Global existence and convergence to the modified Barenblatt solution for the compressible Euler equations with physical vacuum and time-dependent damping

Abstract

In this paper, the smooth solution of the physical vacuum problem for the one dimensional compressible Euler equations with time-dependent damping is considered. Near the vacuum boundary, the sound speed is $C^{1/2}$-Hölder continuous. The coefficient of the damping depends on time, given by this form $\frac{\mu }{(1+t)^\lambda }$, $\lambda ,\ \mu >0$, which decays by order $-\lambda$ in time. Under the assumption that $0<\lambda<1,\ 0<\mu$ or $\lambda =1,\ 2<\mu$, we will prove the global existence of smooth solutions and convergence to the modified Barenblatt solution of the related porous media equation with time-dependent dissipation and the same total mass when the initial data of the Euler equations is a small perturbation of that of the Barenblatt solution. The pointwise convergence rates of the density, velocity and the expanding rate of the physical vacuum boundary are also given. The proof is based on space-time weighted energy estimates, elliptic estimates and Hardy inequality in the Lagrangian coordinates. Our result is an extension of that in Luo–Zeng (Commun Pure Appl Math 69(7):1354–1396, 2016), where the authors considered the physical vacuum free boundary problem of the compressible Euler equations with constant-coefficient damping.

Appendix A. Estimates for \({\tilde{\eta }_x}\) and h

In this appendix, we prove (2.9) and (2.10). The idea of proof follows the line with that in Appendix of [27]. We may write (2.6) as the following system:

$$\begin{aligned} \left\{ \begin{aligned}&h_{t}=z, \\&z_{t}=-\frac{\mu }{(1+t)^\lambda }z-\frac{\mu (\lambda +1)}{\gamma +1}\left[ {\bar{\eta }}_{x}^{-\gamma }-\left( {\bar{\eta }}_{x}+h\right) ^{-\gamma }\right] -{\bar{\eta }}_{x t t}, \\&(h, z)|_{t=0}=(0,0). \end{aligned} \right. \end{aligned}$$

Recalling that ${\bar{\eta }}_{x}(t)=(1+t)^{\frac{\lambda +1}{\gamma +1}},$ we have ${\bar{\eta }}_{x t t}<0 .$ A simple phase plane analysis shows that there exist $0<t_{0}<t_{1}<t_{2}$ such that, starting from $(h, z)=$ (0,0) at $t=0, h$ and z increase in the interval $\left[ 0, t_{0}\right]$ and z reaches its positive maximum at $t_{0} ;$ in the interval $\left[ t_{0}, t_{1}\right] , h$ keeps increasing and reaches its maximum at $t_{1}, z$ decreases from its positive maximum to 0;  in the interval $\left[ t_{1}, t_{2}\right] ,$ both h and z decrease, and z reaches its negative minimum at $t_{2} ;$ in the interval $\left[ t_{2}, \infty \right) , h$ decreases and z increases, and $(h, z) \rightarrow (0,0)$ as $t \rightarrow \infty .$ This can be summarized as follows:

$$\begin{aligned} \begin{array}{l@{\quad }l} z(t) \uparrow _{0}, &{} h(t) \uparrow _{0}, \quad t \in \left[ 0, t_{0}\right] \\ z(t) \downarrow _{0}, &{} h(t) \uparrow , \quad t \in \left[ t_{0}, t_{1}\right] \\ z(t) \downarrow ^{0}, &{} h(t) \downarrow , \quad t \in \left[ t_{1}, t_{2}\right] \\ z(t) \uparrow ^{0}, &{} h(t) \downarrow _{0}, \quad t \in \left[ t_{2}, \infty \right) . \end{array} \end{aligned}$$

