Grow-up rate of solutions for the heat equation with a sublinear source

Abstract

In this paper, we investigate the grow-up rate of solutions for the heat equation with a sublinear source. We find that if the initial value grows fast enough, then it plays a major role in the growing up of solutions, while if the initial value grows slowly, then the sublinear source prevails. As a direct application of these results, we show that the effect of the sublinear source is negligible in the asymptotic behavior of solutions as $t\to \mathrm{\infty }$ if the initial value grows fast enough.

MSC:35K55, 35B40.

1 Introduction

We consider the Cauchy problem of the heat equation with the source

(1.1)

(1.2)

Here $p>0$, $N\ge 1$, and $u_0\in L^{\mathrm{\infty }}({\rho }_{\sigma })\equiv \{\phi :{\rho }_{\sigma }\phi \in L^{\mathrm{\infty }}({\mathbb{R}}^N)\}$ with ${\rho }_{\sigma }(x)={(1+{|x|}^2)}^{-\frac{\sigma }{2}}$.

After the famous work [1], this problem has been widely studied by several authors. It is well known that any positive solutions blow up in finite time if $1<p\le p_F\equiv 1+\frac{2}{N}$ [1–3], while positive global solutions exist if $p>p_F$ [1, 4]. Let

$$p_c=\{\begin{array}{cc}\frac{{(N-2)}^2-4N+8\sqrt{N-1}}{(N-2)(N-10)},\hfill & \text{if }N>10,\hfill \\ \mathrm{\infty },\hfill & \text{if }1\le N\le 10.\hfill \end{array}$$

If $p\ge p_c$, the existence of growing up global solutions, the solutions $u(x,t)$ exist for any $(x,t)\in {\mathbb{R}}^N\times (0,\mathrm{\infty })$ and $u(x,t)\to \mathrm{\infty }$ as $t\to \mathrm{\infty }$ in some senses, has been established by Poláčik and Yanagida [5, 6]. If $p>p_c$ and the initial data $u_0$ satisfy some conditions, Fila, Winkler and Yanagida [7] in 2004 precisely evaluated the grow-up rate of solutions of (1.1)-(1.2) and they found that for large t and some $\ell >0$, the solution $u(x,t)$ satisfies

$$C_1{t}^{\ell }\le {\parallel u(\cdot ,t)\parallel }_{L^{\mathrm{\infty }}({\mathbb{R}}^N)}\le C_2{t}^{\ell },$$

see also [8]. For the Cauchy-Dirichlet problem of (1.1), the existence of growing up global solutions and the grow-up rate of solutions has been investigated by Dold, Galaktionov, Lacey and Vázquez in [9], Galaktionov and King in [10]. If $p>1+\frac{2}{N}$, there are also a lot of papers which intensely investigate the solutions of (1.1)-(1.2) converging to zero at different algebraic rates [11–16].

For the sublinear case ($0<p<1$ in (1.1)), it was Aguirre and Escobedo [17] who first proved that if $0<\sigma <\mathrm{\infty }$, and the initial value $u_0$ satisfies

$$0\le u_0(x)\in L^{\mathrm{\infty }}({\rho }_{\sigma }),$$

then the solutions $u(x,t)$ of (1.1)-(1.2) are global.

Our interest in this paper is to investigate the grow-up rate of solutions for the problem (1.1)-(1.2) with a sublinear source. We first show that if the initial value $u_0$ satisfies

$$0\le u_0\in L^{\mathrm{\infty }}({\rho }_{\sigma })$$

(1.3)

and

$$\underset{|x|\to \mathrm{\infty }}{lim}{|x|}^{-\sigma }{u}_0(x)=A\text{for some }A>0,$$

(1.4)

then the solutions of (1.1)-(1.2) ($0<p<1$) are growing up solutions such that

$$C_1{t}^{\frac{{\ell }_1}{2}}\le {\parallel u(t)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}\le C_2{t}^{\frac{{\ell }_2}{2}}$$

(1.5)

for large t. Here ${\ell }_1={\ell }_2=\sigma$ if $\sigma >\frac{2}{1-p}$, and ${\ell }_1=\frac{2}{1-p}<{\ell }_2\le \frac{2}{1-p}+ϵ$ for any $ϵ>0$ if $0<\sigma \le \frac{2}{1-p}$. Moreover, as an application of these results, we get that if $\frac{2}{1-p}<\sigma <\mathrm{\infty }$ and the initial value $u_0$ satisfies (1.3), (1.4), then the effect of the sublinear source is negligible in the asymptotic behavior of solutions as $t\to \mathrm{\infty }$. While for $\sigma =\frac{2}{1-p}$, Aguirre and Escobedo [17] revealed that the effect of the sublinear source cannot be negligible in the asymptotic behavior of the solutions as $t\to \mathrm{\infty }$. For the absorption case ($u^p$ is replaced by $-u^p$ in (1.1)) and the supercritical case ($p>1+\frac{2}{N}$ in (1.1)), some similar results about the asymptotic behavior of solutions for these problems were established by a lot of papers, see [18–20].

The paper is organized as follows. The next section is devoted to giving the grow-up rate for the solutions of the problem (1.1)-(1.2) with $0<p<1$. In Section 3, we investigate the asymptotic behavior of solutions for the problem (1.1)-(1.2).

