Blowing-up solutions for a nonlinear time-fractional system

Abstract

A nonlinear system with different fractional derivative terms is considered. The existence of positive blowing-up solutions is proved.

1 Introduction

We are concerned with blowing-up solutions of the nonlinear fractional system:

$$\begin{aligned} \left\{ \begin{array}{ll}\displaystyle u'(t)-D^{\alpha }_{t}(u-u(0))(t)=u^{p}(t)v^{q}(t),~t>0, \\ \displaystyle v'(t)-D^{\beta }_{t}(v-v(0))(t)=u^{r}(t)v^{s}(t),~t>0, \\ \displaystyle u(0)=u_0>0,\,\,v(0)=v_0>0, \end{array} \right. \end{aligned}$$

(1.1)

for $u>0, v>0$, where $D^{\sigma }_t$ for $0<\sigma <1$ ($\sigma =\alpha ,\beta$) stands for the Riemann-Liouville fractional derivative defined for an integrable function f by $(D^{\sigma }_tf)(t)=\frac{1}{\Gamma (\sigma -1)}\frac{d}{dt}\int _{0}^{t}\frac{f(\tau )}{(t-\tau )^{\sigma }}\, d\sigma$, p, q, r, s are positive real numbers to be fixed later. There are a couple of physical motivations for considering the system (1.1). Firstly, the type of nonlinearities in the system (1.1) appears in the systems describing processes of heat diffusion and combustion in two component continua with nonlinear heat conduction and volumetric release ($u_t - a \Delta u = u^p v^q, \; v_t -b \Delta u = u^r v^s,$ the subscript t stands for the time derivative, while $\Delta$ stands for the Laplacian operator) [6]. Secondly, as suggested recently [2] one may take $\Delta D^{\alpha }_{t} (u - u(0))$ and $\Delta D^{\beta }_{t} (v - v(0))$ instead of $\Delta u$ and $\Delta v$ if the process takes place in a porous medium. To simplify the analysis one may start by replacing $\Delta (u - u(0))$ and $\Delta (v - v(0))$ by $(u - u(0))$ and $(v - v(0))$, respectively.

Before, we state our results, let us dwell a while on the existing literature. In [3], Furati and Kirane considered blowing-up solutions to the system

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle u'(t)+D^{\alpha }_t(u-u(0))(t)=v^q(t),~t>0, \\ \displaystyle v'(t)+D^{\beta }_t(v-v(0))(t)=u^r(t),~t>0, \\ \displaystyle u(0)=u_0>0,\,\,v(0)=v_0>0, \end{array} \right. \end{aligned}$$

(1.2)

for $u>0, v>0$ and $0<\alpha ,\beta <1.$ Then Kirane and Malik in [4] studied the profile of the blowing-up solutions of system (1.2). The study of the reduced system:

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle u'(t)=u^{p}(t)v^{q}(t),\quad t>0, \\ \displaystyle v'(t)=u^{r}(t)v^{s}(t),\quad t>0, \\ \displaystyle u(0)=u_0>0,\,\,v(0)=v_0>0, \end{array} \right. \end{aligned}$$

(1.3)

for $u>0, \, v>0$ is well documented in the book [6]; in fact, it admits the first integral

$$\begin{aligned} \frac{u(t)^{a_1}}{ a_1} - \frac{v(t)^{a_2}}{ a_2}= C_0 = \frac{u_0^{a_1}}{ a_1} - \frac{v_0^{a_2}}{ a_2} , \; t>0, \end{aligned}$$

(1.4)

where $a_1=r+1-p, \, a_2= q+1-s$. it can then be decoupled in

$$\begin{aligned} u'(t)= & {} u^{p}(t) \left( \frac{a_2}{a_1} u^{a_1} - a_2 C_0 \right) ^{q/ a_2},t>0, \end{aligned}$$

(1.5)

$$\begin{aligned} v'(t)= & {} v^{s}(t) \left( \frac{a_1}{a_2} v^{a_2} - a_1 C_0 \right) ^{r/ a_1},\,t>0. \end{aligned}$$

(1.6)

From here, the occurrence of finite time blow-up in each component can be derived.

