Blow-up for degenerate nonlinear parabolic problem

1.   Introduction

Let $\Gamma\in\left(0, \infty\right]$, $r$ be a nonnegative constant less than $1$, $a$ and $m$ be positive constants, and $p$ be a positive constant greater than $1$. We study the following degenerate nonlinear parabolic first initial-boundary value problem:

where $u_{0}\left(\xi\right)$ is a positive function in $\left(0, a\right)$ such that $u_{0}^{m+1}\left(\xi\right) \in C^{2+\alpha }\left(\bar{D}\right)$ for some $\alpha\in\left(0, 1\right)$ and $u_{0}\left(0\right) = 0 = u_{0}\left(a\right)$.

Problems (1.1)–(1.2) describe the creeping gravity flow of a power-law liquid on a rigid horizontal surface. The solution $u$ is the thickness of the current and $r$ represents the Cartesian symmetry, see ^[5]. It also explains the radial spreading of an axisymmetric current with $\xi$ and $u^{m+1}/\left(m+1\right)$ corresponding respectively to the radial coordinate and the integral of velocity profile of the current, see ^[7]. If $u$ represents the temperature, then it can be interpreted as a nonlinear heat conduction problem with $u^{m}$ being the thermal diffusivity, see [12,pp. 73-74]. When $m = 0$ and $r = 0.5$, it exemplifies heat transfer into one face of a flat cylinder with a small ratio of depth to diameter, see ^[2,15]. Problems (1.1)–(1.2) can illustrate population dynamics when $r = 0$, see ^[6]. (1.1) is a degenerate equation because the thermal diffusivity $u^{m}\rightarrow0$ when $\xi\rightarrow0$ or $\xi\rightarrow a$.

Let $\xi = ax$, $\tau = a^{2}\left(m+1\right) t$, $\Gamma = a^{2}\left(m+1\right) T$, $D = \left(0, 1\right)$, $\Omega = D\times\left(0, T\right)$, $\bar{D} = \left[0, 1\right]$, $\bar{\Omega} = \bar{D}\times\left[0, T\right)$, $\partial D = \left\{ 0, 1\right\}$, and $\partial \Omega = \left(\bar{D}\times\left\{ 0\right\} \right) \cup\left(\partial D\times\left(0, T\right) \right)$. Then, the problems (1.1)–(1.2) are transformed into the degenerate nonlinear parabolic problem below,

When $r = 0$ and $u_{0}\left(x\right) \geq0$ on $\bar{D}$, the multi-dimensional version of the problems (1.3)–(1.4) have been studied by ^{[4,8,11,13,14]}. Let $\mu_{1}$ be the first eigenvalue of the following Sturm-Liouville problem,

When $p = m+1$, Sacks ^[13] proved that if $a^{2}\left(m+1\right) > \mu_{1}$, the solution blows up in a finite time. If $a^{2}\left(m+1\right) \leq\mu_{1}$ (that is, the domain size is sufficiently small), the problems (1.3)–(1.4) have a global solution (also see ^[14]). In the case of $p > m+1$, the solution may or may not exist for all time which depends on the initial condition $u_{0}$, see ^[8,11,13]. Galaktionov ^[4] proved that the problems (1.3)–(1.4) have a global solution if $p < m+1$.

This paper is organized as follows. In section 2, we prove the existence and uniqueness of the classical solution of the problems (1.1)–(1.2). In section 3, we show that $u$ blows up in a finite time when $p\geq m+1$. Then, we prove that there is a global solution when $a$ is sufficiently small. Different from ^[13], our method does not require additional conditions on $p$ and $m$. In section 4, we prove that $u_{t}$ blows up in a finite time when $u$ is unbounded.