2 Growth-up rate of solutions

We take $0<p<1$ in the rest of this paper. For any $0<\sigma <\mathrm{\infty }$, we define a weighted $L^{\mathrm{\infty }}$ space as

$$L^{\mathrm{\infty }}({\rho }_{\sigma })\equiv \{\phi (x);{\rho }_{\sigma }\phi \in L^{\mathrm{\infty }}\left({\mathbb{R}}^N\right)\}$$

with the norm ${\parallel \phi \parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}={\parallel {\rho }_{\sigma }\phi \parallel }_{L^{\mathrm{\infty }}({\mathbb{R}}^N)}$, where ${\rho }_{\sigma }(x)={(1+{|x|}^2)}^{-\frac{\sigma }{2}}$. If ${(1+{|x|}^2)}^{\frac{\sigma }{2}}\le u_0\le C{(1+{|x|}^2)}^{\frac{\sigma }{2}}$, then there exist two subsolutions of the problem (1.1)-(1.2):

$$t\to S(t)u_0(x)={(4\pi t)}^{-\frac{N}{2}}{\int }_{{\mathbb{R}}^N}{exp}^{(-\frac{{|x-y|}^2}{4t})}{u}_0(y)\mathrm{d}y$$

(2.1)

and

$$t\to {((1-p)t)}^{1/(p-1)}.$$

(2.2)

Using a similar method as in [21] (see the Appendix), we can get that there exist constants $C_1,C_2>0$ such that

$$C_1{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}\le S(t)u_0(x)\le C_2{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}.$$

(2.3)

So, for any $x\in {\mathbb{R}}^N$, those two growing up effects given by (2.1) and (2.2) can be compared as $t\to \mathrm{\infty }$. When $0<\sigma <\frac{2}{1-p}$, the one given by (2.2) prevails; when $\frac{2}{1-p}<\sigma <\mathrm{\infty }$, the one given by (2.1) prevails; and they coincide in the critical case $\sigma =\frac{2}{1-p}$.

Inspired by the above discussions, in this paper we first study the grow-up rate of solutions for the problem (1.1)-(1.2). The mild solution $u(x,t)$ of the problem (1.1)-(1.2) is defined as follows:

$$u(x,t)=S(t)u_0(x)+{\int }_0^{t}S(t-s)u^p(x,s)\mathrm{d}s.$$

(2.4)

If the initial value $0\not\equiv u_0\in L^{\mathrm{\infty }}({\rho }_{\sigma })$, the existence and uniqueness of a mild solution for the problem (1.1)-(1.2) has been given in [17].

Lemma 2.1 ([17])

Suppose $0\le u_0\in L^{\mathrm{\infty }}({\rho }_{\sigma })$ and $u_0\not\equiv 0$, then there exists a unique mild global solution u for the problem (1.1)-(1.2) with $0<p<1$ such that

• I. $u\in C^{\mathrm{\infty }}((0,\mathrm{\infty })\times {\mathbb{R}}^N)\cap L_{\mathrm{loc}}^{\mathrm{\infty }}((0,\mathrm{\infty });L^{\mathrm{\infty }}({\rho }_{\sigma }))$;

• II. ${lim}_{t\to 0}u(x,t)=u_0(x)$ for a.e. $x\in {\mathbb{R}}^N$.

Moreover, if $u_0\in C({\mathbb{R}}^N)$, the convergence is uniform on compact subsets of ${\mathbb{R}}^N$.

Our results about the grow-up rate of solutions are the following two theorems.

Theorem 2.1 Let $0<p<1$, $A>0$ and $\frac{2}{1-p}<\sigma <\mathrm{\infty }$. Suppose

$$0\le u_0\in L^{\mathrm{\infty }}({\rho }_{\sigma })$$

(2.5)

and

$$\underset{|x|\to \mathrm{\infty }}{lim}{|x|}^{-\sigma }{u}_0(x)=A.$$

(2.6)

Then there exist constants $T,C_1,C_2>0$, such that

$$C_1{t}^{\frac{\sigma }{2}}\le {\parallel u(t)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}\le C_2{t}^{\frac{\sigma }{2}}\mathit{\text{for}}t>T.$$

(2.7)

Here $u(x,t)$ is the solution of (1.1)-(1.2).

Proof The hypothesis (2.6) clearly implies that there exists a constant $R>0$ such that if $|x|\ge R$, then

$$\frac{A}{2}{|x|}^{\sigma }\le u_0(x)\le 2A{|x|}^{\sigma }.$$

So,

$$u_0(x)\ge \frac{A}{2}{|x|}^{\sigma }-\frac{A}{2}{R}^{\sigma }.$$

From the property of the heat semigroup, we have

$$S(t)u_0(x)\ge S(t)\phi (x)-\frac{A}{2}{R}^{\sigma },$$

where $\phi (x)=\frac{A}{2}{|x|}^{\sigma }$. Using a similar method as [21] (see (A.5)), we obtain that there exists a constant $C_0>0$ such that