If, for example, $a_1>0, a_2>0$, then u blows-up whenever $p+ a_1q/a_2 >1$; this inequality is equivalent to the condition $-rq +(p-1)(s-1)>0$, which is satisfied in this case (since $p<1+r, \, s< 1+q$). Similarly, it can be checked that if $a_1>0, \, a_2>0$, the second component v also blows-up in finite time. From the identity (1.4) it follows that the blow-up times of u(t) and v(t) are the same.

A different situation arises when $a_1a_2 < 0$, for example, if $a_1>0, \, a_2<0$. In this case $C_0 >0$, and since $s>1+q>1$, v blows-up in finite time: $v(t) \, \rightarrow \, +\infty$ as $t \, \rightarrow \, T_0^- < +\infty$. The component u(t) in this case remains bounded: $u(t) \, \rightarrow \, (a_1C_0)^{1/a_1}, \, t \, \rightarrow \, T_0^-$. In the case $a_1<0$, $a_2<0$, the constant $C_0$ can be either sign. For $C_0=0$, both components of the solutions of equations (1.5) and (1.6) lead to finite time blow-up. If $C_0<0$, then u(t) blows-up, while v(t) remains bounded; if $C_0>0$ it is the other way around.

However, such a decoupling and analysis is not directly possible for system (1.1).

2 Results

As argued in [5], (1.1) admits a local solution $(u, v) \in C^1(0,T_{max}) \times C^1(0,T_{max}).$ So,

$$\begin{aligned} D^{\alpha }_t (u-u_0)= & {} ^cD^{\alpha }_t u, \\ \displaystyle D^{\beta }_t (v-v_0)= & {} ^cD^{\beta }_t v, \end{aligned}$$

where $^{c}D^{\alpha }_{t}$ is the so-called Caputo fractional derivative defined, for a differentiable function f, by $(^{c}D^{\alpha }_{t} f)(t)=\frac{1}{\Gamma (\sigma -1)}\displaystyle \int _{0}^{t}\frac{f'(\tau )}{(t-\tau )^{\sigma }}\, d\sigma$.

For the sequel, we need the following lemma.

Lemma 2.1

If $\varphi \in C^1(\mathbb {R})$ is increasing and $u'\ge 0,$ then, $^cD^{\alpha }_t\varphi (u)\le \varphi '(u)^{c}D^{\alpha }_{t}u.$

Proof

We write

$$\begin{aligned} ^cD^{\alpha }_t\varphi (u)= & {} \frac{1}{\Gamma (1-\alpha )}\int _0^t\frac{\frac{d}{ds}\varphi (u(s))}{(t-s)^{\alpha }}ds \\ \displaystyle= & {} \frac{1}{\Gamma (1-\alpha )}\int _0^t\frac{\varphi '(u(s))u'(s)}{(t-s)^{\alpha }}ds \\ \displaystyle\le & {} \frac{\varphi '(u(t))}{\Gamma (1-\alpha )}\int _0^t\frac{u'(s)}{(t-s)^{\alpha }}ds \\ \displaystyle\le & {} \varphi '(u) ^cD^{\alpha }_t u. \end{aligned}$$

Let us recall the Mittag–Leffler function

$$\begin{aligned} e_{1-\alpha }(t)= & {} E_{\alpha ,1}(t^{\alpha })=\sum _{n=0}^{\infty }\frac{t^{\alpha \, n}}{\Gamma (\alpha \, n+1)}, t>0,\\ \displaystyle e'_{1-\alpha }(t)= & {} \sum _{n=1}^{\infty }\frac{t^{\alpha _n-1}}{\Gamma (\alpha _n)}>0,~ t>0. \end{aligned}$$

The solutions of (1.1) satisfies the system of integral equations

$$\begin{aligned} u(t)= & {} u_0+\int _0^t e_{1-\alpha }(t-\tau )u^p(\tau )v^q(\tau )d\tau , \end{aligned}$$

(2.1)

$$\begin{aligned} v(t)= & {} v_0+\int _{0}^{t} e_{1-\beta }(t-\tau )u^r(\tau )v^s(\tau )d\tau . \end{aligned}$$

(2.2)

It is clear then that $u(t)\ge u_0$, $v(t)\ge v_0$.

