A new approach toward boundedness in a two-dimensional parabolic chemotaxis system with singular sensitivity

0.21 MB·English

by  Johannes Lankeit

A new approach toward boundedness in a
two-dimensional parabolic chemotaxis
system with singular sensitivity
Johannes Lankeit
∗
January 22, 2015
5
1
0
2

n
We consider the parabolic chemotaxis model
a
J
u =∆u χ (u v)
1 t − ∇· v∇
2 (vt =∆v v+u
−

]
inasmooth,bounded,convextwo-dimensionaldomainandshowglobalexistenceandbound-
P
A edness of solutions for χ (0,χ0) for some χ0 > 1, thereby proving that the value χ = 1 is
∈
not critical in this regard.
.
h Our main tool is consideration of the energy functional
t
a
m (u,v)= ulnu a ulnv+b √v 2
a,b
F − |∇ |
[ ZΩ ZΩ ZΩ

1 for a>0, b 0, where using nonzero values of b appears to be new in this context.
≥
v Keywords: chemotaxis; singular sensitivity; global existence; boundedness
5 MSC: 35K55 (primary), 35A01, 35A09, 92C17, 35A07, 35B40 (secondary)
7
1
5 1 Introduction
0
.
1 Numerous phenomena in connection with spontaneous aggregation can be described by PDE models
0
incorporating a cross-diffusion mechanism. A prototypical example, which lies at the core of models
5
used for a variety of purposes and to so different aims as the description pattern formation of bacteria
1
: or slime mold in biology [11] or the prediction of burglary in criminology[14], is the following variant of
v
the Keller-Segel system of chemotaxis:
i
X
u =∆u (uS(v) v)
r t
−∇· ∇
a
v =∆v v+u (1)
t
−
∂ u =∂ v =0
ν ∂Ω ν ∂Ω
| |
u(,0)=u ,v(,0)=v
0 0
· ·
in a bounded domain Ω Rn with smooth boundary, with given nonnegative initial data u ,v . We
0 0
⊂
shall be concerned with the case of the singular sensitivity function S given by
χ
S(v)= (2)
v
∗Institut für Mathematik, Universität Paderborn, Warburger Str. 100, 33098 Paderborn, Germany; email:
[email protected]
1
for a constant χ > 0, which is in compliance with the Weber-Fechner law of stimulus perception (see
[12]).
One of the first questions of mathematical interest with respect to this model is that of existence of a
global classical solution, as opposed to blow-up of solutions in finite time. For the vast mathematical
literature on chemotaxis, a large part of which is concerned with this question, see one of the survey
articles [9, 10, 7, 1] and references therein.
According to the standardreasoning in the realm of chemotaxis equations (as e.g. formulated in [1]), in
order to obtain global existence of classical solutions, for the two-dimensional case considered here, it is
sufficient to derive t-independent bounds on the quantities u(t)lnu(t) and v(t)2.
Ω |∇ |
To achieve this in the particular context of (1), it has proven useful to consider the expression
R R
ulnu a ulnv, (3)
−
ZΩ ZΩ
asithasbeendonebyNagai,Senba,Yoshida[15]orBiler[2]. Intheseworks,globalexistenceofsolutions
has been derived for χ 1.
≤
In the present article we shall answer the question whether χ = 1 is a critical value in this regard in
the negative. This question had been left open in [20], where the above-mentioned results have been
generalized to higher dimension n, then obtaining existence in the case χ< 2/n.
Let us mention some more results concerning equation (1): That the classical solutions for χ < 2/n
p
are global-in-time bounded has been shown in [4]. In [20] also weak solutions have been shown to exist
p
for (1), as long as χ< n+2 . In the radially symmetric setting, moreover,certain globalweak “power-
3n 4
−
λ-solutions” exist ([17])q.
Related parabolic-elliptic chemotaxis models are investigated, e.g. in [5], where the presence of terms
describing logistic growth is used to ensure global existence and boundedness of classical solutions. In
[6] global existence and boundedness of classical solutions to the parabolic-elliptic counterpart of (1)
are obtained for even more singular sensitivities of the form 0 < S(v) χ , k 1, under a smallness
≤ vk ≥
condition on χ, which for k =1 and n=2 amounts to χ<1.
Also concerning classical solutions of the fully parabolic system (1), to the best of our knowledge, the
assertions for χ 1 are the best known so far.