$$S(\tau )\phi (x)\ge C_0{(\tau +{|x|}^2)}^{\frac{\sigma }{2}}.$$

So, for $\tau =1+2{(\frac{A}{2C_0})}^{\frac{2}{\sigma }}{R}^2$, there exists a constant $C>0$ (depending on A and σ) such that

$$S(\tau )u_0(x)\ge C{(1+{|x|}^2)}^{\frac{\sigma }{2}}.$$

It follows from the comparison principle that

$$u(x,\tau )\ge S(\tau )u_0(x)\ge C{(1+{|x|}^2)}^{\frac{\sigma }{2}}.$$

From $0\le u_0\in L^{\mathrm{\infty }}({\rho }_{\sigma })$ and I of Lemma 2.1, we obtain that there exists a constant $C>0$ (depending on τ) such that

$$u(x,t){(1+{|x|}^2)}^{-\frac{\sigma }{2}}\le \underset{0\le s\le \tau }{sup}{\parallel u(s)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}\le C\text{for }0\le t\le \tau .$$

Therefore,

$$u(x,t)\le C{(1+{|x|}^2)}^{\frac{\sigma }{2}}\text{for }0\le t\le \tau .$$

So, from (2.3), we have

$$C_1{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}\le S(t)[u(\tau )](x)\le C_2{(1+t+{|x|}^2)}^{\frac{\sigma }{2}},$$

(2.8)

where $C_1$ and $C_2$ are positive constants depending on A, σ and τ. The hypothesis $\frac{2}{1-p}<\sigma <\mathrm{\infty }$ indicates that

$$\sigma (1-p)-2>0.$$

Let

$$a(t)={\left[(1+C_1^{p-1}(1-p){\int }_0^t{(1+s)}^{\frac{\sigma (p-1)}{2}}\mathrm{d}s)\right]}^{\frac{1}{1-p}}.$$

So,

$$\eta \equiv C_1^{p-1}(1-p){\int }_0^{\mathrm{\infty }}{(1+t)}^{\frac{\sigma (p-1)}{2}}\mathrm{d}t=\frac{2C_1^{p-1}(1-p)}{\sigma (1-p)-2}>0.$$

Therefore, $a(t)$ is an increasing function satisfying

$$\{\begin{array}{c}a(0)=1,\hfill \\ a(t)\le {(1+\eta )}^{\frac{1}{1-p}}\hfill & \text{for all }t\ge 0.\hfill \end{array}$$

(2.9)

From (2.8), we have

$$\begin{array}{rcl}{a}^{\prime }(t)& =& C_1^{p-1}a{(t)}^p{(1+t)}^{\frac{\sigma (p-1)}{2}}\\ =& a{(t)}^p{\left[C_1{(1+t)}^{\frac{\sigma }{2}}\right]}^{p-1}\ge a{(t)}^p{[S(t)u(\tau )(x)]}^{p-1}.\end{array}$$

(2.10)

Let $w(x,t)=S(t)u(\tau )(x)$, and assume that

$$\overline{w}(x,t)=a(t)w(x,t).$$

So, from (2.10), one can verify that $\overline{w}(x,t)$ is a supersolution of the following problem:

By (2.8), (2.9) and the comparison principle, we get that

$$\begin{array}{rcl}{C}_1{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}& \le & w(x,t)\le v(x,t)\le \overline{w}(x,t)\\ =& a(t)w(x,t)\le {(1+\eta )}^{\frac{1}{1-p}}w(x,t)\le C_2{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}.\end{array}$$

This means that

$$C_1{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}\le u(x,t+\tau )\le C_2{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}\text{for }t>\tau .$$

Let $T=2\tau +1$. So, there exist two constants, which we still write as $C_1$ and $C_2$, such that

$$C_1{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}\le u(x,t)\le C_2{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}\text{for }t>T.$$

(2.11)

From this, we get (2.7) easily. So we complete the proof of this theorem. □

In the following theorem, we consider the grow-up rate for the solutions of the problem (1.1)-(1.2) when the nonnegative initial value $u_0\in L^{\mathrm{\infty }}({\rho }_{\sigma })$ with $0<\sigma \le \frac{2}{1-p}$.

Theorem 2.2 Let $0<p<1$, and assume that $0<\sigma \le \frac{2}{1-p}$. If the initial value $u_0$ satisfies (2.5) and (2.6), then for any $ϵ>0$, there exist constants $C_1,C_2,T>0$ such that

$$C_1{t}^{\frac{1}{1-p}}\le {\parallel u(t)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}\le C_2{t}^{\frac{1}{1-p}+ϵ}\mathit{\text{for}}t>T.$$

Here $u(x,t)$ is also the solution of (1.1)-(1.2).

