Initial and Boundary Blow-Up Problem for $p$-Laplacian Parabolic Equation with General Absorption

Abstract

In this article, we investigate the initial and boundary blow-up problem for the $p$-Laplacian parabolic equation $u_t-\Delta _p u=-b(x,t)f(u)$ over a smooth bounded domain $\Omega$ of $\mathbb {R}^N$ with $N\ge 2$, where $\Delta _pu=\mathrm{div}(|\nabla u|^{p-2}\nabla u)$ with $p>1$, and $f(u)$ is a function of regular variation at infinity. We study the existence and uniqueness of positive solutions, and their asymptotic behaviors near the parabolic boundary.

Appendix

1.1 Some Basic Results of Regular Variation Theory

In this subsection, we gather some basic results of regular variation theory that are needed in this paper. In most cases, we refer the reader to the basic references (such as [4]) and omit the proofs. However, in certain instances, we feel that we need to provide the proofs as they are not readily available in the literature.

Proposition 5.1

(Representation Theorem) The function $L(u)$ is slowly varying at infinity if and only if it can be written as

$$\begin{aligned} L(u)=M(u)\exp \left\{ \int _b^u\frac{\varphi (t)}{t}\mathrm{d}t\right\} , \ \ \ \forall \ u\ge b \end{aligned}$$

for some $b>0$, where the function $\varphi \in C([b,\infty ))$ satisfies $\displaystyle \lim \limits _{u\rightarrow \infty }\varphi (u)=0$ and $M(u)$ is measurable on $[b,\infty )$ with $\displaystyle \lim \limits _{u\rightarrow \infty }M(u)=M^*\in (0,\infty )$.

Definition 5.1

A function ${\hat{L}}(u)$ is referred to as normalized slowly varying at infinity if it satisfies the requirements in Proposition 5.1 with $M(u)$ replaced by $M^*$. The function $R(u)=M^*u^\rho {\hat{L}}(u)$ is called normalized regularly varying at infinity of index $\rho$ and we write $R(u)\in NRV_\rho$.

We say that $R(u)$ is regularly varying at the origin (from the right) of index $\rho \in \mathbb R$, denoted by $R\in RVZ_\rho$, if $R(1/u)\in RV_{-\rho }$. The set of all normalized regularly varying functions at the origin of index $\rho$ is denoted by $NRVZ_\rho$.

Similar to [7, Remark 2.2], we have

Lemma 5.1

Let $k\in \mathcal {K}_\ell$ with $\ell >0$. Then $k(1/u)$ belongs to $NRV_{(\ell -1)/\ell }$. Furthermore, we have $K(1/u)$ belongs to $NRV_{-1/\ell }$. And hence, $K(u)\in NRVZ_{1/\ell }$ and $k(u)\in NRVZ_{(1-\ell )/\ell }$.

Lemma 5.2

([6, Lemma 2.3]) Suppose that the condition $(F_1)$ holds and the function $\phi$ is given by (1.7). Then

• (i)  $-\phi '(t)=(p'F(\phi (t)))^{1/p}$, and $|\phi '(t)|^{p-2}\phi ''(t)=\frac{p'}{p}f(\phi (t))$, where $p'=\frac{p}{p-1}$;

• (ii)   $-\phi '\in NRVZ_{-r}, \phi \in NRVZ_{1-r}$, where $r=(\rho +1)/(\rho +1-p)>1$.

Following the discussion of [7, Section 2], we can prove

Lemma 5.3

A function $f\in NRV_\rho (\hbox {or }f \in NRVZ_\rho )$ if and only if $f\in C^1[a_1, \infty ) \ ($or $f\in C^1(0, a_1))$ for some $a_1>0$ and $\lim \limits _{s \rightarrow \infty }\frac{sf'(s)}{f(s)}=\rho (\hbox {or }\lim \limits _{s \rightarrow 0^+}\frac{sf'(s)}{f(s)}=\rho )$.

