Simultaneous blow-up of the solution for a singular parabolic system with concentrated sources

1.   Introduction

Let $b$ be a constant such that $b\in (0, 1)$, $L$ is the singular parabolic operator such that $Lu = {u}_{t}-{u}_{xx}-b{u}_{x}/x$, $T$ is a positive real number, $p⩾1$, $q⩾1$, and $\delta \left(x\right)$ is the Dirac delta function. In this paper, we investigate the following system of singular parabolic initial-boundary value problems with concentrated nonlinear source functions:

It is assumed that the initial conditions ${u}_{0}\left(x\right)$ and ${v}_{0}\left(x\right)$ are nonnegative functions on $[0, 1]$ such that ${u}_{0}\left(b\right)$ and ${v}_{0}\left(b\right)$ are positive, and they satisfy conditions below:

The singular operator $L$ is associated with axially symmetric solutions to the heat equation ^[1]. This operator is also related to axially symmetric potentials of elliptic equations ^[2,3,4,5] when ${u}_{t} = 0$. If $b = (N-1)/2$, from ^[6], $Lu$ represents the backward differential equation corresponding to the radial component of N-dimensional Brownian motion. In addition, $Lu$ can describe a stochastic process of a chain of random walks ^[7]. Further, $Lu = 0$ is interpreted as the Fokker-Planck equation of a diffusion problem in which $b$ represents the drift ^[8]. Feller ^[8] discussed the existence and uniqueness of solution of $Lu = 0$ over the domain $(0, \infty)$, Colton ^[9] investigated the Cauchy problem of $Lu = 0$, and Brezis et al. ^[6] studied the differentiability of solution. In addition, Alexiades ^[10] investigated the existence and uniqueness of solution over a domain with moving boundary for $Lu = f(x, t)$. Chan and Chen ^[11] studied the singular quenching problem below:

To a given $b$, they determined the critical length of $u$ numerically through evaluating the approximated solution of the integral solution of (1.3).

For $x\in {\mathbb{R}}^{N}$, Escobedo and Herrero ^[12] studied the blow-up of the Cauchy problem of the following semilinear parabolic system:

They proved that $u$ and $v$ blow up in a finite time when $pq > 1$ and $\left(\mathrm{max}\{p, q\}+1\right)/\left(pq-1\right)⩾N/2$. Levine ^[13] obtained a similar result for the homogeneous Dirichlet boundary-value problem.

In this paper, we prove the existence and uniqueness of solutions $u$ and $v$. Under some assumptions on ${u}_{0}\left(x\right)$ and ${v}_{0}\left(x\right)$, we show that they blow up in a finite time if $p⩾1$, $q⩾1$, and $pq > 1$. We further prove that $u$ and $v$ both blow up simultaneously at $x = b$, then we obtain the result that $u$ and $v$ both blow up totally in $\left(\mathrm{0, 1}\right)$.

2.   Existence and uniqueness of the solution

Let $G\left(x, \xi, t-s\right)$ be the Green's function of $Lu = 0$. For any $x$ and $\xi \in \left[\mathrm{0, 1}\right]$, where $t$ and $s$ are variables and they belong to $(-\infty, \infty)$, $G(x, \xi, t-s)$ satisfies the problem below:

By Chan and Wong ^[14], the Green's function of the operator $L$ is given by

where ${\lambda }_{n}$ for $n = 1, 2, 3, \dots$ are the eigenvalues of the Sturm-Liouville problem

where $\mu$ is an unknown constant. Let ${\phi }_{n}\left(x\right)$ be the corresponding eigenfunction of ${\lambda }_{n}$. The mathematical formula of ${\phi }_{n}\left(x\right)$ is

where ${J}_{\nu }$ is the Bessel function of the first kind of order $\nu = (1-b)/2$. The first eigenvalue ${\lambda }_{1}$ of (2.2) is positive and ${\phi }_{1}\left(x\right)$ satisfies: ${\phi }_{1}\left(x\right) > 0$ in (0, 1) and ${\int }_{0}^{1}{x}^{b}{\phi }_{1}dx = 1$. For ease of computation, let us state the result of Lemma 1 of Chan and Wong.

Lemma 2.1. There exist positive constants ${k}_{1}$ and ${k}_{2}$ such that

(a) $\left|{\phi }_{n}\left(x\right)\right|⩽{k}_{1}{x}^{-b/2}$ for $x$ in $(0, 1]$;

(b) $\left|{\phi }_{n}\left(x\right)\right|⩽{k}_{2}{\lambda }_{n}^{1/4}$ for $x$ on $[0, 1]$.