≤
Since the new possible values for χ are but slightly larger than 1, rather than these values it is the
method that can be consideredthe new contributionof the present article: Key to our approachtoward
the expansion of the interval of values for χ known to yield global solutions, namely, shall be the
employment of an additional summand
b √v 2
|∇ |
Z
in(3). Functionalscontainingthistermhavesuccessfullybeenusedinthecontextofcoupledchemotaxis-
fluid systems (see [21]) or of chemotaxis models with consumption of the chemoattractant [18] (e.g.
obtained from the aforementioned system upon neglection of the fluid).
In the end we will arrive at the following
Theorem 1.1. Let Ω R2 be a convex bounded domain with smooth boundary. Let 0 u C0(Ω),
0
⊂ ≤ ∈
u 0, 0 < v W1,q(Ω). Then there exists χ > 1 such that for any χ (0,χ ) the system (1)
0 6≡ 0 ∈ q>2 0 ∈ 0
has a global classical solution, which is bounded.
S
The plan of the paper is as follows: In the next section we will discuss local existence of and an ex-
tensibility criterion for solutions to (1). Section 3 provides identities and estimates that will facilitate
the usage of the additional term at the center of the proof of Theorem 1.1, to which Section 4 will be
devoted.
2 How to ensure global existence
A general existence theorem for chemotaxis models is the following, taken from [1]:
2
Theorem 2.1. Let n 1 and Ω Rn be a bounded domain with smooth boundary and let q > n. For
some ω ∈(0,1) let S ∈≥Cl1o+cω(Ω×[⊂0,∞)×R2), f ∈C1−(Ω×[0,∞)×R2) and g ∈Cl1o−c(Ω×[0,∞)×R2),
and assume that f(x,t,0,v) 0 for all (x,t,v) Ω [0, )2 and that g(x,t,u,0) 0 for any (x,t,u)
≥ ∈ × ∞ ≥ ∈
Ω [0, )2. Then for all nonnegative u C0(Ω) and v W1,q(Ω) there exist T (0, ] and a
0 0 max
× ∞ ∈ ∈ ∈ ∞
uniquely determined pair of nonnegative functions
u C0(Ω [0,T )) C2,1(Ω (0,T )), (4)
max max
∈ × ∩ ×
v C0(Ω [0,T )) C2,1(Ω (0,T )) L ([0,T );W1,q(Ω)),
∈ × max ∩ × max ∩ ∞loc max
such that (u,v) solves
u =∆u (uS(x,t,u,v) v)+f(x,t,u,v), (5)
t
−∇· ∇
v =∆v v+g(x,t,u,v),
t
−
0=∂ u =∂ v ,
ν ∂Ω ν ∂Ω
| |
u(,0)=u ,v(,0)=v
0 0
· ·
classically in Ω (0,T ) and such that
max
×
if T < , then u(,t) + v(,t) as t T . (6)
max ∞ k · kL∞(Ω) k · kW1,q(Ω) →∞ ր max
Proof. A Banach-typefixedpointargumentprovidesexistence of mildsolutions ona shorttime interval
whose length T depends on u , v . Standard bootstrapping arguments ensure the regularity
k 0k k 0kW1,q
propertieslisted above. It follows∞fromthe dependence of T onthe norms ofu and v thatthe solution
0 0
can be extended to T (0, ] satisfying (6), see [1, Lemma 4.1].
max
∈ ∞
This theoremis not directly applicable to (1), because it does notcoverthe case ofsingularfunctions S.
We will remove this obstruction via use of the following lemma, which is a generalization of Lemma 2.2
of [4].
Lemma 2.2. Let the conditions of Theorem 2.1 be satisfied and let ζ >0.
Then there is η =η(u ,v ,ζ)>0 such that if v and the solution (u,v) to (5) satisfy
0 0 0
infv >0 and inf g(x,s,u(x,s),v(x,s))dx ζ,
0
s∈[0,Tmax)ZΩ ≥
the second component of the solution also fulfils
v(x,t) η for all (x,t) Ω [0,T ).
max
≥ ∈ ×
Proof. Let us fix τ =τ(u ,v )>0 such that
0 0
1
infv(,t) infv for all t [0,τ].
0
Ω · ≥ 2 Ω ∈
Employing the pointwise estimate
(et∆w)(x)≥ (4π1t)n2 e−d42t ZΩw for nonnegative w ∈C0(Ω)
for the Neumann heat semigroup et∆ with d = diamΩ, as provided in [4, Lemma 2.2] following [8,
Lemma 3.1], we can then conclude that
t
v(,t)=et(∆−1)v0+ e(t−s)(∆−1)g(,s,u(,s),v(,s))ds
· · · ·
Z0
≥Z0t (4π(t−1 s))n2 e−4(td−2s)−(t−s)ZΩg(·,s,u(·,s),v(·,s))ds
≥Z0t (4π1r)n2 e−(r+d4r2)drs∈in[0f,t]ZΩg(x,s,u(x,s),v(x,s))dx
3
≥ζZ0τ (4π1r)n2 e−(r+d42r)dr in Ω
for any t∈[τ,Tmax). With η =min{infΩ2v0,ζ 0τ(u0,v0) (4π1r)n2 e−(r+d42r)dr} this proves the claim.
