω ⊂⊂ Ω , (uε )0<ε <ε0 and (vε )0<ε <ε0 converge up to a subsequence to u and v respectively in C2 (ω ) as ε → 0. Passing to the limit with ε → 0
in (10.41) and (10.55)–(10.56) we get
⎧
up
⎪
⎨ Δ u − α u + q + ρ (x) = 0
v
r
u
⎪
⎩Δv − βv + = 0
vs
in Ω ,
in Ω ,
and
m1 ϕ1 ≤ u ≤ M1 ϕ1
in Ω ,
m2Γs,r (ϕ1 ) ≤ v ≤ M2Γs,r (ϕ1 )
in Ω .
(10.57)
(10.58)
Now, we extend u = v = 0 on ∂ Ω . From (10.57) and (10.58) we deduce that u, v ∈
C(Ω ). Hence, the system (10.3) has a classical solution (u, v).
It remains to establish the boundary estimates of the solution to (10.3). This
follows essentially by using the same arguments as above. Let (u, v) be an arbitrary solution of (10.3). Then Δ u − α u + ρ (x) ≤ 0 in Ω which implies that
u ≥ w in Ω , where w is the unique solution of (10.11). By (10.12) it follows that
u ≥ c1 ϕ1 in Ω . Using this inequality in the second equation of (10.3) we deduce
Δ v − β v + c2 ϕ1r v−s ≤ 0 in Ω for some c2 > 0 (we actually have c2 = cr1 > 0). Next,
let ξ be the unique solution of (10.46). A similar argument to that used in before
yields v ≥ c3 ξ in Ω . In view of estimate (10.27) in Proposition 10.3 we derive that
v ≥ c4Γs,r (ϕ1 ) in Ω for some c4 > 0. According to (10.25) it follows that v ≥ c5 ϕ1σ
in Ω . This inequality combined with the first equation in system (10.3) produces
Δ u − α u + c6ϕ1−qσ u p + ρ (x) ≥ 0 in Ω .
Consider the problem
⎧
−qσ
⎨ Δ z − α z + c6ϕ1 z p + ρ (x) = 0 in Ω ,
z>0
in Ω ,
⎩
z=0
on ∂ Ω .
(10.59)
354
10 Pattern Formation and the Gierer–Meinhardt Model in Molecular Biology
Since qσ < p + 1, by Proposition 10.2 there exists a unique solution of (10.59) such
that z ≤ c7 ϕ1 in Ω . Thus, by Theorem 1.2 we get u ≤ z ≤ c7 ϕ1 in Ω . Using this last
inequality in the second equation of (10.3) we finally obtain Δ v − β v + c8 ϕ1r v−s ≥ 0
in Ω for some c8 > 0. By virtue of Proposition 10.3 we have v ≤ c9Γs,r (ϕ1 ) in Ω .
Thus, we have obtained
m1 ϕ1 ≤ u ≤ m2 ϕ1
in Ω ,
m1Γs,r (ϕ1 ) ≤ u ≤ m2Γs,r (ϕ1 )
in Ω ,
for some fixed constants m1 , m2 > 0. Now, the boundary estimates in Theorem
10.4 follows from the above inequalities combined with (10.7). This concludes the
proof.
10.3.2 Further Results on Regularity
Further regularity of the solution to (10.3) can be obtained using the same arguments
as in Gui and Lin [107]. More precisely, it is proved in [107] that if u ∈ C2 (Ω ) ∩
C(Ω ) satisfies −Δ u = u−ν in a smooth bounded domain Ω and u = 0 on ∂ Ω , then
u ∈ C1,1−ν (Ω ). Using the conclusion in Theorem 10.4 we have
Corollary 10.7 Assume that 0 ≤ p < 1 and q, r, s satisfy (10.5).
(i) If q ≤ p and s ≤ r, then the system (10.3) has at least one classical solution.
Moreover, any solution of (10.3) belongs to C2 (Ω ) × C2 (Ω ).
(ii) If −1 < p − q < 0 and −1 < r − s < 0, then the system (10.3) has at least one
classical solution. Moreover, any solution (u, v) of (10.3) satisfies u ∈ C2 (Ω ) ∩
C1,1+p−q(Ω ) and v ∈ C2 (Ω ) ∩C1,1+r−s (Ω ).
Proof. Let (u, v) be a classical solution of (10.3). We rewrite the system (10.3)
in the form
Δ u = f1 (x) in Ω ,
Δ v = f2 (x) in Ω ,
u = v = 0 on ∂ Ω ,
where
f1 (x) = α u(x) −
u p (x)
ur (x)
−
, for all x ∈ Ω .
