Blow-Up of the Solution for a Semilinear Parabolic Problem with a Mixed Source

Abstract

A semilinear parabolic equation with the Dirichlet boundary condition is examined. The reaction source is a mixed nonlinear function. This paper investigates the existence and uniqueness of a solution. A sufficient condition for the solution being blow-up in finite time is found.

1. Introduction

Let L be the parabolic operator such that $Lu=u_t-u_{xx}$, T be a positive real number, $c\in \left(0,1\right)$, and $\delta \left(x\right)$ denote the Dirac delta function. In this paper, we study blow-up of the solution of the following semilinear parabolic initial-boundary value problem with a mixed source: a nonlinear function $u^p$ combined with a concentrated function $\delta \left(x-c\right)u^q$

$$Lu=u^p+\delta \left(x-c\right)u^q\mathrm{in}\left(0,1\right)\times \left(0,T\right),$$

(1)

$$u(x,0)=\psi \left(x\right)\mathrm{for}x\in \left[0,1\right],u\left(0,t\right)=u\left(1,t\right)=0\mathrm{for}t\in \left(0,T\right),$$

(2)

where $p⩾1$ and $q⩾1$, and $\psi \left(x\right)$ is a nontrivial, nonnegative, and differentiable function on $\left[0,1\right]$ such that $\psi \left(0\right)=\psi \left(1\right)=0$. Throughout this paper, we assume that $\psi \left(x\right)$ satisfies the inequality below

$${\psi }^{″}\left(x\right)+{\psi }^p\left(x\right)+\delta (x-c){\psi }^q\left(x\right)⩾0\mathrm{on}[0,1].$$

(3)

We use (3) to show that u is a non-decreasing function in t.

Our study of the problem (1) and (2) is motivated by the papers written by Bimpong-Bota et al. [1] and Ortolvea et al. [2]. In these two articles, the authors investigated instabilities of chemical reactions and transport processes inside a bulk system. Some localized sources, such as electrodes or membranes, are immersed inside this bulk system. They examined the existence of steady state solutions of (1). In (1), $u^p$ can represent the rate of bulk reaction and $\delta \left(x-c\right)u^q$ characterizes the rate of localized heterogeneous reaction at the location $x=c$.

When the source contains only a nonlinear reaction function in the form of $f\left(u\right)$, (1) describes the thermal combustion in a bounded domain (cf. [3,4]). Let $\mathsf{\Omega }$ be a bounded convex domain in ${\mathbb{R}}^n$, and $\overline{\mathsf{\Omega }}$ and $\partial \mathsf{\Omega }$ be its closure and boundary, respectively. Friedman and McLeod [5] studied (1) and (2) with a nonlinear reaction function in the form of

$$\begin{array}{c}{u}_t=\mathsf{\Delta }u+f\left(u\right),(x,t)\in \mathsf{\Omega }\times (0,T),\\ u(x,0)=\psi \left(x\right)\mathrm{for}x\in \overline{\mathsf{\Omega }},u\left(x,t\right)=0\mathrm{for}\left(x,t\right)\in \partial \mathsf{\Omega }\times (0,T).\end{array}$$

They proved that the set of blow-up points is a compact subset of $\mathsf{\Omega }$. When $n=1$ and $f\left(u\right)=u^p$ with $p>1$, Weissler [6] proved that u blows up at a single point if $\psi \left(x\right)$ is symmetric and sufficiently large. Caffarrelli and Friedman [7] proved that the blow-up set consists of one or two points if $\psi \left(x\right)$ has at most two local maxima. When $n⩾1$, Mueller and Weissler [8] proved that u blows up at a single point. Bellout [9] investigated blow-up of u with the Robin boundary condition. If $f\left(u\right)=u^{p\left(x\right)}$, Khelghati and Baghaei [10] showed that u blows up in a finite time when $\psi \left(x\right)$ is sufficiently large.

When the reaction function contains only a concentrated source, (1) and (2) can describe the physical situation of ignition having a highly localized forcing function (cf. [11,12]). Chan and Tian [11] studied the following degenerate parabolic problem

$$x^r{u}_t-u_{xx}=a^{\alpha r+2}\delta (x-b)f\left(u(x,t)\right){\left({\int }_0^1{x}^r\left|u(x,t)\right|dx\right)}^{\alpha }\mathrm{in}\left(0,1\right)\times \left(0,T\right),$$

where a and $\alpha$ are positive constants, r is a non-negative real number, and $b\in (0,1)$. They proved that there exists a unique solution, and u blows up at $x=b$ only.

In Section 2, we prove the existence and uniqueness of the continuous solution u. Then, we prove that $u_t⩾0$. Because of the Dirac delta function, it brings challenges to study them. Our approach is through examining the equivalent integral equation. As (1) and (2) has two different heat sources, and each possesses different strength. Without the nonlinear source $u^p$, intuitively, we expect that the blow-up can occur at $x=c$ due to the singularity of heat source. Our main interest is to investigate the location of blow-up point of this problem if $u^p$ is present. We want to know whether $x=c$ is still a blow-up point. In Section 3, we prove that the solution u blows up in a finite time under the condition: $p+q>2.$ Our method is based on condition (3) alone. Further, an upper bound of blow-up time is obtained. Then, we prove that u blows up at $x=c$. The proof relies on the integral solution.

