# General Asymptotic Supnorm Estimates for Solutions of One-Dimensional Advection-Diffusion Equations in Heterogeneous Media

Authors: José A. Barrionuevo; Lucas S. Oliveira; Paulo R. Zingano
Journal: International Journal of Partial Differential Equations
 (2014)

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## Abstract

We derive general bounds for the large time size of supnorm values∥u(·,t)∥L∞(ℝ) of solutions to one-dimensional advection-diffusion equations ut+(b(x,t)u)x=uxx,x∈ℝ,t>0 with initial data u(·,0)∈Lp0(ℝ)∩L∞(ℝ) for some 1≤p0<∞ and arbitrary bounded advection speeds b(x,t), introducing new techniques based on suitable energy arguments. Some open problems and related results are also given.


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## Body

## 1. Introduction

In this work, we obtain very general large time estimates for supnorm values of solutionsu(·,t) to parabolic initial value problems of the form(1a)ut+(b(x,t)u)x=uxx,x∈ℝ,t>0,(1b)u(·,0)=u0∈Lp0(ℝ)∩L∞(ℝ),1≤p0<∞,for arbitrary continuously differentiable advection fields b∈L∞(ℝ×[0,∞[). Here, bysolution to (1a) and (1b) in some time interval [0,T*[,0<T*≤∞, we mean a function u:ℝ×[0,T*[→ℝ which is bounded in each strip ST=ℝ×[0,T], 0<T<T*, solves (1a) in the classical sense for 0<t<T*, and satisfies u(·,t)→u0 inLloc⁡1(ℝ) as t→0. It follows from the a priori estimates given in Section 2 that all solutions of problem (1a), (1b) are actually globally defined (T*=∞), with u(·,t)∈C0([0,∞[,Lp(ℝ)) for each p≥p0 finite. Given b∈L∞(ℝ×[0,∞[), what then can be said about the size of supnorm values ∥u(·,t)∥L∞(ℝ) for t≫1?When∂b/∂x≥0 for all x∈ℝ,t≥0, it is well known that, for each p0≤p≤∞, ∥u(·,t)∥Lp(ℝ) is monotonically decreasing in t, with
(2)∥u(·,t)∥L∞(ℝ)≤K(p0)∥u0∥Lp0(ℝ)t-1/2p0∀t>0(bx≥0)
for some constant 0<K(p0)<2-1/p0 that depends only on p0; see, for example, [1–5]. For general b(x,t), however, estimating ∥u(·,t)∥L∞(ℝ) is much harder. To see why, let us illustrate with the important case p0=1, where one has
(3)∥u(·,t)∥L1(ℝ)≤∥u0∥L1(ℝ)∀t>0,
as recalled in Theorem 1. Writing (1a) as
(4)ut+b(x,t)ux=uxx-bx(x,t)u,
we observe on the right hand side of (4) that |u(x,t)| is pushed to grow at points (x,t) where bx(x,t)<0. If this condition persists long enough, large values of |u(x,t)| might be generated, particularly at sites where -bx(x,t)≫1. Now, because of constraint (3), any persistent growth in solution size will eventually create long thin structures as shown in Figure 1, which, in turn, tend to be effectively dissipated by viscosity. The final overall behavior that ultimately results from such competition is not immediately clear, either on physical or on mathematical grounds.Figure 1