By theorems of the Bessel-Fourier series expansion of Watson [15, pp. 591 and 594], ${\int }_{0}^{1}G(x, \xi, t){u}_{0}\left(\xi \right)d\xi$ converges to ${u}_{0}\left(x\right)$ for $x\in \left[\mathrm{0, 1}\right]$ when $t\to 0$. By the result of Chan and Wong, the integral solution of (1.1) is given by

We let

then the integral solution represented by (2.3) is equivalent to

In the following, we show that the integral solution represented by (2.4) is a unique continuous solution of (1.1). To achieve this, let us construct two sequences: $\left\{{u}_{m}\right\}$ and $\left\{{v}_{m}\right\}$; they satisfy the iteration process

for $m = 1, 2$, … We assume the initial iteration: ${u}_{0}\left(x, t\right) = {u}_{0}\left(x\right)$ and ${v}_{0}(x, t) = {v}_{0}\left(x\right)$ on $[0, 1]\times [0, T)$. By (2.4), the integral solution of $\left\{{u}_{m}\right\}$ and $\left\{{v}_{m}\right\}$ in (2.5) is given by

By Lemma 4 of Chan and Wong, $G(x, \xi, t-s)$ is a positive function in the set $\left\{\right(x, \xi; t, s):x\mathrm{a}\mathrm{n}\mathrm{d}\xi \mathrm{a}\mathrm{r}\mathrm{e}\; \mathrm{i}\mathrm{n}(0, 1), \mathrm{a}\mathrm{n}\mathrm{d}t > s⩾0\}$. It indicates that ${u}_{m} > 0$ and ${v}_{m} > 0$ in $(0, 1)\times (0, T)$. The following lemma shows that $\left\{{u}_{m}\right\}$ and $\left\{{v}_{m}\right\}$ are nondecreasing sequences and they are continuous functions.

Lemma 2.2. The sequences $\left\{{u}_{m}\right\}$ and $\left\{{v}_{m}\right\}$ described by (2.6) are monotone nondecreasing: ${u}_{0}\left(x\right)⩽{u}_{1}\left(x, t\right)⩽\cdots ⩽{u}_{m}\left(x, t\right)⩽\dots, {v}_{0}\left(x\right)⩽{v}_{1}(x, t)⩽\dots ⩽{v}_{m}(x, t)⩽\dots$, and $\left\{{u}_{m}\right\}$ and $\left\{{v}_{m}\right\}$ are continuous on $\left[\mathrm{0, 1}\right]\times [0, T)$.

Proof. From (1.2), we deduce $H(x, t, {u}_{0}, {v}_{0})⩾0$ and $I(x, t, {v}_{0}, {u}_{0})⩾0$, then by (2.6) with $m = 1$, we obtain

for $x\in \left[\mathrm{0, 1}\right]$. Therefore, ${u}_{1}\left(x, t\right)-{u}_{0}\left(x\right)⩾0$ and ${v}_{1}\left(x, t\right)-{v}_{0}\left(x\right)⩾0$ for $x\in \left[\mathrm{0, 1}\right]$. Suppose that it is true for $m = k$; it gives ${u}_{k}\left(x, t\right)-{u}_{k-1}(x, t)⩾0$ and ${v}_{k}\left(x, t\right)-{v}_{k-1}(x, t)⩾0$ for $(x, t)\in \left[\mathrm{0, 1}\right]\times [0, T)$. Suppose $m = k+1$, and $G(x, \xi, t-s)$ is a positive function for $t > s$

Similarly, we have ${v}_{k+1}\left(x, t\right)-{v}_{k}\left(x, t\right)⩾0$ for $(x, t)\in \left[\mathrm{0, 1}\right]\times [0, T)$. By the mathematical induction, the sequences of solutions $\left\{{u}_{m}\right\}$ and $\left\{{v}_{m}\right\}$ are monotone nondecreasing.

To prove that ${u}_{m}(x, t)$ and ${v}_{m}(x, t)$ are continuous on $\left[\mathrm{0, 1}\right]\times [0, T)$, let us consider (2.6) with $m = 1$. It yields

We show that $G(x, \xi, t){u}_{0}\left(\xi \right)$ and $G(x, b, t-s){v}_{0}^{p}(b, s)$ are integrable. Let us prove that ${\int }_{0}^{t}G(x, b, t-s){v}_{0}^{p}(b, s)ds$ is bounded for $x\in \left[\mathrm{0, 1}\right]$ and $t > 0$. By (2.1), we have

Let ${k}_{3} = \mathrm{m}\mathrm{a}\mathrm{x}\{{max}_{x\in \left[\mathrm{0, 1}\right]}{v}_{0}\left(x\right), {max}_{x\in \left[\mathrm{0, 1}\right]}{u}_{0}\left(x\right)\}$, then