2. Existence and Uniqueness of the Solution

Let $G\left(x,\xi ,t-s\right)$ be the Green’s function corresponding to (1) and (2). For any x and $\xi$ belonging to $[0,1]$, and t and s are variables within the interval $\left(-\infty ,\infty \right)$, $G\left(x,\xi ,t-s\right)$ satisfies the problem below

$$\begin{array}{c}LG=\delta (t-s)\delta (x-\xi )\mathrm{for}s<t,x\in (0,1),\\ G\left(x,\xi ,t-s\right)=0\mathrm{for}t<s,G\left(0,\xi ,t-s\right)=G\left(1,\xi ,t-s\right)=0.\end{array}$$

$G\left(x,\xi ,t-s\right)$ is a positive function in the set $\left\{\right(x,\xi ,t,s):x$ and $\xi$ are in $(0,1)$, and $t>s⩾0\}$. The integral solution of (1) and (2) is represented by

$$\left.\begin{array}{cc}\hfill u(x,t)& ={\int }_0^{1}G\left(x,\xi ,t\right)\psi \left(\xi \right)d\xi +{\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)u^p(\xi ,s)d\xi ds\hfill \\ & +{\int }_0^{t}G(x,c,t-s)u^q\left(c,s\right)ds.\hfill \end{array}\right\}$$

(4)

From Pinsky [13], $G(x,\xi ,t-s)$ is

$$G(x,\xi ,t-s)=2H(t-s)\sum _{j=1}^{\infty }sinj\pi xsinj\pi \xi e^{-j^2{\pi }^2(t-s)},$$

(5)

where $H(t-s)$ is the Heaviside function. The equivalent expression of $G(x,\xi ,t-s)$ is

$$G(x,\xi ,t-s)=\frac{H(t-s)}{\sqrt{4\pi (t-s)}}\sum _{j=-\infty }^{\infty }\left[e^{-\frac{{(x-\xi -2j)}^2}{4(t-s)}}-e^{-\frac{{(x+\xi -2(j+1))}^2}{4(t-s)}}\right].$$

In ${\mathbb{R}}^n$, because of the singularity ${(t-s)}^{-n/2}$ (corresponding to $1/\sqrt{t-s}$ in the above formula for $n=1$) of $G(x,c,t-s)$ in a neighborhood of $x=c$, the integral solution $u(x,t)$ in (4) at $x=c$ does not satisfy (1) and (2) except $n=1$ (cf. [14]).

In the following, we want to prove that the integral solution given by (4) is a unique continuous solution of (1) and (2). To achieve it, we construct a sequence of solution $\left\{u_m\right\}$ satisfying the recursive equation below

$$\left.\begin{array}{c}L{u}_m=u_{m-1}^p+\delta \left(x-c\right)u_{m-1}^q\mathrm{in}(0,1)\times \left(0,T\right),\\ u_m(x,0)=\psi \left(x\right)\mathrm{for}x\in [0,1],u_m\left(0,t\right)=u_m\left(1,t\right)=0\mathrm{for}t\in \left(0,T\right),\end{array}\right\}$$

(6)

for $m=1,2,\dots$ We assume that $u_0\left(x,t\right)=\psi \left(x\right)\mathrm{on}[0,1]\times [0,T)$. The integral solution of (6) is

$$\left.\begin{array}{cc}\hfill u_m(x,t)& ={\int }_0^{1}G\left(x,\xi ,t\right)\psi \left(\xi \right)d\xi +{\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)u_{m-1}^p(\xi ,s)d\xi ds\hfill \\ & +{\int }_0^{t}G(x,c,t-s)u_{m-1}^q\left(c,s\right)ds.\hfill \end{array}\right\}$$

(7)

Since $G\left(x,\xi ,t-s\right)>0$ for $t>s$, $u_m>0$ in $(0,1)\times (0,T)$. The following lemma is going to show that $\left\{u_m\right\}$ is a non-decreasing sequence and $u_m$ is a continuous function.

Lemma 1.

The sequence of solution $\left\{u_m\right\}$ of (7) is monotone non-decreasing: $\psi \left(x\right)⩽u_1\left(x,t\right)⩽\dots ⩽u_m\left(x,t\right)⩽\dots$, and $u_m\left(x,t\right)$ is continuous on $[0,1]\times \left[0,T\right)$.

Proof. 

When $m=1$, from (7) we have

$$\left.\begin{array}{cc}\hfill u_1\left(x,t\right)& ={\int }_0^{1}G(x,\xi ,t)\psi \left(\xi \right)d\xi +{\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)u_0^p(\xi ,s)d\xi ds\hfill \\ & +{\int }_0^{t}G(x,c,t-s)u_0^q(c,s)ds.\hfill \end{array}\right\}$$

(8)

By (3), we obtain the inequality below

$$\begin{array}{ccc}& & u_1\left(x,t\right)-\psi \left(x\right)\hfill \\ & ⩾& {\int }_0^{1}G(x,\xi ,t)(\psi \left(\xi \right)-\psi \left(\xi \right))d\xi +{\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)(u_0^p(\xi ,s)-{\psi }^p\left(\xi \right))d\xi ds\hfill \\ & & +{\int }_0^{t}G(x,c,t-s)(u_0^q(c,s)-{\psi }^q\left(c\right)))ds.\hfill \end{array}$$

Since $G(x,\xi ,t-s)>0$ for $t>s$, $u_1\left(x,t\right)⩾\psi \left(x\right)$ $\mathrm{on}[0,1]\times [0,T)$. To prove that $u_1\left(x,t\right)$ is continuous on $[0,1]\times \left[0,T\right)$, we show that all integrals on the right side of (8) are integrable and ${lim}_{t\to 0}{u}_1\left(x,t\right)=\psi \left(x\right)$ on $[0,1]$. Let us prove that ${\int }_0^{t}G(x,c,t-s)u_0^q(c,s)ds$ is a bounded function. By (5) and switching the summation and integration, the integral becomes

$${\int }_0^{t}G(x,c,t-s)u_0^q(c,s)ds=2\sum _{j=1}^{\infty }sinj\pi xsinj\pi c{\int }_0^t{e}^{-j^2{\pi }^2(t-s)}{\psi }^q\left(c\right)ds.$$

Let $k_1={max}_{x\in [0,1]}\psi \left(x\right)$. We get

$${\int }_0^{t}G(x,c,t-s)u_0^q(c,s)ds⩽\frac{2k_1^q}{{\pi }^2}\sum _{j=1}^{\infty }\frac{1}{j^2}(1-e^{-j^2{\pi }^{2}t}).$$

(9)

It is noticed that the right side of above inequality tends to 0 when t approaches 0. From (9),

$${\int }_0^{t}G(x,c,t-s)u_0^q(c,s)ds⩽\frac{2}{{\pi }^2}{k}_1^q\sum _{j=1}^{\infty }\frac{1}{j^2}=\frac{1}{3}{k}_1^q.$$