Solution profiles showing typical growth in regions withbx<0, where b=5cos⁡x. After reaching maximum height, solution starts decaying very slowly due to its spreading and mass conservation (decay rate is not presently known).As shown in (4), it is not the magnitude of b(x,t) itself but instead itsoscillation that is relevant in determining ∥u(·,t)∥L∞(ℝ). Accordingly, we introduce the quantity B(t) defined by
(5)B(t)=12(sup⁡x∈ℝb(x,t)-inf⁡x∈ℝb(x,t)),t≥0,
which plays a fundamental role in the analysis. Our main result is now easily stated.Main Theorem.For eachp≥p0, one has1(6)limsup⁡t→∞∥u(·,t)∥L∞(ℝ)≤(332πp)1/p·ℬ1/p·limsup⁡t→∞∥u(·,t)∥Lp(ℝ),whereℬ=limsup⁡t→∞B(t).In particular, in the important casep0=1 considered above, we obtain, using (3),
(7)limsup⁡t→∞∥u(·,t)∥L∞(ℝ)≤(332π)·ℬ·∥u0∥L1(ℝ),
so that u(·,t) stays uniformly bounded for all time in this case.2 Estimates similar to (6) can also be shown to hold for the n-dimensional problem
(8)ut+div⁡(b(x,t)u)=Δu,u(·,0)∈Lp(ℝn)∩L∞(ℝn),
but to simplify our discussion we consider here the case n=1 only. Our derivation of (6), which improves some unpublished results by the third author, uses the 1D inequality
(9)∥v∥L∞(ℝ)≤C∞∥v∥L1(ℝ)1/3∥vx∥L2(ℝ)2/3,v∈L1(ℝ)∩H1(ℝ),
where C∞=(3/4)2/3, and can be readily extended to other problems of interest like 1D systems of viscous conservation laws [6, Ch. 9] or the more general equation
(10)ut+(b(x,t,u)u)x=(a(x,t,u)ux)x,a(x,t,u)≥μ(t)>0,
with bounded values b(x,t,u); provided that we assume ∫∞μ(t)dt=∞: using a similar argument, we get the estimate3 [7, Ch. 2]
(11)limsup⁡t→∞∥u(·,t)∥L∞(ℝ)≤(332πp)1/p·ℬμ1/p·limsup⁡t→∞∥u(·,t)∥Lp(ℝ),
for each p≥p0, where(12a)ℬμ=limsup⁡t→∞B(t)μ(t),(12b)B(t)=12(sup⁡x∈ℝb(x,t,u(x,t))-inf⁡x∈ℝb(x,t,u(x,t))).More involving applications, such as problems with superlinear advection or degenerate diffusion, which require considerable extra work, will be studied in the future.


## 2. A Priori Estimates

This section contains some preliminary results on the solutions of problem (1a) and (1b) needed later for our derivation of estimate (6), which is completed in Section 3. (Recall that a solution on some given time interval [0,T*[, 0<T*≤∞, is a functionu(·,t)∈Lloc⁡∞([0,T*[,L∞(ℝ)) which is smooth (C2 in x, C1 in t) in ℝ×]0,T*[ and solves (1a) there, verifying the initial condition in the sense ofLloc⁡1(ℝ), i.e., ∥u(·,t)-u0∥L1(𝕂)→0 as t→0 for each compact 𝕂⊂ℝ. Local existence theory can be found in, e.g., [8, Ch. 6].) We start with a simple Gronwall-type estimate for ∥u(·,t)∥Lq(ℝ), p0≤q<∞. The corresponding result for the supnorm (q=∞) is more difficult to obtain and will be given at the end of Section 2; see Theorem 4.Theorem 1.
Ifu(·,t)∈Lloc⁡∞([0,T*[,L∞(ℝ)) solves problem (1a), (1b), then u(·,t)∈C0([0,T*[,Lq(ℝ)) for each p0≤q<∞, and
(13)∥u(·,t)∥Lq(ℝ)≤∥u(·,0)∥Lq(ℝ)·exp⁡{12(q-1)∫0t‍B(τ)2dτ}
for all 0<t<T*.Proof.
