Rate of decay to 0 of the solutions to a nonlinear parabolic equation
I.Ben Arbi
1Résumé: On étudie l'ordre de convergence vers 0, quand t −→ +∞ de la solution de l'équation ψt−∆ψ+|ψ|p−1ψ = 0 avec les conditions aux limites de Neumann dans un ouvert connexe borné de Rn où p > 1. On montre que soit ψ(t,·)converge vers 0exponentiellement, soit ψ(t,·)décroît comme t−(p−1)1 .
Abstract: We study the decay rate to 0, as t −→ +∞ of the solution of equation ψt−∆ψ+|ψ|p−1ψ = 0 with Neumann boundary conditions in a bounded smooth open connected domain of Rn where p >1. We show that eitherψ(t,·)converges to 0exponentially fast orψ(t,·)decreases liket−(p−1)1 .
Keywords: rate of decay, parabolic equation
1Laboratoire Jacques-Louis Lions, Université Pierre et Marie Curie - Paris 6, boîte courrier 187, 4 place Jussieu, 75252 Paris Cedex 05, FRANCE.
1 Introduction and main results.
In this paper we consider the following nonlinear parabolic equation ( ψt−∆ψ+g(ψ) = 0 in R+×Ω,
∂ψ
∂n = 0 on R+×∂Ω. (1.1)
where Ω is a bounded smooth open connected domain of Rn and g ∈C1(R) satises
g(0) = 0 (1.2)
and for some p > 1
∃c >0,∀s∈R, 0≤g0(s)≤c|s|p−1. (1.3) From (1.2)-(1.3)we deduce that g(s) has the sign ofs and
∀s∈R, |g(s)| ≤ c
p|s|p. (1.4)
We dene the operator A by
D(A) ={ψ ∈H2(Ω),∂ψ
∂n = 0 on ∂Ω}
and
∀ψ ∈D(A), Aψ=−∆ψ On the other hand the operator B dened by
D(B) = {ψ ∈L2(Ω),−∆ψ+g(ψ)∈L2(Ω) and ∂ψ
∂n = 0 on ∂Ω}
and
∀ψ ∈D(B), Bψ =−∆ψ+g(ψ)
is well-known to be maximal monotone in L2(Ω). As a consequence of [1, 2]
for any ψ0 ∈L2(Ω) there exists a unique weak solution of the equation ψ0+Bψ= 0 on R+; ψ(0, x) = ψ0. (1.5) In addition it is well known that if ψ0 ∈L∞(Ω),ψ(t,·)remains inL∞(Ω)for allt >0. Finally [8] contains an estimate of the solution in C( ¯Ω)and C1( ¯Ω) for t >0, which is valid for any suciently regular domain.
Concerning the behaviour for t large, in [5], A.Haraux established in the case of a pure power nonlinearity the exponential convergence to0of the pro- jection on the range of A of the solution of equation (1.1). Moreover in [4], the study of the equation u00+u0−∆u+g(u) = 0 with Neumann boundary conditions and where g satises ∃C, c >0,∀s ∈R, c|s|p−1 ≤g0(s)≤ C|s|p−1 for some p >1, showed that either u(t)converges to 0 exponentially fast, or ku(t)kH1
0(Ω) ≥γt−1/(p−1) with γ >1 for t≥1.
Several authors have treated some variants of equation (1.1), for example in [6] with g(u) = c|u|p−1u−λ1u and with Dirichlet boundary conditions, was studied the decay rate at the innity of solutions to (1.1), where λ1 >0 is the rst eigenvalue of −∆ inH01(Ω). The result obtained there is optimal for positive solutions.
According to La Salle's invariance principle, cf. [3, 7], any solution ψ of (1.1), having a precompact range on R+ with values to L∞(Ω), converges to a continuum of stationary solutions of equation (1.1), which reduces here to the constants of some sub-interval of g−1(0). By monotonicity, it is in fact known that ψ converges to some constant a∈g−1(0).