It follows from the above analysis that there exists a constant $C=C(\lambda ,\mu ,\gamma , M)$ such that

$$\begin{aligned} 0 \le h(t) \le C \quad \text{ for } t \ge 0 . \end{aligned}$$

In view of (2.7), we then see that for some constant $K>0$

$$\begin{aligned} (1+t)^{ \frac{\lambda +1}{\gamma +1}} \le {\tilde{\eta }}_{x} \le K(1+t)^{ \frac{\lambda +1}{\gamma +1}}. \end{aligned}$$

To derive the decay property, we may rewrite (2.8) as

$$\begin{aligned} \left\{ \begin{aligned}&{\tilde{\eta }}_{x t t}+\frac{\mu }{(1+t)^\lambda }{\tilde{\eta }}_{x t}-\frac{\mu (\lambda +1)}{\gamma +1}{\tilde{\eta }}_{x}^{-\gamma }=0,\\&{\tilde{\eta }}_{x}|_{t=0}=1, \quad {\tilde{\eta }}_{x t}|_{t=0}=\frac{\lambda +1}{\gamma +1}. \end{aligned} \right. \end{aligned}$$

(A.1)

Next we give the proof of (2.9) and (2.10) separately. The general ideas are the same.

Case 1: $\varvec{\lambda <1}$:

Then, we have by solving (A.1) that

$$\begin{aligned} \begin{aligned}&{\tilde{\eta }}_{x t}(t)=\frac{\lambda +1}{\gamma +1}e^{-\frac{\mu }{1-\lambda }(1+t)^{1-\lambda }}\\&\quad +\,\frac{(\lambda +1)\mu }{\gamma +1} \int _{0}^{t} e^{\frac{\mu }{1-\lambda }\left[ (1+s)^{1-\lambda }-(1+t)^{1-\lambda }\right] }{\tilde{\eta }}_{x}^{-\gamma }(s) d s\\&\quad \ge 0. \end{aligned} \end{aligned}$$

(A.2)

Next, we use the induction to prove (2.9). First, it follows from (A.2) that

$$\begin{aligned} \begin{aligned} {\tilde{\eta }}_{x t}(t)&\lesssim \, e^{-\frac{\mu }{1-\lambda }(1+t)^{1-\lambda }}+\int _{0}^{t} e^{\frac{\mu }{1-\lambda }\left[ (1+s)^{1-\lambda }-(1+t)^{1-\lambda }\right] }(1+s)^{-\frac{(\lambda +1)\gamma }{\gamma +1}} d s\\&\lesssim \, e^{-\frac{\mu }{1-\lambda }(1+t)^{1-\lambda }}+\left\{ \int _{0}^{t/2}+\int _{t/2}^{t}\right\} e^{\frac{\mu }{1-\lambda }\left[ (1+s)^{1-\lambda }-(1+t)^{1-\lambda }\right] }(1+s)^{-\frac{(\lambda +1)\gamma }{\gamma +1}} d s\\&\lesssim \, e^{-\frac{\mu }{1-\lambda }(1+t)^{1-\lambda }}+e^{-c(1+t)^{1-\lambda }}\int _{0}^{t/2}(1+s)^{-\frac{(\lambda +1)\gamma }{\gamma +1}} ds\\&\quad +\,(1+t)^{-\frac{(\lambda +1)\gamma }{\gamma +1}}\int _{t/2}^{t}e^{\frac{\mu }{1-\lambda }\left[ (1+s)^{1-\lambda }-(1+t)^{1-\lambda }\right] } d s\\&\lesssim \, e^{-\frac{\mu }{1-\lambda }(1+t)^{1-\lambda }}+(1+t)^{-\frac{(\lambda +1)\gamma }{\gamma +1}+\lambda } \\&\lesssim \, (1+t)^{\frac{\lambda +1}{\gamma +1}-1}. \end{aligned} \end{aligned}$$

