# Complicated Asymptotic Behavior of Solutions for Heat Equation in Some Weighted Space

Authors: Liangwei Wang; Jingxue Yin
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We investigate the asymptotic behavior of solutions for the heat equation in the weighted spaceY0σ(ℝN)≡{φ∈C(ℝN):lim|x|→∞(1+|x|2)-σ/2φ(x)=0}. Exactly, we find that the unbounded function space Y0σ(ℝN) with 0<σ<N can provide a setting where complexity occurs in the asymptotic behavior of solutions for the heat equation.


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## Body

## 1. Introduction

In this paper, we consider the asymptotic behavior of solutions to the Cauchy problem of the heat equation(1.1)∂u∂t-Δu=0,(x,t)∈RN×(0,∞),u(x,0)=u0(x),x∈RN,
where N≥1 and the initial value u0∈Y0σ(ℝN).Whether complexity occurs in the asymptotic behavior of solutions for some evolution equations or not mainly depends on the work spaces that one selects [1–9]. In the space Lp(ℝN) with 1≤p<∞, the problem (1.1) under consideration is well posed and the asymptotic behavior of the solutions is rather simple, reflecting the simple structure of the heat equation. Considering, for instance, the problem (1.1) with the initial value u0∈L1(ℝN), it is well-known that the solutions u(x,t) converge as t→∞ toward a multiple of the fundamental solution, the one which has the same integral,
(1.2)u(x,t)=S(t)u0(x)=G(t)*u0(x)⟶MG(x,t),
where G(x,t)=(4πt)-N/2exp(-|x|2/4t) and M=∫ℝN‍u0(x)dx, see [10, 11].It was first found in 2002 [12] by Vázquez and Zuazua that the bounded function space L∞(ℝN) provides a setting where complicated asymptotic behavior of solutions may take place for the heat equation. In fact, they proved that, for any bounded sequence {gj,j=1,2,…} in L∞(ℝN), there exists an initial value u0∈L∞(ℝN) and a sequence tjk→∞ as k→∞ such that
(1.3)u(tjk1/2x,tjk)=S(tjk)u0(tjkx)⟶S(1)gj(x)
uniformly on any compact subset of ℝN as k→∞. Subsequently, Cazenave et al. showed that, in the bounded continuous function space C0(ℝN), the solutions of the heat equation may present more complex asymptotic behavior [13–15]. Meanwhile, considerable attention has also been paid to study the complicated asymptotic behavior of solutions for the porous medium equation and other evolution equations in some bounded function spaces such as C0(ℝN) and L∞(ℝN) (see, e.g., [3, 7, 9, 12, 16] and the references therein).In this paper we find that, even in the unbounded function spaceY0σ(ℝN) with 0<σ<N, the complicated asymptotic behavior of solutions for the heat equation can also occur. For this purpose, we need to establish the Lσp–Lσ∞ smoothing effect and other estimates for the solutions of the problem (1.1) when the initial value u0∈Lσp(ℝN)≡{φ:(1+|·|2)-σ/2φ(·)∈Lp(ℝN)} with 1<p≤∞ or u0∈Y0σ(ℝN).The rest of this paper is organized as follows. In the next section, we give some definitions and some estimates of the solutions to the problem (1.1). Section 3 is devoted to study the complicated asymptotic behavior of the solutions.


## 2. Main Estimates

In this section, we investigate some properties of solutions for the problem (1.1) when the initial value u0 belongs to some weighted spaces. For these purposes, we first introduce the mild solutions u(x,t) of the problem (1.1) which are defined as
(2.1)u(x,t)=S(t)u0(x)=(4πt)-N/2∫RN‍exp⁡(-|x-y|24t)u0(y)dy.
Letting σ≥0 and 1≤p≤∞, we define two weighted spaces Y0σ(ℝN) and Lσp(ℝN) as follows:
(2.2)Y0σ(RN)≡{φ(x)∈C(RN):lim⁡|x|→∞φ(x)(1+|x|2)-σ/2=0},Lσp(RN)≡{φ:φ(⋅)(1+|⋅|2)-σ/2∈Lp(RN)}.