Proof Using the same method as the proof of (2.8), we can get if $0<\sigma \le \frac{2}{1-p}$, then there exist $C_1,C_2,T_1>0$ such that

$$C_1{(t+{|x|}^2)}^{\frac{\sigma }{2}}\le S(t)u_0(x)\le C_2{(t+{|x|}^2)}^{\frac{\sigma }{2}}\text{for }t>T_1\text{ and }x\in {\mathbb{R}}^N.$$

(2.12)

So, by the comparison principle, we can get that there exists a constant $\tau >T_1$ satisfying

$$u(x,\tau )\ge S(\tau )u_0(x)\ge C_1{(1+{|x|}^2)}^{\frac{\sigma }{2}}.$$

We first consider the case of $N>1$. Let

$$B={(1-p)}^{-1}{C}_1^{1-p}\text{and}\underline{w}(x,t)={(1-p)}^{\frac{1}{1-p}}{(B{(1+{|x|}^2)}^{\frac{\sigma (1-p)}{2}}+t)}^{\frac{1}{1-p}}.$$

So, $\underline{w}$ is a subsolution of the following problem:

(2.13)

Here we have used the facts that $\underline{w}(x,0)=C_1{(1+{|x|}^2)}^{\frac{\sigma }{2}}\le u(x,\tau )$ and $N>1$. Therefore, by the comparison principle, for $t>0$, there exists a constant C satisfying

$$C{({(1+{|x|}^2)}^{\frac{\sigma (1-p)}{2}}+t)}^{\frac{1}{1-p}}\le \underline{w}(x,t)\le u(x,t+\tau ).$$

From this, we can get that there exist $C,T>0$ such that

$$C{(1+t)}^{\frac{1}{1-p}}\le {\parallel u(t)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}\text{for }t>T.$$

(2.14)

Now, we consider the case of $N=1$. Let

$$C_3=min(C_1,{(2\sigma )}^{-\frac{1}{1-p}})\text{and}{B}_1={(1-p)}^{-1}{C}_3^{1-p}.$$

Then, we define the function

$$\underline{w_1}(x,t)={(1-p)}^{\frac{1}{1-p}}{(B_1{(1+{|x|}^2)}^{\frac{\sigma (1-p)}{2}}+\frac{1}{2}t)}^{\frac{1}{1-p}}.$$

Therefore, $\underline{w_1}$ is also a subsolution of the problem (2.13). In fact,

$$\frac{\partial \underline{w_1}}{\partial t}-{\underline{w_1}}^p=-\frac{1}{2}{(1-p)}^{\frac{p}{1-p}}{(B_1{(1+{|x|}^2)}^{\frac{\sigma (1-p)}{2}}+\frac{1}{2}t)}^{\frac{p}{1-p}}$$

and

$$\frac{{\partial }^2\underline{w_1}}{\partial x^2}\ge -{(1-p)}^{\frac{1}{1-p}}{B}_1\sigma {(B_1{(1+x^2)}^{\frac{\sigma (1-p)}{2}}+\frac{1}{2}t)}^{\frac{p}{1-p}},$$

so,

$$\frac{\partial \underline{w_1}}{\partial t}-{\underline{w_1}}^p-\frac{{\partial }^2\underline{w_1}}{\partial x^2}\le 0.$$

Using the same method as above, we can get that (2.14) holds for $N=1$. Without loss of generality, we can assume that $t>T$ in the rest of this proof. From the definition of the mild solutions with (2.12), we have

$$\begin{array}{rl}u(x,t)& =S(t)u_0(x)+{\int }_0^{t}S(t-s)u^p(x,s)\mathrm{d}s\\ \le C{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}+{\int }_0^{t}S(t-s){[{(1+{|x|}^2)}^{\frac{\sigma }{2}}u(x,s){(1+{|x|}^2)}^{-\frac{\sigma }{2}}]}^p\mathrm{d}s\\ \le C{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}+C\underset{0\le s\le t}{sup}{\parallel u(s)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}^p{\int }_0^t{(1+{|x|}^2+(t-s))}^{\frac{\sigma p}{2}}\mathrm{d}s\\ \le C{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}+C\underset{0\le s\le t}{sup}{\parallel u(s)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}^p{(1+{|x|}^2+t)}^{\frac{\sigma p}{2}}t\\ \le C{(1+{|x|}^2)}^{\frac{\sigma }{2}}{(1+t)}^{\frac{\sigma }{2}}+C\underset{0\le s\le t}{sup}{\parallel u(s)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}^p{(1+{|x|}^2)}^{\frac{\sigma p}{2}}{(1+t)}^{\frac{\sigma p}{2}+1}\\ \le C{(1+{|x|}^2)}^{\frac{\sigma }{2}}{(1+t)}^{\frac{\sigma }{2}}+C\underset{0\le s\le t}{sup}{\parallel u(s)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}^p{(1+{|x|}^2)}^{\frac{\sigma }{2}}{(1+t)}^{\frac{\sigma p}{2}+1}.\end{array}$$

Here we have used $0<p<1$ and Lemma A.1, see the Appendix. The assumption $0<\sigma \le \frac{2}{1-p}$ implies that

$$\frac{\sigma }{2}\le \frac{\sigma p}{2}+1.$$

Therefore,

$${(1+t)}^{\frac{\sigma }{2}}\le {(1+t)}^{\frac{\sigma p}{2}+1}.$$

By (2.14), we deduce that there exists a constant C such that

$$u(x,t)\le C\underset{0\le s\le t}{sup}{\parallel u(s)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}^p{(1+{|x|}^2)}^{\frac{\sigma }{2}}{(1+t)}^{\frac{\sigma p}{2}+1}.$$

