Blow-Up of Smooth Solution to the Compressible Navier–Stokes–Poisson Equations

Abstract

In this paper, we show the blow-up phenomenon of smooth solutions to the compressible Navier–Stokes–Poisson (N-S-P) equations in $\mathbb {R}^{2}$, under the assumption that the initial density has compact support. The proof is based on some useful physical quantities. In particular, our result is valid for both isentropic and isothermal cases.

1 Introduction

This paper is concerned with the blow-up phenomena of smooth solutions to the Cauchy problem for the following Navier–Stokes–Poisson (N-S-P) system in two-dimensional case:

$$\begin{aligned} \left\{ \begin{array}{lllll} \partial _{t}\rho +\text {div}(\rho u)=0, \\ \partial _{t}(\rho u)+\text {div}(\rho u\otimes u)+\nabla P(\rho )=\mu \Delta u+(\mu +\lambda )\nabla \text {div}u\\ \quad +\kappa \rho \nabla \Phi ,(x,t)\in \mathbb {R}^{2}\times \mathbb {R}_{+} \\ -\Delta \Phi =\rho , \end{array} \right. \end{aligned}$$

(1.1)

with initial data

$$\begin{aligned} \rho |_{t=0}=\rho _{0}(x),u|_{t=0}=u_{0}(x). \end{aligned}$$

(1.2)

The unknown functions $\rho (x,t)$, u(x, t), P, and $\Phi$ denote the density, velocity field, pressure, and potential of the underlying force, respectively. Generally speaking, the pressure P depends on the density and temperature of fluid. However, there are physically relevant situations that we assume the fluid flow is barotropic, i.e., the pressure depends only on the density. This is the case when either the temperature or the entropy is supposed to be constant. The typical expression is

$$\begin{aligned} P(\rho )=\rho ^{\gamma } \ \ \ (\gamma \ge 1), \end{aligned}$$

(1.3)

where $\gamma =1$ stands for the isothermal case and $\gamma >1$ represents the adiabatic constant in the isentropic regime. We introduce the viscous stress tensor T as

$$\begin{aligned} \text {div}T=\mu \Delta u+(\mu +\lambda )\nabla \text {div}u. \end{aligned}$$

The coefficient $\kappa$, which signifies the property of the forcing, is repulsive if $\kappa >0$ and attractive if $\kappa <0$. The coefficients $\mu$ and $\lambda$ represent shear coefficient viscosity of the fluid and the second viscosity coefficient, respectively. As the fluid is assumed to be Newtonian, the two Lamé viscosity coefficients satisfy

$$\begin{aligned} \mu >0,\mu +\lambda \ge 0. \end{aligned}$$

(1.4)

The compressible N-S-P system can be used to describe many models if we consider different potential forces. For example, (1.1) is the self-gravitation model if $\Phi$ is the gravitational potential force, and the semiconductor model if $\Phi$ is the electrostatic potential force. In the literatures, there have been a lot of studies on the N-S-P by physicists and mathematicians because of its physical importance and mathematical challenges. For these results, please refer to [4, 10, 11] and references therein.

The blow-up of smooth solutions to the evolutionary equations arising in the mathematical fluid mechanics has been the subject of many theoretical studies. More precisely, Sideris [6] showed that the lifespan of $C^{1}$ solution to the compressible Euler equations was finite when the initial data are constant outside a bounded set and the initial flow velocity has compact supports. In 1998, Xin [8] used a different method to prove the blow-up result for the compressible Navier–Stokes equations, under two basic hypotheses: the support of the density grows sublinearly in time and the entropy is bounded below. In 2004, Cho [6] extended Xin’s result to the case of fluids with positive heat conduction. Recently, Xin and Yan [9] introduced the concept of isolated mass group to remove the condition that the initial density has compact support and that the smooth solution has finite total energy and obtained a new blow-up result. Inspired by the above pioneering work, many authors study the blow-up phenomena for the N-S-P equations. Jiang and Tan [5] obtained blow-up result of the compressible reactive self-gravitating gas with chemical kinetics equations in $\mathbb {R}^{3}$. Xie [7] showed the blow-up result of smooth solutions to the full compressible N-S-P in $\mathbb {R}^{3}$.