Now, we prove the positivity of $u'$ and $v'$. $\square$

Lemma 2.2

We have $u'>0$ and $v'>0$.

Proof

Let $\displaystyle J^{\alpha }f(t)=\frac{1}{\Gamma (\alpha )}\int _{0}^{t} (t-\tau )^{\alpha -1} f(\tau )d\tau , t>0, \alpha \in \mathbb {R}^{+}.$

As $^cD^{\alpha }f(t)= J^{1-\alpha }f'(t),$ we write the first equation of system (1.1) as:

$$\begin{aligned} u'(t)-J^{1-\alpha }u'(t)=u^p(t)v^q(t), \end{aligned}$$

or

$$\begin{aligned} (I-J^{1-\alpha })u'(t)=u^p(t)v^q(t), \end{aligned}$$

where I is the identity operator; so, formally,

$$\begin{aligned} u'(t)= & {} (I-J^{1-\alpha })^{-1}u^p(t)v^q(t) \\ \displaystyle= & {} u^pv^q+\sum ^{\infty }_{k=1}\frac{t^{(1-\alpha )k-1}}{\Gamma ((1-\alpha )k)}*u^p(t)v^q(t) \\ \displaystyle= & {} u^p(t)v^q(t)+\int _0^{\infty }e'_{1-\alpha }(t-\tau )u^p(\tau )v^q(\tau )d\tau \end{aligned}$$

where $*$ denotes convolution. We clearly have $u'>0.$ Similarly, $v'>0$ can be proved.

As a first consequence, the solution (u, v) of system (1.1) satisfies the system of inequalities

$$\begin{aligned} \left\{ \begin{array}{ll}\displaystyle u'\ge u^pv^q, \quad u(0)=u_0> 0, \\ \displaystyle v'\ge u^rv^s, \quad v(0)=v_0> 0; \end{array} \right. \end{aligned}$$

(2.3)

so it is an upper solution of the system

$$\begin{aligned} \left\{ \begin{array}{ll}\displaystyle {\tilde{u}}'={\tilde{u}}^p {\tilde{v}}^q,~u(0)=u_0> 0, \\ \displaystyle {\tilde{v}}' = {\tilde{u}}^r {\tilde{v}}^s, \,v(0)=v_0> 0. \end{array} \right. \end{aligned}$$

(2.4)

Our first result is the following blow-up result concerning solutions to system (1.1). $\square$

Theorem 2.3

We have the two results:

• If $a_1=r+1-p>0, \, a_2=q+1-s>0$ and $p + (r+1-p)q/(q+1-s)>1$, then the solution $(u, \, v)$ of system (1.1) blows-up in finite time.

• If $C_0=0$, then the solution $(u, \, v)$ of system (1.1) blows-up in finite time.

Proof

The proof is based on the results above concerning system (1.3).

Before we state our next result, let us set $\rho =r/(1-p), \; \gamma =q/(1-s), \, \rho '= r/(r-1+p) \,$ (so $\; \rho +\rho '=\rho \rho '$), $\gamma '=s/(s-1+q)\,$ (so $\gamma + \gamma '=\gamma \gamma '$). Our second result is the following theorem. $\square$

Theorem 2.4

Let (u, v) be the solution of (1.1) associated to the initial condition $(u_0, v_0)$. If $0<p<1, \, 0<s<1, \, r>1-p, \, q>1-s$, and if the condition

$$\begin{aligned} 1- \frac{1}{\rho \gamma } \le \beta + \frac{\alpha }{\gamma } \quad \text {or} \quad 1- \frac{1}{\rho \gamma } \le \alpha + \frac{\beta }{\rho }, \end{aligned}$$

(2.5)

is satisfied, then (u, v) blows-up in a finite time.