Then, we prove ${\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)u_0^p(\xi ,s)d\xi ds$ being bounded above. By (5),

$${\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)u_0^p(\xi ,s)d\xi ds⩽2{\int }_0^t{\int }_0^1\sum _{j=1}^{\infty }|sinj\pi \xi |e^{-j^2{\pi }^2(t-s)}{k}_1^{p}d\xi ds.$$

After interchanging the summation and integration, we obtain

$${\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)u_0^p(\xi ,s)d\xi ds⩽\frac{4}{{\pi }^3}{k}_1^p\sum _{j=1}^{\infty }\frac{1}{j^3}(1-e^{-j^2{\pi }^{2}t}).$$

(10)

It is noticed that the right side of above inequality tends to 0 when t approaches 0. From (10),

$${\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)u_0^p(\xi ,s)d\xi ds⩽\frac{4}{{\pi }^3}{k}_1^p\sum _{j=1}^{\infty }\frac{1}{j^3}.$$

Thus, ${\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)u_0^p(\xi ,s)d\xi ds$ is bounded above on $\left[0,1\right]\times \left[0,T\right)$. By (5) again, we know

$${\int }_0^{1}G(x,\xi ,t)\psi \left(\xi \right)d\xi ⩽2k_1\sum _{j=1}^{\infty }\frac{1}{e^{j^2{\pi }^{2}t}}.$$

The right side is bounded when $t>0$. It is noticed that when $t=0$,

$${\int }_0^{1}G(x,\xi ,0)\psi \left(\xi \right)d\xi ={\int }_0^{1}2\sum _{j=1}^{\infty }sinj\pi xsinj\pi \xi \psi \left(\xi \right)d\xi =\psi \left(x\right).$$

Therefore, ${lim}_{t\to 0}{u}_1\left(x,t\right)=\psi \left(x\right)$ on $[0,1]$. Hence, $u_1\left(x,t\right)$ is continuous on $\left[0,1\right]\times \left[0,T\right)$. Assume that this statement is true for $m=i$, that is, $u_i(x,t)⩾u_{i-1}(x,t)$ and $u_i\left(x,t\right)$ is continuous on $\left[0,1\right]\times \left[0,T\right)$. When $m=i+1$, we have

$$\begin{array}{ccc}& & u_{i+1}\left(x,t\right)-u_i(x,t)\hfill \\ & =& {\int }_0^{1}G(x,\xi ,t)(\psi \left(\xi \right)-\psi \left(\xi \right))d\xi \hfill \\ & & +{\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)(u_i^p(\xi ,s)-u_{i-1}^p(\xi ,s))d\xi ds\hfill \\ & & +{\int }_0^{t}G(x,c,t-s)(u_i^q(c,s)-u_{i-1}^q(c,s)))ds.\hfill \end{array}$$

A similar computation from above yields $u_{i+1}\left(x,t\right)⩾u_i\left(x,t\right)$ and $u_{i+1}\left(x,t\right)$ is continuous on $[0,1]\times \left[0,T\right)$. By the mathematical induction, we conclude that $\psi \left(x\right)⩽u_1\left(x,t\right)⩽\dots ⩽u_m\left(x,t\right)⩽\dots$ and $u_m\left(x,t\right)$ is continuous on $[0,1]\times \left[0,T\right)$ for $m=1,2,\dots$ □

Let h be a positive real number less than T. Our next result is to prove that the sequence $\left\{{\left(u_m\right)}_t\right\}$ is non-negative.

Lemma 2.

${\left(u_m\right)}_t⩾0$ on $[0,1]\times [0,T)$ for $m=1,2,\dots$

Proof. 

To achieve this result, we prove that $u_m(x,t+h)⩾u_m(x,t)$ on $[0,1]\times [0,T-h]$. Let $m=1$. From (6), we get

$$\begin{array}{c}L{u}_1(x,t+h)=u_0^p\left(x,t+h\right)+\delta \left(x-c\right)u_0^q\left(x,t+h\right)\mathrm{in}(0,1)\times (0,T-h],\\ u_1(x,h)=u_1(x,h)\mathrm{for}x\in [0,1],u_1\left(0,t+h\right)=u_1\left(1,t+h\right)=0\mathrm{for}t\in (0,T-h].\end{array}$$

If (6) is subtracted from the above problem, it gives

$$\begin{array}{ccc}& & L\left(u_1(x,t+h)-u_1(x,t)\right)\hfill \\ & =& \left(u_0^p\left(x,t+h\right)-u_0^p(x,t)\right)+\delta (x-c)\left(u_0^q\left(x,t+h\right)-u_0^q(x,t)\right)=0.\hfill \end{array}$$

At $t=0$, $u_1(x,h)-u_1(x,0)=u_1(x,h)-\psi \left(x\right)⩾0$ for $x\in [0,1]$. $u_1\left(x,t+h\right)-u_1\left(x,t\right)=0$ at $x=0$ and 1 for $t\in (0,T-h]$. By the maximum principle, $u_1(x,t+h)⩾u_1(x,t)$ on $[0,1]\times [0,T-h]$. As $h\to 0$, we have ${\left(u_1\right)}_t⩾0$ over $[0,1]\times [0,T)$. Assume that this statement is true for $m=i$. When $m=i+1$, a similar computation yields

$$\begin{array}{ccc}& & L\left(u_{i+1}(x,t+h)-u_{i+1}(x,t)\right)\hfill \\ & =& \left(u_i^p\left(x,t+h\right)-u_i^p(x,t)\right)+\delta (x-c)\left(u_i^q\left(x,t+h\right)-u_i^q(x,t)\right)⩾0.\hfill \end{array}$$

At $t=0$, $u_{i+1}(x,h)-u_{i+1}(x,0)=u_{i+1}(x,h)-\psi \left(x\right)⩾0$ for $x\in [0,1]$. $u_{i+1}\left(x,t+h\right)-u_{i+1}\left(x,t\right)=0$ at $x=0$ and 1 for $t\in (0,T-h]$. By the maximum principle, $u_{i+1}(x,t+h)⩾u_{i+1}(x,t)$ on $[0,1]\times [0,T-h]$. This implies ${\left(u_{i+1}\right)}_t⩾0$ on $[0,1]\times [0,T)$. By the mathematical induction, we have ${\left(u_m\right)}_t⩾0$ on $[0,1]\times [0,T)$ for $m=1,2,\dots$ □

Let $\widehat{t}>0$ and $k_2$ be a positive constant such that $k_2>k_1$. The following result shows that $\left\{u_m\right\}$ is bounded above by $k_2$ on $[0,1]\times [0,\widehat{t}]$.