The proof is standard, so we will only sketch the basic steps. TakingS∈C1(ℝ) such that S′(v)≥0 for all v, S(0)=0,S(v)=sgn⁡(v) for|v|≥1, let (given δ>0)Lδ(u)=∫0u‍S(v/δ)dv, so thatLδ(u)→|u| as δ→0, uniformly inu. LetΦδ(u)=Lδ(u)q. Given R>0, 0<ϵ≤1, let ζR(·) be the cut-off function ζR(x)=0 for |x|≥R, ζR(x)=exp⁡{-ϵ1+x2}-exp⁡{-ϵ1+R2} for |x|<R. Multiplying (1a) by Φδ′(u(x,t))·ζR(x) if q≠2, or u(x,t)·ζR(x) if q=2, and integrating the result on ℝ×[0,t], we obtain, letting δ→0 and then R→∞, since u∈L∞(ℝ×[0,t]),(14a)Uϵ(t)+Vϵ(t)≤Uϵ(0)+∫0t‍Gϵ(τ)Uϵ(τ)dτ,Uϵ(t)=∫ℝ‍|u(x,t)|qwϵ(x)dx,
where wϵ(x)=exp⁡⁡{-ϵ1+x2}, Gϵ(t)=(1/2)q(q-1)B(t)2+ϵ2q·sup⁡0≤τ≤t∥u(·,τ)∥L∞(ℝ)+ϵ, and
(14b)Vϵ(t)={12q(q-1)×∫0t‍∫u≠0‍|u(x,τ)|q-2|ux(x,τ)|2wϵ(x)dxdτ,ifq≠2,∫0t‍∫ℝ‍|ux(x,τ)|2wϵ(x)dxdτ,ifq=2.By Gronwall’s lemma, (14a) and (14b) giveUϵ(t)≤Uϵ(0)·exp⁡⁡{∫0t‍Gϵ(τ)dτ}, from which we obtain (13) by simply letting ϵ→0. This shows, in particular, thatu(·,t)∈Lloc⁡∞([0,T*[,Lq(ℝ)) if p0≤q<∞. Now, to get u(·,t)∈C0([0,T*[,Lq(ℝ)), it is sufficient to show that, given ɛ>0 and 0<T<T* arbitrary, we can find R=R(ɛ,T)≫1 large enough so that we have ∥u(·,t)∥Lq(|x|>R)<ɛ for any 0≤t≤T. Taking ψ∈C2(ℝ) with 0≤ψ≤1 and ψ(x)=0 for all x≤0, ψ(x)=1 for all x≥1, let ΨR,M∈C2(ℝ) be the cut-off function given by ΨR,M(x)=0 if |x|≤R-1, ΨR,M(x)=ψ(|x|-R+1) if R-1<|x|<R, and ΨR,M(x)=1 if R≤|x|≤R+M, ΨR,M(x)=ψ(R+M+1-|x|) if R+M<|x|<R+M+1, ΨR,M(x)=0 if |x|≥R+M+1, where R>1, M>0 are given. Multiplying (1a) by Φδ′(u(x,t))·ΨR,M(x) if q≠2, or u(x,t)·ΨR,M(x) if q=2, and integrating the result on ℝ×[0,t], 0<t≤T, we obtain, as in (14a) and (14b), by letting δ→0, M→∞, that ∥u(·,t)∥Lq(|x|>R)<ɛ/2+∥u(·,0)∥Lq(|x|>R-1) for all 0≤t≤T, provided that we take R>1 sufficiently large. This gives the continuity result, and the proof is complete.An important by-product of the proof above is that we have (lettingϵ→0 in (14a) and (14b), and using (13)), for each 0<T<T* and q≥max⁡{p0,2},
(15)∫0T‍∫ℝ‍|u(x,τ)|q-2|ux(x,τ)|2dxdτ<∞.
Therefore, if we repeat the steps above leading to (14a) and (14b), we obtain (letting δ→0, R→∞, ϵ→0, in this order, taking (13) and (15) into account) the identity
(16)∥u(·,t)∥Lq(ℝ)q+q(q-1)∫0t‍∫ℝ‍|u(x,τ)|q-2|ux(x,τ)|2dxdτ=∥u(·,0)∥Lq(ℝ)q+q(q-1)∫0t∫ℝ‍(b(x,τ)-β(τ))‍|u(x,τ)|q-2000000000000000000×u(x,τ)ux(x,τ)dxdτ
for every 0<t<T* and max⁡{p0,2}≤q<∞, where
(17)β(t)=12(sup⁡x∈ℝb(x,t)+inf⁡x∈ℝb(x,t)),t≥0.