Our rst result is valid without any additional hypothesis ong
Theorem 1.1. Let g satisfy (1.2) and (1.3). Then any solution ψ of (1.1) satises the following alternative as t → ∞: either
kψ(t,·)k∞≤Ce−λ2t, (1.6) or
∃c0 >0,∀t ≥1,
Z
Ω
ψ(t, x)dx
≥c0t−p−11 , (1.7) where λ2 >0 is the second eigenvalue of A in D(A).
Our second result provides a more accurate estimate wheng(ψ) = |ψ|p−1ψ Theorem 1.2. Let us consider the nonlinear parabolic problem (1.1) with g(ψ) =|ψ|p−1ψ, then any solution ψ of (1.1) satises the following alterna- tive as t → ∞: either
kψ(t,·)k∞≤Ce−λ2t, (1.8) or
∀t ≥1, k|ψ(t,·)| −((p−1)t)−p−11 k∞ ≤Kt−(p−1)1 −1, (1.9) where K, C >0, p >1 and λ2 >0 is the second eigenvalue of A in D(A).
In the following Proposition, we consider two special cases showing that both possibilities in the second result in the Theorem 1.1 can actually happen.
Proposition 1.3. Let g satisfy (1.2) and (1.3). Then we have
(i) If Ω is symmetric around 0, g is odd and ψ(0,·) is an odd function in Ω, then any solution of (1.1) satises (1.6).
(ii) Ifψ(0,·)≥0andψ does not vanish a.e inΩ, then any solution of (1.1) satises (1.7).
Finally, our last result shows that the second possibility is sharp for a class of functions g more general than the pure power
Proposition 1.4. Under the additional hypothesis
∃k1 >0,∀s∈R,|g(s)| ≥k1|s|p (1.10) for any solution ψ of (1.1), we have
∀t ≥1, kψ(t,·)k∞ ≤
1 k1(p−1)
p−11
t−p−11 (1.11)
2 Proof of Proposition 1.4.
Proof. If ψ(0,·) = 0 , we have ψ(t,·)≡ 0 and the result is obvious. Other- wise let z be dened by
z(t) =
1
kψ(0,·)k1−p∞ +k1(p−1)t p−11
Then z is a solution of the following nonlinear ODE problem z0+k1zp = 0,
z(0) =kψ(0,·)k∞. (2.1)
Under the additional condition(1.10), we will show that z is a super solution of (1.1). Indeed, we have
zt−∆z+g(z) = −k1(z(0)1−p+k1(p−1)t)−p−1p +g(z)
≥ −k1(z(0)1−p +k1(p−1)t)−p−1p +k1|z|p
≥0.
Since ψ(0,·) ≤ z(0) we deduce, by standard comparison principle, that ψ(t,·)≤z(t) ∀t≥1.
A similar calculation shows that ψ(t,·) ≥ −z(t), ∀t ≥ 1 which concludes the proof.
3 A general result on the range component.
Dening the orthogonal projection P :H −→N, where H =L2(Ω), N = ker(A) and P ψ(t,·) = 1
|Ω|
Z
Ω
ψ(t, x)dx,
as already mentioned in the introduction, [5] showed that forg(ψ) = |ψ|p−1ψ, the following estimate holds
kψ(t)−P ψ(t)kL2(Ω) ≤Ke−λ2t,
for some constant K > 0. In this section, we will show that we have the same result for any function g satisfying (1.3).
Proposition 3.1. Letψ ∈C(R+, L∞)be any solution of (1.1). Assume that g is a locally Lipschitz non-decreasing function. Then we have ,
kψ(t)−P ψ(t)k2 ≤ kψ(0)−P ψ(0)k2 e−λ2t, (3.1) where k.k2 denotes the norm in L2(Ω) and λ2 > 0 is the second eigenvalue of −∆ in L2(Ω) with Neumann boundary conditions .
Proof. We denote by (u, v)the inner product of two functionsu, v ofL2(Ω). Sinceg is a nondecreasing function, for all ψ ∈L∞(Ω), we have a.e. in x∈Ω
(g(ψ)−g(P ψ))(ψ−P ψ)≥0 and then by integrating over Ω
(g(ψ), ψ−P ψ)−(g(P ψ), ψ−P ψ)≥0. (3.2) Since g(P ψ)is a constant and (ψ−P ψ)∈N⊥, we deduce that(g(P ψ), ψ− P ψ) = 0. Hence from (3.2),
(g(ψ), ψ−P ψ)≥0.