R
With this lemma we can weaken the assumptions on the sensitivity S so as to allow for a singularity at
v =0.
Theorem2.3. i)LetS C1+ω(Ω [0, ) R (0, ))forsomeω (0,1)andapart fromthecondition
∈ loc × ∞ × × ∞ ∈
on S let the assumptions of Theorem 2.1 be satisfied.
Additionally, assume that f is nonnegative and g(x,t,u,v) cu for some c > 0 and any (x,t,u,v)
Ω [0, ) R2 and that inf v > 0 and u =: m > 0. T≥hen there is T > 0 such that (1) has∈a
× ∞ × Ω 0 Ω 0 max
unique solution (u,v) as in (4) and such that (6) holds.
R
ii) Furthermore, if there are K ,K >0 such that f(x,t,u,v) K and g(x,t,u,v) K (1+u) for all
1 2 1 2
≤ ≤
(x,t,u,v) Ω (0, )3, and for every η >0, S is bounded on Ω (0, )2 (η, ), and if n=2 and
∈ × ∞ | | × ∞ × ∞
there is M >0 such that
u(,t)lnu(,t) M, and v(,t)2 M for all t [0,T ) (7)
max
· · ≤ |∇ · | ≤ ∈
ZΩ ZΩ
then (u,v) is global and bounded.
Proof. i)Letη :=η(u ,v ,cm)beasinLemma2.2. Letζ: R [0,1]beasmooth,monotonedecreasing
0 0
function with ζ(η)=1 and ζ(η)=0. Define →
2
S(x,t,u,η), (x,t,u,v) Ω [0, ) R ( ,η),
S (x,t,u,v):= 2 ∈ × ∞ × × −∞ 2
η (ζ(v)S(x,t,u, η)+(1 ζ(v))S(x,t,u,v), (x,t,u,v) Ω [0, ) R [η, ).
2 − ∈ × ∞ × × 2 ∞
Then S C1+ω(Ω [0, ) R2) and S and S agree for v η. Let us denote by (5) problem
η ∈ loc × ∞ × η ≥ η
(5) with S replaced by S . Then we can apply Theorem 2.1 to (5) and obtain a solution (u,v) with
η η
the required properties (4) and (6). Nonnegativity of f and integration of the first equation of (5)
η
entail that u(t) m for all t [0,T ) and accordingly g(x,t,u(x,t),v(x,t))dx cm>0 for all
Ω ≥ ∈ max Ω ≥
t [0,T ). Therefore, by Lemma 2.2, v η and hence (u,v) solves (5) as well.
max
∈ R ≥ R
Inorderto carryoverthe uniquenessstatementfromTheorem5,we ensurethat anysolutionof (5) also
solves (5) in Ω [0,T ): Let v be a solution of (5). Let ε (0,η) and define t =inf t:inf v(t)<
η × max ∈ 2 0 { Ω
ε (0, ]. Then (u,v) solves (5) in (0,t ). Assume t < . Then by Lemma 2.2 and continuity of v,
η 0 0
}∈ ∞ ∞
v(x,t ) η >ε=inf v(,t) for all x Ω, a contradiction.
0 Ω
≥ · ∈
ii) Since (u,v) is a solution of (5) , we can apply [1, Lemma 4.3], which turns (7) into a uniform-
η
in-time bound on u(,t) + v(,t) , thus asserting global existence by means of (6) and
k · kL∞(Ω) k · kW1,q(Ω)
boundedness.
Remark 2.4. Throughout the remaining part of the article, we will assume that Ω R2 is a bounded,
⊂
smooth domain, that 0 u C0(Ω), q >2 and v W1,q(Ω), inf v >0 and u =:m>0.
≤ 0 ∈ 0 ∈ Ω 0 Ω 0
Then, in particular, Theorem 2.3 is applicable to (1). Furthermore, any solution (u,v) of (1) satisfies
R
u(t)=m for all t [0,T ). (8)
max
∈
ZΩ
Forthepurposeofusingitinthenextproof,letusrecallthewell-knownGagliardo-Nirenberginequality:
Lemma 2.5. Let Ω Rn be a bounded smooth domain. Let j 0,k 0 be integers and p,q,r,s > 1.