The core of the difficulty in the analysis of (1a) and (1b) is apparent here: under the sole assumption that b is bounded, it is not much clear how one should go about the last term in (16) in order to get more than (13) above. Actually, it will be convenient to consider (16) in the (equivalent) differential form, that is,
(18)ddt∥u(·,t)∥Lq(ℝ)q+q(q-1)∫ℝ‍|u(x,t)|q-2|ux(x,t)|2dx=q(q-1)∫ℝ‍(b(x,t)-β(t))|u(x,t)|q-2000000000000000×u(x,t)ux(x,t)dx
for all t∈[0,T*[∖Eq, where Eq⊂[0,T*[ has zero measure. We then readily obtain, using (9) and the one-dimensional Nash inequality [9]
(19)∥v∥L2(ℝ)≤C2∥v∥L1(ℝ)2/3∥vx∥L2(ℝ)1/3,C2=(334π)1/3,
where the value given above for C2 is optimal [10], the following result.Theorem 2.
Letq≥2p0. If t^∈[0,T*[∖Eq is such that (d/dt)∥u(·,t)∥Lq(ℝ)|t=t^q≥0, then(20a)∥u(·,t^)∥Lq(ℝ)≤(q2C23)1/qB(t^)1/q∥u(·,t^)∥Lq/2(ℝ),(20b)∥u(·,t^)∥L∞(ℝ)≤(q2C2C∞)2/qB(t^)2/q∥u(·,t^)∥Lq/2(ℝ).Proof.
Consider (20a) first. From (5), (17), and (18), we have
(21)∫ℝ‍|u(x,t^)|q-2|ux(x,t^)|2dx≤B(t^)∫ℝ‍|u(x,t^)|q-1|ux(x,t^)|dx.
This gives
(22)∫ℝ‍|u(x,t^)|q-2|ux(x,t^)|2dx≤B(t^)2∥u(·,t^)∥Lq(ℝ)q,
or, in terms of v^∈L1(ℝ)∩L∞(ℝ) defined by v^(x)=|u(x,t^)|q/2 if q>2, v^(x)=u(x,t^) if q=2,
(23)∥v^x∥L2(ℝ)≤q2B(t^)∥v^∥L2(ℝ).
Using (19), we then get ∥v^∥L2(ℝ)2≤(q/2)C23B(t^)∥v^∥L1(ℝ)2, which is equivalent to (20a). Similarly, (20b) can be obtained, using (9).Thus, we can use (20a) and (20b) when ∥u(·,t)∥Lq(ℝ) is not decreasing. If it is decreasing, (18) becomes useless but at least we know in such case that ∥u(·,t)∥Lq(ℝ) is not increasing, which should be useful too. Different values of q have different scenarios, which we will have to piece together in some way. The next result shows us just how. To this end, it is convenient to introduce the quantities 𝔹(t0;t) and 𝕌p(t0;t) defined by
(24)𝔹(t0;t)=sup⁡{B(τ):t0≤τ≤t},(25)𝕌p(t0;t)=sup⁡{∥u(·,τ)∥Lp(ℝ):t0≤τ≤t},
given p≥p0,0≤t0≤t<T* arbitrary.Theorem 3.
Letq≥2p0. For each 0≤t0<T*, we have
(26)𝕌q(t0;t)≤max⁡{∥u(·,t0)∥Lq(ℝ);(q2C23)1/q𝔹(t0;t)1/q𝕌q/2(t0;t)}
for all t0≤t<T*.Proof.
Setλq(t)=((q/2)C23)1/q𝔹(t0;t)1/q𝕌q/2(t0;t). There are three cases to consider.
Case I.
∥
u
(
·
,
τ
)
∥
L
q
(
ℝ
)
>
λ
q
(
t
) for all t0≤τ≤t. By (20a), Theorem 2, we must then have (d/dτ)∥u(·,τ)∥Lq(ℝ)q<0 for all τ∈[t0,t]∖Eq, so that ∥u(·,τ)∥Lq(ℝ) is monotonically decreasing in [t0,t]. In particular, 𝕌q(t0;t)=∥u(·,t0)∥Lq(ℝ) in this case, and (26) holds.
Case II.