This implies that

$$\underset{0\le s\le t}{sup}{\parallel u(s)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}\le C{(1+t)}^{\frac{\sigma p}{2(1-p)}+\frac{1}{1-p}}.$$

Using the integral expression (2.4) again, we have

$$\begin{array}{rl}u(x,t)& \le C{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}+C\underset{0\le s\le t}{sup}{\parallel u(s)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}^p{\int }_0^t{(1+t-s+{|x|}^2)}^{\frac{\sigma p}{2}}\mathrm{d}s\\ \le C{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}+C{(1+t)}^{\frac{\sigma p^2}{2(1-p)}+\frac{p}{1-p}}{(1+t+{|x|}^2)}^{\frac{\sigma p}{2}}t.\end{array}$$

Here we have used the fact that

$${(1+{|x|}^2+t)}^{\alpha }\le {(1+t)}^{\alpha }{(1+{|x|}^2)}^{\alpha }\text{for }\alpha >0.$$

Notice that for $t>s$ and $m>0$,

$$S(t-s)\varphi (x)\le C{(1+t+{|x|}^2)}^{\frac{m}{2}},$$

where $\varphi (x)={(1+s+{|x|}^2)}^{\frac{m}{2}}$. So,

$$\begin{array}{rcl}u(x,t)& =& S(t)u_0(x)+{\int }_0^{t}S(t-s)u^p(x,s)\mathrm{d}s\le C{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}\\ +C{\int }_0^t[{(1+t+{|x|}^2)}^{\frac{p\sigma }{2}}+{(1+s)}^{\frac{\sigma p^3}{2(1-p)}+\frac{p^2}{1-p}}{(1+t+{|x|}^2)}^{\frac{\sigma p^2}{2}}{s}^p]\mathrm{d}s\\ \le & C[{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}+{(1+t+{|x|}^2)}^{\frac{p\sigma }{2}}t\\ +{(1+t)}^{\frac{\sigma p^3}{2(1-p)}+\frac{p^2}{1-p}}{(1+t+{|x|}^2)}^{\frac{\sigma p^2}{2}}{t}^{1+p}].\end{array}$$

(2.15)

Iterating (2.15) $n-1$ times, we get that

$$\begin{array}{rcl}u(x,t)& \le & C[{(1+t+{|x|}^2)}^{\frac{\sigma }{2}}+{(1+t+{|x|}^2)}^{\frac{p\sigma }{2}}t+{(1+t+{|x|}^2)}^{\frac{p^2\sigma }{2}}{t}^{1+p}+\cdots \\ +{(1+t+{|x|}^2)}^{\frac{p^{n-1}\sigma }{2}}{t}^{\frac{1-p^{n-1}}{1-p}}\\ +{(1+t)}^{[\frac{\sigma p}{2(1-p)}+\frac{1}{1-p}]p^n}{(1+t+{|x|}^2)}^{\frac{\sigma p^n}{2}}{t}^{\frac{1}{1-p}-\frac{p^n}{1-p}}]\\ \le & C[{(1+{|x|}^2)}^{\frac{\sigma }{2}}{(1+t)}^{\frac{\sigma }{2}}+{(1+{|x|}^2)}^{\frac{p\sigma }{2}}{(1+t)}^{1+\frac{p\sigma }{2}}\\ +{(1+{|x|}^2)}^{\frac{p^2\sigma }{2}}{(1+t)}^{\frac{1-p^2}{1-p}+\frac{p^2\sigma }{2}}+\cdots +{(1+{|x|}^2)}^{\frac{p^{n-1}\sigma }{2}}{(1+t)}^{\frac{1-p^{n-1}}{1-p}+\frac{p^{n-1}\sigma }{2}}\\ +{(1+t)}^{\frac{\sigma p^n}{2(1-p)}+\frac{1}{1-p}}{(1+{|x|}^2)}^{\frac{\sigma p^n}{2}}]\\ \le & C(n){(1+{|x|}^2)}^{\frac{\sigma }{2}}[{(1+t)}^{\frac{1}{1-p}+\frac{\sigma p^n}{2(1-p)}}+{(1+t)}^{\frac{1}{1-p}+\frac{\sigma p^{n-1}}{2}}].\end{array}$$

(2.16)

Here we have used the facts that

$$\frac{\sigma }{2}\le \frac{p\sigma }{2}+1\le \frac{p^2\sigma }{2}+\frac{1-p^2}{1-p}\le \cdots \le \frac{p^{n-1}\sigma }{2}+\frac{1-p^{n-1}}{1-p}\le \frac{p^{n-1}\sigma }{2}+\frac{1}{1-p}$$

and

$${(1+{|x|}^2)}^{\frac{\sigma p^m}{2}}\le {(1+{|x|}^2)}^{\frac{\sigma }{2}}\text{for }m>0.$$

So, for any $ϵ>0$, we can select n large enough to satisfy

$$ϵ>max(\frac{p^n\sigma }{2(1-p)},\frac{p^{n-1}\sigma }{2}).$$

From (2.16), we thus have

$${\parallel u(t)\parallel }_{L^{\mathrm{\infty }}({\rho }_{\sigma })}\le C{(1+t)}^{\frac{1}{1-p}+ϵ}.$$

Combining this with (2.14), we can get the desired results. So we complete the proof of this theorem. □

Remark 2.1 From Theorem 2.1 and Theorem 2.2, we find that if $\sigma >\frac{2}{1-p}$, then the main effect on the growing up of solutions comes from the initial value; while if $0<\sigma \le \frac{2}{1-p}$, then the sublinear source has a major effect.