By Lemma 2.1, it gives

By Watson [15, p. 506], $O\left({\lambda }_{n}\right) = O\left({n}^{2}\right)$ for $n$ large. This shows that the above series converges for $t > 0$. The upper bound of the series is independent of $x$ for each fixed $t > 0$. For each positive integer $j$ greater than 1, the finite series $\sum_{n = 1}^{j}{\int }_{0}^{t}{b}^{b}{\phi }_{n}\left(b\right){\phi }_{n}\left(x\right){e}^{-{\lambda }_{n}\left(t-s\right)}{v}_{0}^{p}(b, s)ds$ is continuous for $x\in \left[\mathrm{0, 1}\right]$ and $t > 0$. Further, by (2.7), this finite sum satisfies the following inequality:

Therefore, $\sum_{n = 1}^{j}{\int }_{0}^{t}{b}^{b}{\phi }_{n}\left(b\right){\phi }_{n}\left(x\right){e}^{-{\lambda }_{n}\left(t-s\right)}{v}_{0}^{p}(b, s)ds$ converges uniformly. As $j\to \infty$, ${\int }_{0}^{t}G(x, b, t-s){v}_{0}^{p}(b, s)ds$ is continuous for $x\in \left[\mathrm{0, 1}\right]$ and $t > 0$. Through a similar computation, it yields

By Lemma 2.1(a) and ${\phi }_{n}\left(0\right) = {\phi }_{n}\left(1\right) = 0$, the righthand series converges for $x\in \left[\mathrm{0, 1}\right]$ when $t > 0$. By the similar computation above, we obtain ${\int }_{0}^{1}G(x, \xi, t){u}_{0}\left(\xi \right)d\xi$ as continuous for $x\in \left[\mathrm{0, 1}\right]$ and $t > 0$. By (2.7), ${\int }_{0}^{t}G(x, b, t-s){v}_{0}^{p}(b, s)ds\to 0$ as $t\to 0$. Since ${\int }_{0}^{1}G(x, \xi, t){u}_{0}\left(\xi \right)d\xi$ converges to ${u}_{0}\left(x\right)$ for $x\in \left[\mathrm{0, 1}\right]$ when $t\to 0$, it follows that $\underset{t\to 0}{\mathrm{lim}}{u}_{1}\left(x, t\right) = {u}_{0}\left(x\right)$. Therefore, ${u}_{1}\left(x, t\right) = H\left(x, t, {u}_{0}, {v}_{0}\right)$ is continuous on $\left[\mathrm{0, 1}\right]\times [0, T)$. By the mathematical induction and computation above, ${u}_{m}\left(x, t\right)$ is continuous on $\left[\mathrm{0, 1}\right]\times [0, T)$ for $m = 1, 2, \dots$ A similar computation proves that ${v}_{m}(x, t)$ is continuous on $\left[\mathrm{0, 1}\right]\times [0, T)$ for $m = 1, 2, \dots$ This completes the proof.

Let $h$ be a positive real number less than $T$. We show that sequences $\left\{{\left({u}_{m}\right)}_{t}\right\}$ and $\left\{{\left({v}_{m}\right)}_{t}\right\}$ are nonnegative functions over $\left[\mathrm{0, 1}\right]$.

Lemma 2.3. ${\left({u}_{m}\right)}_{t}⩾0$ and ${\left({v}_{m}\right)}_{t}⩾0$ on $\left[\mathrm{0, 1}\right]\times (0, T)$ for $m = 1, 2, \dots$

Proof. To establish this result, we prove that ${u}_{m}\left(x, t+h\right)⩾{u}_{m}\left(x, t\right)$ on $\left[\mathrm{0, 1}\right]\times [0, T-h]$ through the mathematical induction. When $m = 1$, we follow Lemma 2.2 to get ${u}_{1}\left(x, h\right)⩾{u}_{0}\left(x\right)$ and ${v}_{1}\left(x, h\right)⩾{v}_{0}\left(x\right)$. From (2.5), we deduce

and

When (2.5) is subtracted from the above two problems, we deduce

and

By (2.3),

Therefore, ${u}_{1}\left(x, t+h\right)⩾{u}_{1}(x, t)$ and ${v}_{1}\left(x, t+h\right)⩾{v}_{1}(x, t)$ on $\left[\mathrm{0, 1}\right]\times [0, T-h]$. As $h\to 0,$ it implies ${\left({u}_{1}\right)}_{t}⩾0$ and ${\left({v}_{1}\right)}_{t}⩾0$ over $\left[\mathrm{0, 1}\right]\times \left(0, T\right)$. Assume that this statement is true for $m = k$. When $m = k+1$, it follows a similar computation and Lemma 2.2 to achieve