⊂ ≥ ≥
There are constants c ,c >0 such that for any function w Lq(Ω) Ls(Ω) with Dkw Lr(Ω),
1 2
∈ ∩ ∈
Djw p ≤c1 Dkw αr kwk1q−α+c2kwks,
(cid:13) (cid:13) (cid:13) (cid:13)
whenever 1 = j + 1 k α+(cid:13)1 α a(cid:13)nd j (cid:13)α<1(cid:13).
p n r − n −q k ≤
Proof. Cf. [16, p. 1(cid:0)26] (cid:1)
4
Lemma 2.6. Let (u,v) be a solution to (1), let τ =min 1,Tmax , and assume there exists C >0 such
{ 2 }
that
t+τ u2
|∇ | C for any t (0,T τ)
max
u ≤ ∈ −
Zt ZΩ
and that
u(t)lnu(t) C for any t (0,T )
max
≤ ∈
ZΩ
Then T = and (u,v) is bounded.
max
∞
Proof. Letc ,c >0 be the constantsyieldedby Lemma 2.5for j =0, k =1, q =2, r=2,α= 1. Then
1 2 2
with m from (8),
t+τ u2 = t+τ √u 4L4(Ω) ≤ t+τ c1 ∇√u αL2(Ω) √u 1L−2(αΩ)+c2 √u L2(Ω) 4
Zt ZΩ Ztt+τ(cid:13)(cid:13) (cid:13)(cid:13) u2Zt 14 (cid:16) (cid:13)(cid:13) (cid:13)(cid:13) 4 c(cid:13)(cid:13)4mC(cid:13)(cid:13) (cid:13)(cid:13) (cid:13)(cid:13) (cid:17)
≤Zt c1(cid:18)ZΩ |∇4u| (cid:19) m41 +c2m12! ≤ 14 +c42m2
holds for any t (0,T τ).
max
∈ −
Multiplying the second equation of (1) by ∆v and integrating, from Young’s inequality we obtain
−
t t 1 t 1 t
v(t)2 v 2 ∆v 2 v 2+ u2+ ∆v 2 on (0,T ),
0 max
|∇ | ≤ |∇ | − | | − |∇ | 2 2 | |
ZΩ ZΩ Z0 ZΩ Z0 ZΩ Z0 ZΩ Z0 ZΩ
that is, y(t):= v(t)2 satisfies the differential inequality y +y f on [0,T ), where f = 1 u2
Ω|∇ | ′ ≤ max 2 Ω
t+τ
satisfies f(s)ds C for all t (0,T τ) and with some constant C > 0. Let z be a solution
t | R | ≤ ∈ max− R
to z +z =f, z(0)=z = v 2 and observe that the variation-of-constantsformula entails
′ R 0 Ω|∇ 0|
R
t ⌊t/τ⌋−1 (k+1)τ t
z(t) e−tz0 = e−sf(t s)ds e−s f(t s)ds+ e−s f(t s)ds
− − ≤ | − | | − |
Z0 Xk=0 Zkτ Zτ⌊t/τ⌋
t/τ 1
⌊ ⌋− 1
e kτC+C C 1+ for t (0,T ),
≤ − ≤ 1 e τ ∈ max
k=0 (cid:18) − − (cid:19)
X
so that an ODE comparison yields boundedness of y = v(t)2.
Ω|∇ |
Togetherwiththesecondassumption,theboundon ulnu,thisissufficienttoconcludeglobalexistence
Ω R
and boundedness of solutions by Theorem 2.3 ii).
R
3 Some useful general estimates and identities
Lemma 3.1. Let Ω be convex and let w C2(Ω) satisfy ∂ w = 0. Then for all x ∂Ω also
ν ∂Ω
∈ | ∈
∂ w(x)2 0.
ν
|∇ | ≤
Proof. This is Lemme 2.I.1 of [13].
Lemma 3.2. For all positive w C2(Ω) satisfying ∂ w =0
ν ∂Ω
∈ |
1 1 w4
wD2lnw2 = D2w2+ w2∆w |∇ |
| | w| | w2|∇ | − w3
ZΩ ZΩ ZΩ ZΩ
Proof. This proof is also contained in the proof of [21, Lemma 3.2]. The equality rests on the pointwise
identity
1 1 2 w4 1 1
wD2lnw2 =w w2+ D2w = |∇ | + D2w2 w2 w
| | −w2|∇ | w w3 w| | − w2∇|∇ | ·∇
(cid:18) (cid:19)
5
and integration by parts in the last term giving
1 1 w4
w2 w = w2∆w 2 |∇ | .