Endowed with the obvious norms,
(2.3)‖φ‖Y0σ(RN)=‖φ(⋅)(1+|⋅|2)-σ/2‖L∞(RN),‖φ‖Lσp(RN)=‖φ(⋅)(1+|⋅|2)-σ/2‖Lp(RN),
the spaces Y0σ(ℝN) and Lσp(ℝN) are both Banach spaces. Notice that if σ=0, then
(2.4)Y00(RN)=C0(RN),L0∞(RN)=L∞(RN),L0p(RN)=Lp(RN).
Next we give the definition of the ω-limit set ωσμ,β(u0) which is our main study object in this paper.Definition 2.1.
Letσ≥0, μ,  β>0, and suppose that u0∈Y0σ(ℝN). The ω-limit set ωσμ,β(u0) is given by
(2.5)ωσμ,β(u0)≡{f∈Y0σ(RN);∃tn⟶∞s.t.Dtnμ,β[S(tn)u0]→n→∞finY0σ(RN)}.
HereDλμ,βφ(x)≡λμφ(λ2βx) for φ∈Lloc1(ℝN) and λ>0.
In the rest of this section, we will consider the properties of the solutions for the problem (1.1) when the initial value u0∈Lσp(ℝN) or u0∈Y0σ(ℝN).
The following theorem can be seen as some extension of the maximum principle for the problem (1.1).Theorem 2.2.
Let0≤σ<∞. Suppose that
(2.6)u0∈Lσ∞(RN)
and that u(x,t)=S(t)u0(x) are the mild solutions of the problem (1.1). Then
(2.7)u(t)=S(t)u0∈Lσ∞(RN)fort>0.
Moreover, if t>1, then
(2.8)‖S(t)u0‖Lσ∞(RN)≤Ctσ‖u0‖Lσ∞(RN),
or if 0<t≤1, then
(2.9)‖S(t)u0‖Lσ∞(RN)≤C‖u0‖Lσ∞(RN).Remark 2.3.
Letσ=0. From Theorem 2.2, we can obtain the well-known result (maximum principle) that if u0∈L0∞(ℝN)=L∞(ℝN), then
(2.10)‖S(t)u0‖L∞(RN)≤C‖u0‖L∞(RN).Proof.
To prove this theorem, we need the fact that if(2.11)φ(x)=M(1+|x|2)σ/2forsomeM>0,
then there exists a constant C such that
(2.12)S(t)φ(x)≤C(1+t+|x|2)σ/2,
which proof can be found in [17]; we give the proof here for completeness. Consider the following problem:
(2.13)∂v∂t-Δv=0,inRN×(0,∞),v(x,0)=v0(x)=M|x|σ,inRN.
For λ>0, from (2.1), we can get that
(2.14)Dλ-σ/2,1/2S(λt)v0(x)=λ-σ/2[S(λt)v0](λ1/2x)=S(t)[Dλ-σ/2,1/2v0](x)=S(t)v0(x).
By the existence and the regularity theories of the solutions, we can obtain that, for t>0,
(2.15)0<S(t)v0∈C∞((0,∞)×RN),
see [10, 18]. Now taking t=1, λ=s and g(x)=S(1)v0(x) in the expression (2.14), we have
(2.16)S(s)v0(x)=sσ/2g(s-1/2x).
The fact that S(s)v0(x)∈C([0,∞)×ℝN∖(0,0)) clearly implies that, for |x|=1,
(2.17)sσ/2g(s-1/2x)=S(s)v0(x)⟶v0(x)=M|x|σ=M
as s→0. Let
(2.18)y=s-1/2x.
So,
(2.19)|y|⟶∞ass⟶0.
Therefore,
(2.20)|y|-σg(y)-M⟶0
as |y|→∞. So, there exists constant 0<C<∞ such that
(2.21)0≤g(x)≤C(1+|x|2)σ/2.
By (2.16), we thus have
(2.22)S(s)v0(x)≤C(s+|x|2)σ/2.