Lemma 3.

For any $k_2>k_1$, there exists $\widehat{t}>0$ such that $k_2>u_m$ on $[0,1]\times [0,\widehat{t}]$ for $m=1,2,\dots$

Proof. 

From Lemma 1, we know that ${\int }_0^{1}G\left(x,\xi ,t\right)\psi \left(\xi \right)d\xi$, ${\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)u_{m-1}^p(\xi ,s)d\xi ds$, and ${\int }_0^{t}G(x,c,t-s)u_{m-1}^q\left(c,s\right)ds$ are bounded above for $m=1,2,\dots$ By (9), (10), and ${\int }_0^{1}G(x,\xi ,0)\psi \left(\xi \right)d\xi =\psi \left(x\right)$ on $[0,1]$, there exists $\widehat{t}>0$ such that

$$\left.\begin{array}{c}{k}_2>{\int }_0^{1}G\left(x,\xi ,\widehat{t}\right)\psi \left(\xi \right)d\xi +{\int }_0^{\widehat{t}}{\int }_0^{1}G(x,\xi ,\widehat{t}-s)k_2^{p}d\xi ds\hfill \\ +{\int }_0^{\widehat{t}}G(x,c,\widehat{t}-s)k_2^{q}ds.\hfill \end{array}\right\}$$

(11)

Let $m=1$. Then, by (7)

$$\begin{array}{cc}\hfill u_1(x,\widehat{t})& ={\int }_0^{1}G\left(x,\xi ,\widehat{t}\right)\psi \left(\xi \right)d\xi +{\int }_0^{\widehat{t}}{\int }_0^{1}G(x,\xi ,\widehat{t}-s)u_0^p(\xi ,s)d\xi ds\hfill \\ \hfill & +{\int }_0^{\widehat{t}}G(x,c,\widehat{t}-s)u_0^q\left(c,s\right)ds.\hfill \end{array}$$

As $k_2>k_1$, the above equation is subtracted from (11) to give

$$\begin{array}{ccc}& & k_2-u_1(x,\widehat{t})\hfill \\ & >& {\int }_0^{\widehat{t}}{\int }_0^{1}G(x,\xi ,\widehat{t}-s)[k_2^p-u_0^p(\xi ,s)]d\xi ds+{\int }_0^{\widehat{t}}G(x,c,\widehat{t}-s)[k_2^q-u_0^q\left(c,s\right)]ds>0,\hfill \end{array}$$

for $x\in [0,1]$. By Lemma 2, $u_1$ is non-decreasing in t for $0⩽t⩽\widehat{t}$. Therefore, $k_2>u_1(x,t)$ on $[0,1]\times [0,\widehat{t}]$. Assume that this statement is true for $m=i$. That is, $k_2>u_i(x,t)$ on $[0,1]\times [0,\widehat{t}]$. When $m=i+1$, a similar computation yields

$$\begin{array}{ccc}& & k_2-u_{i+1}(x,\widehat{t})\hfill \\ & >& {\int }_0^{\widehat{t}}{\int }_0^{1}G(x,\xi ,\widehat{t}-s)[k_2^p-u_i^p(\xi ,s)]d\xi ds+{\int }_0^{\widehat{t}}G(x,c,\widehat{t}-s)[k_2^q-u_i^q\left(c,s\right)]ds>0,\hfill \end{array}$$

for $x\in [0,1]$. Therefore, $k_2>u_{i+1}$ on $[0,1]\times [0,\widehat{t}]$. By the mathematical induction, we have $k_2>u_m$ on $[0,1]\times [0,\widehat{t}]$ for $m=1,2,\dots$ □

Let $u\left(x,t\right)={lim}_{m\to \infty }{u}_m(x,t)$. Let us show that (4) has a continuous solution.

Theorem 1.

The integral Equation (4) has a continuous solution $u\left(x,t\right)⩾\psi \left(x\right)$ on $[0,1]\times [0,\widehat{t}]$.

Proof. 

To obtain this result, we are going to prove $\left\{u_m\right\}$ to converge uniformly. Let $S_{m+1}={max}_{[0,1]\times [0,\widehat{t}]}[u_{m+1}(x,t)-u_m(x,t)]$. By (7),

$$\begin{array}{ccc}& & u_{m+1}(x,t)-u_m(x,t)\hfill \\ & =& {\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)\left[u_m^p(\xi ,s)-u_{m-1}^p(\xi ,s)\right]d\xi ds\hfill \\ & & +{\int }_0^{t}G(x,c,t-s)\left[u_m^q\left(c,s\right)-u_{m-1}^q\left(c,s\right)\right]ds.\hfill \end{array}$$

By $p⩾1$ and $q⩾1$, the mean value theorem, and Lemma 3, we have

$$u_m^p(\xi ,s)-u_{m-1}^p(\xi ,s)⩽p{k}_2^{p-1}[u_m(\xi ,s)-u_{m-1}(\xi ,s)],$$

and

$$u_m^q(c,s)-u_{m-1}^q(c,s)⩽q{k}_2^{q-1}[u_m(c,s)-u_{m-1}(c,s)].$$

Then, we follow a similar computation of (9) and (10) to get

$$\begin{array}{ccc}\hfill S_{m+1}& ⩽& p{k}_2^{p-1}{S}_m{\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)d\xi ds+q{k}_2^{q-1}{S}_m{\int }_0^{t}G(x,c,t-s)ds\hfill \\ & ⩽& \left[\frac{4p{k}_2^{p-1}}{{\pi }^3}\sum _{j=1}^{\infty }\frac{1}{j^3}(1-e^{-j^2{\pi }^{2}t})+\frac{2q{k}_2^{q-1}}{{\pi }^2}\sum _{j=1}^{\infty }\frac{1}{j^2}(1-e^{-j^2{\pi }^{2}t})\right]S_m.\hfill \end{array}$$