∥
u
(
·
,
t
0
)
∥
L
q
(
ℝ
)
>
λ
q
(
t
) and ∥u(·,t1)∥Lq(ℝ)≤λq(t) for some t1∈]t0,t]. In this case, let t2∈]t0,t] be such that we have ∥u(·,τ)∥Lq(ℝ)>λq(t) for all t0≤τ<t2, while ∥u(·,t2)∥Lq(ℝ)=λq(t). We claim that ∥u(·,τ)∥Lq(ℝ)≤λq(t) for every t2≤τ≤t: in fact, if this were not true, we could then find t3,t4 with t2≤t3<t4≤t such that ∥u(·,τ)∥Lq(ℝ)>λq(t) for all t3<τ≤t4, ∥u(·,t3)∥Lq(ℝ)=λq(t). By (20a), Theorem 2, this would require (d/dτ)∥u(·,τ)∥Lq(ℝ)q<0 for all τ∈]t3,t4]∖Eq, so that ∥u(·,τ)∥Lq(ℝ) could not increase anywhere on [t3,t4]. This contradicts ∥u(·,t3)∥Lq(ℝ)<∥u(·,t4)∥Lq(ℝ), and so we have ∥u(·,τ)∥Lq(ℝ)≤λq(t) for every t2≤τ≤t, as claimed. On the other hand, by (20a), ∥u(·,τ)∥Lq(ℝ) has to be monotonically decreasing on [t0,t2], just as in Case I. Therefore, we have 𝕌q(t0;t)=∥u(·,t0)∥Lq(ℝ) in this case again, which shows (26).
Case III.Consider ∥u(·,t0)∥Lq(ℝ)≤λq(t). This gives ∥u(·,τ)∥Lq(ℝ)≤λq(t) for every t0≤τ≤t, by repeating the argument used on the interval [t2,t] in Case II. It follows that we must have 𝕌q(t0;t)≤λq(t) in this case, and the proof of Theorem 3 is complete.An important application of Theorem3 is the following result.Theorem 4.
Letp0≤p<∞, 0≤t0<T*. Then
(27)∥u(·,t)∥L∞(ℝ)≤(2p)1/p·max⁡{∥u(·,t0)∥L∞(ℝ);𝔹(t0;t)1/p𝕌p(t0;t)}
for any t0≤t<T*, where 𝔹(t0;t) and 𝕌p(t0;t) are given in (24) and (25) above.Proof.
Letk∈ℤ, k≥2. Applying (26) successively with q=2p,4p,…,2kp, we obtain(28a)∥u(·,t)∥L2kp(ℝ)≤max⁡{∥u(·,t0)∥L2kp(ℝ);K(k,ℓ)1/p·𝔹(t0;t)(1/p)(2-ℓ-2-k)·∥u(·,t0)∥L2ℓp(ℝ),1≤ℓ≤k-1;K(k,0)1/p·𝔹(t0;t)(1/p)(1-2-k)·𝕌p(t0;t)},
where
(28b)K(k,ℓ)=∏j=ℓ+1k(2j-1pC23)2-j,0≤ℓ≤k-1.Now, for 1≤ℓ≤k-1,
(29)𝔹(t0;t)(1/p)(2-ℓ-2-k)·∥u(·,t0)∥L2ℓp(ℝ)≤𝔹(t0;t)(1/p)(2-ℓ-2-k)·∥u(·,t0)∥Lp(ℝ)(2-ℓ-2-k)/(1-2-k)·∥u(·,t0)∥L2kp(ℝ)(1-2-ℓ)/(1-2-k)≤max⁡{∂y∂x∥u(·,t0)∥L2kp(ℝ);000000000𝔹(t0;t)(1/p)(1-2-k)·∥u(·,t0)∥Lp(ℝ)∂y∂x}
by Young’s inequality (see, e.g., [11, page 622]); in particular, we get, from (28a) and (28b),
(30)∥u(·,t)∥L2kp(ℝ)≤(2p)1/p·max⁡{∥u(·,t0)∥L2kp(ℝ);𝔹(t0;t)(1/p)(1-2-k)·𝕌p(t0;t)},
since K(k,ℓ)≤2p for all 0≤ℓ≤k-1. Letting k→∞, (27) is obtained.It follows from Theorems1 and 4 that u(·,t) is globally defined (T*=∞). Now, from (27), we immediately obtain, letting t→∞,
(31)limsup⁡t→∞∥u(·,t)∥L∞(ℝ)≤(2p)1/p·max⁡{∥u(·,t0)∥L∞(ℝ);𝔹(t0)1/p𝕌p(t0)}
for any t0≥0, where 𝔹(t0) and 𝕌p(t0) are given by
(32)𝔹(t0)=sup⁡{B(t):t≥t0},(33)𝕌p(t0)=sup⁡{∥u(·,t)∥Lp(ℝ):t≥t0}.