3 Asymptotic behavior

In this section, we will use the fact that the mild solutions of the problem (1.1)-(1.2) given by Lemma 2.1 also satisfy the following integral identity:

$${\int }_0^T{\int }_{{\mathbb{R}}^N}[(\frac{\partial \xi }{\partial t}+\mathrm{\Delta }\xi )u+\xi u^p]\mathrm{d}x\mathrm{d}t+{\int }_{{\mathbb{R}}^N}\xi (x,0)u_0(x)\mathrm{d}x=0,$$

(3.1)

for any $\xi \in C^{1,2}([0,T]\times {\mathbb{R}}^N)$ which vanishes for large $|x|$ and at $t=T$.

The following result gives the fact that if $\sigma >\frac{2}{1-p}$, then the sublinear source is negligible in the asymptotic behavior of the rescaled solution $t^{-\frac{\sigma }{2}}u(x,t)$ as $t\to \mathrm{\infty }$. Similar to [19, 22, 23], we follow the framework by Kamin and Peletier [19] to give the proof of our result.

Theorem 3.1 Let $0<p<1$ and $\frac{2}{1-p}<\sigma <\mathrm{\infty }$. If the initial value $u_0$ satisfies (2.5) and (2.6), then

$$\underset{t\to \mathrm{\infty }}{lim}{t}^{-\frac{\sigma }{2}}|u(x,t)-S(t)\phi (x)|=0$$

(3.2)

uniformly on sets $\{(y,s);|y|\le \gamma s^{\frac{1}{2}}\}$, $\gamma >0$. Here $u(x,t)$ is the mild solution of (1.1)-(1.2) and $\phi (x)=A{|x|}^{\sigma }$.

Proof We first define the functions

and

$$w_{\lambda }(x,t)={\lambda }^{-\sigma }\left[S\left({\lambda }^{2}t\right)u_0\right](\lambda x).$$

Using the comparison principle, we know that

$$w(x,t)\le u(x,t),$$

and

$$w_{\lambda }(x,t)\le u_{\lambda }(x,t)\text{for all }\lambda \ge 1.$$

For $t>0$, without loss of generality, we can assume that λ is large enough to satisfy $\lambda t>T$, where T is the constant given by Theorem 2.1. So, from (2.11), we have

$$\begin{array}{rcl}{u}_{\lambda }(x,t)& \le & C{\lambda }^{-\sigma }{[1+{\lambda }^{2}t+{\lambda }^2{|x|}^2]}^{\frac{\sigma }{2}}\le C{({\lambda }^{-2}+t+{|x|}^2)}^{\frac{\sigma }{2}}\\ \le & C{({\lambda }^{-2}+t)}^{\frac{\sigma }{2}}{(1+{({\lambda }^{-2}+t)}^{-1}{|x|}^2)}^{\frac{\sigma }{2}}\\ \le & C{({\lambda }^{-2}+t)}^{\frac{\sigma }{2}}{(1+{({\lambda }^{-2}+t)}^{-\frac{1}{2}}|x|)}^{\sigma }.\end{array}$$

(3.3)

So, if $\lambda >2$ and $0<\tau <\frac{1}{4}$, then

$$\begin{array}{rcl}{\int }_0^{\tau }{\int }_{B_1}{u}_{\lambda }(x,t)\mathrm{d}x\mathrm{d}t& \le & C{\int }_{{\lambda }^{-2}}^{\tau +{\lambda }^{-2}}{s}^{\frac{N+\sigma }{2}}{\int }_0^{s^{-\frac{1}{2}}}{(1+r)}^{N+\sigma -1}\mathrm{d}r\mathrm{d}s\\ \le & C{\int }_{{\lambda }^{-2}}^{\tau +{\lambda }^{-2}}{s}^{\frac{N+\sigma }{2}}[{(1+s^{-\frac{1}{2}})}^{N+\sigma }-1]\mathrm{d}s\\ \le & C{\int }_{{\lambda }^{-2}}^{\tau +{\lambda }^{-2}}{s}^{\frac{N+\sigma }{2}}{\left(2s^{-\frac{1}{2}}\right)}^{N+\sigma }\mathrm{d}s\le C\tau .\end{array}$$

Similarly, for any $q>0$, from (3.3), we can obtain the following integral estimates:

$$\begin{array}{rcl}{\int }_0^{\tau }{\int }_{B_1}{u}_{\lambda }{(x,t)}^q\mathrm{d}x\mathrm{d}t& \le & C{\int }_{{\lambda }^{-2}}^{\tau +{\lambda }^{-2}}{s}^{\frac{N+q\sigma }{2}}[{(1+s^{-\frac{1}{2}})}^{N+q\sigma }-1]\mathrm{d}s\\ \le & C{\int }_{{\lambda }^{-2}}^{\tau +{\lambda }^{-2}}{(1+s^{\frac{1}{2}})}^{\frac{N+\sigma }{2}}\mathrm{d}s\le C{(1+\tau +{\lambda }^{-2})}^{N+q\sigma }{\int }_{{\lambda }^{-2}}^{\tau +{\lambda }^{-2}}\mathrm{d}s\\ \le & C{(1+\tau +{\lambda }^{-2})}^{N+q\sigma }\tau .\end{array}$$

