We study the global (in time) existence of nonnegative solutions of the Gierer-Meinhardt system with mixed boundary conditions. In the research, the Robin boundary and Neumann boundary conditions were used on the activator and the inhibitor conditions respectively. Based on the priori estimates of solutions, the considerable results were obtained.
Biological spatial pattern formation is one area in applied mathematics under- going vivid investigations in recent years. Most models involved in biological phenomena are of the general reaction-diffusion type considered by Turing [
One of the famous studied models in biological spatial pattern formation is the Gierer-Meinhardt system which has received numerous attention and has been extensively studied [
( A t = ϵ 2 Δ A − A + A p H q + b , in Ω × ( 0 , T ) τ H t = D Δ H − H + A r H s , in Ω × ( 0 , T ) ϵ ∂ A ∂ ν + a A = 0 = ∂ H ∂ ν , on ∂ Ω × ( 0 , T ) H ( x , 0 ) = H 0 ( x ) > 0 , A ( x , 0 ) = A 0 ( x ) ≥ 0 in Ω ¯ (1)
where a ≥ 0 , b ≥ 0 and Ω ⊂ ℝ N is a bounded smooth domain; Δ is the Laplace or diffusion operator in ℝ N ; ν ( x ) is the unit outer normal at x ∈ ∂ Ω , ∂ / ∂ ν : = ∇ ⋅ ν is the directional derivative in the direction of the vector ν . We assume that the reaction exponents ( p > 1 , q > 0 , r > 0 , s ≥ 0 ) satisfy
0 < p − 1 r < q s + 1 . (2)
The diffusion constants are ϵ > 0 and D > 0 for the activator and inhibitor respectively. The time relaxation constant τ > 0 was mathematically intro- duced due to its usefulness on the stability of the system. The constant b provides additional support to the inhibitor and may be thought of as a measure of the effectiveness of the inhibitor in suppressing the production of the activator and that of its own. In [
In [
Special case was considered for the Neumann boundary condition (i.e. a = 0 ) in [
In this paper we consider the Robin boundary condition (a ≠ 0) on the activator and Neumann boundary condition on the inhibitor and study the global (in time) existence of solutions for the Gierer-Meinhardt system in (1). The theorem and lemmas in this current manuscript are inspired by [
Theorem 1. Suppose Ω is a smooth bounded domain with a smooth
boundary ∂ Ω in ℝ N . Assume that p − 1 r < min { 1 , q s + 1 } . Let A 0 and
H 0 ∈ W 2, l ( Ω ) , l > max { N , 2 } . Then every solution ( A ( x , t ) , H ( x , t ) ) of (1) exists globally in time.
The local existence and uniqueness of (1) is standard and more details can be found in [
u ( t ) = inf x ∈ Ω H ( x , t ) , t ∈ [ 0 , T )
then
u ( 0 ) = inf x ∈ Ω H 0 ( x ) > 0
Lemma 1. u ( t ) ≥ u ( 0 ) e − t τ for all 0 ≤ t < T .
Proof. Let
H * ( x , t ) = H ( x , t ) − u ( 0 ) e − t τ (3)
then H * ( x , t ) satisfies
H t * = H t − [ u ( 0 ) e − t τ ] ′ = 1 τ [ D Δ H − H + A r H s ] + 1 τ u ( 0 ) e − t τ = 1 τ [ D Δ H * − ( H * + u ( 0 ) e − t τ ) + A r H s ] + 1 τ u ( 0 ) e − t τ = 1 τ [ D Δ H * − H * − u ( 0 ) e − t τ + A r H s ] + 1 τ u ( 0 ) e − t τ = 1 τ [ D Δ H * − H * + A r H s ] − 1 τ u ( 0 ) e − t τ + 1 τ u ( 0 ) e − t τ = 1 τ [ D Δ H * − H * + A r H s ] = 1 τ [ D Δ H * − H * ] + A r τ H s ≥ 1 τ [ D Δ H * − H * ] in Ω × [ 0 , T )
and
∂ H * ∂ ν = ∂ ∂ ν [ H ( x , t ) − u ( 0 ) e − t τ ] = ∂ ∂ ν H ( x , t )
but
∂ H ∂ ν = 0 on ∂ Ω × [ 0 , T )
thus
∂ H * ∂ ν = 0 on ∂ Ω × [ 0 , T ) .
Additionally at t = 0 , from (3)
H * ( x , 0 ) = H ( x , 0 ) − u ( 0 ) = H ( x , 0 ) − inf x ∈ Ω H ( x , 0 ) ≥ 0
So H * ( x ,0 ) ≥ 0 for any x ∈ Ω .