ρ
(x),
f
(x)
=
β
v(x)
−
2
vq (x)
vs (x)
10.3 Case 0 ≤ p < 1
355
Note that in our settings we have σ = 1 in (10.6) and by virtue of Theorem 10.4
there exist c1 , c2 > 0 such that c1 d(x) ≤ u, v ≤ c2 d(x) in Ω . Hence,
| f1 (x)| ≤ m1 d p−q(x) + ρ (x)
in Ω
in Ω . (10.60)
and | f2 (x)| ≤ m2 d r−s (x)
(i) Since 0 ≤ p−q and 0 ≤ r −s, by (10.60) we get f1 , f2 ∈ L∞ (Ω ). Next, standard
elliptic arguments lead us to u, v ∈ C2 (Ω ).
(ii) Let us assume that −1 < p − q < 0 and −1 < r − s < 0. From (10.60) we
derive
| f1 (x)| ≤ cd p−q (x)
in Ω
and | f2 (x)| ≤ cd r−s (x)
in Ω ,
for some positive constant c > 0.
If N = 1 then, for all x1 , x2 ∈ Ω we have
|u (x1 ) − u (x2 )| ≤
x2
x1
| f1 (t)|dt ≤ c
x2
x1
˜ 1 − x2 |1+p−q,
d p−q(t)dt ≤ c|x
where c˜ > 0 does not depend on x1 , x2 . This yields u ∈ C1,1+p−q(Ω ) and similarly
v ∈ C1,1+r−s (Ω ).
If N ≥ 2, the conclusion follows exactly in the same way as in [107]. More precisely, let G denote the Green’s function for the Laplace operator. Then for all x ∈ Ω
we have
u(x) =
Ω
G (x, y) f1 (y)dy,
v(x) =
Gx (x, y) f1 (y)dy,
∇v(x) =
Ω
G (x, y) f2 (y)dy,
and
∇u(x) =
Ω
Ω
Gx (x, y) f2 (y)dy.
Then, for all x1 , x2 ∈ Ω , x1 = x2 we have
|∇u(x1 ) − ∇u(x2)| ≤
≤c
Ω
|Gx (x1 , y) − Gx (x2 , y)|| f1 (y)|dy
Ω
|Gx (x1 , y) − Gx (x2 , y)|d p−q (y)dy,
and similarly
|∇v(x1 ) − ∇v(x2)| ≤ c
Ω
|Gx (x1 , y) − Gx (x2 , y)|d r−s (y)dy.
356
10 Pattern Formation and the Gierer–Meinhardt Model in Molecular Biology
From now on, we need only to employ the sharp estimates given in [107, Theorem
1.1] in order to obtain u ∈ C1,1+p−q(Ω ) and v ∈ C1,1+r−s (Ω ). This finishes the proof
of Corollary 10.7.
10.3.3 Uniqueness of a Solution
The issue of uniqueness is a delicate matter even in one dimension. In this case the
system (10.3) reads
⎧
up
⎪
⎪
u − α u + q + ρ (x) = 0
⎪
⎪
v
⎨
ur
v −βv+ s = 0
⎪
⎪
v
⎪
⎪
⎩
u(0) = u(1) = 0, v(0) = v(1) = 0.
in (0, 1),
in (0, 1),
(10.61)
In [38] it is proved that the system (10.61) has a unique solution provided that
p = q = r = s = 1. The main idea is to write (10.61) as a linear system with smooth
coefficients and then to use the C2 [0, 1] × C2 [0, 1] regularity of the solution. This
approach has been used in [94] (see also [95] or [93, Theorem 2.7]) in the case
β ≤ α , 0 < q ≤ p ≤ 1 and r − p = s − q ≥ 0.
We are able to show that the uniqueness of the solution to (10.61) still holds
provided that
− 1 < p − q < 1,
−1 < r − s < 1.
(10.62)
Note that for the above range of exponents, the solutions of (10.61) do not necessarily belong to C2 [0, 1] ×C2[0, 1]. We prove that a C1+δ -regularity up to the boundary
of the solution suffices in order to have uniqueness. Therefore, we prove
Theorem 10.8 Let Ω = (0, 1), 0 ≤ p < 1 and q, r, s > 0 verify (10.62). Then the
system (10.61) has a unique classical solution.
Unlike the Neumann boundary condition, in which large multiplicities of solutions are observed, the uniqueness in the above result seems to be a particular feature
of the Dirichlet boundary condition together with the sublinear character of the first
equation in the system (10.61).
Proof. Let (u, v) be a classical solution of (10.61). Then, by virtue of Corollary
10.7, we have
10.3 Case 0 ≤ p < 1
357
u, v ∈ C2 [0, 1] × C2[0, 1]
if 0 ≤ p − q, 0 ≤ r − s,
and
u ∈ C2 (0, 1) ∩C1,1+p−q[0, 1], v ∈ C2 (0, 1) ∩C1,1+r−s [0, 1],
if −1 < p − q < 0, −1 < r − s < 0. Furthermore, by Hopf’s maximum principle we
also have that u (0) > 0, v (0) > 0, u (1) < 0 and v (1) < 0.