However, all the previous works are concerned with the non-isentropic case, in which the energy equation plays an important role. By virtue of the energy equation and the adiabatic exponent $\gamma >1$, we can easily obtain that the density $\rho$ is compactly supported all the time; while the methods used above cannot be applied to our N-S-P model straightforwardly. On one hand, it is difficult to obtain the fact that compact support of the initial data will not change in time. On the other hand, it is hard to construct the function as the total energy has a negative part when $\kappa >0$. To overcome the above-mentioned difficulties, we use some useful physical quantities in the radially symmetric cases, proving the blow-up for both isothermal (i.e., $\gamma =1$) and isentropic (i.e., $\gamma >1$) case whether $\kappa >0$ or $\kappa <0$. In some senses, we improve the corresponding previous result [2], which proves the isothermal compressible Navier-Stokes equations for two-dimensional case. For simplicity of presentation, we introduce the following physical quantities:

$$\begin{aligned} m(t)= & {} \int _{\mathbb {R}^{2}}\rho (x,t)\mathrm{d}x, \end{aligned}$$

(1.5)

$$\begin{aligned} M(t)= & {} \int _{\mathbb {R}^{2}}\rho (x,t)|x|^{2}\mathrm{d}x, \end{aligned}$$

(1.6)

$$\begin{aligned} F(t)= & {} \int _{\mathbb {R}^{2}}\rho (x,t)u(x,t)\cdot x\mathrm{d}x, \end{aligned}$$

(1.7)

which represent the total mass, second moment, and radial component of momentum, respectively.

Throughout this paper, we always assume $m(0)>0$ and $F(0)>0$. Moreover, we assume that the initial density $\rho _{0}$ has compact support, i.e., there exists a positive constant R such that

$$\begin{aligned} \text {supp}\rho _{0}\subset B_{R}, \end{aligned}$$

(1.8)

where $B_{R}$ denotes the ball in $\mathbb {R}^{2}$ centered at origin with radius R.

Our main results can be summarized as follows.

Theorem 1.1

Let $(\mu ,\lambda )$ satisfy (1.4), $\kappa \ne 0$, $\gamma \ge 1$ and $(\rho ,u)\in C^{1}\big ([0,T],H^{m}(\mathbb {R}^{2})\big )(m>2)$ is a spherically symmetric solution to the compressible N-S-P system (1.1) with initial data $\rho _{0}(x)$ having compact support. Assume that the initial data are spherically symmetric, i.e.,

$$\begin{aligned} \rho _{0}(x)=\rho _{0}\big (|x|\big ), u_{0}(x)=\frac{x}{|x|}u_{0}\big (|x|\big ). \end{aligned}$$

Then, the lifespan of the solution $(\rho ,u)$ is finite.

In the following theorem, we point out that the function constructed by Xin in [8] can be applied to the N-S-P model under the special coefficient $\kappa$.

Theorem 1.2

Let $(\mu ,\lambda )$ satisfy (1.4), $\kappa \!=\!-1$, $\gamma \!>\!2$ and $(\rho ,u)\!\in \! C^{1}\big ([0,T],H^{m}(\mathbb {R}^{2})\big )$ $(m>2)$ is a spherically symmetric solution to the compressible N-S-P system (1.1) with initial data $\rho _{0}(x)$ having compact support. The initial data are spherically symmetric as Theorem 1.1. Then the lifespan of the solution $(\rho ,u)$ is finite.

Remark 1.1

By changing the condition on u, such as $u\in C^{1}\big ([0,T],W^{m,1}(\mathbb {R}^{N})\big )(m>2)$, we can extend our result to high-dimensional case.

2 Proof of Theorem 1.1

Before the proof of Theorem 1.1, we give the following key lemma which plays an important role in the proof.

Lemma 2.1

Assume that $(\mu ,\lambda )$ satisfies (1.4), and $(\rho ,u)\in C^{1}\big ([0,T],H^{m}(\mathbb {R}^{2})\big )(m>2)$ is a spherically symmetric solution to the compressible N-S-P system. Then

$$\begin{aligned} u(x,t)\equiv 0, \quad x\in B^{c}_{R}. \end{aligned}$$

(2.1)

Moreover, the support of solution will not change in time.

Proof

First, let $X(t,\alpha )$ denote the particle path starting at $\alpha$ when $t=0$, i.e.,

$$\begin{aligned} \frac{d}{{d}t}X(t,\alpha )=u(X(t,\alpha ),t), \quad X(0,\alpha )=\alpha . \end{aligned}$$

(2.2)

And we denote by $\Omega (t)$ the closed region that is the image of $B_{R}$ under the flow map (2.2):

$$\begin{aligned} \Omega (t)=\big \{x=X(t,\alpha ), \alpha \in B_{R}\big \}. \end{aligned}$$

(2.3)

In the sequel, we shall prove that $\Omega (t)=\Omega (0)$.