In view of Lemma 5.3 we can prove

Lemma 5.4

Assume that $f\in NRV_\rho \ (\hbox {or }f \in NRVZ_\rho )$ and $\rho \not =0$. If $f'(s)>0$, then the inverse function $f^{-1}(y)$ of $f(s)$ belongs to $NRV_{1/\rho }\ ($or $f^{-1}(y)\in NRVZ_{1/\rho })$. If $f'(s)<0$, then the inverse function $f^{-1}(y)$ of $f(s)$ belongs to $NRVZ_{1/\rho }\ ($or $f^{-1}(y)\in NRV_{1/\rho })$.

Since $\phi \in NRVZ_{1-r}$ and $r>1, K\in RVZ_{1/\ell }$, by Lemma 5.4 we have $\phi ^{-1}\in RV_{1/(1-r)},\, K^{-1}\in RVZ_{\ell }$. Thanks to $k\in RVZ_{(1-\ell )/\ell }$, it follows that

$$\begin{aligned}&k\circ K^{-1}\circ \phi ^{-1}(s)\in RV_{\frac{1-\ell }{1-r}},&\end{aligned}$$

(5.1)

$$\begin{aligned}&f^*(s)=\left( k\circ K^{-1}\circ \phi ^{-1}(s)\right) ^pf(s)\in RV_{\frac{p(1-\ell )}{1-r}+\rho }=RV_{q}.&\end{aligned}$$

(5.2)

In the following, $C$ represents a generic positive constant which can differ from line to line.

Lemma 5.5

Let $\varrho >0$ be a constant. Assume that $f$ is a positive continuous function in $(0, \infty )$, and $f\in RV_{\gamma }$ with $\gamma >\max \{1,\,\varrho \}$. Let $w(x,t)$ be a positive function with positive lower bound $w_0$. Then for any given constant $C\ge 1$, there is a constant $\Lambda >0$, which depends on $w_0, C$ and $\gamma$, such that, for all $(x,t)$,

$$\begin{aligned} C(\Lambda +\Lambda ^\varrho )f(w(x,t))<f(\Lambda w(x,t)). \end{aligned}$$

Proof

Denote $\alpha =\max \{1,\,\varrho \}$ and choose $\sigma >0$ satisfying $\gamma -\sigma >\alpha$. Choose $\Lambda _0\ge 2$ and $0<\varepsilon \ll 1$ with $(1-\varepsilon )\Lambda _0^\sigma >2C$. By the definition, there is a constant $z_0=z_0(\Lambda _0)>0$ such that

$$\begin{aligned} f(\Lambda _0z)\ge (1-\varepsilon )\Lambda _0^\gamma f(z), \ \ \ \forall \ z\ge z_0. \end{aligned}$$

(5.3)

For any positive integer $j\ge 2$ and $z\ge z_0$, we have $\Lambda _0^{j-i}z\ge z\ge z_0$ for all $1\le i\le j-1$, and by inductively

$$\begin{aligned} f(\Lambda _0^jz)&= f(\Lambda _0\Lambda _0^{j-1}z)\ge (1-\varepsilon )\Lambda _0^\gamma f(\Lambda _0^{j-1}z)\ge \cdots \nonumber \\&\ge [(1-\varepsilon )\Lambda _0^\gamma ]^jf(z)=[(1-\varepsilon )\Lambda _0^\sigma ]^jf(z) \Lambda _0^{(\gamma -\sigma )j}\nonumber \\&> 2C\Lambda _0^{(\gamma -\sigma )j}f(z). \end{aligned}$$

(5.4)

As $f(z)\rightarrow \infty$ as $z\rightarrow \infty$, we can choose $z_0$ so large that $f(z)\le f(z_0)$ for all $w_0\le z\le z_0$.