− w2∇|∇ | ·∇ w2|∇ | − w3
ZΩ ZΩ ZΩ
Lemma 3.3. Let w C2(Ω) be positive and satisfy ∂ w =0. Then
ν ∂Ω
∈ |
1 1 3 w2∆w 1 w4 1 1
∆w2 = D2w2 |∇ | + |∇ | + ∂ w2.
− w| | − w| | − 2 w2 2 w3 2 w ν|∇ |
ZΩ ZΩ ZΩ ZΩ Z∂Ω
Proof. This results from [21, Lemma 3.1] upon the choice of h(w) = 1. The proof can be found in [3,
w
Lemma 2.3].
Lemma 3.4. (i) For all positive w C2(Ω) satisfying ∂ w =0,
ν ∂Ω
∈ |
1 1 1 1 1
∆w2 = wD2lnw2 w2∆w+ ∂ w2.
− w| | − | | − 2 w2|∇ | 2 w ν|∇ |
ZΩ ZΩ ZΩ Z∂Ω
(ii) If furthermore Ω is convex, then
1 1 1
∆w2 wD2lnw2 w2∆w.
− w| | ≤− | | − 2 w2|∇ |
ZΩ ZΩ ZΩ
Proof. This is a direct consequence of the previous three lemmata.
Lemma 3.5. There is c > 0 such that for all positive w C2(Ω) fulfilling ∂ w = 0 the following
0 ν ∂Ω
∈ |
estimate holds:
w4
wD2lnw2 c |∇ | .
| | ≥ 0 w3
ZΩ ZΩ
Proof. An even more general version of this lemma and its proof can be found in [21, Lemma 3.3].
Remark 3.6. As can be seen from the referenced lemma, the constant in the above statement can be
chosen to be 1 .
(2+√2)2
4 The energy functional. Proof of Theorem 1.1
In this section let us investigate the energy functional defined by
(u(t),v(t))= u(t)lnu(t) a u(t)lnv(t)+b v(t)2, t [0,T ), (9)
a,b max
F − |∇ | ∈
ZΩ ZΩ ZΩ
p
for nonnegative parameters a,b.
If we want to gain useful information from this functional, the upper bounds on its derivative that we
will derive, should be accompanied by bounds for from below. In order to ensure those, let us first
a,b
F
provide the following estimate for solutions of (1).
Lemma 4.1. Let (u,v) be a solution to (1). For any p>0 there is C >0 such that
p
vp(t) C for any t [0,T ).
p max
≤ ∈
ZΩ
Proof. Since t u(t) is constant by (8), for p 1 this is a consequence of Duhamel’s formula
7→ k kL1(Ω) ≥
for the solutionofthe secondequationof (1) andestimatesfor the Neumann heatsemigroup,whichcan
e.g. be found in [19, Lemma 1.3]: They provide C >0 such that for all t (0,T ),
max
∈
t
kv(t)kLp(Ω) ≤(cid:13)et(∆−1)v0(cid:13)Lp(Ω)+Z0 (cid:13)e(t−s)(∆−1)(u(s)−m)(cid:13)Lp(Ω)+(cid:13)e−(t−s)m(cid:13)Lp(Ω)ds
(cid:13) (cid:13) t (cid:13) (cid:13) (cid:13) (cid:13)
≤k(cid:13)v0kLp(Ω)+(cid:13) C(1+(cid:13)(t−s)−n2(1−p1))e−(t−s)(cid:13)ku(s)−m(cid:13)kL1(Ω)+(cid:13)e−(t−s)m|Ω|p1 ds.
Z0 (cid:16) (cid:17)
The case p (0,1) then follows from vp 1+v.
∈ ≤
6
The following lemma gives bounds from below as well as means to turn boundedness of (u,v) into
a,b
F
boundedness of ulnu.
Ω
Lemma 4.2. LeRt a,b 0. For any solution (u,v) of (1), there is γ R such that
≥ ∈
1
(u,v) ulnu γ on (0,T ).
a,b max
F ≥ 2 −
ZΩ
Proof. Denoting m= u(t) as in (8), we have
Ω
R 1 u21 1 va u
(u,v) ulnu+ uln = ulnu+m ln ,
Fa,b ≥ 2ZΩ ZΩ va 2ZΩ ZΩ(cid:18)− u21(cid:19)m
similar as in the proof of [2, Thm. 3]. Hence, following an idea from the proof of [15, Lemma 3.3] in
applying Jensen’s inequality with the probability measure udλ and the convexfunction ln, we obtain
m −
1 vau21
(u,v) ulnu mln
a,b
F ≥2 − m
ZΩ ZΩ
1
1 1 2
ulnu mln v2a u
≥2ZΩ − m(cid:18)ZΩ ZΩ (cid:19) !