In fact, it follows from the continuity Eq. (1.1)$_1$ that the density is simply transported along particle paths, so that

$$\begin{aligned} \text {supp}_{x}\rho (x,t)\subset \Omega (t). \end{aligned}$$

Consequently, from momentum equation, one has

$$\begin{aligned} \mu \Delta u+(\mu +\lambda )\nabla \text {div}u=0, \quad \ \text {on}\{t\}\times \mathbb {R}^{2}\backslash \Omega (t). \end{aligned}$$

(2.4)

Since $u(x,t)=\frac{x}{|x|}\overline{u}\big (|x|,t\big )$ for some radially symmetric function $\overline{u}$, we obtain from (2.4) that

$$\begin{aligned} \overline{u}_{rr}+\bigg (\frac{\overline{u}}{r}\bigg )_{r}=0 \quad \text {on}\{t\}\times \mathbb {R}^{2}\backslash \Omega (t). \end{aligned}$$

Using the condition $u\in H^{m}(\mathbb {R}^{2})$, we have

$$\begin{aligned} \overline{u}_{r}+\frac{\overline{u}}{r}=0 \quad \text {on} \{t\}\times \mathbb {R}^{2}\backslash \Omega (t). \end{aligned}$$

Solving the ODE gives

$$\begin{aligned} \overline{u}=\frac{C(t)}{r}, \end{aligned}$$

where C(t) is a constant dependent on t. Then using $u\in C\big ([0,T],L^{2}(\mathbb {R}^{2})\big )$ and adopting the method in [2], we get

$$\begin{aligned} \overline{u}\equiv 0 \quad \text {on}\{t\}\times \mathbb {R}^{2}\backslash \Omega (t). \end{aligned}$$

It follows from the definition of $\Omega (t)$ that if $\alpha \in \Omega ^{c}(0)$, then

$$\begin{aligned} u\big (X(t,\alpha ),t\big )=0. \end{aligned}$$

Thus

$$\begin{aligned} X(t,\alpha )=\alpha +\int _0^tu\big (X(s,\alpha ),s\big )\mathrm{d}s=\alpha , \end{aligned}$$

and

$$\begin{aligned} \Omega (0)=\Omega (t) \quad \forall 0\le t\le T. \end{aligned}$$

This completes the proof of Lemma 2.1. $\square$

By virtue of Lemma 2.1, we are in a position to prove Theorem 1.1 in the following.

Proof of Theorem 1.1

Firstly multiplying the momentum equation (1.1)$_2$ by x and integrating over $\mathbb {R}^{2}$, we have

$$\begin{aligned} \frac{d}{dt}{F(t)}&=\frac{d}{{d}t}\int _{\mathbb {R}^{2}}\rho u\cdot x\mathrm{d}x =\int _{\mathbb {R}^{2}}(\rho u)_{t}\cdot x\mathrm{d}x\nonumber \\&=-\int _{\mathbb {R}^{2}}\text {div}(\rho u\otimes u)\cdot x\mathrm{d}x -\int _{\mathbb {R}^{2}}\nabla \rho ^{\gamma }\cdot x\mathrm{d}x\nonumber \\&\quad +\int _{\mathbb {R}^{2}}\text {div}T\cdot x\mathrm{d}x +\kappa \int _{\mathbb {R}^{2}}\rho \nabla \Phi \cdot x\mathrm{d}x\nonumber \\&=\sum ^{4}_{k=1}I_{k}. \end{aligned}$$

(2.5)

We calculate the integrals $I_{k}$ one by one. Starting with $I_{1}$ and utilizing mass equation, we get

$$\begin{aligned} I_{1}=\int _{\mathbb {R}^{2}}\rho |u|^{2}\mathrm{d}x. \end{aligned}$$

(2.6)

Similarly, we have

$$\begin{aligned} I_{2}=2\int _{\mathbb {R}^{2}}\rho ^{\gamma }\mathrm{d}x. \end{aligned}$$

(2.7)

Using Lemma 2.1 and the definition of viscous stress tensor T, we can easily obtain

$$\begin{aligned} I_{3}=0. \end{aligned}$$

(2.8)