Setting
w=ψ−P ψ, we have since ∆P ψ =P∆ψ = 0
w0−∆w=ψ0 −∆ψ−P ψ0+ ∆P ψ = (I−P)(ψ0−∆ψ) =−(I−P)g(ψ).
Thus , since(w,(I−P)g(ψ)) = ((I−P)w, g(ψ)) = (w, g(ψ)) = (g(ψ), ψ−P ψ) we nd
1 2
d
dtkw(t)k22 = (w,∆w)−(g(ψ), ψ−P ψ)≤ −λ2|w|2. By integrating we obtain (3.1).
4 Proof of Theorem 1.1.
We set ψ =u+w, where u=P ψ and w= (I−P)ψ.By projecting (1.1) on N we obtain
u0+P(g(ψ)) = 0, (4.1)
where we have used that P(Aψ) = 0, sinceR(A)⊂N⊥. Noticing that u0+P(g(u)) +P(g(ψ)−g(u)) =u0+g(u) +P(g(ψ)−g(u)), we can rewrite the equation (4.1) as
u0+g(u) =−P(g(ψ)−g(u)).
By the assumption (1.3), we deduce that
|P(g(ψ)−g(u))| ≤ 1
|Ω|kg(ψ)−g(u)k1 ≤ c
|Ω|(kψkp−12p−2+kukp−12p−2)kwk2. But ψ and u are uniformly bounded and from Proposition 3.1 we have the estimate kw(t)k2 ≤Ke−λ2t. Therefore
|P(g(ψ)−g(u))| ≤K0e−λ2t, with K0 >0. Which leads us to study the equation:
u0 +g(u) = f(t) in R+, (4.2)
where
f(t) =P(g(ψ)−g(u)) and
|f(t)| ≤K0e−λ2t.
Using the same method as in [5], we show the following result:
Lemma 4.1. Let c > 0, γ > 0, p > 1 and g satisfying (1.2) and (1.3) Let M > 0 such that
M ≤γ 2c
p−11
, (4.3)
c1 >0 with
c1 ≤ γ 2M, then for every function f satisfying
|f(t)| ≤c1e−γt,
there exists a unique function v ∈C1(R+) satisfying
∀t≥0, v0+g(v) =f(t) (4.4) and
sup
t∈(0,+∞)
eγt|v(t)| ≤M . (4.5) Proof. Since any solution of (4.4).(4.5) satises the integral equation
v(t) =− Z +∞
t
(f(s)−g(v(s))ds. (4.6) We look for a solution of (4.6). It is then natural to introduce the following function space :
X ={v ∈C(0,+∞); sup
t∈(0,+∞)
eγt|v(t)| ≤M}, equipped with the distance associated to the norm
kvkγ = sup
t∈(0,+∞)
eγt|v(t)|. (4.7) We consider the operator T :X →C(0,+∞)dened by
Tv(t) = − Z +∞
t
(f(s)−g(v(s))ds.
From (1.4), we have the estimate
∀s∈R+; |g(v(s))| ≤ c
p|v(s)|p.
First we will show that T(X)⊂X. Let v ∈X, then for all t≥0,
|Tv(t)| ≤ Z +∞
t
|f(s)|ds+ Z +∞
t
|g(v(s))|ds
≤ Z +∞
t
|f(s)|ds+ c p
Z +∞
t
|v(s)|pds
≤ c1
γe−γt+ c pMp
Z +∞
t
e−pγsds
≤ c1
γ + cMp p2γ
e−γt
≤ M
2 +M cMp−1 p2γ
e−γt
≤ M
2 + M 2p2
e−γt.
Since p > 1, it follows that
|Tv(t)| ≤M e−γt. Hence by (4.5), we obtain that Tv ∈X, with
kTv(t)kγ ≤M.
Secondly, we will prove that T is a contraction on X. In fact, for x,x¯∈ X and for all t≥0
|Tx(t)− Tx(t)| ≤¯ Z +∞
t
|g(x(s))−g(¯x(s))|ds
≤cMp−1 Z +∞
t
e−pγseγs|x(s)−x(s)|ds¯
≤ cMp−1
pγ kx−xk¯ γe−γt. Then we have
|Tx(t)− Tx(t)|e¯ γt≤ cMp−1
pγ kx−xk¯ γ.