Hence from maximum principle, H * ( x , t ) ≥ 0 in Ω ¯ × [ 0, T ) and thus
H ( x , t ) − u ( 0 ) e − t τ ≥ 0,
u ( t ) = inf x ∈ Ω H ( x , t ) ≥ u ( 0 ) e − t τ , t ∈ [ 0 , T )
□
Lemma 2. For any two constants α > 1 , β ≥ 0 , let p − 1 r < min { 1 , q s + 1 } .
Define
h α , β ( t ) = ∫ Ω A α ( x , t ) H β ( x , t ) d x , 0 ≤ t < T
Suppose
2 ε τ D ( α − 1 ) ( β + 1 ) ≥ ( τ ϵ 2 + D ) α β , (4)
then
h ˙ α , β ( t ) ≤ ( β τ − α ) h α , β ( t ) + C u − δ h α , β 1 − ϑ ( t ) . (5)
Here
ϑ = σ α ( r r − ( p − 1 ) − σ ) (6)
and
δ = r r − ( p − 1 ) − σ [ q − ( s + 1 ) ( p − 1 ) r − ( s + 1 r − β α ) σ ] (7)
where δ > 0 , σ > 0 , ϑ ∈ ( 0,1 ) and
C = [ α ( β / τ ) − p − 1 + σ r ] r r − ( p − 1 ) − σ | Ω | ϑ
Proof. Let α > 1 and β ≥ 0 ,
h ˙ α , β ( t ) = d d t ( ∫ Ω A α H β d x ) = ∫ Ω ( A α H β ) t d x = ∫ Ω [ α A α − 1 H β A t − β A α H β + 1 H t ] d x = ∫ Ω [ α A α − 1 H β ( ϵ 2 Δ A − A + A p H q + b ) − β A α τ H β + 1 ( D Δ H − H + A r H s ) ] d x = ∫ Ω α ϵ 2 A α − 1 H β Δ A d x − ∫ Ω α A α H β d x + ∫ Ω α A α + p − 1 H β ( H q + b ) d x − ∫ Ω β D A α τ H β + 1 Δ H d x + ∫ Ω β A α τ H β d x − ∫ Ω β A α + r τ H β + s + 1 d x = ( β τ − α ) ∫ Ω A α H β d x + α ϵ 2 ∫ Ω A α − 1 H β Δ A d x − β D τ ∫ Ω A α H β + 1 Δ H d x + α ∫ Ω A α + p − 1 H β ( H q + b ) d x − β τ ∫ Ω A α + r H β + s + 1 d x
But
α ϵ 2 ∫ Ω A α − 1 H β Δ A d x = α ϵ 2 ∫ ∂ Ω A α − 1 H β ∇ A ⋅ ν d S − α ϵ 2 ( α − 1 ) ∫ Ω A α − 2 H β | ∇ A | 2 d x + α ϵ 2 β ∫ Ω A α − 1 H β + 1 ∇ A ∇ H d x = α ϵ 2 ∫ ∂ Ω A α − 1 H β ( − a A ϵ ) d S − α ε 2 ( α − 1 ) ∫ Ω A α − 2 H β | ∇ A | 2 d x + α ϵ 2 β ∫ Ω A α − 1 H β + 1 ∇ A ∇ H d x = − α ϵ a ∫ ∂ Ω A α H β d S − α ϵ 2 ( α − 1 ) ∫ Ω A α − 2 H β | ∇ A | 2 d x + α ϵ 2 β ∫ Ω A α − 1 H β + 1 ∇ A ∇ H d x
Additionally,