(3.4)

Using the same method as above and the comparison principle, we can get the similar integral estimates for $w_{\lambda }(x,t)$. For any $T_1>t>0$, from (3.1), we have

$${\iint }_{S_{\tau }+S_{T_1}^{\tau }}[{\xi }_t(u_{\lambda }-w_{\lambda })+\mathrm{\Delta }\xi (u_{\lambda }-w_{\lambda })]\mathrm{d}t\mathrm{d}x={\iint }_{\overline{S_{T_1}}}{\lambda }^{-\kappa }\xi u_{\lambda }^p\mathrm{d}x\mathrm{d}t,$$

(3.5)

where $S_{\tau }\equiv (0,\tau ]\times {\mathbb{R}}^N$, $S_{T_1}^{\tau }\equiv (\tau ,T_1]\times {\mathbb{R}}^N$, $\kappa =\sigma (1-p)-2>0$ and $\xi \in C^{1,2}(\overline{S_{T_1}})$ which vanishes for large $|x|$ and at $t=T_1$. For any $ϵ>0$, by the integral estimates of $u_{\lambda }(x,t)$ and $w_{\lambda }(x,t)$, there exists $\tau >0$ such that

$${\iint }_{S_{\tau }}[{\xi }_t(w_{\lambda }-u_{\lambda })+\mathrm{\Delta }\xi (w_{\lambda }-u_{\lambda })]\mathrm{d}x\mathrm{d}t<\frac{ϵ}{3}.$$

(3.6)

From the fact that $\kappa >0$ and (3.4), we can get that there exists ${\lambda }_1$ such that if $\lambda >{\lambda }_1$, then

$${\iint }_{S_{T_1}}{\lambda }^{-\kappa }\xi u_{\lambda }^p\mathrm{d}x\mathrm{d}t<\frac{ϵ}{3}.$$

(3.7)

Let $z_{\lambda }(x,t)=u_{\lambda }(x,t)-w_{\lambda }(x,t)$. From (3.1), we can get that for any compact subset K of $S_{T_1}$, $u_{\lambda }(x,t)$ and $w_{\lambda }(x,t)$ have a uniform upper bound, which means that the sequence $z_{\lambda }(x,t)$ is equicontinuous on K (see [17, 24, 25]). So, we can get that there exist a subsequence $z_{{\lambda }^{\prime }}(x,t)$ and a function $z(x,t)\in C(S_{T_1})$ such that

$$z_{{\lambda }^{\prime }}(x,t)\to z(x,t)$$

as ${\lambda }^{\prime }\to \mathrm{\infty }$ uniformly on K. Therefore, we have as ${\lambda }^{\prime }\to \mathrm{\infty }$, omitting the primes,

$${\iint }_{S_{T_1}^{\tau }}[{\xi }_t(u_{\lambda }-w_{\lambda })+\mathrm{\Delta }\xi (u_{\lambda }-w_{\lambda })]\mathrm{d}t\mathrm{d}x\to {\iint }_{S_{T_1}^{\tau }}[{\xi }_t+\mathrm{\Delta }\xi ]z\mathrm{d}t\mathrm{d}x.$$

Combining this with (3.5)-(3.7), we obtain that

$${\iint }_{S_{T_1}}[{\xi }_t+\mathrm{\Delta }\xi ]z\mathrm{d}t\mathrm{d}x=0.$$

Therefore, it follows from the uniqueness of the solutions of the heat equation that

$$z(x,t)=0\text{for all }(x,t)\in S_{T_1}.$$

Thus the entire sequence $z_{\lambda }$ converges to $z=0$. Therefore, we have proved that for any $0<t<T<\mathrm{\infty }$,

$$u_{\lambda }(x,t)-w_{\lambda }(x,t)\to 0\text{as }\lambda \to \mathrm{\infty }$$

uniformly on any compact subset of ${\mathbb{R}}^N$. Thus, taking $t=1$ and $\lambda =s^{\frac{1}{2}}$, we obtain

$$\underset{s\to \mathrm{\infty }}{lim}{s}^{-\frac{\sigma }{2}}{\parallel u(s^{\frac{1}{2}}\cdot ,s)-w(s^{\frac{1}{2}}\cdot ,s)\parallel }_{L_{\mathrm{loc}}^{\mathrm{\infty }}({\mathbb{R}}^N)}=0.$$

(3.8)

From (2.8) and $0\le u_0\in L^{\mathrm{\infty }}({\rho }_{\sigma })$, we have

$$w_{t^{\frac{1}{2}}}(x,1)=t^{-\frac{\sigma }{2}}S(t)u_0\left(t^{\frac{1}{2}}x\right)\le C{(t^{-\frac{1}{2}}+1+{|x|}^2)}^{\frac{\sigma }{2}}.$$