Next, we will calculate $I_{4}(t)$. By the Poisson equation (1.1)$_{3}$, we get

$$\begin{aligned} I_{4}&=\int _{\mathbb {R}^{2}}\rho \nabla \Phi \cdot x\mathrm{d}x =\int _{\mathbb {R}^{2}}(-\Delta \Phi )\nabla \Phi \cdot x\mathrm{d}x\nonumber \\&=-\sum ^{2}_{i,j=1}\int _{\mathbb {R}^{2}}\partial _{ii}\Phi (\partial _{j}\Phi \cdot x_{j})\mathrm{d}x\nonumber \\&=\sum ^{2}_{i,j=1}\int _{\mathbb {R}^{2}}\partial _{i}\Phi \partial _{ji}\Phi \cdot x_{j}\mathrm{d}x +\sum ^{2}_{i,j=1}\int _{\mathbb {R}^{2}}\partial _{i}\Phi \partial _{j}\Phi \cdot \partial _{i}x_{j}\mathrm{d}x\nonumber \\&=\sum ^{2}_{i,j=1}\frac{1}{2}\int _{\mathbb {R}^{2}}\partial _{j}|\partial _{i}\Phi |^2\cdot x_{j}\mathrm{d}x +\sum ^{2}_{i,j=1}\int _{\mathbb {R}^{2}}\partial _{i}\Phi \partial _{j}\Phi \delta _{ij}\mathrm{d}x\nonumber \\&=-\int _{\mathbb {R}^{2}}|\nabla \Phi |^{2}\mathrm{d}x+\int _{\mathbb {R}^{2}}|\nabla \Phi |^{2}\mathrm{d}x=0. \end{aligned}$$

(2.9)

Gathering the identities (2.5)–(2.9), we obtain

$$\begin{aligned} \frac{d}{{d}t}{F(t)}=\int _{\mathbb {R}^{2}}\rho |u|^{2}\mathrm{d}x +2\int _{\mathbb {R}^{2}}\rho ^{\gamma }\mathrm{d}x. \end{aligned}$$

Integrating the above identity with respect to t, we get

$$\begin{aligned} F(t)=F(0)+\int ^t_0\int _{\mathbb {R}^{2}}\rho |u|^{2}\mathrm{d}x\mathrm{d}t +2\int ^t_0\int _{\mathbb {R}^{2}}\rho ^{\gamma }\mathrm{d}x\mathrm{d}t, \end{aligned}$$

which implies

$$\begin{aligned} F(t)\ge F(0). \end{aligned}$$

(2.10)

By virtue of the continuity equation and integrating by parts formula, we obtain

$$\begin{aligned} \frac{d}{{d}t}{M(t)}=\frac{d}{{d}t}\int _{\mathbb {R}^{2}}\rho |x|^{2}\mathrm{d}x=2F(t). \end{aligned}$$

(2.11)

Integrating (2.11), we get

$$\begin{aligned} M(t)=M(0)+2\int ^t_0F(s)\mathrm{d}s. \end{aligned}$$

(2.12)

From (2.10) and (2.12), we deduce

$$\begin{aligned} M(t)\ge M(0)+2F(0)t. \end{aligned}$$

(2.13)

Obviously, the right-hand side of (2.9) grows linearly in t. We shall show that the left-hand side is bounded. On the other hand, from the continuity equation, we obtain

$$\begin{aligned} \frac{d}{{d}t}{m(t)}=\frac{d}{{d}t}\int _{\mathbb {R}^{2}}\rho \mathrm{d}x=0, \end{aligned}$$

which implies

$$\begin{aligned} \int _{\mathbb {R}^{2}}\rho \mathrm{d}x=\int _{\mathbb {R}^{2}}\rho _{0}\mathrm{d}x=m(0). \end{aligned}$$

Thus, from (1.8) and mass conservation, we have

$$\begin{aligned} M(t)=\int _{\mathbb {R}^{2}}\rho |x|^{2}\mathrm{d}x =\int _{B_{R}}\rho |x|^{2}\mathrm{d}x\le R^{2}\int _{B_{R}}\rho \mathrm{d}x=R^{2}m(0). \end{aligned}$$

(2.14)

Putting (2.13) and (2.14) together, we conclude that

$$\begin{aligned} R^{2}m(0)\ge M(0)+2F(0)t. \end{aligned}$$

Hence, the lifespan of the classical solution of N-S-P is finite and we finish the proof of Theorem 1.1. $\square$

3 Proof of Theorem 1.2

In this section, we will give the proof of Theorem 1.2 and show that the blow-up result can be obtained by the nonlinear function introduced in [8], only when $\kappa =-1$ and $\gamma >2$.