This proves (2.9) for $k=1 .$ For $k\ge 2$, we make the induction hypothesis that (2.9) holds for all $\ell =1,2, \ldots , k-1,$ that is

$$\begin{aligned} \left| \frac{d^{\ell } {\tilde{\eta }}_{x}(t)}{d t^{\ell }}\right| \le C_k (1+t)^{\frac{\lambda +1}{\gamma +1}-\ell }, \quad \ell =1,2, \ldots , k-1. \end{aligned}$$

(A.3)

It suffices to prove (A.3) holds for $\ell =k .$ We derive from (A.1) that

$$\begin{aligned} \begin{aligned}&\frac{d^{k+1} {\tilde{\eta }}_{x}}{d t^{k+1}}(t)+\frac{\mu }{(1+t)^\lambda }\frac{d^{k} {\tilde{\eta }}_{x}}{d t^{k}}(t)\\&\quad =\frac{(\lambda +1)\mu }{\gamma +1} \frac{d^{k-1} {\tilde{\eta }}_{x}^{-\gamma }}{d t^{k-1}}(t)-\mu \sum ^{k-1}_{\ell =1}C^\ell _{k-1}\frac{d^\ell (1+t)^{-\lambda }}{dt^\ell }\frac{d^{k-\ell } {\tilde{\eta }}_{x}}{dt^{k-\ell }}. \end{aligned} \end{aligned}$$

Solving this ODE gives that

$$\begin{aligned} \begin{aligned} \frac{d^{k} {\tilde{\eta }}_{x}}{d t^{k}}(t)&=e^{-\frac{\mu }{1-\lambda }(1+t)^{1-\lambda }}\frac{d^{k} {\tilde{\eta }}_{x}}{d t^{k}}(0)\\&\qquad +\,\frac{(\lambda +1)\mu }{\gamma +1}\int ^t_0 e^{\frac{\mu }{1-\lambda }\left[ (1+s)^{1-\lambda }-(1+t)^{1-\lambda }\right] }\frac{d^{k-1} {\tilde{\eta }}_{x}^{-\gamma }}{d t^{k-1}}(s)ds\\&\qquad -\mu \int ^t_0 e^{\frac{\mu }{1-\lambda }\left[ (1+s)^{1-\lambda }-(1+t)^{1-\lambda }\right] }\sum ^{k-1}_{\ell =1}C^\ell _{k-1}\frac{d^\ell (1+s)^{-\lambda }}{dt^\ell }\frac{d^{k-\ell } {\tilde{\eta }}_{x}}{dt^{k-\ell }}(s)ds \end{aligned} \end{aligned}$$

(A.4)

where $\frac{d^{k} {\tilde{\eta }}_{x}}{d t^{k}}(0)$ can be determined by the equation inductively. We need to bound the second and the third term on the righthand of (A.4).

Using (A.3), we can deduce by induction that

$$\begin{aligned} \left| \frac{d^{\ell } {\tilde{\eta }}^{-\gamma }_{x}(t)}{d t^{\ell }}\right| \le C_k (1+t)^{-\frac{(\lambda +1)\gamma }{\gamma +1}-\ell }, \quad \ell =1,2, \ldots , k-1. \end{aligned}$$

(A.5)