2.   Existence and uniqueness of the solution

We assume that the initial data $u_{0}\left(x\right)$ satisfies the condition below,

We note that $u_{0} = \left[Kx\sin\left(\pi\left(1-x\right) ^{2}/2\right) \right] ^{1/\left(m+1\right) }$, where $K$ is a positive constant, satisfies (2.1) and $u_{0}\left(x\right) = 0$ on $\partial D$. Let $v = u^{m+1}$, the problems (1.3)–(1.4) become

where $v_{0}\left(x\right) = u_{0}^{m+1}\left(x\right)$. To prove the existence of a solution, Chan and Chan ^[1] consider the following nonlinear parabolic problem with $\varepsilon$ being a small positive number less than $1$,

They prove that $v_{\varepsilon}\in C\left(\bar{\Omega}\right) \cap C^{2+\alpha, 1+\alpha/2}\left(D\times\left[0, T\right) \right)$, and the sequence of solutions: $\left\{ v_{\varepsilon}\right\}$ converges to $v\in C\left(\bar{\Omega}\right) \cap C^{2+\alpha, 1+\alpha/2}\left(D\times\left[0, T\right) \right)$ when $\varepsilon\rightarrow0$. They also show that $v > 0$ in $D\times\left[0, T\right)$ and $v\left(x, t\right) \geq v_{0}\left(x\right)$ on $\bar{D}\times\left[0, T\right)$. Using these results, they prove that the problems (1.3)–(1.4) have a solution $u\in C\left(\bar{\Omega}\right) \cap C^{2+\alpha, 1+\alpha/2}\left(D\times\left[0, T\right) \right)$, $u > 0$ in $D\times\left[0, T\right)$, and $u\left(x, t\right) \geq u_{0}\left(x\right)$ on $\bar{D} \times\left[0, T\right)$. By (2.1), they show that $u_{t}\geq0$ and $v_{t}\geq0$ in $D\times\left[0, T\right)$. Further, they prove that $u$ is unbounded in $D\times\left(0, T\right)$ if $T < \infty$. For ease of reference, let us state their Theorem 2.8 below.

Theorem 2.1. Problems (1.3)–(1.4) have a solution $u\in C\left(\bar{\Omega}\right) \cap C^{2+\alpha, 1+\alpha/2}\left(D\times\left[0, T\right) \right)$. If $T < \infty$, then $u$ is unbounded in $D\times\left(0, T\right)$.

Let $Lv = v^{-m/\left(m+1\right) }v_{t}/\left(m+1\right) -v_{xx}-rv_{x}/x$ and $\beta\left(x, t\right)$ be a bounded function on $\bar{\Omega}$. Here is a comparison theorem.

Lemma 2.2. Suppose that $y$ and $s\in C\left(\bar{\Omega}\right) \cap C^{2, 1}\left(\Omega\right)$, and

Then, $y\geq s$ on $\bar{\Omega}$.

Proof. If not, let us assume that $s > y$ somewhere, say, $\left(\bar{x}, \bar{t}\right)$ $\in\Omega$. By the continuity of $s$ and $y$ over $\bar{\Omega}$, there exists an interval $\left(a_{1}, a_{2}\right) \subset D$ such that $\bar{x}\in\left(a_{1}, a_{2}\right)$, $s\left(a_{1}, \bar {t}\right) -y\left(a_{1}, \bar{t}\right) = 0$, $s\left(a_{2}, \bar {t}\right) -y\left(a_{2}, \bar{t}\right) = 0$, $s\left(x, \bar{t}\right) > y\left(x, \bar{t}\right)$ for $x\in\left(a_{1}, a_{2}\right)$, and $s\leq y$ in $\left[a_{1}, a_{2}\right] \times\left[0, \bar{t}\right)$. Then,

Let $\tilde{\phi}\left(x\right)$ and $\tilde{\lambda}$ be the first eigenfunction and eigenvalue of the following Sturm-Liouville problem,

By Theorem 3.1.2 of Pao [9,p. 97], $\tilde{\phi}\left(x\right)$ exists and $\tilde{\lambda} > 0$. Further, $\tilde{\phi}\left(x\right) > 0$ in $\left(a_{1}, a_{2}\right)$. Let $\gamma$ be a positive real number to be determined. By the above equation, we have