Since $f\in RV_\gamma$, we have $\displaystyle \lim _{s\rightarrow \infty }s^{-\gamma +\sigma }f(s)=\infty$. For the given constant $A> 2Cf(z_0)/w_0^{\gamma -\sigma }$, there is a constant $S_0>1$ such that $f(s)\ge As^{\gamma -\sigma }$ for all $s\ge S_0$. It is obvious that there is a constant $\Lambda ^*>1$ such that $\Lambda z>S_0$ for all $\Lambda \ge \Lambda ^*$ and $w_0\le z\le z_0$. Therefore,

$$\begin{aligned}&f(\Lambda z)\ge A(\Lambda z)^{\gamma -\sigma }\ge Aw_0^{\gamma -\sigma }\Lambda ^{\gamma -\sigma } \frac{f(z)}{f(z_0)}\ge 2C\Lambda ^{\gamma -\sigma }f(z), \nonumber \\&\quad \forall \ \Lambda \ge \Lambda ^*, \ \, w_0\le z\le z_0.\qquad \end{aligned}$$

(5.5)

Note that $\Lambda _0\ge 2$, we can choose an integer $j\ge 2$ such that $\Lambda _0^j\ge \Lambda ^*$. Take $\Lambda =\Lambda _0^j$, then our conclusion is true. In fact, for those $(x,t)$ with $w(x,t)\le z_0$, by (5.5)

$$\begin{aligned} f(\Lambda w(x,t))>2C\Lambda ^{\gamma -\sigma }f(w(z,t))>2C\Lambda ^\alpha f(w(x,t)). \end{aligned}$$

For those $(x,t)$ with $w(x,t)>z_0$, by (5.4),

$$\begin{aligned} f(\Lambda w(x,t))&= f(\Lambda _0^jw(x,t))>2C\Lambda _0^{(\gamma -\sigma )j}f(w(x,t))\\&= 2C\Lambda ^{\gamma -\sigma } f(w(x,t))>2C\Lambda ^\alpha f(w(x,t)). \end{aligned}$$

The proof is complete. $\square$

When $f$ is increasing, the following better result can be obtained:

Lemma 5.6

Under the conditions of Lemma 5.5, we further assume that $f$ is increasing in $(0, \infty )$. Then for any given constant $C\ge 1$, there is a constant $\Lambda ^*>0$, which depends on $w_0, C$ and $\gamma$, such that for all $(x,t)$ and all $\Lambda \ge \Lambda ^*$,

$$\begin{aligned} C(\Lambda +\Lambda ^\varrho )f(w(x,t))<f(\Lambda w(x,t)). \end{aligned}$$

Lemma 5.7

Suppose that $f\in RV_{\gamma }$ with $\gamma \in \mathbb {R}$, is continuous, increasing and positive in $(0,\infty )$. Then for any given $\tau >0$, there is a constant $c=c(\tau )>0$, such that

$$\begin{aligned} f(a)+f(b)>cf(a+b),\ \ \forall \ a,b\ge \tau . \end{aligned}$$

Proof

We assume by contradiction that there exist two sequences $\{a_n\}$ and $\{b_n\}$, such that $f(a_n)+f(b_n)\le \frac{1}{n}f(a_n+b_n)$. Since $f$ is continuous, increasing and positive in $[\tau ,\infty )$, it follows $a_n+b_n\rightarrow \infty$ as $n\rightarrow \infty$. Suppose that $a_n\le b_n$ without loss of generality. Since $f$ is increasing on $(0,\infty )$, we have $f(b_n)\le \frac{1}{n}f(2b_n)$, i.e., $f(2b_n)/f(b_n)\ge n$. On the other hand, since $f\in RV_{\gamma }$, we have $\displaystyle \lim _{n\rightarrow \infty }[f(2b_n)/f(b_n)]=2^\gamma$, which is a contradiction. $\square$

Lemma 5.8

Assume that $f$ is a positive continuous function and $f\in RV_{\gamma }$ with $\gamma >0$. Let $\tau >0$ be a given constant. Then there exist a positive, continuous and increasing function $g\in RV_{\gamma }$ and a constant $\sigma$, such that

$$\begin{aligned} \sigma g(u)\le f(u)\le g(u), \ \ \ \forall \ u\ge \tau . \end{aligned}$$