Notice that
(2.23)0≤φ(x)≤C(1+v0(x)).
Therefore, by comparison principle and (2.22), we can get that, for all t≥0, there exists constant C>0 such that
(2.24)0≤S(t)φ(x)≤C(1+t+|x|2)σ/2.
So we complete the proof of (2.12). For any t>0, from (2.1) and (2.12), we thus obtain that
(2.25)|u(x,t)|=|S(t)u0(x)|=|(4πt)-N/2∫RN‍exp⁡(-|x-y|2/4t)u0(y)dy|=|(4πt)-N/2∫RN‍exp⁡(-|x-y|2/4t)(1+|y|2)σ/2u0(y)(1+|y|2)-σ/2dy|≤‖u0‖Y∞σ(RN)(4πt)-N/2∫RN‍exp⁡(-|x-y|2/4t)(1+|y|2)σ/2dy≤C(1+t+|x|2)σ/2‖u0‖Y∞σ(RN).
Therefore, if t>1, then
(2.26)|u(x,t)|≤C(1+|x|2)σ/2tσ‖u0‖Y∞σ(RN).
This clearly illustrates (2.8). If 0<t≤1, then
(2.27)|u(x,t)|≤C(1+|x|2)σ/2‖u0‖Y∞σ(RN).
From this, we can get (2.9). So we complete the proof of this theorem.Theorem 2.4 (Lσp–Lσ∞ smoothing effect).
Let1<p<∞ and 0≤σ<∞. Suppose u0∈Lσp(ℝN) and that u(x,t)=S(t)u0(x) are the solutions of the problem (1.1). Then
(2.28)u(t)=S(t)u0∈Lσ∞(RN)fort>0.
Moreover, if t>1, then
(2.29)‖S(t)u0‖Lσ∞(RN)≤Ctσ-N/2p‖u0‖Lσp(RN),
or if 0<t≤1, then
(2.30)‖S(t)u0‖Lσ∞(RN)≤Ct-N/2p‖u0‖Lσp(RN).Remark 2.5.
Ifσ=0, then Theorem 2.4 captures the result Lp–L∞ smoothing effect for the heat equation.Proof.
For anyt>0, from (2.1) and Theorem 2.2, we thus obtain that
(2.31)|u(x,t)|=|S(t)u0(x)|=|(4πt)-N/2∫RN‍exp⁡(-|x-y|2/4t)u0(y)dy|=|(4πt)-N/2∫RN‍exp⁡(-|x-y|2/4t)(1+|y|2)σ/2u0(y)(1+|y|2)-σ/2dy|≤[(4πt)-N/2∫RN‍exp⁡(-|x-y|2/4t)(1+|y|2)p′σ/2dy]1/p′×[(4πt)-N/2∫RN‍exp⁡(-|x-y|2/4t)(1+|y|2)-pσ/2u0p(y)dy]1/p≤C(1+t+|x|2)σ/2t-N/2p‖u0‖Lσp(RN).
Here 1/p+1/p′=1. From this, we can get that, if t≥1, then
(2.32)‖u(t)‖Lσ∞(RN)≤Ctσ/2-N/2p‖u0‖Lσp(RN),
or if 0<t<1, then
(2.33)‖u(t)‖Lσ∞(RN)≤Ct-N/2p‖u0‖Lσp(RN).
So we complete the proof of this theorem.In the following theorem, we consider the property of the solutionsu(x,t) of (1.1) with the initial data u0∈Y0σ(ℝN).Theorem 2.6.
Let0≤σ<∞. If u0∈Y0σ(ℝN), then
(2.34)S(t)u0∈Y0σ(RN)fort≥0.Proof.
Forσ=0, the above theorem is a well-known result. So, in the rest of this proof, we assume that 0<σ<∞. From (2.1), we have
(2.35)(1+|x|2)-σ/2|u(x,t)|=(1+|x|2)-σ/2|(4πt)-N/2∫RN‍exp⁡(-|x-y|2/4t)u0(y)dy|=(1+|x|2)-σ/2|(4πt)-N/2∫|y|≤M‍exp⁡(-|x-y|2/4t)u0(y)dy|+(1+|x|2)-σ/2|(4πt)-N/2∫|y|≤M‍exp⁡(-|x-y|2/4t)u0(y)dy|=I1(x)+I2(x).