Proof

Multiplying the first equation of (1.1) by $\theta u^{\theta -1}$, we obtain

$$\begin{aligned} \theta u^{\theta -1}u^pv^q=\theta u^{\theta -1}u' - \theta u^{\theta -1} D_t^{\alpha }(u-u_0)= \theta u^{\theta -1}u' - \theta u^{\theta -1} (^cD_t^{\alpha }u). \end{aligned}$$

(2.6)

Using Lemma 2.2 with $\varphi (u)=u^{\theta }$, we have

$$\begin{aligned} ^{c}D_{t}^{\alpha }u^{\theta }\le (\theta u^{\theta -1}) (^{c}D_t^{\alpha }u), \end{aligned}$$

so,

$$\begin{aligned} -\theta u^{\theta -1}D_t^{\alpha }(u-u_0)\le - ^cD_t^{\alpha }u^{\theta }. \end{aligned}$$

Whereupon,

$$\begin{aligned} \theta u^{\theta +p-1}v^q \le (u^{\theta })'- ^{c}D_{t}^{\alpha }u^{\theta }. \end{aligned}$$

(2.7)

In a similar manner, multiplying the second equation of (1.1) by $\lambda v^{\lambda -1}$ and using Lemma 2.1, we obtain

$$\begin{aligned} \lambda u^r v^{s-1+\lambda }\le (v^{\lambda })'-^{c}D_t^{\beta }v^{\lambda }. \end{aligned}$$

(2.8)

Now, setting $\displaystyle u^{\theta }=:U, v^{\lambda }=:V$, the inequalities (2.7) and (2.8) take the form

$$\begin{aligned} \displaystyle \theta U^{\frac{\theta +p-1}{\theta }}V^{\frac{q}{\lambda }}\le & {} U'- ^{c}D_{t}^{\alpha }U, \end{aligned}$$

(2.9)

$$\begin{aligned} \displaystyle \lambda U^{\frac{r}{\theta }} V^{\frac{s+\lambda -1}{\lambda }}\le & {} V'- ^{c}D_{t}^{\beta }V. \end{aligned}$$

(2.10)

Let us choose

$$\begin{aligned} \left\{ \begin{array}{ll}\displaystyle \theta +p-1=0 \quad \Leftrightarrow \quad \theta =1-p,\\ \displaystyle \lambda +s-1=0 \quad \Leftrightarrow \quad \lambda =1-s. \end{array} \right. \end{aligned}$$

Consequently, the system of Eqs. (2.9) and (2.10) can be written

$$\begin{aligned} \theta V^{\gamma }\le & {} U'-^{c}D_{t}^{\alpha }U, \end{aligned}$$

(2.11)

$$\begin{aligned} \lambda U^{\rho }\le & {} V'-^{c}D_{t}^{\beta }V, \end{aligned}$$

(2.12)

where $q/ \lambda =\gamma , r/\theta =\rho .$

Now we multiply (2.11) by

$$\begin{aligned} \varphi (t)=\left\{ \begin{array}{ll} \displaystyle \Big (1-\frac{t}{T}\Big )^{\sigma }, &{} t\le T, \; \sigma>>1,\\ \displaystyle 0,\,\,&{} t \ge T, \end{array} \right. \end{aligned}$$

and integrate over (0, T); we obtain

$$\begin{aligned} \theta \int _0^{T} V^{\gamma }\varphi (\tau ) d\tau \le \int _0^{T} U'(\tau )\varphi (\tau )d\tau -\int _0^{T} (^{c}D_{t}^{\alpha }U(\tau ))\varphi (\tau )d\tau . \end{aligned}$$

(2.13)