Substituting (A.3) and (A.5) into (A.4), we can get

$$\begin{aligned} \begin{aligned} \frac{d^{k} {\tilde{\eta }}_{x}}{d t^{k}}(t)&\lesssim \,e^{-\frac{\mu }{1-\lambda }(1+t)^{1-\lambda }}\\&\quad +\int ^t_0 e^{\frac{\mu }{1-\lambda }\left[ (1+s)^{1-\lambda }-(1+t)^{1-\lambda }\right] }(1+s)^{-\frac{(\lambda +1)\gamma }{\gamma +1}-(k-1)}ds\\&\quad +\sum ^{k-1}_{\ell =1}\int ^t_0 e^{\frac{\mu }{1-\lambda }\left[ (1+s)^{1-\lambda }-(1+t)^{1-\lambda }\right] }(1+s)^{-\lambda -\ell }(1+s)^{-\frac{(\lambda +1)\gamma }{\gamma +1}-(k-\ell )}(s)ds\\&\lesssim e^{-\frac{\mu }{1-\lambda }(1+t)^{1-\lambda }}\\&\quad +\int ^t_0 e^{\frac{\mu }{1-\lambda }\left[ (1+s)^{1-\lambda }-(1+t)^{1-\lambda }\right] }(1+s)^{-\frac{(\lambda +1)\gamma }{\gamma +1}-(k-1)}ds\\&\lesssim \,e^{-\frac{\mu }{1-\lambda }(1+t)^{1-\lambda }}+(1+t)^{-\frac{(\lambda +1)\gamma }{\gamma +1}-(k-1)+\lambda }\\&\lesssim (1+t)^{\frac{(\lambda +1)}{\gamma +1}-k}. \end{aligned} \end{aligned}$$

This finishes the proof of (2.9).

Case 2: $\varvec{\lambda =1}$:

We have by solving (A.1) that

$$\begin{aligned} {\tilde{\eta }}_{x t}(t)=\frac{2}{\gamma +1}(1+t)^{-\mu }+\frac{2\mu }{\gamma +1} (1+t)^{-\mu }\int _{0}^{t} (1+s)^\mu {\tilde{\eta }}_{x}^{-\gamma }(s) d s \ge 0. \end{aligned}$$

(A.6)

The same as case $0<\lambda <1$, we use the induction to prove (2.10) for $k<\mu +\frac{2}{\gamma +1}$. First, it follows from (A.6) that

$$\begin{aligned} \begin{aligned} {\tilde{\eta }}_{x t}(t)&\lesssim (1+t)^{-\mu }+ (1+t)^{-\mu }\int _{0}^{t} (1+s)^\mu (1+s)^{-\frac{2\gamma }{\gamma +1}} d s\\&\lesssim (1+t)^{-\mu }+ (1+t)^{-\mu } (1+t)^{\mu -\frac{2\gamma }{\gamma +1}+1} \\&=(1+t)^{-\mu }+ (1+t)^{\frac{2}{\gamma +1}-1}\\&\lesssim (1+t)^{\frac{2}{\gamma +1}-1}. \end{aligned} \end{aligned}$$

This proves (2.10) for $k=1 .$ For $2 \le k <\mu +\frac{2}{\gamma +1}$, we make the induction hypothesis that (2.10) holds for all $\ell =1,2, \ldots , k-1,$ that is

$$\begin{aligned} \left| \frac{d^{\ell } {\tilde{\eta }}_{x}(t)}{d t^{\ell }}\right| \le C_k (1+t)^{\frac{2}{\gamma +1}-\ell }, \quad \ell =1,2, \ldots , k-1. \end{aligned}$$

(A.7)

It suffices to prove (A.7) holds for $\ell =k .$ We derive from (A.1) that

$$\begin{aligned} \begin{aligned}&\frac{d^{k+1} {\tilde{\eta }}_{x}}{d t^{k+1}}(t)+\frac{\mu }{1+t}\frac{d^{k} {\tilde{\eta }}_{x}}{d t^{k}}(t)\\&\quad =\frac{2\mu }{\gamma +1} \frac{d^{k-1} {\tilde{\eta }}_{x}^{-\gamma }}{d t^{k-1}}(t)-\mu \sum ^{k-1}_{\ell =1}C^\ell _{k-1}\frac{d^\ell (1+t)^{-1}}{ds^\ell }\frac{d^{k-\ell } {\tilde{\eta }}_{x}}{ds^{k-\ell }}. \end{aligned} \end{aligned}$$