For any ε>0, from u0∈Y0σ(ℝN), we obtain that there exists an M>0 such that if |y|>M, then
(2.36)|u0(y)(1+|y|2)-σ/2|<ε.
So, from (2.12), we have
(2.37)I2(x)=(1+|x|2)-σ/2|(4πt)-N/2∫|y|>M‍exp⁡(-|x-y|2/4t)u0(y)(1+|y|2)-σ/2(1+|y|2)σ/2dy|≤(1+|x|2)-σ/2sup⁡|y|>M|u0(y)(1+|y|2)-σ/2|(4πt)-N/2∫|y|>M‍exp⁡(-|x-y|2/4t)(1+|y|2)σ/2dy≤C(1+|x|2)-σ/2(1+t+|x|2)σ/2ε.
So, if t>1, then
(2.38)I2(x)≤C(1+|x|2)-σ/2(1+t+|x|2)σ/2ε≤Ctσ/2ε,
or if 0≤t≤1, then
(2.39)I2(x)≤C(1+|x|2)-σ/2(1+t+|x|2)σ/2ε≤Cε.
Notice also that
(2.40)u0∈Y0σ(RN)⊂C(RN).
So, there exists a constant C such that
(2.41)sup⁡|y|≤M|u0(y)|≤C.
This means that
(2.42)I1(x)=(1+|x|2)-σ/2|(4πt)-N/2∫|y|≤M‍exp⁡(-|x-y|2/4t)u0(y)dy|≤C(1+|x|2)-σ/2.
So, there exists an M1>0 such that if |x|>M1, then
(2.43)I1(x)<ε.
Combining this with (2.35), (2.38), and (2.39), we can obtain that, for t≥0,
(2.44)u(t)=S(t)u0∈Y0σ(RN).
So we complete the proof of this theorem.


## 3. Complicated Asymptotic Behavior

In this section, we investigate the asymptotic behavior of solutions for the problem (1.1) and give the fact that the weighted space Y0σ(ℝN) with 0≤σ<N can provide a setting where complexity occurs in the asymptotic behavior of solutions.Theorem 3.1.
Letμ>0, σ≥0, p>1, and β>1/2. If
(3.1)μ+2βσ<Np,f∈Y0σ(RN)⋂‍Lσp(RN),
then there exists an initial value u0∈Y0σ(ℝN) and a sequence tn→∞ as n→∞ such that
(3.2)Dtnμ,βS(tn)u0→n→∞finY0σ(RN).
That is, f∈ωσμ,β(u0).Proof.
Suppose thata>2 is a constant. Then let
(3.3)λ1=a,λj=max⁡(j2p/(N-p(μ+2βσ))λj-1(2βN-p(μ+2βσ))/(N-p(μ+2βσ)),(2jλj-1)1/μ)forj>1.
Now we define the initial-value u0 as
(3.4)u0(x)=∑j=1∞‍λj-μf(xλj2β)=∑j=1∞‍Dλj-1μ,βf(x).
Let
(3.5)l=max⁡(‖f‖Y0σ(RN),‖f‖Lσp(RN)).
So,
(3.6)‖u0‖Y0σ(RN)≤∑j=1∞‍λj-μ‖f(1λj2β⋅)‖Y0σ(RN)≤∑j=1∞‍2-j‖f(x)‖Y0σ(RN)≤l.
Here we have used the fact that if 0<λ≤1, then
(3.7)‖f(λ⋅)‖Y0σ(RN)=sup⁡x∈RN|(1+x2)-σ/2f(λx)|=sup⁡x∈RN|f(λx)(1+|λx|2)-σ/2(1+|λx|21+|x|2)σ/2|≤sup⁡x∈RN|f(λx)(1+|λx|2)-σ/2|=‖f‖Y0σ(RN).