So, for any $x\in {\mathbb{R}}^N$, by Lebesgue’s dominated convergence theorem, we have

$$\begin{array}{rcl}{w}_{t^{\frac{1}{2}}}(x,1)& =& t^{-\frac{\sigma }{2}}S(t)u_0\left(t^{\frac{1}{2}}x\right)\\ =& {(4\pi )}^{-\frac{N}{2}}{\int }_{{\mathbb{R}}^N}exp(-\frac{{|x-y|}^2}{4})t^{-\frac{\sigma }{2}}{u}_0\left(t^{\frac{1}{2}}y\right)\mathrm{d}y\to S(1)\phi (x)\end{array}$$

(3.9)

as $t\to \mathrm{\infty }$. The uniform upper bound of $w_{t^{\frac{1}{2}}}(x,1)$ on any compact subset M of ${\mathbb{R}}^N$ implies that the sequence $w_{t^{\frac{1}{2}}}(x,1)$ is equicontinuous on M, see [17, 24, 25]. Therefore, from (3.9), we have

$$w_{t^{1/2}}(x,1)=t^{-\frac{\sigma }{2}}S(t)u_0\left(t^{\frac{1}{2}}x\right)\to S(1)\phi (x)$$

uniformly on any compact sets of ${\mathbb{R}}^N$ as $t\to \mathrm{\infty }$. By (3.8), we thus have (3.2). So we complete the proof of this theorem. □

Appendix

Lemma A.1 Let $M_1,M_2>0$ and $0<\sigma <\mathrm{\infty }$. If

$$M_1{(1+{|x|}^2)}^{\frac{\sigma }{2}}\le u_0\le M_2{(1+{|x|}^2)}^{\frac{\sigma }{2}},$$

(A.1)

then there exist two constants $C(M_1,\sigma ),C(M_2,\sigma )>0$ such that

$$C(M_1,\sigma ){(1+t+{|x|}^2)}^{\frac{\sigma }{2}}\le S(t)u_0(x)\le C(M_2,\sigma ){(1+t+{|x|}^2)}^{\frac{\sigma }{2}}.$$

(A.2)

Proof Consider the following problem:

where $M>0$ is a constant. For $\lambda >1$, from (2.1), we can get

$${\lambda }^{-\frac{\sigma }{2}}[S(\lambda t)v_0]\left({\lambda }^{\frac{1}{2}}x\right)=S(t)\left[{\lambda }^{-\frac{\sigma }{2}}{v}_0({\lambda }^{\frac{1}{2}}\cdot )\right](x)=S(t)v_0(x).$$

(A.3)

By existence and regularity theories for solutions, we can obtain that for $t>0$,

$$0<S(t)v_0\in C^{\mathrm{\infty }}((0,\mathrm{\infty })\times {\mathbb{R}}^N),$$

see [24, 25]. Now taking $t=1$, $\lambda =s$ and $g(x)=S(1)v_0(x)$ in the expression (A.3), we have

$$S(s)v_0(x)=s^{\frac{\sigma }{2}}g\left(s^{-\frac{1}{2}}x\right).$$

(A.4)

The fact that $S(s)v_0(x)\in C([0,\mathrm{\infty })\times {\mathbb{R}}^N)$ clearly implies that for $|x|=1$,

$$s^{\frac{\sigma }{2}}g\left(s^{-\frac{1}{2}}x\right)=S(s)v_0(x)\to v_0(x)=M{|x|}^{\sigma }=M$$

as $s\to 0$. Let

$$y=s^{-\frac{1}{2}}x.$$

So

$$|y|\to \mathrm{\infty }\text{as }s\to 0.$$

Therefore,

$${|y|}^{-\sigma }g(y)-M\to 0$$

as $|y|\to \mathrm{\infty }$. So, there exist constants $0<C_1(M)\le C_2(M)<\mathrm{\infty }$ satisfying

$$C_1(M){(1+{|x|}^2)}^{\frac{\sigma }{2}}\le g(x)\le C_2(M){(1+{|x|}^2)}^{\frac{\sigma }{2}}.$$

By (A.4), we thus have

$$C_1(M){(s+{|x|}^2)}^{\frac{\sigma }{2}}\le S(s)v_0(x)\le C_2(M){(s+{|x|}^2)}^{\frac{\sigma }{2}}.$$

(A.5)

Let $\phi (x)=M{(1+{|x|}^2)}^{\frac{\sigma }{2}}$. So there exist two constants $C_1(M,\sigma ),C_2(M,\sigma )>0$ such that

$$C_1(M,\sigma )(1+v_0(x))\le \phi (x)\le C_2(M,\sigma )(1+v_0(x)).$$

Therefore, by the comparison principle and (A.5), we can get that for all $t\ge 0$, there exist constants $C_1(M,\sigma ),C_2(M,\sigma )>0$ such that

$$C_1(M,\sigma ){(1+s+{|x|}^2)}^{\frac{\sigma }{2}}\le S(t)\phi (x)\le C_2(M,\sigma ){(1+s+{|x|}^2)}^{\frac{\sigma }{2}}.$$

By (A.1) and the comparison principle, we have (A.2). So we complete the proof of this lemma. □