1 m m
ulnu+ lnm lnC
2a
≥2 2 − 2
ZΩ
after applying Hölder’s inequality and with C as in 4.1.
2a
Lemma 4.3. Let a,b 0. For any solution (u,v) of (1),
≥
i) (u,v) is bounded below.
a,b
F
ii) If sup (u(t),v(t))< then sup u(t)lnu(t)< .
t∈[0,Tmax)Fa,b ∞ t∈[0,Tmax) Ω ∞
Proof. Both statements are immediate consequences of LeRmma 4.2.
Lemma 4.1 as well enables us to control the first two summands of (u,v) from above by u2.
Fa,b Ω v
Lemma 4.4. Let (u,v) be a solution to (1) and let a>0. Then for any δ >0 there is c >0Rsuch that
δ
u2
ulnu a ulnv δ +c on [0,T ).
δ max
− ≤ v
ZΩ ZΩ ZΩ
Proof. Given a>0 let ε (0,1) be so small that 1+ε 2aε >0. There is C >0 such that for any x>0
∈ 1−ε ε
we have lnx C xε. Therefore for any δ > 0 Young−’s inequality and Lemma 4.1 provide C > 0 and
ε δ
≤
c >0 satisfying
δ
ulnu−a ulnv = ulnvua ≤Cε uv1a+εε ≤δ (u1+εv−1+2ε)1+2ε +Cδ (v1+ε−22aε)1−2ε
ZΩ ZΩ ZΩ ZΩ ZΩ ZΩ
u2
δ +c .
δ
≤ v
ZΩ
With these preparations, we turn to the time derivative of (u,v), beginning with the already inves-
a,b
F
tigated first part:
Lemma 4.5. For any a 0 and any solution (u,v) of (1),
≥
d u2 u v u v 2 u2
(u,v)(t)= |∇ | +(χ+2a) ∇ ·∇ a(χ+1) |∇ | +a u a
dtFa,0 − u v − v2 − v
ZΩ ZΩ ZΩ ZΩ ZΩ
holds on (0,T ).
max
7
Proof. Using the first equation of (1) in d ulnu a ulnv and integrating by parts we obtain:
dt Ω − Ω
(cid:0)R R (cid:1)
d u
ulnu a ulnv = u lnu+ u a u lnv a v
t t t t
dt − − − v
(cid:18)ZΩ ZΩ (cid:19) ZΩ ZΩ ZΩ ZΩ
u u v u u
= ∇ u χ v +a ∇ u χ v a (∆v v+u)
− u ∇ − v∇ v ∇ − v∇ − v −
ZΩ (cid:16) (cid:17) ZΩ (cid:16) (cid:17) ZΩ
u2 u v u v u v 2
= |∇ | +χ ∇ ·∇ +a ∇ ·∇ aχ |∇ |
− u v v − v2
ZΩ ZΩ ZΩ ZΩ
u v u v 2 u2
+a ∇ ·∇ a |∇ | +a u a .
v − v2 − v
ZΩ ZΩ ZΩ ZΩ
Since we do not know the sign of Ω ∇uv·∇v and, in this situation, cannot control Ω u|∇v2v|2, we are left
with Young’s inequality, hoping thRat the resulting coefficient (χ+42a)2 −a(χ+1) oRf Ω u|∇v2v|2 turns out
to be negative. This can be achieved if χ<1.
However,itbecomespossibletocopewithlargerparametersif Ω u|∇v2v|2 canbecontroRlled,e.g. byhaving
control over Ω |∇vv3|4 and Ω uv2. The second term already beinRg in place, fortunately, the first is one of
the terms arising from the following:
R R
Lemma 4.6. Let Ω be convex. For any solution (u,v) of (1),
d v 4 v 2 u v v 2u
4 √v 2 2c |∇ | |∇ | +2 ∇ ·∇ |∇ | (10)
dt |∇ | ≤− 0 v3 − v v − v2
(cid:18)ZΩ (cid:19) ZΩ ZΩ ZΩ ZΩ
holds on (0,T ), where c is the constant provided by Lemma 3.5.