Thus, ${u}_{k+1}\left(x, t+h\right)⩾{u}_{k+1}(x, t)$. Similarly, we obtain ${v}_{k+1}\left(x, t+h\right)⩾{v}_{k+1}(x, t)$. When $h\to 0$, ${\left({u}_{k+1}\right)}_{t}⩾0$ and ${\left({v}_{k+1}\right)}_{t}⩾0$ over $\left[\mathrm{0, 1}\right]\times \left(0, T\right)$. By the mathematical induction, we conclude that ${\left({u}_{m}\right)}_{t}⩾0$ and ${\left({v}_{m}\right)}_{t}⩾0$ on $\left[\mathrm{0, 1}\right]\times \left(0, T\right)$ for $m = 1, 2, \dots$ This completes the proof.

Lemma 2.4. Given $\stackrel{-}{t} > 0$, there exist positive constants ${\alpha }_{1}$ and ${\alpha }_{2}$ (depending on $\stackrel{-}{t}$) both less than $1$ such that

and

Proof. By (2.6), we get

Since $p⩾1$, the mean value theorem, and Lemma 2.2, there exists a positive constant ${k}_{4}$ such that

for $s\in [0, t]$. Based on the above inequality, we obtain the expression

We follow a similar method of Lemma 2.2 to obtain an upper bound of the integral of the Green's function

The righthand series converges and tends to zero for positive $t$ being close to 0. Let us choose $t = \stackrel{-}{t}$ sufficiently close to 0 such that

Hence, ${\mathrm{m}\mathrm{a}\mathrm{x}}_{\left[\mathrm{0, 1}\right]\times [0, \stackrel{-}{t}]}\left|H\left(x, t, {u}_{0}, {v}_{m}\right)-H\left(x, t, {u}_{0}, {v}_{m-1}\right)\right| < {\alpha }_{1}{\mathrm{m}\mathrm{a}\mathrm{x}}_{\left[\mathrm{0, 1}\right]\times [0, \stackrel{-}{t}]}\left|{v}_{m}-{v}_{m-1}\right|$. Similarly, we have ${\alpha }_{2} < 1$ such that ${\mathrm{m}\mathrm{a}\mathrm{x}}_{\left[\mathrm{0, 1}\right]\times [0, \stackrel{-}{t}]}\left|I\left(x, t, {v}_{0}, {u}_{m}\right)-I\left(x, t, {v}_{0}, {u}_{m-1}\right)\right| < {\alpha }_{2}{\mathrm{m}\mathrm{a}\mathrm{x}}_{\left[\mathrm{0, 1}\right]\times [0, \stackrel{-}{t}]}\left|{u}_{m}-{u}_{m-1}\right|$. The proof is complete.

Let ${k}_{5}$ and ${k}_{6}$ be positive constants greater than ${k}_{3}$. We prove that $\left\{{u}_{m}\right\}$ and $\left\{{v}_{m}\right\}$ are bounded above by ${k}_{5}$ and ${k}_{6}$, respectively, on $\left[0, 1\right]\times \left[0, \stackrel{-}{t}\right]$.

Lemma 2.5. For any ${k}_{5} > {k}_{3}$ and ${k}_{6} > {k}_{3}$, there exists $\stackrel{-}{t} > 0$ such that ${k}_{5} > {u}_{m}$ and ${k}_{6} > {v}_{m}$ on $\left[0, 1\right]\times \left[0, \stackrel{-}{t}\right]$ for $m = 1, 2, \dots$

Proof. Based on Lemma 2.2, we know that ${\int }_{0}^{1}G(x, \xi, t){u}_{0}\left(\xi \right)d\xi$, ${\int }_{0}^{t}G(x, b, t-s){v}_{m-1}^{p}(b, s)ds$, ${\int }_{0}^{1}G(x, \xi, t){v}_{0}\left(\xi \right)d\xi$, and ${\int }_{0}^{t}G(x, b, t-s){u}_{m-1}^{q}(b, s)ds$ are bounded above when $m = 1, 2, \dots$ By Lemma 2.2 again, $\underset{t\to 0}{\mathrm{lim}}{\int }_{0}^{1}G(x, \xi, t){v}_{0}\left(\xi \right)d\xi = {v}_{0}\left(x\right)$ and $\underset{t\to 0}{\mathrm{lim}}{\int }_{0}^{1}G(x, \xi, t){u}_{0}\left(\xi \right)d\xi = {u}_{0}\left(x\right)$ on $\left[0, 1\right]$, and ${k}_{5} > {k}_{3}$ and ${k}_{6} > {k}_{3}$. Thus, there exists $\stackrel{-}{t} > 0$ such that