First, we compute:

$$\begin{aligned} \int _0^T U'(\tau )\varphi (\tau )d\tau= & {} U\varphi \Big |_0^T-\int _0^T U(\tau )\varphi '(\tau )d\tau \\= & {} -U(0)-\int _0^T U(\tau )\varphi '(\tau )d\tau \\\le & {} -U(0)+\int _0^T U(\tau )|\varphi '(\tau )|d\tau . \end{aligned}$$

Writing

$$\begin{aligned} \int _0^T U(\tau )|\varphi '(\tau )| \, d\tau =\int _0^T U(\tau )\varphi ^{\frac{1}{\rho }}(\tau )\varphi ^{-\frac{1}{\rho }}(\tau )|\varphi '(\tau )| \, d\tau , \end{aligned}$$

and then using Holder’s inequality, we obtain the estimate

$$\begin{aligned} \int _0^T U(\tau )|\varphi '(\tau )| \, d\tau \le \left( \int _0^T U^{\rho }(\tau )\varphi (\tau ) \, d\tau \right) ^{\frac{1}{\rho }} \left( \int _0^T\varphi ^{-\frac{\rho '}{\rho }}(\tau )|\varphi '(\tau )|^{\rho '} \, d\tau \right) ^{\frac{1}{\rho '}}, \end{aligned}$$

(2.14)

where $\rho '+\rho =\rho \rho '.$

Alike, we obtain the estimate

$$\begin{aligned} \int _0^T (^cD_\tau ^{\alpha }U(\tau ))\varphi (\tau ) \, d\tau= & {} \int _0^T U(\tau ) (^cD_{\tau |T}^{\alpha }\varphi (\tau )) \, d\tau \nonumber \\\le & {} \left( \int _0^T U^{\rho }(\tau )\varphi (\tau ) \, d\tau \right) ^{\frac{1}{\rho }}\nonumber \\&\times \left( \int _0^T\varphi ^{-\frac{\rho '}{\rho }}(\tau )|^cD_{\tau |T}^{\alpha }\varphi (\tau )|^{\rho '} \, d\tau \right) ^{\frac{1}{\rho '}}. \end{aligned}$$

(2.15)

Setting

$$\begin{aligned} A(\varphi ,\rho )=\left( \int _0^T\varphi ^{-\frac{\rho '}{\rho }} (\tau )|\varphi '(\tau )|^{\rho '}d\tau \right) ^{\frac{1}{\rho '}}, \end{aligned}$$

and

$$\begin{aligned} B(\varphi ,\rho )=\left( \int _0^T\varphi ^{-\frac{\rho '}{\rho }} (\tau )|^cD_{\tau |T}^{\alpha }\varphi (\tau )|^{\rho '}d\tau \right) ^{\frac{1}{\rho '}}, \end{aligned}$$

in (2.14) and (2.15) and inserting the resulting expressions in (2.13), we get

$$\begin{aligned} \theta \int _0^T V^{\gamma }\varphi \, d\tau +U(0)\le \left( \int _0^T U^{\rho }\varphi d\tau \right) ^{\frac{1}{\rho }}A(\varphi ,\rho )+\left( \int _0^T U^{\rho }\varphi \, d\tau \right) ^{\frac{1}{\rho }}B(\varphi ,\rho ). \end{aligned}$$

Similarly, we obtain

$$\begin{aligned} \lambda \int _0^T U^{\rho }\varphi +V(0)\le \left( \int _0^T V^{\gamma }\varphi \, d\tau \right) ^{\frac{1}{\gamma }}A(\varphi ,\gamma )+\left( \int _0^T V^{\gamma }\varphi \, d\tau \right) ^{\frac{1}{\gamma }}B(\varphi ,\gamma ). \end{aligned}$$

Setting $I=\displaystyle \int _0^T U^{\rho }\varphi \, d\tau$ and $J=\displaystyle \int _0^T V^{\gamma }\varphi \, d\tau ,$ we have

$$\begin{aligned} \theta J+U(0)\le I^{\frac{1}{\rho }}\Big (A(\varphi ,\rho )+B_{\alpha }(\varphi ,\rho )\Big ), \end{aligned}$$