max 0
Proof. From the second equation of (1), we obtain
d d v 2 2 v v v 2v
4 √v 2 = |∇ | = ∇ ·∇ t |∇ | t
dt |∇ | dt v v − v2
(cid:18)ZΩ (cid:19) ZΩ ZΩ ZΩ
2 v ∆v 2 v 2 2 v u v 2∆v v 2v v 2u
= ∇ ·∇ |∇ | + ∇ ·∇ |∇ | + |∇ | |∇ | .
v − v v − v2 v2 − v2
ZΩ ZΩ ZΩ ZΩ ZΩ ZΩ
Integrationby parts in the first integraland merging the secondand second to last summand lead us to
d ∆v 2 v 2∆v v 2 u v v 2u
4 √v 2 = 2 | | + |∇ | |∇ | +2 ∇ ·∇ |∇ | . (11)
dt |∇ | − v v2 − v v − v2
(cid:18)ZΩ (cid:19) ZΩ ZΩ ZΩ ZΩ ZΩ
By Lemma 3.4 and due to the convexity of Ω we can transform the first summand according to
∆v 2 1
2 | | 2 v D2lnv 2 v 2∆v,
− v ≤− | | − v2|∇ |
ZΩ ZΩ ZΩ
making the second term in the right hand side of (11) vanish:
d u v v 2u
4 √v 2 2 v D2lnv 2+2 ∇ ·∇ |∇ | . (12)
dt |∇ | ≤− | | v − v2
(cid:18)ZΩ (cid:19) ZΩ ZΩ ZΩ
We are left with a term we can estimate with the help of Lemma 3.5:
v 4
2 v D2lnv 2 2c |∇ | ,
− | | ≤− 0 v3
ZΩ ZΩ
thereby gaining the term which will make the crucialdifference in the estimates to come and arrivingat
d v 4 v 2 u v v 2u
4 √v 2 2c |∇ | |∇ | +2 ∇ ·∇ |∇ | .
dt |∇ | ≤− 0 v3 − v v − v2
(cid:18)ZΩ (cid:19) ZΩ ZΩ ZΩ ZΩ
8
If we combine the previous two lemmata, we are led to:
Lemma 4.7. Let Ω R2 be a convex, bounded, smooth domain and let a,b 0, δ (0,1). Then for
⊂ ≥ ∈
any solution (u,v) of (1),
2
d 1 (χ+2a+ b)2 b bc v 4
(u,v)(t) 2 aχ a 0 |∇ |
dtFa,b ≤4a(1−δ) 4(1−δ) − − − 4!+− 2 ZΩ v3
 u2 u2 b v 2 
δ |∇ | δ +a u |∇ | on (0,T ). (13)
max
− u − v − 4 v
ZΩ ZΩ ZΩ ZΩ
Proof. An estimate for d (u(t),v(t)) is given by the sum of the terms from Lemma 4.5 and Lemma
dtFa,b
4.6:
d u2 b u v b u v 2
(u,v)(t) |∇ | + χ+2a+ ∇ ·∇ aχ+a+ |∇ |
dtFa,b ≤− u 2 v − 4 v2
ZΩ (cid:18) (cid:19)ZΩ (cid:18) (cid:19)ZΩ
u2 b v 4 b v 2
+a u a c |∇ | |∇ | .
− v − 2 0 v3 − 4 v
ZΩ ZΩ ZΩ ZΩ
In order to finally still have some control over |∇u|2, as required for Lemma 2.6, we retain a small
u
portion of this term when applying Young’s inequality:
R
u2 b u v u2 (χ+2a+ b)2 u v 2
|∇ | + χ+2a+ ∇ ·∇ 1+(1 δ) |∇ | + 2 |∇ | ,
− u 2 v ≤ − − u 4(1 δ) v2
ZΩ (cid:18) (cid:19)ZΩ ZΩ − ZΩ
(cid:0) (cid:1)
so that
d u2 (χ+2a+ b)2 b u v 2
(u,v)(t) δ |∇ | + 2 aχ a |∇ |
dtFa,b ≤− ZΩ u 4(1−δ) − − − 4!ZΩ v2
u2 b v 4 b v 2
+a u a c |∇ | |∇ | .
− v − 2 0 v3 − 4 v
ZΩ ZΩ ZΩ ZΩ
By virtue of the presence of − Ω |∇vv3|4, which originates from the additional summand of the energy
functionalandthe preparationsRofSection3,wecancontinueestimating Ω u|∇v2v|2 by Ω uv2 and Ω |∇vv3|4
andstillhopefornegativecoefficientsinfrontoftheintegrals,incontrasttothe situationofLemma4.5.