for $x\in [0, 1]$. By Lemma 2.3, ${\left({u}_{m}\right)}_{t}⩾0$ and ${\left({v}_{m}\right)}_{t}⩾0$, which leads to ${k}_{5} > {u}_{1}$ and ${k}_{6} > {v}_{1}$ on $\left[0, 1\right]\times [0, \stackrel{-}{t}]$. Assume that it is true for $m = i$, we follow a similar calculation to obtain ${k}_{5} > {u}_{i+1}$ and ${k}_{6} > {v}_{i+1}$ on $\left[0, 1\right]\times [0, \stackrel{-}{t}]$. By the mathematical induction, we have ${k}_{5} > {u}_{m}$ and ${k}_{6} > {v}_{m}$ on $\left[0, 1\right]\times [0, \stackrel{-}{t}]$ for $m = 1, 2, \dots$

Let $u\left(x, t\right) = \underset{m\to \infty }{\mathrm{lim}}{u}_{m}(x, t)$ and $v\left(x, t\right) = \underset{m\to \infty }{\mathrm{lim}}{v}_{m}(x, t)$. We have the result below.

Theorem 2.6. The integral equation (2.3) has a unique continuous solution ${k}_{5} > u(x, t)⩾{u}_{0}\left(x\right)$ and ${k}_{6} > v(x, t)⩾{v}_{0}\left(x\right)$ on $\left[\mathrm{0, 1}\right]\times [0, \stackrel{-}{t}]$.

Proof. Based on Lemma 2.2 and Dini's theorem, the sequences $\left\{{u}_{m}\right\}$ and $\left\{{v}_{m}\right\}$ converge uniformly to continuous solutions $u(x, t)$ and $v(x, t)$ such that ${k}_{5} > u(x, t)⩾{u}_{0}\left(x\right)$ and ${k}_{6} > v(x, t)⩾{v}_{0}\left(x\right)$ on $\left[\mathrm{0, 1}\right]\times [0, \stackrel{-}{t}]$. By Lemma 2.4, $u(x, t)$ and $v(x, t)$ are unique. The proof is complete.

Based on Lemma 2.3, we obtain that ${u}_{t}$ and ${v}_{t}$ are nonnegative.

Lemma 2.7. ${u}_{t}⩾0$ and ${v}_{t}⩾0$ on $\left[\mathrm{0, 1}\right]\times (0, \stackrel{-}{t}]$.

Let ${t}_{b}$ be the supremum of $\stackrel{-}{t}$ such that the integral solution (2.3) has a unique continuous solution on $\left[\mathrm{0, 1}\right]\times [0, \stackrel{-}{t}]$. We follow Theorem 3 of Chan ^[16] to obtain the result below.

Theorem 2.8. If ${t}_{b}$ is finite, then $u(x, t)$ and $v(x, t)$ are unbounded somewhere on $\left[\mathrm{0, 1}\right]$ when $t\to {t}_{b}$.

3.   Blow-up of the solution

In this section, let us assume that the initial data ${u}_{0}\left(x\right)$ and ${v}_{0}\left(x\right)$ both attain positive maximum at $x = b$ only. By Theorem 2.6, $u\left(b, t\right) > 0$ and $v\left(b, t\right) > 0$ for $t > 0$. Our first result is to prove that $u$ and $v$ both reach their maximum at $x = b$, then we show that $u$ and $v$ blow up in a finite time.

Lemma 3.1. $u$ and $v$ both attain their maximum at $x = b$ for $t⩾0$.

Proof. We prove this result using contradiction. From (1.1), it gives

Let $t = {t}_{1}$ be the infimum of $t$ such that $u(x, t)$ attains another maximum at ${x}_{1}$ for some ${x}_{1}\in (0, b)\cup (b, 1)$. As $u\left(x, 0\right) = {u}_{0}\left(x\right)$ reaches its positive maximum at $x = b$, ${t}_{1} > 0$, then ${u}_{x}\left({x}_{1}, {t}_{1}\right) = 0$ and ${u}_{xx}\left({x}_{1}, {t}_{1}\right) < 0$. By Lemma 2.7, it yields ${u}_{t}\left({x}_{1}, {t}_{1}\right)⩾0$. Therefore, $Lu\left({x}_{1}, {t}_{1}\right) > 0$. It contradicts $Lu = 0$ at $x\ne b$ for $t > 0$. Thus, $u(x, t)$ attains its maximum at $x = b$ for $t⩾0$. Through a similar computation, we prove that $v(x, t)$ attains its maximum at $x = b$ for $t⩾0$. This completes the proof.