(2.16)

$$\begin{aligned} \lambda I+V(0)\le J^{\frac{1}{\gamma }}\Big (A(\varphi ,\gamma )+B_{\beta }(\varphi ,\gamma )\Big ). \end{aligned}$$

(2.17)

Using $I\le C J^{\frac{1}{\gamma }}\Big (A(\varphi ,\gamma )+B_{\beta }(\varphi ,\gamma )\Big )$ in (2.16), we obtain

$$\begin{aligned} J+U(0)\le C J^{\frac{1}{\rho \gamma }}\Big (A(\varphi ,\gamma )+B_{\beta }(\varphi ,\gamma )\Big )^{\frac{1}{\rho }}\Big (A(\varphi ,\rho )+B_{\alpha }(\varphi ,\rho )\Big ), \end{aligned}$$

which implies

$$\begin{aligned} J^{1-\frac{1}{\rho \gamma }}\le C \Big (A(\varphi ,\gamma )+B_{\beta }(\varphi ,\gamma )\Big )^{\frac{1}{\rho }}\Big (A(\varphi ,\rho )+B_{\alpha }(\varphi ,\rho )\Big ). \end{aligned}$$

(2.18)

Similarly, we have $J \le C I^{\frac{1}{\rho }}\Big (A(\varphi ,\rho )+B_{\alpha }(\varphi ,\rho )\Big )$, and

$$\begin{aligned} I+V(0) \le C I^{\frac{1}{\rho \gamma }}\Big (A(\varphi ,\rho )+B_{\beta }(\varphi ,\rho )\Big )^{\frac{1}{\gamma }}\Big (A(\varphi ,\gamma )+B_{\alpha }(\varphi ,\gamma )\Big ). \end{aligned}$$

Thus

$$\begin{aligned} I^{1-\frac{1}{\rho \gamma }}\le C \Big (A(\varphi ,\rho )+B_{\beta }(\varphi ,\rho )\Big )^{\frac{1}{\gamma }}\Big (A(\varphi ,\gamma )+B_{\alpha }(\varphi ,\gamma )\Big ). \end{aligned}$$

(2.19)

As $\int _0^T \varphi ^{\vartheta } |^cD^{\alpha }_{t|T}\varphi |^{\widetilde{\rho }}dt=C_{\widetilde{\rho },\alpha }T^{1-\alpha \widetilde{\rho }},$ where $\vartheta$ is any exponent, we have:

$$\begin{aligned} B(\varphi ,\rho )=C T^{ \frac{1-\alpha \rho '}{\rho '} },\quad A(\varphi ,\rho )=CT^{ \frac{1- \rho '}{\rho '} } . \end{aligned}$$

(2.20)

Inserting (2.20) into (2.19), we obtain the estimate

$$\begin{aligned} I^{ 1- \frac{1}{\rho \gamma }}\le & {} C \left( T^{ \frac{1 - \rho '}{\rho '} } +T^{ \frac{1-\alpha \rho '}{\rho '}}\right) ^{\frac{1}{\gamma }} \left( T^{ \frac{1-\gamma '}{\gamma '} }+T^{ \frac{1-\beta \gamma '}{\gamma '} } \right) \nonumber \\\le & {} C T^{\frac{1-\alpha \rho '}{\gamma \rho '}} T^{ \frac{1-\beta \gamma '}{\gamma '}},\quad for~~ T>>1 \nonumber \\\le & {} C T^{ - \left( \frac{1}{\rho \gamma } + \frac{\alpha }{\gamma } \right) +1-\beta }. \end{aligned}$$

(2.21)