Proof

As $u^{-\gamma }f(u)$ is a slow variation function, by Proposition 5.1,

$$\begin{aligned} u^{-\gamma }f(u)=M(u)\psi (u), \ \ \ \text{ with } \ \psi (u)=\exp \left\{ \int _b^u\frac{\varphi (t)}{t}\mathrm{d}t\right\} , \ \ \ u\ge b>0. \end{aligned}$$

Since $\displaystyle \lim \limits _{u\rightarrow \infty }M(u)=M^*\in (0,\infty )$, there is a constant $u_1>0$ such that $M^*/2<M(u)<2M^*$ for all $u\ge u_1$. Hence,

$$\begin{aligned} \frac{1}{2}M^*u^\gamma \psi (u)\le f(u)\le 2M^*u^\gamma \psi (u), \ \ \ \forall \ u\ge u_1. \end{aligned}$$

The direct computation gives $\left( u^\gamma \psi (u)\right) '=u^{\gamma -1}[\gamma +\varphi (u)]\psi (u)$. By use of $\displaystyle \lim \limits _{u\rightarrow \infty }\varphi (u)=0$, it follows that there is a $u_2>0$ such that $\gamma +\varphi (u)>0$ when $u\ge u_2$. That is, the function $u^\gamma \psi (u)$ is increasing for $u\ge u_2$. Take a positive, continuous and increasing function $g_1(u)$ such that $g_1(u)=u^\gamma \psi (u)$ when $u\ge u_0=\max \{u_1, u_2\}$. It is obvious that $g_1(u)\in RV_\gamma$ and $\frac{1}{2} M^* g_1(u)\le f(u)\le 2M^*g_1(u)$ for all $u\ge u_0$. Note that both $f$ and $g_1$ are continuous and positive in $[\tau , u_0]$, there is a constant $C>0$ such that $C^{-1}g_1(u)\le f(u)\le Cg_1(u)$ for all $\tau \le u\le u_0$. Take $g(u)=(2M^*+C)g_1(u)$, then our conclusion holds. $\square$

1.2 Some Results on the Unique Solution of (1.5)

Lemma 5.9

Assume that $g(u)$ and $h(u)$ are continuous functions and $h(u)$ is positive in $[a, \infty )$ for some constant $a>0$, and that $h(u)\in NRV_\gamma$ with $\gamma >1$. Let $v(t)$ and $w(t)$ be the positive solutions of the problems, respectively:

$$\begin{aligned}&v'(t)=- g(v(t)), \ \ t>0; \qquad v(0)=\infty ,&\\&w'(t)=- h(w(t)), \ \ t>0; \qquad w(0)=\infty .&\end{aligned}$$

If $\lim \limits _{u\rightarrow \infty }\frac{g(u)}{h(u)}=1$, then $\lim \limits _{t\rightarrow 0^+}\frac{v(t)}{w(t)}=1$.

Proof

Note that $h(u)\in NRV_\gamma$ and $\gamma >1+\nu$, similar to the proof of Lemma 5.8 we have that $h(u)/u^{1+\nu }$ is increasing when $u\ge u_0$ for some large constant $u_0$. Choose $\varepsilon _n$ with $0<\varepsilon _n<1/2$ and $\varepsilon _n\rightarrow 0$ as $n\rightarrow \infty$. In view of $g(u)/h(u)\rightarrow 1$ as $u\rightarrow \infty$, there is $u_n\ge 2^{1/\nu }u_0$ such that $(1-\varepsilon _n)h(u)\le g(u)\le (1+\varepsilon _n)h(u)$ when $u\ge u_n$.