We multiply (1.1) by ${x}^{b}{\phi }_{1}\left(x\right)$, and then integrate the expression with respect to $x$ over $\left[\mathrm{0, 1}\right]$,

Using integration by parts and (2.2), it yields

By Lemma 3.1, $u(x, t)⩽u(b, t)$ and $v(x, t)⩽v(b, t)$ on $\left[\mathrm{0, 1}\right]$ for $t⩾0$, we have

Let ${k}_{7} = {b}^{b}{\phi }_{1}\left(b\right)$. The system of differential equations of (3.1) and (3.2) becomes

Let $\mathrm{\Psi }\left(t\right) = {\int }_{0}^{1}{x}^{b}{\phi }_{1}udx$ and $\mathrm{\Lambda }\left(t\right) = {\int }_{0}^{1}{x}^{b}{\phi }_{1}vdx$, then (3.3) and (3.4) become

Let $\mathrm{\Gamma } > 0$ and $\text{B}$ be a positive constant such that

To obtain the result if either $u$ or $v$ blows up somewhere on $\left[\mathrm{0, 1}\right]$ in a finite time, we study the blow-up of system of differential inequalities (3.3) and (3.4) when $p⩾1$, $q⩾1$, and $pq > 1$. We examine (3.3) and (3.4) in two different cases: (I) $q = 1$, $p > 1$, or $p = 1$, $q > 1$; (II) $q\;\mathrm{a}\mathrm{n}\mathrm{d}\;p > 1$. To case (I), we prove either $u$ or $v$ to blow up in a finite time through constructing a lower solution. We then compare the solutions of case (I) with case (II) to show either $u$ or $v$ to blow up in a finite time by the Picard iterates.

Lemma 3.2. (I) Suppose that $q = 1$ and $p > 1$, or $p = 1$ and $q > 1$, then either $u$ or $v$ blows up somewhere on $\left[\mathrm{0, 1}\right]$ in a finite time.

(II) Suppose that $p > 1$ and $q > 1$, and assume that

(a) ${\left(\mathrm{\Lambda }\left(0\right)\right)}^{p}{k}_{7}/{\lambda }_{1}⩾\mathrm{\Psi }\left(0\right)$ and ${\left(\mathrm{\Psi }\left(0\right)\right)}^{q}{k}_{7}/{\lambda }_{1}⩾\mathrm{\Lambda }\left(0\right)$,

(b) $\mathrm{\Psi }\left(0\right)⩾1$ or $\mathrm{\Lambda }\left(0\right)⩾1$,

then either $u$ or $v$ blows up somewhere on $\left[\mathrm{0, 1}\right]$ in a finite time.

Proof. Case (I): Suppose $q = 1$ and $p > 1$ and let $U\left(t\right) = {\int }_{0}^{1}{x}^{b}{\phi }_{1}udx$ and $V\left(t\right) = {\int }_{0}^{1}{x}^{b}{\phi }_{1}vdx$. Let us consider the following system of ordinary differential equations:

with the initial conditions: $U\left(0\right) = \mathrm{\Psi }\left(0\right)$ and $V\left(0\right) = \mathrm{\Lambda }\left(0\right)$. We differentiate the second equation of (3.6) with respect to $t$ and then substitute it into the first one to obtain

with the initial conditions: $V\left(0\right) = \mathrm{\Lambda }\left(0\right) > 0$ and ${V}^{{'}}\left(0\right) = \mathrm{\Lambda }{'}\left(0\right) > 0.$ For $t\in [0, \mathrm{\Gamma })$, we construct a lower solution of the above equation as $E\left(t\right) = {(\mathrm{\Gamma }-t)}^{-1/(p-1)}/B$, where $B$ satisfies $V\left(0\right) > {\mathrm{\Gamma }}^{-1/(p-1)}/B = E\left(0\right)$ and ${V}^{{'}}\left(0\right) > {\mathrm{\Gamma }}^{-p/(p-1)}/[B(p-1)]$. The first and second derivatives of $E\left(t\right)$ are

We evaluate the expression ${E}^{{'}{'}}/{k}_{7}+2{\lambda }_{1}{E}^{{'}}/{k}_{7}+{\lambda }_{1}^{2}E/{k}_{7}-{k}_{7}{E}^{p}$ to have