Therefore, since Mp−1 satises (4.3), we conclude that∀x,x¯∈X kTx− Txk¯ γ ≤ 1
2kx−xk¯ γ.
Thus T is a 12-Lipschitz functional on the complete metric spaceX and the result follows from the Banach xed point theorem. From (4.6) it follows eas- ily thatvsatises (4.4). Then the uniqueness ofvfollows from the uniqueness of the solution of (4.6) (T is a contraction) and the fact that any solution of (4.4) satises (4.6). The existence comes from the fact that conversely any solution of (4.6) satises (4.4).
Proof of Theorem 1.1 continued. Consequently, we have a solution v that satises the equation (4.4) and for all t ≥T0
|v(t)| ≤M e−λ2t, (4.8) where M =M0eγT0 and M0 >0. If we subtract (4.4) from (4.2) we obtain
(u−v)0+g(u)−g(v) = 0.
Settingz =u−v,we complete the proof analyzing two cases.
1st case: If z(T0) = 0, then for all t ≥ T0, z(t) = 0. Hence u ≡ v and from (4.8) it follows that
|u(t)| ≤M e−λ2t. Then, using (3.1), we obtain
kψ(t)k2 ≤M0e−λ2t. Finally by reasoning as in [5], [6] we obtain (1.6).
2nd case: If z(T0)6= 0 then ∀t≥T0, z(t)6= 0 and we have z0(t) + g(u(t))−g(v(t))
u(t)−v(t) z(t) = 0.
Since g is a monotonic function,
α(t) := g(u(t))−g(v(t)) u(t)−v(t) is a strictly positive function. Moreover, ∃θ ∈]0,1[,
α(t) = g0(θu(t) + (1−θ)v(t))≤c|θu(t) + (1−θ)v(t)|p−1 (4.9) We distinguish 2 cases:
p >2 then by convexity of the pth power we have
|θu(t) + (1−θ)v(t)|p−1 ≤θ|u|p−1+ (1−θ)|v|p−1 ≤ |u|p−1+|v|p−1.
1< p <2we study the function(x+y)a−xa for 0< a < 1 and x, y >0 we prove that X −→ (1 +X)a −Xa is a decreasing function on (0,+∞), we deduce 0 < (1 + X)a − Xa < 1 and it follows by homogeneity that (x+y)a < xa+ya by letting X = xy. Then we conclude that |θu(t) + (1− θ)v(t)|p−1 ≤ |u|p−1 +|v|p−1.Consequently we obtain
∀p >1, α(t)≤c(|u|p−1 +|v|p−1). (4.10) Setting y=|z|, we have that
y0 +α(t)y ≥0.
Then the estimate (4.10) implies that
y0 ≥ −c(|u|p−1+|v|p−1)y≥ −c(|z+v|p−1+|v|p−1)y.
Hence there exists some constants c2, c3 >0 such that y0 ≥ −c2(|z|p−1)y−c3|v|p−1y.
Since y=|z|,
y0 ≥ −c2yp−c3|v|p−1y.
Putting a(t) = c3|v|p−1, we deduce that
y0 +a(t)y ≥ −c2yp. (4.11) We set
A(t) =−c3 Z +∞
t
|v|p−1ds, ω(t) =eA(t)y and by replacing ω in (4.11), we obtain
ω(t)≥
1
ω(0)1−p+ (p−1)c4t p−11
with c4 = R+∞
0 a(s)ds
−(p−1)
. Then for t large enough we have ω(t)≥Kt−p−11 .
Since t −→eA(t) is a bounded function, we conclude (1.7) by observing that u=z+v and v tends to0 exponentially at innity.
5 Proof of Theorem 1.2.
Considering g(ψ) = |ψ|p−1ψ , g satises (1.2) and (1.3) with c = p. Hence Lemma 4.1 is applicable with c=p, therefore we assume
M ≤ λ2
2p p−11
.