Taking (t0(n))n such that t0(n)→∞ and ∥u(·,t0(n))∥L∞(ℝ)→liminf⁡t→∞∥u(·,t)∥L∞(ℝ), and applying (31) with t0=t0(n) for each n, we then obtain, letting n→∞,
(34)limsup⁡t→∞⁡∥u(·,t)∥L∞(ℝ)≤(2p)1/p·max⁡{liminf⁡t→∞⁡∥u(·,t)∥L∞(ℝ);ℬ1/p·𝒰p},
where ℬ and 𝒰p are given by
(35)ℬ=limsup⁡t→∞B(t),𝒰p=limsup⁡t→∞∥u(·,t)∥Lp(ℝ).


## 3. Large Time Estimates

In this section, we use the results obtained above to derive two basic large time estimates (given in Theorems5 and 6) for solutions u(·,t) of problem (1a), (1b), which represent important intermediate steps that will ultimately lead to the main result stated in Theorem 7.Theorem 5.
Letq≥2p0, and ℬ≥0 be as defined in (35). Then
(36)limsup⁡t→∞∥u(·,t)∥Lq(ℝ),≤(q2C23)1/q·ℬ1/q·limsup⁡t→∞∥u(·,t)∥Lq/2(ℝ),
where C2=(33/(4π))1/3 is the constant in the Nash inequality (19).Proof.
We setp=q/2 and assume that 𝒰p is finite. As in the proof of Theorem 2, we take v∈L∞(ℝ×[0,∞[) given by v(x,t)=|u(x,t)|p if p>1, v(x,t)=u(x,t) if p=1. It follows that
(37)∥v(·,t)∥L2(ℝ)2=∥u(·,t)∥L2p(ℝ)2p,∥vx(·,t)∥L2(ℝ)2=p2∫ℝ‍|u(x,t)|2p-2|ux(x,t)|2dx.
Therefore, from (18), we have, for some null set E2p⊂[0,∞[,
(38)ddt∥v(·,t)∥L2(ℝ)2+4(1-12p)∥vx(·,t)∥L2(ℝ)2≤4p(1-12p)B(t)∥v(·,t)∥L2(ℝ)∥vx(·,t)∥L2(ℝ)
for all t∈[0,∞[∖E2p, and so, by (19),
(39)ddt∥v(·,t)∥L2(ℝ)2+4(1-12p)∥vx(·,t)∥L2(ℝ)2≤4pC2(1-12p)B(t)∥v(·,t)∥L1(ℝ)2/3∥vx(·,t)∥L2(ℝ)4/3.
This gives, by Young’s inequality ([11, page 622]), for all t∈[0,∞[∖E2p,
(40)ddt∥v(·,t)∥L2(ℝ)2+43(1-12p)∥vx(·,t)∥L2(ℝ)2≤43(1-12p)(pC2)3B(t)3∥v(·,t)∥L1(ℝ)2.
Setting
(41)λp=limsup⁡t→∞⁡g(t),g(t)=(pC23)1/2B(t)1/2∥v(·,t)∥L1(ℝ),
we claim that
(42)limsup⁡t→∞∥v(·,t)∥L2(ℝ)≤λp.
In fact, let us argue by contradiction. If (42) is false, we can pick 0<η≪1 and a sequence (tj)j≥0, tj→∞, such that ∥v(·,tj)∥L2(ℝ)>λp+η (for all j≥0) and g(t)≤λp+η/2 for all t≥t0. From (20a), Theorem 2, it will then follow that
(43)∥v(·,t)∥L2(ℝ)>λp+η,∀t≥t0.