By (3.5), we have ${E}^{{'}{'}}/{k}_{7}+2{\lambda }_{1}{E}^{{'}}/{k}_{7}+{\lambda }_{1}^{2}E/{k}_{7}-{k}_{7}{E}^{p} > 0$, then by the comparison theorem [17, p. 96], we have $V\left(t\right)⩾E\left(t\right)$ for $t\in [0, \mathrm{\Gamma })$. As $E\left(t\right)\to \mathrm{\infty }$ when $t\to \mathrm{\Gamma }$, $V\left(t\right)$ blows up in a finite time less than or equal to $\mathrm{\Gamma }$. Thus, $U+V$ blows up in a finite time. By the comparison theorem, $\mathrm{\Psi }+\mathrm{\Lambda }$ blows up in a finite time. Hence, either $u$ or $v$ in (3.3) and (3.4) blows up on $\left[\mathrm{0, 1}\right]$ in a finite time. We follow a similar computation to obtain the same result when $p = 1$ and $q > 1$, with $p$ and $\mathrm{\Lambda }\left(0\right)$ being replaced by $q$ and $\mathrm{\Psi }\left(0\right)$, respectively, in (3.5).

Case (II): Suppose that $p > 1$ and $q > 1$. Let $W\left(t\right) = {\int }_{0}^{1}{x}^{b}{\phi }_{1}udx$ and $Y\left(t\right) = {\int }_{0}^{1}{x}^{b}{\phi }_{1}vdx$ be the solution of the following system of ordinary differential equations:

with initial conditions: $\text{W}\left(0\right) = \mathrm{\Psi }\left(0\right)$ and $\text{Y}\left(0\right) = \mathrm{\Lambda }\left(0\right)$. Multiplying the integrating factor ${e}^{{\lambda }_{1}t}$ on both sides, this system becomes

Using the Picard iterates, the approximated solutions of $W$ and $Y$ are given by

with ${W}_{0} = W\left(0\right)$ and ${Y}_{0} = Y\left(0\right)$, $m = 1, \; \; \; \; 2, \; \dots$ By assumption (a), it yields $W\left(0\right)⩽{W}_{1}\left(t\right)⩽{W}_{2}\left(t\right)\dots ⩽{W}_{m}\left(\mathrm{t}\right)⩽\dots$ and $Y\left(0\right)⩽{Y}_{1}\left(t\right)⩽{Y}_{2}\left(t\right)\dots ⩽{Y}_{m}\left(\mathrm{t}\right)⩽\dots$ Similarly, the representation of the approximated solutions of $U$ and $V$ are

with ${U}_{0} = U\left(0\right)$ and ${V}_{0} = V\left(0\right)$, $m = 1, 2, \dots$ The system (3.9) and (3.10) is subtracted from (3.7) and (3.8) to yield

We show that ${W}_{m}\left(t\right)⩾{U}_{m}\left(t\right)$ and ${Y}_{m}\left(t\right)⩾{V}_{m}\left(t\right)$ for $m = 1, 2, \dots$ through the mathematical induction. When $m = 1$,

By assumption (b), ${W}_{0}⩾1$. It concludes that ${W}_{1}\left(t\right) = {U}_{1}\left(t\right)$ and ${Y}_{1}\left(t\right)⩾{V}_{1}\left(t\right)$. Assume that it is true for $m = i$, so that ${W}_{i}\left(t\right)⩾{U}_{i}\left(t\right)$ and ${Y}_{i}\left(t\right)⩾{V}_{i}\left(t\right)$. When $m = i+1$, we have

Thus, ${W}_{i+1}\left(t\right)⩾{U}_{i+1}\left(t\right)$ and ${Y}_{i+1}\left(t\right)⩾{V}_{i+1}\left(t\right)$. By the mathematical induction, ${W}_{m}\left(t\right)⩾{U}_{m}\left(t\right)$ and ${Y}_{m}\left(t\right)⩾{V}_{m}\left(t\right)$ for $m = 1, 2, \dots$ By the convergence of the Picard iterates [18, pp. 71-74], ${U}_{m}\left(t\right)\to U\left(t\right)$, ${V}_{m}\left(t\right)\to V\left(t\right)$, ${W}_{m}\left(t\right)\to W\left(t\right)$, and ${Y}_{m}\left(t\right)\to Y\left(t\right)$ as $m\to \infty$. Thus, $W+Y⩾U+V$. As $U+V$ blows up in a finite time, $W+Y$ blows up in a finite time. In case $\mathrm{\Lambda }\left(0\right)⩾1$ in assumption (b), we repeat the above procedure to compare the solution of (3.7) and (3.8) to Case (I) for $p = 1$ and $q > 1$ to obtain the same result. Hence, either $u$ or $v$ in (3.3) and (3.4) blows up on $\left[\mathrm{0, 1}\right]$ in a finite time. This completes the proof.