Proof of Theorem 1.2

First, we introduce the function

$$\begin{aligned} I(t)&=\int _{\mathbb {R}^{2}}x^{2}\rho \mathrm{d}x-2(1+t)\int _{\mathbb {R}^{2}}xu\rho \mathrm{d}x+(1+t)^{2}\nonumber \\&\quad \int _{\mathbb {R}^{2}}\left( \rho u^{2}+\frac{2}{\gamma -1}\rho ^{\gamma }+|\nabla \Phi |^{2}\right) \mathrm{d}x. \end{aligned}$$

(3.1)

Lemma 2.1 is valid, since in the radially symmetric case

$$\begin{aligned} \Delta u=\nabla \text {div}u =\partial _{r}\left( \overline{u}_{r}+\frac{\overline{u}}{r}\right) \frac{x}{r}. \end{aligned}$$

Consequently, one has

$$\begin{aligned} \Omega (0)=\Omega (t). \end{aligned}$$

Then

$$\begin{aligned} I'(t)&=\int _{\mathbb {R}^{2}}(x^{2}\rho _t-2x\rho u)\mathrm{d}x -2(1+t)\\&\quad \int _{\mathbb {R}^{2}}\left[ x(\rho u)_{t} -\rho u^{2}-\frac{2}{\gamma -1}\rho ^{\gamma }-|\nabla \Phi |^{2}\right] \mathrm{d}x\\&\quad +(1+t)^{2}\int _{\mathbb {R}^{2}}\left( \rho u^{2}+\frac{2}{\gamma -1}\rho ^{\gamma }+|\nabla \Phi |^{2}\right) _t\mathrm{d}x\\&=I_1+I_2+I_3. \end{aligned}$$

Direct calculation by using the mass equation and $u\in C^{1}\big ([0,T],H^{m}(\mathbb {R}^{2})\big )(m>2)$ implies that $I_1=0$. As to the second term

$$\begin{aligned} I_2&=2(1+t)\int _{\mathbb {R}^{2}}x\cdot \left[ \text {div}(\rho u\otimes u)+\nabla \rho ^{\gamma }-\mu \Delta u-(\mu +\lambda )\nabla \text {div}u+\rho \nabla \Phi \right] \mathrm{d}x\\&\quad +2(1+t)\int _{\mathbb {R}^{2}}\left( \rho u^{2}+\frac{2}{\gamma -1}\rho ^{\gamma }+|\nabla \Phi |^{2}\right) \mathrm{d}x\\&=2(1+t)\left[ \int _{\mathbb {R}^{2}}\left( \frac{2(2-\gamma )}{\gamma -1}\rho ^{\gamma } +|\nabla \Phi |^{2}\right) \mathrm{d}x\right. \\&\quad \left. +(2\mu +\lambda )\int ^{\infty }_{0}\partial _{r}\left( \frac{\overline{u}}{r}+\overline{u}_r\right) \frac{|x|^2}{r}r\mathrm{d}r\right] \\&=2(1+t)\int _{\mathbb {R}^{2}}\left[ \frac{2(2-\gamma )}{\gamma -1}\rho ^{\gamma } +|\nabla \Phi |^{2} +2(2\mu +\lambda )\left( \frac{\overline{u}}{r}+\overline{u}_r\right) \right] \mathrm{d}x, \end{aligned}$$

and

$$\begin{aligned} I_3&=(1+t)^2\int _{\mathbb {R}^{2}}2\big [-\nabla \rho ^{\gamma }\cdot u+\mu \Delta u\cdot u+(\mu +\lambda )(\nabla \text {div}u)\cdot u\big ]\mathrm{d}x\\&+(1+t)^2\frac{2}{\gamma -1}\int _{\mathbb {R}^{2}}\big [-\text {div}(\rho ^\gamma u)-\big ((\rho ^\gamma )'\rho -\rho \big )\text {div}u\big ]\mathrm{d}x\\&=2(1+t)^2\int ^\infty _0(2\mu +\lambda )\partial _r\left( \frac{\overline{u}}{r}+\overline{u}_r\right) \frac{x}{r}\cdot \left( \overline{u}\frac{x}{r}\right) r\mathrm{d}r\\&=-2(1+t)^2(2\mu +\lambda )\int _{\mathbb {R}^{2}} \left( \frac{\overline{u}}{r}+\overline{u}_r\right) ^2\mathrm{d}x. \end{aligned}$$