In fact, suppose that (43) were false, so that we had ∥v(·,t~)∥L2(ℝ)≤λp+η for some t~>t0. Taking j≫1 with tj>t~, we could then find t^∈[t~,tj[ such that ∥v(·,t)∥L2(ℝ)>λp+η for all t∈]t^,tj], while ∥v(·,t^)∥L2(ℝ)=λp+η, and so there would exist t*∈[t^,tj]∖E2p with (d/dτ)∥v(·,t)∥L2(ℝ)2 positive at t=t*. By (20a), we would have ∥v(·,t*)∥L2(ℝ)≤λp, but this would contradict the fact that ∥v(·,t)∥L2(ℝ)≥λp+η everywhere on [t^,tj]. Thus, we conclude that (43) cannot be false, as claimed. We then obtain, from (19), (40), and (43),
(44)∥v(·,t)∥L2(ℝ)6≤C26∥v(·,t)∥L1(ℝ)4∥vx(·,t)∥L2(ℝ)2≤g(t)6+2p2p-1∥v(·,t)∥L1(ℝ)4(-ddt∥v(·,t)∥L2(ℝ)2)
for all t∈[t0,∞[∖E2p. Recalling that ∥v(·,t)∥L2(ℝ)>λp+η, g(t)≤λp+η/2, for  all t≥t0, this gives
(45)-ddt∥v(·,t)∥L2(ℝ)2≥K(η),∀t∈[t0,∞[∖E2p
for some constant K(η)>0 independent of t, which cannot be, since this implies
(46)∥v(·,t0)∥L2(ℝ)2≥K(η)·(t-t0)∀t>t0.
This contradiction shows (42), which is equivalent to (36), and the proof is complete.Applying (36) successively with q=2p,4p,…,2kp, we get
(47)limsup⁡t→∞∥u(·,t)∥L2kp(ℝ)≤[∏j=1k‍(2j-1pC23)2-j]1/p·ℬ(1/p)(1-2-k)·𝒰p
for k≥1 arbitrary, where 𝒰p=limsup⁡t→∞∥u(·,t)∥Lp(ℝ). Letting k→∞, this suggests(48a)limsup⁡t→∞∥u(·,t)∥L∞(ℝ)≤K(p)·ℬ1/p·limsup⁡t→∞∥u(·,t)∥Lp(ℝ),
where
(48b)K(p)=[∏j=1∞‍(2j-1pC23)2-j]1/p=(332πp)1/p(cf. (6) above), as long as the limit processes k→∞, t→∞ can be interchanged. That this is indeed the case is a consequence of (34) and the following result.Theorem 6.
Letp≥p0. Then
(49)liminf⁡t→∞∥u(·,t)∥L∞(ℝ)≤(pC2C∞)1/p·ℬ1/p·limsup⁡t→∞∥u(·,t)∥Lp(ℝ),
where C2, C∞ are the constants given in (19) and (9).Proof.
Again, assuming𝒰p finite (otherwise, (49) is obvious; cf. endnote4), we introduce, as in the previous proof, v∈L∞(ℝ×[0,∞[) given by v(x,t)=|u(x,t)|p if p>1, and v(x,t)=u(x,t) if p=1. Thus, (40) is valid, and setting λp∈ℝ, g∈L∞([0,∞[) by
(50)λp=limsup⁡t→∞g(t),g(t)=pC2B(t)∥v(·,t)∥L1(ℝ),
we have that (49) is obtained if we show that
(51)liminf⁡t→∞∥v(·,t)∥L∞(ℝ)≤C∞·λp.