Therefore, the sequence of (3.4) is convergent in Y0σ(ℝN). This means that
(3.8)u0∈Y0σ(RN).
As a result of (2.1), we see that, for 0<t<T<∞,(3.9)Dλnμ,β[S(λn2t)u0](x)=S(tλn2-4β)(un+vn+wn)=S(tλn2-4β)un+S(tλn2-4β)vn+S(tλn2-4β)wn,
where
(3.10)un(x)=∑j=1n-1‍Dλnμ,β[Dλj-1μ,βf(x)]=λnμ∑j=1n-1‍λj-μf(xλn2βλj2β),vn(x)=Dλnμ,β[Dλn-1μ,βf(x)]=f(x),wn=∑j=n+1∞‍Dλnμ,β[Dλj-1μ,βf(x)]=λnμ∑j=n+1∞‍λj-μf(xλn2βλj2β).
Notice that, if λ≥1, then
(3.11)‖f(λ⋅)‖Ypσ(RN)=(∫RN‍|(1+|x|2)-σ/2f(λx)|pdx)1/p≤(∫RN‍|f(λx)(1+|λx|2)-σ/2(1+|λx|21+|x|2)σ/2|pdx)1/p≤λσ(∫RN‍|f(λx)(1+|λx|2)-σ/2|pdx)1/p=λσ-N/p‖f‖Ypσ(RN),
we thus obtain that
(3.12)‖un‖Ypσ(RN)≤∑j=1n-1‍(λnλj)μ-2βN/p+2βσ‖f‖Ypσ(RN)≤n(λnλn-1)μ-2βN/p+2βσ‖f‖Ypσ(RN).
Consequently, for any t>0, we can select N large enough to satisfy that if n>N, then
(3.13)0<λn2-4βt<1.
Here we have used the hypothesis β>1/2. So, (2.30), (3.11), and (3.12) indicate
(3.14)‖S(λn2-4βt)un(x)‖Y0σ(RN)≤C(λn2-4βt)-N/2p‖un(x)‖Lσp(RN)≤Cλn-12βN/p-2βσ-μt-2N/p(nl)λnμ-N/p+2βσ.
From (3.3), we thus obtain that
(3.15)‖S(λn2-4βt)un(x)‖Y0σ(RN)≤Ct-2N/pln-1⟶0
as n→∞. From (3.7) and the definition of wn(x), we have
(3.16)‖wn‖Y0σ(RN)≤‖∑j=n+1∞‍(λnλj)μf((λnλj)2β⋅)‖Y0σ(RN)≤∑j=n+1∞‍(λnλj)μl.
Applying (2.9) to S(tλn2-4β)wn, we thus have
(3.17)‖S(tλn2-4β)wn‖Y0σ(RN)≤Cλnμ∑j=n+1∞‍λj+1-μ≤Cl∑j=n+1∞‍2-j⟶0
as n→∞. Here we have used (3.3) and the fact that 0<λn2-2βt<1 for n>N. At present, we want to verify the claim that, for 0≤t<∞,
(3.18)(S(λn2-4βt)vn-f)Y0σ(RN)→n→∞0.
In fact, if 0<t<∞, then
(3.19)‖S(λn2-4βt)vn-f‖Y0σ(RN)≤sup⁡|x|≤M(1+|x|2)-σ/2|S(λn2-4βt)vn(x)-f(x)|+sup⁡|x|>M(1+|x|2)-σ/2|S(λn2-4βt)vn(x)|+sup⁡|x|>M(1+|x|2)-σ/2|f(x)|.
Notice that
(3.20)vn=f∈Y0σ(RN).
Therefore, from Theorem 2.6, we have
(3.21)S(λn2-4βt)vn∈Y0σ(RN).
So, for any ε>0, from (2.35), (2.39), and (2.43), we get that if n>N, there exists a constant M independing on n and t such that
(3.22)sup⁡|x|>M(1+|x|2)-σ/2|S(λn2-4βt)vn(x)|<ε3,sup⁡|x|>M(1+|x|2)-σ/2|f(x)|<ε3.