By the mean-value theorem, there exists ${\eta }_1$ between 0 and t such that

$$\sum _{j=1}^{\infty }\frac{1}{j^3}(1-e^{-j^2{\pi }^{2}t})={\pi }^2\sum _{j=1}^{\infty }\frac{1}{j{e}^{j^2{\pi }^2{\eta }_1}}t.$$

Similarly, there exists ${\eta }_2$ between 0 and t such that

$$\sum _{j=1}^{\infty }\frac{1}{j^2}(1-e^{-j^2{\pi }^{2}t})={\pi }^2\sum _{j=1}^{\infty }\frac{1}{e^{j^2{\pi }^2{\eta }_2}}t.$$

Thus,

$$S_{m+1}⩽\left(\frac{4p{k}_2^{p-1}}{\pi }\sum _{j=1}^{\infty }\frac{1}{j{e}^{j^2{\pi }^2{\eta }_1}}+2q{k}_2^{q-1}\sum _{j=1}^{\infty }\frac{1}{e^{j^2{\pi }^2{\eta }_2}}\right)t{S}_m.$$

Since ${\sum }_{j=1}^{\infty }1/\left(j{e}^{j^2{\pi }^2{\eta }_1}\right)$ and ${\sum }_{j=1}^{\infty }1/e^{j^2{\pi }^2{\eta }_2}$ are convergent, we choose $t={\sigma }_1>0$ sufficiently close to 0 such that

$$\left(\frac{4p{k}_2^{p-1}}{\pi }\sum _{j=1}^{\infty }\frac{1}{j{e}^{j^2{\pi }^2{\eta }_1}}+2q{k}_2^{q-1}\sum _{j=1}^{\infty }\frac{1}{e^{j^2{\pi }^2{\eta }_2}}\right){\sigma }_1<1.$$

Then, we have $S_{m+1}<S_m$. Therefore, ${\left\{S_m\right\}}_{m=1}^{\infty }$ is a decreasing sequence. By the Dini’s theorem, $\left\{u_m\right\}$ converges uniformly. As $u\left(x,t\right)={lim}_{m\to \infty }{u}_m(x,t)$, $u\left(x,t\right)$ is continuous on $[0,1]\times [0,{\sigma }_1]$. When $t>{\sigma }_1$, we repeat the above process through replacing $\psi \left(x\right)$ by $u\left(x,{\sigma }_1\right)$ to obtain that $u\left(x,t\right)$ is continuous on $[0,1]\times [0,\widehat{t}]$. By the Monotone Convergence Theorem,

$$\begin{array}{ccc}\hfill u(x,t)& =& {\int }_0^{1}G(x,\xi ,t)\psi \left(\xi \right)d\xi +{\int }_0^t{\int }_0^{1}G(x,\xi ,t-s)u^p(\xi ,s)d\xi ds\hfill \\ & & +{\int }_0^{t}G(x,c,t-s)u^q\left(c,s\right)ds,\hfill \end{array}$$

is a solution of (1) and (2) on $[0,1]\times [0,\widehat{t}]$. Since $u_m(x,t)⩾\psi \left(x\right)$ on $[0,1]\times [0,\widehat{t}]$ for $m=0,1,2,\dots$, we have $u(x,t)⩾\psi \left(x\right)$ on $[0,1]\times [0,\widehat{t}]$. □

Based on a similar proof of Lemma 2, we show that $u_t$ is non-negative.

Lemma 4.

$u_t⩾0$ on $[0,1]\times [0,\widehat{t}]$.

Proof. 

By Lemma 2, we have $u_k\left(x,t+h\right)⩾u_k(x,t)\mathrm{on}[0,1]\times [0,T)$ for $k=0,1,2,\dots$ When $k\to \infty$, we have $u(x,t+h)⩾u(x,t)$ on $[0,1]\times [0,T)$. Hence, $u_t⩾0$ on $[0,1]\times [0,T)$. □

The result below shows the uniqueness of u.

Theorem 2.

The integral solution u in (4) is unique on $[0,1]\times [0,\widehat{t}]$.

Proof. 

Suppose that $\widehat{u}\left(x,t\right)$ and $\tilde{u}\left(x,t\right)$ are two different solutions of (4). $\widehat{u}\left(x,t\right)-\tilde{u}\left(x,t\right)$ satisfies

$$\begin{array}{ccc}& & \widehat{u}\left(x,t\right)-\tilde{u}\left(x,t\right)\hfill \\ & =& {\int }_0^t{\int }_0^{1}G\left(x,\xi ,t-s\right)\left({\widehat{u}}^p(\xi ,s)-{\tilde{u}}^p(\xi ,s)\right)d\xi ds+{\int }_0^{t}G(x,c,t-s)\left({\widehat{u}}^q\left(c,s\right)-{\tilde{u}}^q(c,s)\right)ds,\hfill \end{array}$$

and $|\widehat{u}\left(x,t\right)-\tilde{u}\left(x,t\right)|>0$ for some $(x,t)\in [0,1]\times [0,\widehat{t}]$. By the mean value theorem and (5), we have the inequality below

$$\begin{array}{ccc}& & |\widehat{u}\left(x,t\right)-\tilde{u}\left(x,t\right)|\hfill \\ & ⩽& 2p\underset{\left[0,1\right]\times \left[0,t\right]}{max}{v}_1^{p-1}\underset{\left[0,1\right]\times \left[0,t\right]}{max}\left|\widehat{u}-\tilde{u}\right|\sum _{j=1}^{\infty }{\int }_0^t{\int }_0^1\left|sinj\pi \xi \right|e^{-j^2{\pi }^2(t-s)}d\xi ds\hfill \\ & & +2q\underset{\left[0,1\right]\times \left[0,t\right]}{max}{v}_2^{q-1}\underset{\left[0,1\right]\times \left[0,t\right]}{max}\left|\widehat{u}-\tilde{u}\right|\sum _{j=1}^{\infty }{\int }_0^t{e}^{-j^2{\pi }^2(t-s)}ds\hfill \end{array}$$

where $v_1$ is between $\widehat{u}(\xi ,s)$ and $\tilde{u}(\xi ,s)$ and $v_2$ is between $\widehat{u}\left(c,s\right)$ and $\tilde{u}(c,s)$. Let

$$M=max\{\underset{\left[0,1\right]\times \left[0,t\right]}{max}{v}_1^{p-1},\underset{\left[0,1\right]\times \left[0,t\right]}{max}{v}_2^{q-1}\}.$$