Using integration by parts, $\tilde{\phi}^{\prime}\left(a_{1}\right) \geq 0$, and $\tilde{\phi}^{\prime}\left(a_{2}\right) \leq0$, we have

By (2.4), we get

From this, we have

By (2.6), we obtain

The above expression is equivalent to

By the mean value theorem, there exists an $\zeta$ between $s^{1/\left(m+1\right) }$ and $y^{1/\left(m+1\right) }$ such that

By the Gronwall inequality (cf. Walter [16,pp. 14-15]),

As $\beta$ is bounded, we choose $\gamma$ such that $\gamma\geq\left(m+1\right) \left(\tilde{\lambda}-\beta\right) \zeta^{m}$. By $y\geq s\ $in $\left[a_{1}, a_{2}\right] \times\left[0, \tilde{t}\right)$, we have

Since $x^{r}\tilde{\phi}e^{\gamma\bar{t}} > 0$ in $\left(a_{1}, a_{2}\right)$, the above inequality contradicts (2.5). Therefore, $y\geq s$ in$\Omega$. As $y\geq s$ on $\partial\Omega$, $y\geq s$ on $\bar{\Omega}$. The proof is complete.

Let $\mathcal{L}y = v^{-m/\left(m+1\right) }y_{t}/\left(m+1\right) -y_{xx}-ry_{x}/x$. Based on a similar computation of Lemma 2.2, we have the following result.

Lemma 2.3. Suppose that $y$ and $s\in C\left(\bar{\Omega}\right) \cap C^{2, 1}\left(\Omega\right)$, and

Then, $y\geq s$ on $\bar{\Omega}$.

By Theorem 2.1 and Lemma 2.2, we obtain the result of the existence and uniqueness of solution.

Theorem 2.4. Problems (1.3)–(1.4) and (2.2)–(2.3) have the unique classical solution.

3.   Blow-up of the solution and global existence

Instead of using condition (2.1), let us assume that $u_{0}$ satisfies the inequality below in the following two sections:

Then, by (1.3) and $u\in C\left(\bar{\Omega}\right) \cap C^{2+\alpha, 1+\alpha/2}\left(D\times\left[0, T\right) \right)$, we have $u_{t}\left(x, 0\right) > 0$ ($v_{t}\left(x, 0\right) > 0$) in $D$. We want to prove that $v_{t}\left(x, t\right) > 0$ in $D$ for $t > 0$. To achieve it, we have the following two results.

Lemma 3.1. $v\left(x, t\right) > v_{0}\left(x\right)$ in $\Omega$.

Proof. From (3.1), we obtain

As stated in section 2, we have $v\left(x, t\right) \geq v_{0}\left(x\right)$ on $\bar{D}\times\left[0, T\right)$. Subtract the above inequality from (2.2), it gives

Further, we know that $v\left(x, t\right) = v_{0}\left(x\right) = 0$ on $\partial D\times\left(0, T\right)$ and $v\left(x, 0\right) = v_{0}\left(x\right)$ on $\bar{D}$. Suppose that $v\left(\tilde{x}, t\right) = v_{0}\left(\tilde{x}\right)$ for some $\tilde{x}\in D$ and $t > 0$. Then, the set

is non-empty. Let $\tilde{t}$ denote its infimum. Suppose that $\tilde{t} > 0$. Then, $v\left(\tilde{x}, \tilde{t}\right) = v_{0}\left(\tilde{x}\right)$ and $v\left(x, t\right) > v_{0}$ in $D\times\left(0, \tilde{t}\right)$. Therefore, $\left(v\left(\tilde{x}, \tilde{t}\right) -v_{0}\left(\tilde{x}\right) \right) _{t}\leq0$. From section 2, we have $v_{t}\left(\tilde{x}, \tilde{t}\right) \geq0$. Thus, $\left(v\left(\tilde{x}, \tilde{t}\right) -v_{0}\left(\tilde{x}\right) \right) _{t} = 0$. Further, $v\left(x, t\right) -v_{0}\left(x\right)$ attains its local minimum at $\left(\tilde{x}, \tilde{t}\right)$. This implies that $\left(v\left(\tilde{x}, \tilde{t}\right) -v_{0}\left(\tilde{x}\right) \right) _{x} = 0$ and $\left(v\left(\tilde{x}, \tilde{t}\right) -v_{0}\left(\tilde {x}\right) \right) _{xx} > 0$. Since $v\left(\tilde{x}, \tilde{t}\right) > 0$, we have