Combining all the identities, we get

$$\begin{aligned} I'(t)&=4(1+t)\frac{2-\gamma }{\gamma -1}\int _{\mathbb {R}^{2}}\rho ^{\gamma }\mathrm{d }x +4(1+t)(2\mu +\lambda )\int _{\mathbb {R}^{2}}\left( \overline{u}_{r}+\frac{\overline{u}}{r}\right) \mathrm{d}x\nonumber \\&-2(1+t)^{2}(2\mu +\lambda )\int _{\mathbb {R}^{2}}\left( \overline{u}_{r}+\frac{\overline{u}}{r}\right) ^{2}\mathrm{d}x +2(1+t)\int _{\mathbb {R}^{2}}|\nabla \Phi |^{2}\mathrm{d}x\nonumber \\&\le 4(1+t)\frac{2-\gamma }{\gamma -1}\int _{\mathbb {R}^{2}}\rho ^{\gamma }\mathrm{d}x +2(2\mu +\lambda )|\Omega (0)| +2(1+t)\int _{\mathbb {R}^{2}}|\nabla \Phi |^{2}\mathrm{d}x\nonumber \\&\le \frac{6-2\gamma }{1+t}I(t)+2(2\mu +\lambda )|\Omega (0)|. \end{aligned}$$

(3.2)

In the following, we adopt a similar method as in [3].

Case 1 If $\gamma \ge 3$, we have from (3.2)

$$\begin{aligned} I'(t)\le 2(2\mu +\lambda )|\Omega (0)|, \end{aligned}$$

which implies

$$\begin{aligned} I(t)\le I(0)+2(2\mu +\lambda )|\Omega (0)|t. \end{aligned}$$

(3.3)

From (3.1) and (3.3), we get

$$\begin{aligned} \int _{\mathbb {R}^{2}}\rho ^{\gamma }\mathrm{d}x\le \frac{\gamma -1}{2}I(0)(1+t)^{-2} +(2\mu +\lambda )(\gamma -1)|\Omega (0)|(1+t)^{-1}. \end{aligned}$$

(3.4)

By conservation of mass and the Hölder inequality, we have

$$\begin{aligned} \int _{\Omega (0)}\rho _0&=\int _{\Omega (t)}\rho \mathrm{d}x\\&\le \left( \int _{\Omega (t)}\rho ^{\gamma }\mathrm{d}x)^{\frac{1}{\gamma }}(\Omega (t)\right) ^\frac{\gamma -1}{\gamma }\\&\le \big (\Omega (0)\big )^\frac{\gamma -1}{\gamma } \bigg (\frac{\gamma -1}{2}I(0)(1\!+\!t)^{-2} +(2\mu \!+\!\lambda )(\gamma -1)|\Omega (0)|(1+t)^{-1}\bigg )^{\frac{1}{\gamma }}, \end{aligned}$$

which yields that T must be finite.

Case 2 If $2<\gamma <3$, we get

$$\begin{aligned} \big ((1+t)^{2\gamma -6}I(t)\big )' \le 2(2\mu +\lambda )|\Omega (0)|(1+t)^{2\gamma -6}, \end{aligned}$$

which gives

$$\begin{aligned} (1+t)^{2\gamma -6}I(t) \le I(0)+2(2\mu +\lambda )|\Omega (0)|H(t), \end{aligned}$$

(3.5)

where

$$\begin{aligned} H(t)=\left\{ \begin{array}{lllll} \frac{1}{2\gamma -5}(1+t)^{2\gamma -5} \qquad \quad &{}\text {if} \, \, \, 2\gamma -5\ne 0;\\ \ln (1+t) \qquad \quad &{}\text {if} \, \, \, 2\gamma -5=0. \end{array} \right. \end{aligned}$$

Thus we obtain that

$$\begin{aligned} \int _{\mathbb {R}^{2}}\rho ^{\gamma }\mathrm{d}x\le \frac{\gamma -1}{2}I(0)(1+t)^{4-2\gamma } +(2\mu +\lambda )(\gamma -1)|\Omega (0)|(1+t)^{4-2\gamma }H(t). \end{aligned}$$

(3.6)

Similar to the estimates (3.3), (3.4), (3.6) also implies that T must be finite. So we finish the proof. $\square$