In (4.2), we replaceg(ψ) = |ψ|p−1ψ , we can subtract (4.4) from (4.2), we deduce:
(u−v)0+|u|p−1u− |v|p−1v = 0. (5.1) We will study two cases.
1st case: Ifz(T0) = 0 then for all t≥T0, z(t) = 0. Hence u≡v and from (4.8) it follows that
|u(t)| ≤M e−λ2t. Moreover, using (3.1), we obtain (1.8).
2nd case: if z(T0)6= 0 then for all t≥T0 z(t)6= 0 and we have z0(t) +|u(t)|p−1u(t)− |v(t)|p−1v(t) =z0(t) +α(t)z(t) = 0.
with
α(t) = |u(t)|p−1u(t)− |v(t)|p−1v(t) u(t)−v(t)
= |z(t) +v(t)|p−1(z(t) +v(t))− |v(t)|p−1v(t) z(t)
=
|z(t)|p−1|1 + v(t)z(t)|p−1(z(t) +v(t))− |v(t)|p−1v(t)
z(t) .
In that case α(t) > 0, indeed t 7→ |u(t)|p−1u(t) is non decreasing function.
By applying Taylor's formula, we obtain
α(t) =|z(t)|p−1+β(t), (5.2) with|β(t)| ≤Be−ηt,whereη >0is any positive number smaller than(p−1)λ2 and B >0. Replacing α by its expression in (5.2), equation (5.1) becomes
z0+|z|p−1z+β(t)z = 0. (5.3)
Let y=|z|,we obtain
y0+yp+β(t)y = 0. (5.4)
Setting ξ(t) = eA(t)y(t), with A(t) = −R+∞
t β(s)ds we nd
|A(t)| ≤ Z +∞
t
|β(s)|ds≤ B
ηe−ηt. (5.5)
By Taylor's formula we have
∀h∈[−1,1], |eh−1| ≤2|h|, (5.6) we can give the following estimate
|ξ(t)−y(t)| ≤y(t)|eA(t)−1| ≤2|A(t)|y(t)≤ 2B
η kyk∞e−ηt, we conclude
|ξ(t)−y(t)| ≤ke−δt, (5.7) where δ=η and k = 2Bη kyk∞ .
Replacing ξ(t) in (5.3), we have
−e−A(t)ξ0(t) =e−pA(t)ξp(t)
⇔ −ξ0(t)
ξp(t) =e(1−p)A(t)
⇔( 1
ξp−1(t))0 = (p−1)e(1−p)A(t).
t7→e−(p−1)A(t) is bounded and tends to 1at innity, ξ is given by
ξ(t)p−1 = 1
ξ(t0)1−p+ (p−1)Rt
t0e−(p−1)A(s)ds. we set
D(t) =ξ(t0)1−p+ (p−1)Rt
t0e−(p−1)A(s)ds and h(t) = −(p−1)A(t), then we show thatD(t)−(p−1)t is bounded.
From (5.5), we know that t 7→ h(t) is an integrable function, then by (5.6), (e−(p−1)A(t)−1) is also integrable. In order to show (1.9), we proceed
as follows
|D(t)−(p−1)t|=|ξ(t0)1−p+ (p−1) Z t
t0
e−(p−1)A(s)ds−(p−1)t|
=|ξ(t0)1−p+ (p−1) Z t
t0
(e−(p−1)A(s)−1)ds−(p−1)t0|
Using (5.6) we obtain
|D(t)−(p−1)t| ≤ |K|+ 2(p−1)2 Z t
t0
|A(s)|ds
≤M1 =|K|+ 2(p−1)2B η2
with K =ξ(t0)1−p−(p−1)t0. Setting d(t) = D(t)−(p−1)t, we obtain
ξ(t)−
1 (p−1)t
(p−1)1
=
1
D(t) (p−1)1
−
1 (p−1)t
(p−1)1
=
1
D(t)−(p−1)t+ (p−1)t (p−1)1
−
1 (p−1)t
(p−1)1
=
1 d(t) + (p−1)t
(p−1)1
−
1 (p−1)t
(p−1)1
=
1 (p−1)t
(p−1)1
1
d(t) (p−1)t+ 1
!(p−1)1
−
1 (p−1)t
(p−1)1
=
1 (p−1)t
(p−1)1
d(t) (p−1)t + 1
−(p−1)1
−1
=
1 (p−1)t
(p−1)1 1−
d(t) (p−1)t + 1
−(p−1)1 ! .