It leads to a contradiction. If $\tilde{t} = 0$, we have $v\left(x, 0\right) = v_{0}\left(x\right)$ on $\bar{D}$ and $v\left(x, t\right) > v_{0}\left(x\right)$ for $t > 0$ in $D$. Hence, $v\left(x, t\right) > v_{0}\left(x\right)$ in $\Omega$.

Let $h$ be a small positive real number and $q\left(x, t\right) = v\left(x, t+h\right)$. Further, $q$ is the solution of the following problem:

We follow a similar calculation of Lemma 3.1 to obtain the corollary below.

Corollary 3.2. $q\left(x, t\right) > v\left(x, t\right)$ in $\Omega$.

Having these two results, we prove $v_{t}$ being positive in the domain.

Lemma 3.3. $v_{t} > 0$ in $\Omega$.

Proof. From the result of section 2, $v_{t}\geq0$ in $D\times\left[0, T\right)$. Let us assume that $v_{t}\left(\rho, \omega\right) = 0$ for some $\left(\rho, \omega\right) \in\Omega$. Then, there exists a neighborhood $\left(a_{3}, a_{4}\right) \times\left(t_{1}, t_{2}\right) \subset\Omega$ such that $\left(\rho, \omega\right) \in\left(a_{3}, a_{4}\right) \times\left(t_{1}, t_{2}\right)$. We differentiate (2.2) with respect to $t$ to obtain

Since $v > 0$ in $\left(a_{3}, a_{4}\right) \times\left(t_{1}, t_{2}\right)$, it gives

By the strong maximum principle (cf. Protter and Weinberger [10,pp. 168-169]), $v_{t}\equiv0$ in $\left(a_{3}, a_{4}\right) \times\left(t_{1}, t_{2}\right)$. This contradicts Corollary 3.2 that $v$ is strictly increasing in $t$ in $\Omega$. Therefore, $v_{t} > 0$ in $\left(a_{3}, a_{4}\right) \times\left(t_{1}, t_{2}\right)$. Since $\left(\rho, \omega\right)$ is arbitrary in $\Omega$, $v_{t} > 0$ in $\Omega$.

To study the blow-up of the solution $u$, we let $z^{1/\left(1-r\right) } = x$. By a direct computation,

Then, the problems (2.2)–(2.3) are transformed into

Let

Since $v > 0$ in $D\times\left[0, T\right)$, $F\left(t\right) > 0$ over $\left[0, T\right)$. We modify Lemma 4.3 of Deng, Duan and Xie ^[3] to obtain the result below.

Lemma 3.4. If $p\geq m+1$, then

Proof. By a direct computation, the derivative of $F\left(t\right)$ is given by

By $v_{t}\left(x, 0\right) > 0$ in $D$ and Lemma 3.3 $v_{t} > 0$ in $\Omega$, we have $F^{\prime}\left(t\right) > 0$ over $\left[0, T\right)$. By (3.5), (3.8) is rewritten as

Differentiating $F^{\prime}\left(t\right)$ with respect to $t$ and by (3.5), we have

Using integration by parts and $p\geq m+1$, we obtain

By (3.5), the above expression becomes

By assumption $p\geq m+1$, it yields

By (3.8) and the Cauchy-Schwartz inequality, we obtain

Then, by (3.7) and (3.9), we have

This completes the proof.

Lemma 3.5. If $p\geq m+1$, then the solution $u$ blows up somewhere on $\bar{D}$ in a finite time $T$.

