Lobachevskii Journal of Mathematics http://ljm.ksu.ru ISSN 1818-9962 Vol. 24, 2006, 63–72

© Changchun Liu

Abstract. In this paper, the author investigates the initial boundary value problem for a nonlinear degenerate parabolic equation, which comes from a compressible fluid flowing in a homogeneous isotropic rigid porous medium. We establish the existence of nonnegative self-similar solutions.

1. Introduction

During the past years, many authors have paid much attention to the following equation

in ${\mathbb{ℝ}}^n$. The equation is a continuous model, which comes from a compressible fluid flowing in a homogeneous isotropic rigid porous medium. The equation is also called non-newtonian polypropic filtration equations. Zhao and Yuan [4], Zhao and Xu [3] consider the Cauchy problem for $\alpha =0$. They proved the existence of weak solution; see also [6], [7]. After introducing the radial variable $r=\mid x\mid$, we see that the radial symmetric solution satisfies

On the basis of physical consideration, as usual the equation (1.1) is supplemented with the natural boundary value conditions

and the initial value condition

In this paper, we study the self-similar solution of the problem (1.1)-(1.3). We are seeking solutions of the form

A direct calculation shows that $y\left(x\right)$ should satisfy the following ordinary differential equation

where $m>1,N>1$, $\beta =\frac{\gamma \left(mN-1\right)+1}{N-\alpha +1}$, $C^{N+1}=u_0^{mn-1}\beta ,\lambda =\frac{\gamma }{\beta }$, $u_0>0$, $\gamma >0$ with the following initial value conditions

where

In recent years, the equation (1.4) with $\alpha =0,m=1$ has also caused much of attention [1, 2, 5].

Definition A function $y\left(x\right)$ is said to be a solution of the problem (1.4)–(1.5), if the following conditions are satisfied:

1) $y\left(x\right)\in C^1\left[0,+\infty \right)$,

2) the function $x^{\alpha +n-1}\mid {\left(\mid y{\mid }^{m-1}y\right)}^{\prime }{\mid }^{N-1}{\left(\mid y{\mid }^{m-1}y\right)}^{\prime }$ is continuously differentiable in $\left[0,+\infty \right)$, and (1.4) a.e. holds in $\left[0,+\infty \right)$.

3) (1.5) holds.

2. The two-point boundary value problem

In this section, we first study the properties of the solution for the problem (1.4)-(1.5).

Proof. Otherwise, if there exists a point $x^{\ast }\in \left(0,+\infty \right)$ such that $y\left(x^{\ast }\right)<0$. By $y\left(0\right)=1$ and $y\left(+\infty \right)=0$, we know that there would exist a point $a_1\in \left(0,+\infty \right)$ and $y\left(a_1\right)$ is a minimum value. Hence, we have $y^{\prime }\left(a_1\right)=0$. Again by $y\left(+\infty \right)=0$, $y\left(a_1\right)<0$, we know that there exists a point $a_2\in \left[a_1,+\infty \right)$, such that $y\left(a_2\right)=\frac{1}{2}y\left(a_1\right)$. In addition, the mean value theorem yields,

where $b\in \left[a_1,a_2\right]$. Since $y\left(a_2\right)-y\left(a_1\right)>0$ and $a_2-a_1>0$, we obtain $y^{\prime }\left(b\right)>0$. Taking $\left[c_1,c_2\right]\subset \left[a_1,a_2\right]$ such that $y^{\prime }\left(c_1\right)=0$ and $y\left(x\right)<0$, $y^{\prime }\left(x\right)>0$ in $\left(c_1,c_2\right)$. Integrating the equation (1.4) with respect to $x$ over $\left(c_1,c_2\right)$, we see that

Observing that $\lambda x^{n-1}y\left(x\right)\le 0$ and $x^n{y}^{\prime }\left(x\right)>0$, we have $\lambda x^{n-1}y\left(x\right)-x^n{y}^{\prime }\left(x\right)<0$. Hence

The contradiction shows $y\ge 0$.