β D τ ∫ Ω A α H β + 1 Δ H d x = β D τ ∫ ∂ Ω A α H β + 1 ∇ H ⋅ ν d S + β ( β + 1 ) D τ ∫ Ω A α H β + 2 | ∇ H | 2 d x − α β D τ ∫ Ω A α − 1 H β + 1 ∇ A ∇ H d x = β D τ ∫ ∂ Ω A α H β + 1 ( 0 ) d S + β ( β + 1 ) D τ ∫ Ω A α H β + 2 | ∇ H | 2 d x − α β D τ ∫ Ω A α − 1 H β + 1 ∇ A ∇ H d x = β ( β + 1 ) D τ ∫ Ω A α H β + 2 | ∇ H | 2 d x − α β D τ ∫ Ω A α − 1 H β + 1 ∇ A ∇ H d x
now we have,
h ˙ α , β ( t ) = ( β τ − α ) ∫ Ω A α H β d x − α ϵ a ∫ ∂ Ω A α H β d S − α ϵ 2 ( α − 1 ) ∫ Ω A α − 2 H β | ∇ A | 2 d x + α β ( ϵ 2 + D τ ) ∫ Ω A α − 1 H β + 1 ∇ A ∇ H d x − β ( β + 1 ) D τ ∫ Ω A α H β + 2 | ∇ H | 2 d x + α ∫ Ω A α + p − 1 H β + q d x − β τ ∫ Ω A α + r H β + s + 1 d x . = ( β τ − α ) ∫ Ω A α H β d x − α ϵ a ∫ ∂ Ω A α H β d S − α ϵ 2 ( α − 1 ) ∫ Ω A α H β | ∇ A A | 2 d x + α β ( ϵ 2 + D τ ) ∫ Ω A α H β ∇ H H ∇ A A d x − β ( β + 1 ) D τ ∫ Ω A α H β | ∇ H H | 2 d x + α ∫ Ω A α + p − 1 H β + q d x − β τ ∫ Ω A α + r H β + s + 1 d x . = ( β τ − α ) ∫ Ω A α H β d x − α ϵ a ∫ ∂ Ω A α H β d S + α ∫ Ω A α + p − 1 H β + q d x − β τ ∫ Ω A α + r H β + s + 1 d x + ∫ Ω A α H β [ − α ϵ 2 ( α − 1 ) | ∇ A A | 2 d x + α β ( ϵ 2 + D τ ) ∇ H H ∇ A A d x − β ( β + 1 ) D τ | ∇ H H | 2 d x ] .
We deduce from above a quadratic equation involving ∇ A A and ∇ H H . Let us
fix α > 1 and choose β ≥ 0 , we have
2 ϵ τ D ( α − 1 ) ( β + 1 ) ≥ ( τ ϵ 2 + D ) α β ,
therefore the quadratic form involving ∇ A A and ∇ H H in the inequality above
is non-positive since its determinant
[ α β ( ϵ 2 + D τ ) ∇ H H ] 2 ≤ 4 [ − ϵ 2 α ( α − 1 ) ] [ − β D τ ( β + 1 ) | ∇ H H | 2 ] .
Thus
h ˙ α , β ( t ) ≤ ( β τ − α ) ∫ Ω A α H β d x − α ϵ a ∫ ∂ Ω A α H β d S + α ∫ Ω A α + p − 1 H β + q d x − β τ ∫ Ω A α + r H β + s + 1 d x ≤ ( β τ − α ) ∫ Ω A α H β d x + α ∫ Ω A α + p − 1 H β + q d x − β τ ∫ Ω A α + r H β + s + 1 d x = ( β τ − α ) h α , β ( t ) + α ∫ Ω A α + p − 1 H β + q d x − β τ ∫ Ω A α + r H β + s + 1 d x .
We have
p − 1 r < 1 ⇒ r − ( p − 1 ) > 0
and
p − 1 r < q s + 1 ⇒ q > ( p − 1 ) ( s + 1 ) r
we choose σ > 0 sufficiently small such that
r − ( p − 1 ) > σ
and
q > ( p − 1 ) ( s + 1 ) r + ( s + 1 r − β α ) σ .