Since | (p−1)td(t) |< 1 for t large enough. Let η(t) =
1 + (p−1)td(t) −(p−1)1
, by the mean value Theorem and if we suppose that |(p−1)td(t) | ≤ 12, we obtain
|η0(t)| ≤ 1
p−1 ×21+p−11 .
Therefore
|1−
1 + d(t) (p−1)t
−(p−1)1
| ≤ 1
p−1 ×21+p−11 | d(t) (p−1)t|.
As we have seen above, d(t) is bounded byM1 then we conclude
ξ(t)−
1 (p−1)t
(p−1)1
≤Ct−1−p−11 .
With C = (p−11 )p−1p ×21+p−11 M1.
We recall that z has a constant sign on [T0,+∞[ and z and u have the same sign. As in Section 4, we set u=v+z and ψ =u+w, we distinguish two cases
- If z >0, this implies that for t large enoughu >0and |ψ|=ψ.
Then
||ψ| −
1 (p−1)t
(p−1)1
≤ |u−
1 (p−1)t
(p−1)1
+|w| ≤Kt−1−p−11 . Indeed,
|u−
1 (p−1)t
(p−1)1
≤ |u−z|+|z−ξ|+|ξ−
1 (p−1)t
(p−1)1
≤M e−λ2t+ke−δt+Ct−1−p−11 .
Since we have (5.7), we obtain (1.9).
- If we suppose that z <0, then implies that u <0. By similar calcula- tions we obtain the same result.
Indeed, |ψ|=−ψ, and we have
||ψ| −
1 (p−1)t
(p−1)1
≤ | −u−
1 (p−1)t
(p−1)1
+|w|
≤Kt−1−p−11
since
| −u−
1 (p−1)t
(p−1)1
≤ |u−z|+| −z−ξ|+|ξ−
1 (p−1)t
(p−1)1
≤M e−λ2t+ke−δt+Ct−1−p−11 .
Also since we have (5.7), nally we obtain (1.9).
6 Proof of Proposition 1.3
(i) If ψ is an odd function, then ψ(0,−x) = −ψ(0, x). It implies for all t >0, ψ(t,−x) = −ψ(t, x).
In that case
u(t) = P ψ(t, x) = 1
|Ω|
Z
Ω
ψ(t, x)dx= 0.
Moreover, we know that ψ =u+w, wherew(t)∈N⊥ for all t≥0and (cf. Proposition 3.1) we have
kw(t)kL∞(Ω) ≤Ke−λ2t, hence the solution ψ satises (1.8).
(ii) If ψ(0, x) ≥ 0 and ψ does not vanish a.e in Ω, then for all t ≥ 0 ψ(t, x)>0, it implies that
Z
Ω
ψ(t, x)dx >0. (6.1)
We suppose that we have kψ(t,·)k∞ ≤ Ce−λ2t and we consider the problem (1.1) then we integrate on Ω, we obtain
Z
Ω
ψt(t, x)dx=− Z
Ω
g(ψ(t, x))dx. (6.2)
An elementary calculation shows that we have Z
Ω
g(ψ(t, x))dx≤ c p
Z
Ω
|ψ(t, x)|pdx
≤ c p
Z
Ω
|ψ(t, x)|p−1ψ(t, x)dx
≤ c p
Z
Ω
kψ(t, x)kp−1∞ ψ(t, x)dx
≤ c
pCp−1e−(p−1)λ2t Z
Ω
ψ(t, x)dx Now we set y(t) =R
Ωψ(t, x)dx.From (6.2), we deduce
y0(t)≥ −M e−δty(t) (6.3) with M = pcCp−1 and δ= (p−1)λ2.
Sincey(t)>0by (6.1), we can integrate in the interval[0, t], we obtain y(t)≥y(0) exp
−M Z t
0
e−δsds
≥y(0) exp
−M δ
>0. (6.4) Hence y does not tend to0 for t large, this contradicts our hypothesis and we conclude that y satises (1.7).
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