We follow a similar computation of Theorem 1 to obtain

$$\underset{\left[0,1\right]\times \left[0,t\right]}{max}\left|\widehat{u}-\tilde{u}\right|⩽\left(\frac{4p}{\pi }\sum _{j=1}^{\infty }\frac{1}{j{e}^{j^2{\pi }^2{\eta }_3}}+2q\sum _{j=1}^{\infty }\frac{1}{e^{j^2{\pi }^2{\eta }_4}}\right)Mt\underset{\left[0,1\right]\times \left[0,t\right]}{max}\left|\widehat{u}-\tilde{u}\right|,$$

where ${\eta }_3$ and ${\eta }_4$ are between 0 and t. We choose $t={\sigma }_2>0$ sufficiently close to 0 such that

$$\left(\frac{4p}{\pi }\sum _{j=1}^{\infty }\frac{1}{j{e}^{j^2{\pi }^2{\eta }_3}}+2q\sum _{j=1}^{\infty }\frac{1}{e^{j^2{\pi }^2{\eta }_4}}\right)M{\sigma }_2<1.$$

Then, we have

$$\underset{\left[0,1\right]\times \left[0,{\sigma }_2\right]}{max}\left|\widehat{u}\left(x,t\right)-\tilde{u}\left(x,t\right)\right|<\underset{\left[0,1\right]\times \left[0,{\sigma }_2\right]}{max}\left|\widehat{u}\left(x,t\right)-\tilde{u}\left(x,t\right)\right|.$$

This contradiction shows that $\widehat{u}\left(x,t\right)=\tilde{u}\left(x,t\right)$ in $\left[0,1\right]\times \left[0,{\sigma }_2\right]$. We repeat the above process through replacing $\psi \left(x\right)$ by $\widehat{u}\left(x,{\sigma }_2\right)$ to conclude that $\widehat{u}\left(x,t\right)=\tilde{u}\left(x,t\right)$ on $\left[0,1\right]\times [0,\widehat{t}]$. □

Let $t_b$ be the supremum of $\widehat{t}$ such that the integral Equation (4) has a unique continuous solution over $[0,1]\times [0,\widehat{t}]$. The following result shows that u is unbounded if $t_b$ is finite.

Theorem 3.

If $t_b$ is finite, then $u\left(x,t\right)$ is unbounded somewhere in $[0,1]$ when $t\to t_b$.

Proof. 

Suppose that $u\left(x,t\right)$ exists at $t_b$ for all $x\in [0,1]$. We prove this result through showing that $u\left(x,t\right)$ continues to exist for $t>t_b$. Let us consider the following problem

$$\begin{array}{c}Lw=w^p+\delta \left(x-c\right)w^q\mathrm{in}\left(0,1\right)\times (t_b,t_b+h),\\ w(x,t_b)=u\left(x,t_b\right)\mathrm{for}x\in \left[0,1\right],w\left(0,t\right)=w\left(1,t\right)=0\mathrm{for}t\in \left(t_b,t_b+h\right),\end{array}$$

for some $h>0$. We want to prove that the solution $w(x,t)$ exists on $\left[0,1\right]\times \left[t_b,t_b+h\right)$. Similar to (6) and (7), let us construct a sequence of solution $\left\{w_m\right\}$ on $[0,1]\times [t_b,t_b+h)$. The integral solution of $w_m$ is given by

$$\begin{array}{cc}\hfill w_m(x,t)& ={\int }_0^{1}G\left(x,\xi ,t\right)u\left(\xi ,t_b\right)d\xi +{\int }_{t_b}^t{\int }_0^{1}G(x,\xi ,t-s)w_{m-1}^p(\xi ,s)d\xi ds\hfill \\ \hfill & +{\int }_{t_b}^{t}G(x,c,t-s)w_{m-1}^q\left(c,s\right)ds,\hfill \end{array}$$

where $m=1,2,\dots$ and $w_0\left(x,t\right)=u\left(x,t_b\right)\mathrm{on}[0,1]\times [t_b,t_b+h)$. We follow a similar proof of Lemma 1 to obtain a continuous monotone non-decreasing sequence solution $\left\{w_m\right\}$ on $[0,1]\times [t_b,t_b+h)$. Let $w={lim}_{m\to \infty }{w}_m$. We follow from the proof of Theorems 1 and 2 to show that w is a unique continuous solution on $[0,1]\times [t_b,t_b+h)$. Let us define $z(x,t)$ as

$$z(x,t)=\left\{\begin{array}{c}u(x,t)\mathrm{for}[0,1]\times [0,t_b),\hfill \\ w(x,t)\mathrm{for}[0,1]\times [t_b,t_b+h).\hfill \end{array}\right.$$

Then, $z(x,t)$ is a unique continuous solution of (1) and (2) on $[0,1]\times [0,t_b+h)$. It leads to a contradiction. Hence, $u\left(x,t\right)$ is unbounded somewhere in $[0,1]$ when $t\to t_b$. □

3. Blow-Up of the Solution

In this section, we prove that u blows up in a finite time under the condition (3). Our proof does not rely on the first eigenfunction method [15] which requires $\psi \left(x\right)$ being large enough in the domain. Instead, we modify the method of Lemma 3.4 of [16] to show the finite time blow-up of u.

In the sequel, let us define $T=t_b$, $\beta$ be a positive real number depending on p and q such that $\beta >2$, and F be a function in t defined as

$$F\left(t\right)=\frac{1}{\beta }{\int }_0^1{u}^{\beta }dx.$$

Clearly, $F\left(t\right)>0$ and $F^{\prime }\left(t\right)>0$ for $t⩾0$. Further, $F\left(t\right)$ has the following property.

Lemma 5.