Thanks to the fact that $v(t), w(t)\rightarrow \infty$ as $t\rightarrow 0$, there is $t_n>0$ such that $v(t)\ge u_n$ for all $0<t\le t_n$. Therefore,

$$\begin{aligned} (1-\varepsilon _n)h(v(t))\le g(v(t))\le (1+\varepsilon _n)h(v(t)), \ \ \forall \ 0<t\le t_n. \end{aligned}$$

Hence

$$\begin{aligned} v'(t)\le -(1-\varepsilon _n)h(v(t)), \ \ v(t)\ge 2^{1/\nu }u_0, \ \ \forall \ 0<t\le t_n; \ \ \ v(0)=0. \end{aligned}$$

Denote $\sigma _n=(1-\varepsilon _n)^{1/\nu }$. In view of $\varepsilon _n<1/2$, we see that $\sigma _nv(t)\ge u_0$ for all $n$ and $0<t\le t_n$. Notice that $h(u)/u^{1+\nu }$ is increasing for $u\ge u_0$ and $\sigma _n<1$, it is easily seen that $\sigma _n(1-\varepsilon _n)h(v(t))=\sigma _n^{1+\nu }h(v(t))\ge h(\sigma _nv(t))$ in $(0, t_n]$. Therefore, $y_n(t)=\sigma _nv(t)$ satisfies

$$\begin{aligned} y'_n(t)\le -\sigma _n(1-\varepsilon _n)h(v(t))\le -h(y_n(t)), \ \ \forall \ 0<t\le t_n. \end{aligned}$$

The comparison principle gives that $y_n(t)=\sigma _nv(t)\le w(t)$ for all $0<t\le t_n$. Hence $\lim \limits _{t\rightarrow 0^+}\frac{v(t)}{w(t)}\le 1/\sigma _n$, and consequently $\lim \limits _{t\rightarrow 0^+}\frac{v(t)}{w(t)}\le 1$ by letting $n\rightarrow \infty$.

Similarly, we can prove that $\lim \limits _{t\rightarrow 0^+}\frac{v(t)}{w(t)}\ge 1$. $\square$

The following corollary can be drawn; we shall omit the proof:

Corollary 1

In Lemma 5.9, if we replace the condition $h(u)\in NRV_\gamma$ by $h(u)\in RV_\gamma$ with $\gamma >1$, the conclusion is also true.

Following the same line of the above argument, we can also establish:

Lemma 5.10

Assume that $g(u)$ and $h(u)$ are continuous functions and $h(u)$ is positive in $[a, \infty )$ for some constant $a>0$. Suppose further that $h(u)\in RV_\gamma$ with $\gamma >1$. Let $v(t)$ and $w(t)$ be the positive solutions of the following problems, respectively:

$$\begin{aligned}&v'(t)=-g(v(t)), \ \ t>0; \qquad v(0)=\infty ,&\\&w'(t)=-h(w(t)), \ \ t>0; \qquad w(0)=\infty .&\end{aligned}$$

If $\lim \limits _{u\rightarrow \infty }\frac{g(u)}{h(u)}=c$ for some constant $c>0$, then for any $\nu :0<\nu <\gamma -1$,

$$\begin{aligned} c^{1/\nu }\le \liminf _{t\rightarrow 0^+}\frac{w(t)}{v(t)}\le \limsup _{t\rightarrow 0^+}\frac{w(t)}{v(t)}\le 1 \ \ \ \text{ if } \ c\le 1,\\ 1\le \liminf _{t\rightarrow 0^+}\frac{w(t)}{v(t)}\le \limsup _{t\rightarrow 0^+}\frac{w(t)}{v(t)}\le c^{1/\nu }\ \ \ \text{ if } \ c>1. \end{aligned}$$

Using Lemma 5.10, we can also obtain:

Lemma 5.11

Assume that $g(u)$ and $h(u)$ are positive continuous differentiable functions, and $g(u)\in RV_\theta , h(u)\in RV_\gamma$ with $\gamma +\theta >1$. Let $c>0$ be a given constant. Denote by $v(t)$ and $w(t)$ solutions of the following problems, respectively:

$$\begin{aligned}&v'(t)=-g(c v(t))h(v(t)), \ \ t>0; \qquad v(0)=\infty ,&\\&w'(t)=-g(w(t))h(w(t)), \ \ t>0; \qquad w(0)=\infty .&\end{aligned}$$

Then there exists a constant $C>1$ such that $C^{-1}v(t)\le w(t)\le Cv(t)$ in $(0, T]$.