So that

$$\begin{aligned} \begin{aligned} \frac{d^{k} {\tilde{\eta }}_{x}}{d t^{k}}(t)&=(1+t)^{-\mu }\frac{d^{k} {\tilde{\eta }}_{x}}{d t^{k}}(0)\\&\qquad +\,\frac{2\mu }{\gamma +1}(1+t)^{-\mu }\int ^t_0 (1+s)^{\mu }\frac{d^{k-1} {\tilde{\eta }}_{x}^{-\gamma }}{d t^{k-1}}(s)ds\\&\quad -\,\mu (1+t)^{-\mu }\int ^t_0 \sum ^{k-1}_{\ell =1}(1+s)^{\mu }C^\ell _{k-1}\frac{d^\ell (1+s)^{-1}}{dt^\ell }\frac{d^{k-\ell } {\tilde{\eta }}_{x}}{dt^{k-\ell }}(s)ds. \end{aligned} \end{aligned}$$

(A.8)

We need to bound the second and the third terms on the righthand of (A.8).

Using (A.7), we can deduce by induction that

$$\begin{aligned} \left| \frac{d^{\ell } {\tilde{\eta }}^{-\gamma }_{x}(t)}{d t^{\ell }}\right| \le C_k (1+t)^{-\frac{2\gamma }{\gamma +1}-\ell }, \quad \ell =1,2, \ldots , k-1. \end{aligned}$$

(A.9)

Substituting (A.7) and (A.9) into (A.8), we can get

$$\begin{aligned} \frac{d^{k} {\tilde{\eta }}_{x}}{d t^{k}}(t)\lesssim&\,(1+t)^{-\mu }+(1+t)^{-\mu }\int ^t_0 (1+s)^{\mu }(1+s)^{-\frac{2\gamma }{\gamma +1}-(k-1)}ds\nonumber \\&+(1+t)^{-\mu }\sum ^{k-1}_{\ell =1}\int ^t_0 (1+s)^{\mu }(1+s)^{-1-\ell }(1+s)^{\frac{2}{\gamma +1}-(k-\ell )}ds \nonumber \\ \lesssim&\,(1+t)^{-\mu }+(1+t)^{-\mu }\int ^t_0(1+s)^{\mu +\frac{2}{\gamma +1}-k-1}ds\nonumber \\ \lesssim&\,(1+t)^{-\mu }+(1+t)^{-\mu }(1+t)^{\mu +\frac{2}{\gamma +1}-k}\nonumber \\ \lesssim&\,(1+t)^{\frac{2}{\gamma +1}-k}. \end{aligned}$$

(A.10)

This finishes the proof of (A.7) for $k<\mu +\frac{2}{\gamma +1}$. If $k=\mu +\frac{2}{\gamma +1}$, from the third line of (A.10), we have

$$\begin{aligned} \begin{aligned} \left| \frac{d^{k} {\tilde{\eta }}_{x}}{d t^{k}}(t)\right| \lesssim&(1+t)^{-\mu }+(1+t)^{-\mu }\int ^t_0(1+s)^{-1}ds\\ \lesssim&(1+t)^{-\mu }\ln (1+t). \end{aligned} \end{aligned}$$

When $k>\mu +\frac{2}{\gamma +1}$, it is a routine work to prove that

$$\begin{aligned} \left| \frac{d^{k} {\tilde{\eta }}_{x}}{d t^{k}}(t)\right| \le (1+t)^{-\mu }\ln (1+t) . \end{aligned}$$

by again using induction. Since in this case, the terms

$$\begin{aligned} \int ^t_0 (1+s)^{\mu }\frac{d^{k-1} {\tilde{\eta }}_{x}^{-\gamma }}{d t^{k-1}}(s)ds,\quad \int ^t_0 (1+s)^{\mu }\frac{d^\ell (1+s)^{-1}}{dt^\ell }\frac{d^{k-\ell } {\tilde{\eta }}_{x}}{dt^{k-\ell }}(s)ds \end{aligned}$$

in (A.8), actually are bounded by $\ln (1+t)$. This finishes the proof of (2.10). $\square$