If $p+q>2$, then

$${\left(F^{\prime }\left(t\right)\right)}^2<\frac{\beta }{2(\beta -1)}F\left(t\right)F^{″}\left(t\right),$$

(12)

for $t⩾0$.

Proof. 

The derivative of $F\left(t\right)$ is

$$F^{\prime }\left(t\right)={\int }_0^1{u}^{\beta -1}{u}_{t}dx.$$

(13)

By (1), we get

$$F^{\prime }\left(t\right)={\int }_0^1[u^{\beta -1}{u}_{xx}+u^{p+\beta -1}+\delta (x-c)u^{q+\beta -1}]dx.$$

The derivative of $F^{\prime }\left(t\right)$ is

$$\begin{array}{ccc}\hfill F^{″}\left(t\right)& =& {\int }_0^1[(\beta -1)u^{\beta -2}{u}_t{u}_{xx}+u^{\beta -1}{u}_{xxt}]dx\hfill \\ & & +{\int }_0^1(p+\beta -1)u^{p+\beta -2}{u}_{t}dx+(q+\beta -1)u^{q+\beta -2}(c,t)u_t(c,t).\hfill \end{array}$$

By (1) and a direct computation, we have

$$\begin{array}{ccc}\hfill F^{″}\left(t\right)& =& {\int }_0^1(\beta -1)u^{\beta -2}{u}_t[u_t-u^p-\delta (x-c)u^q]dx\hfill \\ & & +{\int }_0^1{u}^{\beta -1}{u}_{xxt}dx+{\int }_0^1(p+\beta -1)u^{p+\beta -2}{u}_{t}dx\hfill \\ & & +(q+\beta -1)u^{q+\beta -2}(c,t)u_t(c,t)\hfill \\ & =& (\beta -1){\int }_0^1{u}^{\beta -2}{\left(u_t\right)}^{2}dx+{\int }_0^1{u}^{\beta -1}{u}_{xxt}dx\hfill \\ & & +p{\int }_0^1{u}^{p+\beta -2}{u}_{t}dx+q{u}^{q+\beta -2}(c,t)u_t(c,t).\hfill \end{array}$$

Using integration by parts and $u(0,t)=u(1,t)=0$, it yields

$$\begin{array}{ccc}\hfill F^{″}\left(t\right)& =& (\beta -1){\int }_0^1{u}^{\beta -2}{\left(u_t\right)}^{2}dx+p{\int }_0^1{u}^{p+\beta -2}{u}_{t}dx+q{u}^{q+\beta -2}(c,t)u_t(c,t)\hfill \\ & & +(\beta -1)(\beta -2){\int }_0^1{u}^{\beta -3}{\left(u_x\right)}^2{u}_{t}dx+(\beta -1){\int }_0^1{u}^{\beta -2}{u}_t{u}_{xx}dx.\hfill \end{array}$$

By (1) and a direct computation, it gives

$$\left.\begin{array}{c}{F}^{″}\left(t\right)=2(\beta -1){\int }_0^1{u}^{\beta -2}{\left(u_t\right)}^{2}dx+(\beta -1)(\beta -2){\int }_0^1{u}^{\beta -3}{\left(u_x\right)}^2{u}_{t}dx\\ +[p-(\beta -1)]{\int }_0^1{u}^{p+\beta -2}{u}_{t}dx+[q-(\beta -1)]u^{q+\beta -2}(c,t)u_t(c,t).\end{array}\right\}$$

(14)

We want to prove that

$$F^{″}\left(t\right)>2(\beta -1){\int }_0^1{u}^{\beta -2}{\left(u_t\right)}^{2}dx$$

(15)

through investigating three different cases: (i) $q>p⩾1$, (ii) $p>q⩾1$, (iii) $p=q>1$.

Case (i): If $q>p⩾1$, then $p+q>2$. We let $\beta =p+1$ when $p>1$. Then, by (14)

$$F^{″}\left(t\right)=2p{\int }_0^1{u}^{p-1}{\left(u_t\right)}^{2}dx+p(p-1){\int }_0^1{u}^{p-2}{\left(u_x\right)}^2{u}_{t}dx+(q-p)u^{q+p-1}(c,t)u_t(c,t).$$

(16)

By Lemma 4, $u_t⩾0$ for $t⩾0$ when $x\in [0,1]$. Then,

$$F^{″}\left(t\right)>2p{\int }_0^1{u}^{p-1}{\left(u_t\right)}^{2}dx.$$

That is, (15). When $p=1$, we set

$$\beta =\frac{2{\int }_0^1{u}^{\beta -1}{u}_{t}dx+(q+1)u^{q+\beta -2}(c,t)u_t(c,t)}{{\int }_0^1{u}^{\beta -1}{u}_{t}dx+u^{q+\beta -2}(c,t)u_t(c,t)}$$

to obtain (15).

Case (ii): If $p>q⩾1$, we let $\beta =q+1$ when $q>1$. We follow the computation of case (i) to obtain (15). When $q=1$, we set

$$\beta =\frac{(p+1){\int }_0^1{u}^{p+\beta -2}{u}_{t}dx+2u^{\beta -1}(c,t)u_t(c,t)}{{\int }_0^1{u}^{p+\beta -2}{u}_{t}dx+u^{\beta -1}(c,t)u_t(c,t)}$$

to establish (15).

Case (iii): If $p=q>1$, we let $\beta =p+1$. By (16), we obtain

$$F^{″}\left(t\right)=2p{\int }_0^1{u}^{p-1}{\left(u_t\right)}^{2}dx+p(p-1){\int }_0^1{u}^{p-2}{\left(u_x\right)}^2{u}_{t}dx>2(\beta -1){\int }_0^1{u}^{\beta -2}{\left(u_t\right)}^{2}dx.$$

By above three cases, (13), and the Cauchy-Schwartz inequality, we have

$${\left(F^{\prime }\left(t\right)\right)}^2⩽{\int }_0^1{u}^{\beta }dx{\int }_0^1{u}^{\beta -2}{\left(u_t\right)}^{2}dx<\frac{\beta }{2(\beta -1)}F\left(t\right)F^{″}\left(t\right),$$

for $t⩾0$. The proof is complete. □

The result below provides an estimate of the upper bound of blow-up time of u.