Similarly, we can prove $y^{\prime }\left(x\right)\le 0$. Otherwise, if there exists $x^{\ast }\in \left(0,+\infty \right)$ such that $y^{\prime }\left(x^{\ast }\right)>0$. By $y\left(0\right)=1$, $y\left(+\infty \right)=0$, we know that there would exist a $x_1\in \left(0,+\infty \right)$ with $y\left(x_1\right)$ being a maximum value. Hence, we have $y^{\prime }\left(x_1\right)=0$. By $y\left(x_1\right)>0$ which is a maximum value, there exists a point $x_2\in \left[x_1,+\infty \right)$, satisfying $y\left(x_2\right)=\frac{1}{2}y\left(x_1\right)$. The mean value theorem yields,

where $b\in \left[x_1,x_2\right]$. By $y\left(x_2\right)-y\left(x_1\right)<0$ and $x_2-x_1>0$, therefore we have $y^{\prime }\left(b\right)<0$. we take $\left[c_1,c_2\right]\subset \left[a_1,a_2\right]$ such that $y^{\prime }\left(c_1\right)=0$ and $y\left(x\right)>0$, $y^{\prime }\left(x\right)<0$ in $\left(c_1,c_2\right)$. Integrating the equation (1.4) with respect to $x$ over $\left(c_1,c_2\right)$, we see that

Observing that $\lambda x^{n-1}y\left(x\right)>0$ and $-x^n{y}^{\prime }\left(x\right)>0$, then $\lambda x^{n-1}y\left(x\right)-x^n{y}^{\prime }\left(x\right)>0$. Thus we have

The contradiction shows $y^{\prime }\le 0$. $\square$

To prove the proposition 2.2, we need some lemmas. We first have

Proof. We first prove $y\left(x\right)\ge 0$. Otherwise, if there exists a point $a_1$ such that $y\left(a_1\right)<0$. By $y\left(+\infty \right)=0$, we know that there would exist a point $x_1$ and $y\left(x_1\right)<0$ is a minimum value. Hence, we have $y^{\prime }\left(x_1\right)=0$. Integrating the equation (1.4) with respect to $x$ over $\left(x_1,x_2\right)$, and using $y^{\prime }\left(x_0\right)=0$, $y^{\prime }\left(x_1\right)=0$ and $y\left(x_0\right)>0$, $y\left(x_1\right)<0$, we see that

and

The contradiction shows $y\ge 0$. Similarly, we can prove $y^{\prime }\left(x\right)\le 0$. $\square$

Proof. For simplicity, we only prove the first inequality $y\left(x\right)\le 0$, since the other can be shown similarly. Otherwise, if there exists $a_1$ such that $y\left(a_1\right)>0$, by $y\left(+\infty \right)=0$, we know that there would exist a $x_1$ and $y\left(x_1\right)<0$ is a maximum value. Hence, we have $y^{\prime }\left(x_1\right)=0$. By $y^{\prime }\left(x_0\right)=0$, $y^{\prime }\left(x_1\right)=0$ and $y\left(x_0\right)>0$, $y\left(x_1\right)<0$. Integrating the equation (1.4) with respect to $x$ over $\left(x_1,x_2\right)$, we see that

and

The contradiction shows $y\le 0$. $\square$

Proof of Proposition 2.2. We first prove that $y\left(x\right)\ge 0$. Otherwise, if there exists $x^{\ast }\in \left(0,+\infty \right)$ such that $y\left(x^{\ast }\right)<0$, by $y\left(0\right)=1$, $y\left(+\infty \right)=0$, we know that there would exist a $a_1\in \left(0,+\infty \right)$ and $y\left(a_1\right)$ is a minimum value. Therefore, we have $y^{\prime }\left(a_1\right)=0$. By lemma 2.2, we have $y^{\prime }\left(x\right)\ge 0$, $y\left(x\right)\le 0$ in $\left[a_1,+\infty \right)$ . Multiplying both sides of the equation (1.4) by $-x^{-\lambda -1}$, we have

Then, integrating the resulting relation with respect to $x$ over $\left(a_1,M\right)$, we have

However, the right hand side

By $lim{x}^{-\lambda }y\left(x\right)=0$, we have the right hand side $>0$, as $M\to +\infty$. The contradiction shows $y\le 0$.