1.3 Some Results on the Corresponding Elliptic Boundary Blow-Up Problem

In this final subsection, we recall, for the sake of ease of reference for the reader, some results about the boundary blow-up solutions of the $p-$Laplacian elliptic equation

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} \Delta _p u= b(x)f(u),\ \ &{}x\in \Omega ,\\ u=\infty , \ \ &{}x\in \partial \Omega , \end{array}\right. \end{aligned}$$

(5.6)

where $b(x)\in C^\alpha (\Omega )$ for some $0<\alpha <1$ with $b(x)\ge 0$ and $b(x)\not \equiv 0$ in $\Omega$. Set $\Omega _0=\{x\in \Omega :\, b(x)=0\}$ and assume that ${\bar{\Omega }}_0\subset \Omega$ and $b(x)>0$ in $\Omega \backslash \bar{\Omega }_0$.

Theorem 5.1

([6, Theorem 1.1]) Assume that $f$ satisfies $(F_2)$ and

$(F_3)\ \ \displaystyle \int _1^\infty F^{-1/p}(t)\mathrm{d}t<\infty$.

Then the problem (5.6) has at least one positive solution.

In fact, $(F_1)$ implies $(F_3)$, see [6, Lemmas 2.1 and 2.2, and Remark 2.4].

Theorem 5.2

([6, Theorem 1.2]) Assume that $(F_1)-(F_2)$ hold, the function $\phi$ is defined by (1.7).

• (i) If there exist a function $k\in \mathcal {K}_\ell$ and a positive continuous function $\beta (y)$ defined on $\partial \Omega$ such that

  $$\begin{aligned} \lim _{\Omega \ni x\rightarrow y} \frac{b(x)}{k^p(d(x))}=\beta (y) \ \ \mathrm{uniformly \ for } \ y\in \partial \Omega , \end{aligned}$$

  then (5.6) has a unique positive solution and the blow-up rate is given by

  $$\begin{aligned} \lim _{\Omega \ni x\rightarrow y}\frac{u(x)}{\phi (K(d(x)))} =\left( \frac{r+\ell -1}{r\beta (y)}\right) ^{\frac{r-1}{p}} \ \ \ \mathrm{uniformly\ for}\ \ y\in \partial \Omega . \end{aligned}$$

• (ii) Suppose that there exist a function $k\in \mathcal {K}_\ell$ and constants $0<\beta _1\le \beta _2, \delta >0$, such that

  $$\begin{aligned} \beta _1k^p(d(x))\le b(x)\le \beta _2k^p(d(x))\ \ \mathrm{for \ all}\ \ x\in \Omega \ \ \mathrm{with}\ \ d(x)\le \delta . \end{aligned}$$

  Then the problem (5.6) has a positive solution $u$ satisfying

  $$\begin{aligned} \displaystyle \liminf _{d(x)\rightarrow 0}\frac{u(x)}{\phi ( K(d(x)))} \ge \left( \frac{r+\ell -1}{r\beta _2}\right) ^{\frac{r-1}{p}},\ \ \ \displaystyle \limsup _{d(x)\rightarrow 0}\frac{u(x)}{\phi (K(d(x)))} \le \left( \frac{r+\ell -1}{r\beta _1}\right) ^{\frac{r-1}{p}}. \end{aligned}$$

• (iii) When $p=2$ and $\ell \not =0$, under the condition of (ii), the positive solution of (5.6) is also unique.

Remark 5.1

From the proof of [6, Theorem 1.2] it can be seen that, for any given $y\in \partial \Omega$, if the limit

$$\begin{aligned} \lim _{\Omega \ni x\rightarrow y} \frac{b(x)}{k^p(d(x))}=\beta (y) \end{aligned}$$

exists, then any positive solution $u(x)$ of (5.6) satisfies

$$\begin{aligned} \lim _{\Omega \ni x\rightarrow y}\frac{u(x)}{\phi (K(d(x)))} =\left( \frac{r+\ell -1}{r\beta (y)}\right) ^{\frac{r-1}{p}}. \end{aligned}$$