Now, we write
A α + p − 1 H β ( H q + b ) ≤ A α + p − 1 H β H q = A α + p − 1 H β H δ ( r − ( p − 1 ) − σ r ) + ( p − 1 ) ( s + 1 ) r + ( s + 1 r − β α ) σ = A α + p − 1 H β H δ ( r − ( p − 1 ) − σ r ) H ( p − 1 ) ( s + 1 ) r H ( s + 1 r − β α ) σ = A α + p − 1 H β σ α H β H δ ( r − ( p − 1 ) − σ r ) H ( p − 1 ) ( s + 1 ) r H s + 1 r σ = A α + p − 1 H β σ α H β H δ ( r − ( p − 1 ) − σ r ) H ( s + 1 ) p − 1 + σ r = A α + p − 1 H β σ α H β H δ ( r − ( p − 1 ) − σ r ) H ( s + 1 ) p − 1 + σ r = A α + p − 1 H β ϑ ( 1 − p − 1 + σ r ) H β − β p − 1 + σ r + β p − 1 + σ r H δ ( 1 − p − 1 + σ r ) H ( s + 1 ) p − 1 + σ r = A α + p − 1 H β ϑ ( 1 − p − 1 + σ r ) H β ( 1 − p − 1 + σ r ) H δ ( 1 − p − 1 + σ r ) H ( β + s + 1 ) p − 1 + σ r
but
A α + p − 1 = A α A p − 1 A σ A σ = A α A p − 1 + σ A α ϑ ( 1 − p − 1 + σ r ) = A α ( 1 − p − 1 + σ r + p − 1 + σ r ) A r p − 1 + σ r A α ϑ ( 1 − p − 1 + σ r ) = A α ( 1 − p − 1 + σ r ) A α p − 1 + σ r A r p − 1 + σ r A α ϑ ( 1 − p − 1 + σ r ) = A α ( 1 − ϑ ) ( 1 − p − 1 + σ r ) A ( α + r ) p − 1 + σ r
thus
A α + p − 1 H β + q = A α ( 1 − ϑ ) ( 1 − p − 1 + σ r ) A ( α + r ) p − 1 + σ r H β ϑ ( 1 − p − 1 + σ r ) H β ( 1 − p − 1 + σ r ) H δ ( 1 − p − 1 + σ r ) H ( β + s + 1 ) p − 1 + σ r = H − δ ( 1 − p − 1 + σ r ) A α ( 1 − ϑ ) ( 1 − p − 1 + σ r ) A ( α + r ) p − 1 + σ r H β ( 1 − ϑ ) ( 1 − p − 1 + σ r ) H ( β + s + 1 ) p − 1 + σ r = [ H − δ ( A α H β ) 1 − ϑ ] 1 − p − 1 + σ r ( A α + r H β + s + 1 ) p − 1 + σ r
where ϑ and δ are defined by (6) and (7)
α A α + p − 1 H β ( H q + b ) ≤ α [ H − δ ( A α H β ) 1 − ϑ ] 1 − p − 1 + σ r ( A α + r H β + s + 1 ) p − 1 + σ r ≤ α [ u − δ ( A α H β ) 1 − ϑ ] 1 − p − 1 + σ r ( A α + r H β + s + 1 ) p − 1 + σ r = α [ u − δ ( A α H β ) 1 − ϑ ] 1 − p − 1 + σ r ( β τ − 1 β τ − 1 A α + r H β + s + 1 ) p − 1 + σ r = α ( β τ ) − p − 1 + σ r [ u − δ ( A α H β ) 1 − ϑ ] 1 − p − 1 + σ r ( β τ A α + r H β + s + 1 ) p − 1 + σ r
by Young’s inequality, we obtain
α A α + p − 1 H β ( H q + b ) ≤ [ α ( β τ ) − p − 1 + σ r ] r r − ( p − 1 ) − σ u − δ ( A α H β ) 1 − ϑ + β τ A α + r H β + s + 1
α A α + p − 1 H β ( H q + b ) − β τ A α + r H β + s + 1 ≤ [ α ( β τ ) − p − 1 + σ r ] r r − ( p − 1 ) − σ u − δ ( A α H β ) 1 − ϑ .
Therefore
α ∫ Ω A α + p − 1 H β ( H q + b ) d x − β τ ∫ Ω A α + r H β + s + 1 d x ≤ [ α ( β τ ) − p − 1 + σ r ] r r − ( p − 1 ) − σ u − δ ∫ Ω ( A α H β ) 1 − ϑ d x
but by Hölder’s inequality
∫ Ω ( A α H β ) 1 − ϑ d x ≤ ( ∫ Ω d x ) ϑ ( ∫ Ω A α H β d x ) 1 − ϑ = | Ω | ϑ h α , β 1 − ϑ .
Thus
α ∫ Ω A α + p − 1 H β ( H q + b ) d x − β τ ∫ Ω A α + r H β + s + 1 d x ≤ [ α ( β τ ) − p − 1 + σ r ] r r − ( p − 1 ) − σ u − δ | Ω | ϑ h α , β 1 − ϑ = C u − δ h α , β 1 − ϑ .
Finally,
h ˙ α , β ( t ) ≤ ( β τ − α ) h α , β + C u − δ h α , β 1 − ϑ .
□
Remark 1. The condition in (4) is true for any
α ≥ 2 and 0 < β ≤ 1 2 K
where K ≥ max { τ ϵ 2 D , D τ ϵ 2 }
Lemma 3. Let 0 ≤ δ < 1 , θ > 0 and ζ > 0 on ( 0, T ) be an integrable function. Let h α , β = h α , β ( t ) be a nonnegative function on [ 0, T ) satisfying the differential inequality
h ˙ α , β ( t ) ≤ − θ h α , β + ζ h α , β δ , 0 ≤ t < T . (8)
Then
h α , β ( t ) ≤ κ , 0 ≤ t < T (9)
where κ is the maximal root of the algebraic equation
x − G ( ζ ) x δ = h α , β ( 0 ) .