Proof. By a direct computation,

By (3.10), $p > 1$, and $F > 0$ over $\left[0, T\right)$, we have

We integrate the above inequality over $\left(0, t\right)$ to get

Equivalently,

Then, we integrate this inequality over $\left(0, t\right)$ to obtain

Since $F\left(0\right) > 0$, $F^{\prime}\left(0\right) > 0$, and $p > 1$, the right side of the above inequality is a decreasing function in $t$ and is equal to zero in a finite time. Therefore, there exists some finite $T$ such that $F^{-\left(p-1\right) /\left(p+1\right) }\left(T\right) = 0$. Hence, $F\left(T\right) = \infty$. It implies that $v\left(z, t\right) \rightarrow\infty$ when $t\rightarrow T$ for some $z\in\bar{D}$. Thus, $u\left(x, t\right)$ blows up somewhere on $\bar{D}$ in a finite time $T$.

Now, we prove that $u$ exists globally if $a$ is sufficiently small. This can be achieved through constructing a global-exist upper solution of the problems (2.2)–(2.3). In this proof, we do not have additional conditions on $p$ and $m$.

Theorem 3.6. If $a$ is small enough, then $u$ exists globally.

Proof. It suffices to prove that $v(x, t)$ exists globally. Let $V\left(x\right) = kx^{1-r}\left(1-x\right)$ where $k$ is a positive constant. Then, $V\left(x\right) \in C\left(\bar{D}\right) \cap C^{2}\left(D\right)$. We choose $k$ such that $V\left(x\right) \geq v_{0}\left(x\right)$. Clearly, $V\left(x\right) = 0$ at $x = 0$ and $x = 1$. The expression of $V_{x}$ and $V_{xx}$ is below

By a direct computation,

If $a$ is sufficiently small, then $V_{xx}+rV_{x}/x+a^{2}\left(m+1\right) V^{p/\left(m+1\right) }\leq0\left( = V_{t}\right)$. By Lemma 2.2, $V\left(x\right) \geq v\left(x, t\right)$ on $\bar{D}\times\left[0, \infty\right)$. Therefore, $v$ exists globally which implies $u$ exists globally.

4.   Blow-up of $u_{t}$

In this section, we want to prove that $u_{t}$ tends to infinity if $u$ blows up. From Lemma 3.3, $v_{t} > 0$ in $\Omega$. Let $J\left(x, t\right) = v_{t}\left(x, t\right) -\varepsilon v\left(x, t\right)$ where $\varepsilon$ is a small positive number. Then, $J = 0$ on $\partial D\times\left[0, T\right)$. Let $t_{3}\in\left(0, T\right)$. We choose $\varepsilon$ such that $J\left(x, t_{3}\right) \geq0$ on $\bar{D}$.

Lemma 4.1. If $p\geq m+1$, then $J\geq0$ on $\bar{D}\times\left[t_{3}, T\right)$.

Proof. By a direct computation, $J_{t} = v_{tt}-\varepsilon v_{t}$, $J_{x} = v_{tx}-\varepsilon v_{x}$, and $J_{xx} = v_{txx}-\varepsilon v_{xx}$. From (3.4), we have

By Lemma 3.3, $J_{t}+\varepsilon v_{t} = v_{tt}$, and (2.2), we have

Simplifying the above inequality and by $p\geq m+1$, it gives

By Lemma 2.3, we have $J\geq0$ on $\bar{D}\times\left[t_{3}, T\right)$.

Our main result below is immediately followed by Lemma 3.5 and Lemma 4.1.

Theorem 4.2. If $p\geq m+1$ and $u$ is unbounded somewhere on $\bar{D}$ in a finite time $T$, then $u_{t}$ blows up at $T$.

5.   Conclusion

In this paper, we prove the existence and uniqueness of the solution of a degenerate nonlinear parabolic problem. This solution blows up in a finite time if $p\geq m+1$. Then, we show that $u_{t}$ blows up somewhere in the domain in a finite time.

Acknowledgments

The author thanks the anonymous referee for careful reading. This research did not receive any specific grant funding agencies in the public, commercial, or not-for-profit sectors.

Conflict of interest

The author declares that there are no conflicts of interest in this paper.