The fact vn=f∈C(ℝN) means that
(3.23)S(t)vn∈C([0,∞)×RN).
Therefore, for any ε>0, there exists an N1>0 such that if n>N1, then
(3.24)sup⁡|x|≤M(1+|x|2)-σ/2|S(λn2-4βt)vn(x)-f(x)|<ε3.
So, from (3.19), (3.22), and (3.24), we thus obtain that
(3.25)‖S(λn2-4βt)vn-f‖Y0σ(RN)⟶0
as n→∞. From (3.9), (3.15), (3.17), and (3.18), we obtain that for any fixed 0<t<∞,
(3.26)Dλnμ,βS(λn2t)u0=S(λn2-2βt)(un+vn+wn)→n→∞finY0σ(RN).
Equation (3.2) follows from (3.26) by setting t=1 and tn=λn2. The proof of this theorem is complete.Theorem 3.2.
Letμ>0, σ≥0, and β>1/2. If
(3.27)0<μ+2βσ<N,
then there exists an initial value u0∈Y0σ(ℝN) such that
(3.28)ωσμ,β(u0)=Y0σ(RN).Remark 3.3.
Ifσ=0, the above results had been given by Cazenave et al. (see [15, Corollary 6.3]). So our results capture their results. Here we have used some of their ideas.Proof.
By (3.27), there exists a constant p>1 such that
(3.29)μ+2βσ<Np.
Therefore, there exists a countable dense subset F of Y0σ(ℝN) such that
(3.30)F⊂Y0σ(RN)⋂‍Lσp(RN).
So, there exists a sequence {φj}j≥1⊂F such that (i)
for anyϕ∈F, there exists a subsequence {φjk}k≥1 of the sequence {φj}j≥1 satisfying
(3.31)φjk(x)=ϕ∀k≥1,(ii)
there exists a constantC>0 satisfying
(3.32)max⁡(‖φj‖Y0σ(RN),‖φj‖Lσp(RN))≤Cjforj≥1.Now we select a constant
(3.33)a>2
and then take
(3.34)λj={aifj=1,max⁡(j4p/(N-p(μ+2βσ))λj-1(2βN-p(μ+2βσ))/(N-p(μ+2βσ)),(2jλj-1j)1/μ)ifj>1.
The initial-value u0 is given by
(3.35)u0(x)=∑j=1∞‍λj-μφj(xλj2β)=∑j=1∞‍Dλj-1μ,βφj(x).
From (3.7), (3.32), and (3.34), we have
(3.36)‖u0‖Y0σ(RN)≤∑j=1∞‍λj-μ‖φj‖Y0σ(RN)≤C∑j=1∞‍2-j≤C.
This means that the sequence (3.35) is convergent in Y0σ(ℝN). So,
(3.37)u0∈Y0σ(RN).
Similar to the proof of Theorem 2.2, we can prove that, for any ϕ∈F and 0<t<∞, there exists a sequence λjk→∞ as k→∞ such that
(3.38)Dλjkμ,βS(λjk2t)u0→ϕinY0σ(RN)
as k→∞. Let tnk=λnk1/2 and t=1 in (2.1); we thus have
(3.39)Dtjkμ,βS(tjk)u0⟶ϕinY0σ(RN).
Therefore,
(3.40)F⊂ωσμ,β(u0)⊂Y0σ(RN).
Notice that F is dense in Y0σ(ℝN) and that ωσμ,β(u0) is a closed subset of Y0σ(ℝN). We thus obtain from (3.40) that
(3.41)ωσμ,β(u0)=Y0σ(RN).
So we complete the proof of this theorem.Remark 3.4.
For any0≤σ<N, there exist constants μ>0 and β>1/2 such that
(3.42)0<μ+2βσ<N.
Therefore, from Theorem3.2, we can get that the weighted space Y0σ(ℝN) provides a setting where complexity occurs in the asymptotic behavior of solutions for the problem (1.1).



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