Moreover, if T = ∞ , we have
lim sup t → ∞ h α , β ( t ) ≤ κ ∞ , (10)
where κ ∞ is the maximal root of the algebraic equation
x − G ∞ ( ζ ) x δ = 0.
Proof.
h ˙ α , β ( t ) ≤ − θ h α , β + ζ h α , β δ
h ˙ α , β ( t ) + θ h α , β ≤ ζ h α , β δ
e θ t [ h ˙ α , β ( t ) + θ h α , β ] ≤ e θ t ζ h α , β δ
d d t [ e θ t h α , β ] ≤ ζ e θ t h α , β δ
∫ 0 t d d t [ e θ χ h α , β ] d χ ≤ ∫ 0 t e θ χ ζ ( χ ) h α , β δ ( χ ) d χ
e θ t h α , β ( t ) − h α , β ( 0 ) ≤ ∫ 0 t e θ χ ζ ( χ ) h α , β δ ( χ ) d χ
e θ t h α , β ( t ) ≤ h α , β ( 0 ) + ∫ 0 t e θ χ ζ ( χ ) h α , β δ ( χ ) d χ
h α , β ( t ) ≤ e − θ t h α , β ( 0 ) + ∫ 0 t e − θ ( t − χ ) ζ ( χ ) h α , β δ ( χ ) d χ . (11)
Let
h α , β ¯ ( t ) = sup 0 < χ < t h α , β ( χ ) ,
and
G ( ζ ) : = sup 0 < t < T ∫ 0 t e − θ ( t − χ ) ζ ( χ ) d χ
in particular, at T = ∞
G ∞ ( ζ ) : = lim sup t → ∞ ∫ 0 t e − θ ( t − χ ) ζ ( χ ) d χ
we obtain now that
h α , β ¯ ( t ) ≤ h α , β ( 0 ) + G ( ζ ) h α , β δ ¯ ( t ) (12)
Notice that the quantity G ( ζ ) is finite and hence (9) follows from (12). As t → ∞ in (11), we ascertain
lim sup t → ∞ h α , β ( t ) ≤ G ∞ ( ζ ) lim sup t → ∞ h α , β δ ( t )
thus (10) follows since G ∞ ( ζ ) is finite. □
The next Lemma follows after applying Lemma 3 to (5).
Lemma 4. For any α , β > 0 such that β < τ α , and all conditions in Lemma 2 hold true. Then there exists a constant C ( T ) = C α , β ( T ) ≤ ∞ such that
h α , β ( t ) ≤ C ( T ) (13)
for all t ∈ [ 0, T ) .
Proof. For sufficiently small β > 0 , such that β < τ α with α > 1 , we obtain
2 ϵ τ D ( α − 1 ) ( β + 1 ) ≥ ( τ ϵ 2 + D ) α β , β τ − α < 0 ,
therefore we deduce from Lemma 2 that h α , β satisfies
h ˙ α , β ( t ) ≤ ( β τ − α ) h α , β + C u − δ h α , β 1 − ϑ .
Since
β τ − α < 0
and
u ( t ) ≥ u ( 0 ) e − t τ
for all t ∈ [ 0, T ) , then from Lemma 3, (13) is true for α > 1 and sufficiently small β such that β < τ α . Since H is bounded away from zero, then (13) is true for any α , β > 0 . □
From Lemma 1 and Lemma 4, we deduce the Corollary below.
Corollary 1. Let l ≥ 1 and all other assumptions in Theorem 1, Lemma 2, Lemma 3, and Lemma 4 hold true. Define
g 1 ( A , H ) = A p H q + b
g 2 ( A , H ) = A r H s
then there exist positive constant C l ( T ) , such that
‖ g j ( A , H ) ‖ L l ( Ω ) ≤ C l ( T ) , j = 1 , 2
for all 0 ≤ t < T .
Proof. The proof to this Corollary follows from Lemma 3 and Lemma 4. □
In this paper, we have studied the Gierer-Meinhardt system with Robin boundary conditions and Neumann boundary conditions on the activator and inhibitor respectively. Global existence of solutions have been obtained under the mixed boundary conditions using a priori estimates of solutions.
Antwi-Fordjour, K. and Nkashama, M. (2017) Global Existence of Solutions of the Gierer-Meinhardt System with Mixed Boundary Conditions. Applied Mathematics, 8, 857-867. https://doi.org/10.4236/am.2017.86067