Theorem 3.3. If either $u$ or $v$ blows up at $t = {t}_{b}$, then $u$ and $v$ blow up simultaneously at the point $x = b$ at $t = {t}_{b}$.

Proof. Without loss of generality, we suppose that $u$ blows up somewhere at ${x}^{*}$ where ${x}^{*}\in \left(\mathrm{0, 1}\right)$ and ${x}^{*}\ne b$. Let ${a}_{1}$ and ${a}_{2}$ be positive real numbers such that ${x}^{*}$ and $b\in [{a}_{1}, {a}_{2}]⊊(0, \; 1)$. By (2.3), the integral solution of $u\left(x, t\right)$ at $x = b$ is given by

By the mean value theorem, there exists $\theta \in (0, t)$ such that

As $G\left(b, b, t-\theta \right) > 0$ for $t > \theta$, there exists a positive constant ${k}_{8}$ (depending on $b$ and $t$) such that $G\left(b, b, t-\theta \right)⩾{k}_{8}$ and

Since $u$ does not blow up at $b$, ${\int }_{0}^{t}{v}^{p}\left(b, s\right)ds$ is bounded when $t\to {t}_{b}$. Further, there exists a positive constant ${k}_{9}$ (depending on $x$ and $t$) such that $G(x, \xi, t)⩽{k}_{9}$ for $x\in [{a}_{1}, {a}_{2}]$. Let us consider the integral solution of $u\left(x, t\right)$ at $x = {x}^{*}$, which satisfies the following inequality:

Because ${\int }_{0}^{t}{v}^{p}\left(b, s\right)ds$ is bounded when $t\to {t}_{b}$, we have $u\left({x}^{*}, t\right)$ being bounded when $t\to {t}_{b}$. It leads to a contradiction. Thus, $b$ is a blow-up point of $u\left(x, t\right)$. By (3.11), if we replace ${x}^{*}$ by $b$, then we have ${\int }_{0}^{t}{v}^{p}\left(b, s\right)ds\to \infty$ as $t\to {t}_{b}$. Thus, $v(x, t)$ blows up at $x = b.$ Hence, $u\left(x, t\right)$ and $v(x, t)$ blow up simultaneously at $x = b$.

Corollary 3.4. If u and v blow up at the point $x = b$ at $t = {t}_{b}$, then the blow-up set of u and v is $\left(\mathrm{0, 1}\right)$.

Proof. By Theorem 3.3, $u\left(b, t\right)\to \infty$ as $t\to {t}_{b}$. It implies ${\int }_{0}^{t}{v}^{p}\left(b, s\right)ds\to \infty$ as $t\to {t}_{b}$. For $x\in [{a}_{1}, {a}_{2}]⊊(0, \; 1),$ we know ${\int }_{0}^{t}G(x, b, t-s){v}^{p}\left(b, s\right)ds⩾{\int }_{0}^{t}{k}_{8}{v}^{p}\left(b, s\right)ds\to \infty$ as $t\to {t}_{b}$. By (2.3), we have $u\left(x, t\right)\to \infty$ as $t\to {t}_{b}$ for $x\in [{a}_{1}, {a}_{2}]$. Similarly, it yields $v\left(x, t\right)\to \infty$ as $t\to {t}_{b}$ for $x\in [{a}_{1}, {a}_{2}]$ if $v\left(b, t\right)\to \infty$ as $t\to {t}_{b}$. Since $[{a}_{1}, {a}_{2}]$ is any subset of $(0, \; 1)$, the blow-up set of $u$ and $v$ is $(0, \; \; \; \; 1)$. This completes the proof.

4.   Conclusions

In this paper, the existence and uniqueness of the solution of a system of singular parabolic problems with concentrated nonlinear reaction sources and homogeneous Dirichlet boundary condition: ${u}_{t}-{u}_{xx}-b{u}_{x}/x = \delta (x-b){v}^{p}$, ${v}_{t}-{v}_{xx}-b{v}_{x}/x = \delta (x-b){u}^{q}$ are established. The result is obtained by investigating the corresponding integral solutions. If $p⩾1$, $q⩾1$, and $pq > 1$, we prove that either $u$ or $v$ blows up in a finite time through constructing a lower solution. Further, we show that $u$ and $v$ blow up simultaneously in a finite time at $b$, and they blow up everywhere in the domain except the boundary.

Use of AI tools declaration

The author declares he has not used Artificial Intelligence (AI) tools in the creation of this article.

Conflict of interest

The author declares no conflict of interest in this paper.