We argue by contradiction and assume that (51) is false. Taking then 0<η≪1,t0≫1 so that ∥v(·,t)∥L∞(ℝ)≥C∞·(λp+η) and g(t)≤λp+η/2 hold for all t≥t0, we get, by (9) and (40),
(52)∥v(·,t)∥L∞(ℝ)3≤C∞3∥v(·,t)∥L1(ℝ)∥vx(·,t)∥L2(ℝ)2≤C∞3g(t)3+C∞32p2p-1∥v(·,t)∥L1(ℝ)×(-ddt∥v(·,t)∥L2(ℝ)2)
for all t∈[t0,∞[∖E2p. Since ∥v(·,t)∥L∞(ℝ)≥C∞·(λp+η), g(t)≤λp+η/2, this gives
(53)-ddt∥v(·,t)∥L2(ℝ)2≥K(η),∀t∈[t0,∞[∖E2p
for some constant K(η)>0 independent of t. As before, this implies that ∥v(·,t0)∥L2(ℝ)2≥K(η)·(t-t0) for all t≥t0, which is impossible because ∥v(·,t0)∥L2(ℝ) is finite. This contradiction establishes (51) above, completing the proof of Theorem 6.We are finally in good position to derive (6), (48a), and (48b). Combining (34) and (49) above, we obtain
(54)limsup⁡t→∞∥u(·,t)∥L∞(ℝ)≤(2p2)1/p·ℬ1/p·𝒰p
for each p≥p0, so that we have, in particular,
(55)limsup⁡t→∞∥u(·,t)∥L∞(ℝ)≤(22k+1p2)1/2kp·ℬ1/2kp·𝒰2kp
for each k≥0. By (47), we then get
(56)limsup⁡t→∞∥u(·,t)∥L∞(ℝ)≤{(22k+1p2)2-k·∏j=1k‍(2j-1pC23)2-j}1/p·ℬ1/p·𝒰p
for all k. Letting k→∞, Theorem 7 is obtained, and our argument is complete.Theorem 7.
Letp≥p0. Assuming b∈L∞(ℝ×[0,∞[), then (6), (48a), and (48b) hold.It is worth noticing that the corresponding estimate for then-dimensional problem (8), namely,
(57)limsup⁡t→∞∥u(·,t)∥L∞(ℝn)≤K(n,p)·ℬn/p·limsup⁡t→∞∥u(·,t)∥Lp(ℝn),
where ℬ≥0 is similarly defined, can be also derived in arbitrary dimension n>1.


## 4. Concluding Remarks

We close our discussion of the problem (1a), (1b), given b∈L∞(ℝ×[0,∞[), 1≤p0<∞, indicating a few questions which were not answered by our analysis:(a)
characterize allb∈L∞(ℝ×[0,∞[) for which it is true that ∥u(·,t)∥L∞(ℝ)→0 (as t→∞) for every solution u(·,t) of problem (1a) and (1b);(b)
same question as (a) above, but requiring only thatlimsup⁡∥u(·,t)∥L∞(ℝ)<∞ (as t→∞) for every solution u(·,t) of problem (1a) and (1b), in case p0>1;5(c)
givenp0>1, characterize all b∈L∞(ℝ×[0,∞[) such that ∥u(·,t)∥Lp0(ℝ)→0 (as t→∞) for every solution u(·,t) of problem (1a) and (1b);(d)
same question as (c) above, but requiring only thatlimsup⁡∥u(·,t)∥Lp0(ℝ)<∞ (as t→∞) for every solution u(·,t) of problem (1a) and (1b);(e)
forp0=1, characterize all b∈L∞(ℝ×[0,∞[) such that ∥u(·,t)∥L1(ℝ)→|m| (as t→∞) for every solution u(·,t), where m=∫ℝ‍u0(x)dx is the solution mass;(f)
forp0=1, and b∈L∞(ℝ×[0,∞[) not satisfying property (e), what are the values of lim⁡t→∞∥u(·,t)∥L1(ℝ) in case of initial states that change sign?These questions can be similarly posed for solutionsu(·,t) of autonomous problems
(58)ut+(b(x)u)x=uxx,u(·,0)∈Lp0(ℝ)∩L∞(ℝ),
where b∈L∞(ℝ) does not depend on the time variable. For (58), question (e) has been answered in [12] (see also [13]). Another interesting question is the following: (g)
when (58) admits no stationary solutions other than the trivial solution u=0, is it true that lim⁡t→∞∥u(·,t)∥L∞(ℝ)=0 for every solution u(·,t)?Moreover, for solutionsu(·,t) of (1a) and (1b) or (58) with ∥u(·,t)∥L∞(ℝ)→0 as t→∞, there is the question of determining the proper decay rate.6 As suggested by Figure 1, solution decay may sometimes happen at remarkably slow rates.



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