Assume by contradiction that the solution is global and bounded. Then, by letting $T\rightarrow +\infty$ in (2.21), we obtain a contradiction if

$$\begin{aligned} - \left( \frac{\alpha }{\gamma } + \frac{1}{\rho \gamma } \right) +1 - \beta <0 \quad \Leftrightarrow \quad 1- \frac{1}{\rho \gamma } \le \beta + \frac{\alpha }{\gamma }. \end{aligned}$$

Alike, similar estimates along J leads to a contradiction if

$$\begin{aligned} 1- \frac{1}{\rho \gamma } \le \alpha + \frac{\beta }{\rho } . \end{aligned}$$

The case for $1- \frac{1}{\rho \gamma } = \beta + \frac{\alpha }{\gamma }$ or $1- \frac{1}{\rho \gamma } =\alpha + \frac{\beta }{\rho }$ can be treated as in [4]. $\square$

3 Estimate of the blow-up time and profile of the solution

We consider only solutions under conditions of Theorem 2.4.

Without loss of generality, we assume that $\theta>\lambda \Leftrightarrow s>p.$ Then we have:

$$\begin{aligned} U'\ge & {} \lambda V^{\sigma }, \; \sigma = \frac{q}{\lambda }>1,\\ V'\ge & {} \lambda U^{\delta }, \; \delta \frac{r}{\rho } >1; \end{aligned}$$

so $U\ge \overset{\sim }{U}$ and $V\ge \overset{\sim }{V}$ where

$$\begin{aligned} \overset{\sim }{U}'(\tau )= & {} \overset{\sim }{V}^{\sigma }(\tau ), \; \; \overset{\sim }{U_0}= U_0, \end{aligned}$$

(3.1)

$$\begin{aligned} \overset{\sim }{V}'(\tau )= & {} \overset{\sim }{U}^{\delta }(\tau ) \; \; \overset{\sim }{V_0}= V_0. \end{aligned}$$

(3.2)

Solutions of (3.1) and (3.2) are explicitly given by

$$\begin{aligned} \overset{\sim }{U}(\tau )= \overset{\sim }{C_1} ( \overset{\sim }{T_{max}} - \tau )^{ - \frac{1+\delta }{\sigma \delta -1 } } \end{aligned}$$

and

$$\begin{aligned} \overset{\sim }{V}(\tau )= \overset{\sim }{C_2}(\overset{\sim }{T_{max}} - \tau )^{ -\frac{1+\sigma }{\sigma \delta -1} }, \end{aligned}$$

where

$$\begin{aligned} \overset{\sim }{C_1}= & {} \left( \frac{(1+\sigma )(1+\delta )^{\sigma }}{ (\sigma \delta -1)^{ \delta +1}} \right) ^{\frac{1}{\sigma \delta -1}},\\ \overset{\sim }{C_2}= & {} \left( \frac{(1+\delta )(1+\sigma )^{\delta }}{(\sigma \delta -1)^{ \sigma +1} } \right) ^{ \frac{1}{\sigma \delta -1}} \end{aligned}$$

and

$$\begin{aligned} \overset{\sim }{T_{max}} = \left( \frac{U_0\,\overset{\sim }{C_2}}{V_0\,\overset{\sim }{C_2}}\right) ^{\frac{1}{\delta -\sigma }}. \end{aligned}$$

Whereupon,

$$\begin{aligned} U(t)\ge & {} C_{1}({T_{max}}-t)^{-\frac{1+\delta }{\sigma \delta -1}}\\ V(t)\ge & {} C_{2}({T_{max}}-t)^{-\frac{1+\sigma }{\sigma \delta -1}}, \end{aligned}$$

where

$$\begin{aligned} C_1=\overset{\sim }{C_1} \lambda ^{ - \frac{1+\sigma }{\sigma \delta -1 } }, \qquad C_{2}=\overset{\sim }{C_2} \lambda ^{ - \frac{1+\delta }{\sigma \delta -1 } }, \end{aligned}$$

and

$$\begin{aligned} T_{max} = \overset{\sim }{T_{max}}/\lambda . \end{aligned}$$