Assume that there exist (u1 , v1 ) and (u2 , v2 ) two different solutions of (10.61).
First we claim that we cannot have u2 ≥ u1 or v2 ≥ v1 in [0,1]. Assume by contradiction that u2 ≥ u1 in [0,1]. Then
v2 − β v2 +
ur2
ur1
=
0
=
v
−
β
v
+
1
1
vs2
vs1
in (0, 1),
and by Corollary 1.3 we get v2 ≥ v1 in [0, 1]. This implies that
u1 − α u1 +
u1p
u2p
q + ρ (x) ≤ 0 = u2 − α u2 + q + ρ (x) in (0, 1).
v2
v2
(10.63)
p
On the other hand, the mapping Ψ (x,t) = −α t + vqt(x) + ρ (x), (x,t) ∈ (0, 1) ×
2
(0, ∞) satisfies the hypotheses in Theorem 1.2. Hence u2 ≤ u1 in [0, 1], that is u1 ≡
u2 . This also implies v1 ≡ v2 , which is a contradiction. Replacing u1 by u2 and v1
by v2 , we also get that the situation u1 ≥ u2 or v1 ≥ v2 in [0,1] is not possible.
Set U = u2 − u1 and V = v2 − v1 . From the above arguments, both U and V
change sign in (0, 1). The key result in our approach is the following.
Proposition 10.9 U and V vanish only at finitely many points in the interval [0, 1].
Proof. Subtracting the corresponding equations for (u1 , v1 ) and (u2 , v2 ) we obtain
the following linear problem
W (x) + A(x)W(x) = 0 in (0, 1),
W(0) = W(1) = 0,
where W = (U,V )T and A(x) = (Ai j (x))1≤i, j≤2 is a 2 × 2 matrix defined as
A11 (x) = −α +
⎧
⎪
⎪
⎨
⎪
⎪
⎩
1
vq2 (x)
u2p (x) − u1p(x)
, u1 (x) = u2 (x)
u2 (x) − u1(x)
u p−1(x)
,
u1 (x) = u2 (x)
p 1q
v1 (x)
·
(10.64)
358
10 Pattern Formation and the Gierer–Meinhardt Model in Molecular Biology
⎧
⎪
⎪
⎨−
q
q
u1p (x)
v (x) − v1 (x)
· 2
, v1 (x) = v2 (x)
q
q
v1 (x)v2 (x) v2 (x) − v1 (x)
p
A12 (x) =
u1 (x)
⎪
⎪
−q q+1
,
v1 (x) = v2 (x)
⎩
v1 (x)
⎧
ur (x) − ur1 (x)
1
⎪
⎪
· 2
, u1 (x) = u2 (x)
⎨ s
v2 (x) u2 (x) − u1 (x)
A21 (x) =
r−1
u (x)
⎪
⎪
⎩
,
u1 (x) = u2 (x)
r 1s
v1 (x)
⎧
ur1 (x)
vs2 (x) − vs1 (x)
⎪
⎪
·
, v1 (x) = v2 (x)
⎨ s
v1 (x)vs2 (x) v2 (x) − v1 (x)
A22 (x) = −β −
r
u1 (x)
⎪
⎪
s s+1
,
v1 (x) = v2 (x).
⎩
v1 (x)
Lemma 10.10 We have
(i) Ai j ∈ C(0, 1), for all 1 ≤ i, j ≤ 2.
(ii) A12 (x) = 0 and A21 (x) = 0 for all x ∈ (0, 1).
(iii) d 1−(p−q)(x)A1 j ∈ L∞ (0, 1) and d 1−(r−s) (x)A2 j ∈ L∞ (0, 1), for j = 1, 2.
Proof. The claims in (i) and (ii) are easy to verify. We prove only the statement
in (iii). To this aim, let us notice first that by the regularity of solutions, there exist
c1 , c2 > 0 such that
c1 d(x) ≤ ui , vi ≤ c2 d(x) in (0, 1), 1 ≤ i ≤ 2.
(10.65)
By (10.65) and the fact that
|aq − bq| ≤ q|a − b| max{aq−1, bq−1 }
for all a, b > 0,
we have
d(x)|A12 (x)| ≤ qd(x)
u1p (x)
q
v1 (x)vq2 (x)
q−1
max{vq−1
1 (x), v2 (x)}
p
≤ qd p−q(x)
u1 (x)
d(x)
≤ cd p−q(x)
for all 0 < x < 1.
max
d(x)
v1 (x)
q+1
,
d(x)
v2 (x)
q+1
Hence d 1−(p−q)(x)A12 ∈ L∞ (0, 1). We obtain similar estimates for d 1−(p−q)(x)A11
and d 1−(r−s) (x)A2 j , j = 1, 2. This concludes the proof.