Theorem 4.

If $p+q>2$, then there exists T given by

$$T=\frac{\beta }{(\beta -2)}\frac{F\left(0\right)}{F^{\prime }\left(0\right)}$$

such that u blows up somewhere in $(0,1)$ for a finite time τ where $\tau ⩽T$.

Proof. 

By a direct computation, we obtain

$$\frac{d}{dt}{F}^{-(\beta -2)/\beta }=-\frac{(\beta -2)}{\beta }{F}^{-2(\beta -1)/\beta }{F}^{\prime },$$

and

$$\frac{d^2}{d{t}^2}{F}^{-(\beta -2)/\beta }=\frac{2(\beta -1)(\beta -2)}{{\beta }^2}{F}^{(-3\beta +2)/\beta }\left[{\left(F^{\prime }\right)}^2-\frac{\beta }{2(\beta -1)}F{F}^{″}\right].$$

By (12) and $\beta >2$, it yields

$$\frac{d^2}{d{t}^2}{F}^{-(\beta -2)/\beta }<0.$$

After integrating this expression over $[0,t]$, we have

$$\frac{d{F}^{-(\beta -2)/\beta }\left(t\right)}{dt}-\frac{d{F}^{-(\beta -2)/\beta }\left(0\right)}{dt}<0.$$

Equivalently,

$$\frac{d{F}^{-(\beta -2)/\beta }\left(t\right)}{dt}<-\frac{(\beta -2)}{\beta }{F}^{-2(\beta -1)/\beta }\left(0\right)F^{\prime }\left(0\right).$$

Then, we integrate the above expression over $[0,t]$ to obtain

$$F^{-(\beta -2)/\beta }\left(t\right)<-\frac{(\beta -2)}{\beta }{F}^{-2(\beta -1)/\beta }\left(0\right)F^{\prime }\left(0\right)t+F^{-(\beta -2)/\beta }\left(0\right).$$

Since $F\left(0\right)>0$ and $F^{\prime }\left(0\right)>0$, the right side of above inequality is a decreasing function in t. Thus, there exists a finite time T such that

$$-\frac{(\beta -2)}{\beta }{F}^{-2(\beta -1)/\beta }\left(0\right)F^{\prime }\left(0\right)T+F^{-(\beta -2)/\beta }\left(0\right)=0.$$

Then, we solve for T to get

$$T=\frac{\beta }{(\beta -2)}\frac{F\left(0\right)}{F^{\prime }\left(0\right)}.$$

Therefore, there exists $\tau ⩽T$ such that $F^{-(\beta -2)/\beta }\left(\tau \right)=0$. This implies $F\left(\tau \right)=\infty$. Hence, u blows up somewhere in $(0,1)$ at $\tau$. The proof is complete. □

The result below is to show that the solution u blows up at $x=c$.

Theorem 5.

If u blows up in a finite time, then $x=c$ is a blow-up point.

Proof. 

Suppose that u blows up at $x^{*}\in (0,1)$ where $x^{*}\ne c$. Let $a_1$ and $a_2$ be positive real numbers such that $x^{*}$ and $c\in [a_1,a_2]⊊(0,1)$. By the integral solution (4) at $x=x^{*}$, it gives

$$\left.\begin{array}{c}u(x^{*},t)={\int }_0^{1}G(x^{*},\xi ,t)\psi \left(\xi \right)d\xi +{\int }_0^t{\int }_0^{1}G(x^{*},\xi ,t-s)u^p(\xi ,s)d\xi ds\hfill \\ +{\int }_0^{t}G(x^{*},c,t-s)u^q(c,s)ds.\hfill \end{array}\right\}$$

By the mean value theorem, there exist ${\eta }_5\in (0,t)$ and $\widehat{\xi }\in (0,1)$ such that

$$\left.\begin{array}{c}u(x^{*},t)={\int }_0^{1}G(x^{*},\xi ,t)\psi \left(\xi \right)d\xi +G(x^{*},\widehat{\xi },t-{\eta }_5){\int }_0^t{\int }_0^1{u}^p(\xi ,s)d\xi ds\hfill \\ +{\int }_0^{t}G(x^{*},c,t-s)u^q(c,s)ds.\hfill \end{array}\right\}$$

By (5), there exists a positive constant $k_3$ such that $G(x^{*},\widehat{\xi },t-{\eta }_5)⩽k_3$ (depends on $x^{*}$ and t). Thus,

$$u(x^{*},t)⩽{\int }_0^{1}G(x^{*},\xi ,t)\psi \left(\xi \right)d\xi +k_3{\int }_0^t{\int }_0^1{u}^p(\xi ,s)d\xi ds+{\int }_0^{t}G(x^{*},c,t-s)u^q(c,s)ds.$$

Since $x^{*}$ is a blow-up point, $u(x^{*},t)\to \infty$ when $t\to T$ but $u(c,t)$ remains bounded above when $t\to T$. This implies that ${\int }_0^t{\int }_0^1{u}^p(\xi ,s)d\xi ds\to \infty$ when $t\to T$. Since $G\left(x,\xi ,t\right)>0$ for $t>0$, there exists a positive constant $k_4$ (depends on x and t) such that $G\left(x,\xi ,t\right)⩾k_4$ for $x\in [a_1,a_2]$. If we consider (4) again at $x=c$, it yields

$$u(c,t)⩾k_4[{\int }_0^t{\int }_0^1{u}^p(\xi ,s)d\xi ds+{\int }_0^t{u}^q(c,s)ds].$$

Because ${\int }_0^t{\int }_0^1{u}^p(\xi ,s)d\xi ds\to \infty$ when $t\to T$, we have $u(c,t)\to \infty$ when $t\to T$. It leads to a contradiction. Thus, c is a blow-up point. □

4. Conclusions

In this paper, the existence and uniqueness of solution of a semilinear parabolic problem with a mixed reaction source: $u_t-u_{xx}=u^p+\delta (x-c)u^q$ is studied. We establish this result through investigating the corresponding integral solution. A sufficient condition for finite time blow-up of u is provided. Then, an upper bound of the blow-up time is obtained. Further, we prove that u blows up at c.