In doing so we keep some part of u2 for the sake of a later applicationRof Lemma 4R.4 and arrRive at
Ω v
R 2
d u2 u2 1 (χ+2a+ b)2 b v 4
(u,v)(t) δ |∇ | +a(1 δ) + 2 aχ a |∇ |
dtFa,b ≤− ZΩ u − ZΩ v 4a(1−δ) 4(1−δ) − − − 4!+ZΩ v3
u2 b v 4 b v 2
+a u a c |∇ | |∇ | ,
− v − 2 0 v3 − 4 v
ZΩ ZΩ ZΩ ZΩ
which amounts to (13).
Lemma 4.8. Let a>0, b 0, χ>0 be such that
≥
2
1 (χ+2a+ b)2 b bc
ϕ(a,b;χ):= 2 aχ a 0 <0, (14)
4a 4 − − − 4! − 2 
+
 
and let (u,v) be a solution of (1). Then there are κ,δ >0 and c>0 such that for any t (0,T ),
max
∈
d u(t)2
(u,v)(t)+κ (u,v)(t)+δ |∇ | c.
a,b a,b
dtF F u(t) ≤
ZΩ
9
Proof. By continuity of
2
1 (χ+2a+ b)2 b bc
δ ϕ (a,b;χ):= 2 aχ a 0
δ
7→ 4a(1 δ) 4(1 δ) − − − 4! − 2 
− − +
 
in δ = 0, for fixed a,b,χ, negativity of ϕ(a,b;χ) entails the existence of δ > 0 so that ϕ (a,b;χ) is
δ
negative as well. Therefore, by Lemma 4.7,
d u2 u2
(u,v)+δ |∇ | +δ +b √v 2 a u on (0,T ).
a,b max
dtF u v |∇ | ≤
ZΩ ZΩ ZΩ ZΩ
Since u is constant in time by (8), Lemma 4.4 implies the assertion.
Ω
LemmRa 4.9. If
χ χ>0;there are a>0 and b 0 such that ϕ(a,b;χ)<0 =:M,
0
∈ ≥
n o
then
(0,χ ) M.
0
⊂
Proof. Since, for any fixed a>0, b 0,
≥
1 b2 2
χ ϕ(a,b;χ)= χ2+4a2+ +bχ+2ab 4a b 32abc
0
7→ 64a (cid:18) 4 − − (cid:19)+− !
is monotone, for any a>0, b 0
≥
ϕ(a,b;χ )<0 implies ϕ(a,b;χ)<0 for any 0<χ<χ .
0 0
Lemma 4.10. There is χ >1 such that ϕ(a,b;χ )<0 for some a>0,b>0.
0 0
Proof. Since ϕ(1,0,1)=0 and
2
d 1 d 1 b2 2 1 1 b2 b 1 c
0
ϕ ,b,1 = +b c b = ( +b)( +1) c = <0,
0 0
db (cid:18)2 (cid:19)(cid:12)(cid:12)b=0 db 32(cid:18) 4 (cid:19) − 2 !(cid:12)(cid:12)b=0 (cid:20)16 4 2 − 2 (cid:21)(cid:12)(cid:12)b=0 − 2
(cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12)
thereis b>0 su(cid:12)chthatϕ(1,b,1)<0 andby continu(cid:12)ityofϕwith respectto χ,the asse(cid:12)rtionfollows.
2
Proof of Theorem 1.1. By Lemma 4.10, there are a,b > 0, χ > 1 such that ϕ(a,b,χ ) < 0 and hence,
0 0
by Lemma 4.9, also ϕ(a,b,χ)<0 for χ (0,χ ). An application of Lemma 4.8 thus reveals that for all
0
∈
t>0
d u2
(u,v)(t)+κ (u,v)(t)+δ |∇ | c (15)
a,b a,b
dtF F u ≤
ZΩ
forsomeκ,δ,c>0. Togetherwiththeboundednessof (u,v)frombelowbyLemma4.3i)thisensures
a,b
F
that Fa,b(u,v) is bounded so that an integration of (15) also shows the boundedness of tt+1 Ω |∇uu|2.
Since (u,v) is bounded, by Lemma 4.3 ii) the same holds true for ulnu and so the conditions of
Fa,b Ω R R
Lemma 2.6 are met and Theorem 1.1 follows.
R
Remark 4.11. Assuming c = 1 , as permitted by Remark 3.6,
0 (2+√2)2
1.1 10 5 ϕ(0.49,0.001;1.015)<0,
−
− · ≈
i.e. χ >1.015.
0
10