Similarly, we can prove $y^{\prime }\left(x\right)\ge 0$. $\square$

Proof. If there exists a point $x_0\in \left[0,+\infty \right)$, such that $y\left(x_0\right)>0$ and $y^{\prime }\left(x_0\right)=0$. Without loss of generality, we assume there exists a strictly monotone sequence ${\left\{x_j\right\}}_{j=1}^{\infty }$ such that ${lim}_{j\to +\infty }{x}_j=x_0$, $y\left(x_j\right)>0$, $y^{\prime }\left(x_j\right)=0$. Set $f\left(x_j\right)=x_j^{\alpha +n-1}{\left|{\left(\mid y{\mid }^{m-1}y\left(x_j\right)\right)}^{\prime }\right|}^{N-1}{\left(\mid y{\mid }^{m-1}y\left(x_j\right)\right)}^{\prime }$. Hence

where ${\xi }_j\in \left[x_j,x_{j+1}\right]$. Letting $j\to +\infty$, we have $y\left(x_0\right)=0$. The contradiction shows $y^{\prime }<0$. $\square$

3. Self-similar solution

To prove the existence of solutions of the problem (1.4)-(1.5), we set

and consider the following problem

To prove the existence of solution of the problem (3.1)-(3.3), we consider the problem (3.1), (3.2) and

Proof. If $v\left(t\right)\in C\left[0,1\right]$ satisfies

and let

It is seen that $\left(v\left(t\right),W\left(t\right)\right)$ is a solution of the problem (3.1), (3.2), (3.4).

On the other hand, clearly, for any $a\in \left[0,1\right]$, we have

Define the map $\varphi :\Omega \to \Omega$,

where $\Omega =\left\{v\left(t\right)\in C\left[0,1\right];h\le v\left(t\right)\le \left(\varphi h\right)\left(t\right)\right\}$. It is seen that the operator $\varphi$ is $\Omega$ to $\Omega$ continuous and compact. By Leray–Schauder principle of fixed point, the operator $\varphi$ has a fixed point $v\left(t,h\right)$ in $\Omega$, which is the desired solution of the (3.5). Hence it is the solutions of the problem (3.1), (3.2), (3.4). $\square$

Proof. We first show the left inequality. If this were not true, then there will be a point $t_0\in \left[0,1\right]$, such that

By $v\left(0,h_1\right)-v\left(0,h_2\right)=h_1-h_2>0$, hence $t_0⁄=0$.

Since $t_0⁄=0$, then there exists a interval $\left(a,t_0\right]$, where $0<a<t_0\le 1$, such that

Using (3.7), we obtain $v\left(t_0,h_1\right)-v\left(t_0,h_2\right)=0$, this yields a contradiction.

We can obtain the right inequality, by the left inequality and (3.5). $\square$

By the lemma 3.1, we have the following conclusion.

Proof. By the lemma 3.1, we know that

Substituting $v\left(t,h\right)$ into (3.5), then letting $h\to 0$, using lemma 3.2, we have the problem (3.1)-(3.3) admits one and only one nonnegative continuous solution $v\left(t\right)$. $\square$

Proof. Now, we construct a solution of the problem (1.4)-(1.5) by the solution of the problem (3.1)-(3.3). Suppose $\left(v\left(t\right),W\left(t\right)\right)$ is a solution of the problem (3.1)-(3.3), then $W\left(t\right)$ is a strictly decreasing function in $\left[0,1\right]$. Hence the inverse function $t=y\left(x\right)$ of $x=W\left(t\right)$ exists in $\left[0,W\left(0\right)\right)$. If $W\left(0\right)<+\infty$, we define the $y\left(x\right)=0$, as $x\in \left[W\left(0\right),+\infty \right)$. Hence the $y\left(x\right)$ is a nonnegative function with $y\left(0\right)=1$, ${lim}_{x\to +\infty }{x}^{{\left[-\lambda \right]}_{+}}y\left(x\right)=0$, in $\left[0,+\infty \right)$. We set $x_0=W\left(0\right)$. Observing that $x=W\left(t\right)$ in $\left[0,x_0\right)$ and $y^{\prime }\left(x\right)=\frac{1}{W^{\prime }\left(t\right)}<0$ a.e. in $\left(0,x_0\right)$. As $x_0<+\infty$, we have $y_{-}^{\prime }\left(x_0\right)=\frac{1}{W_{+}^{\prime }\left(0\right)}=0$. Again as $x\in \left[x_0,+\infty \right)$, $y\left(x\right)=0$, hence, we know that $y^{\prime }\left(x\right)$ continuous at $x_0$ and $y^{\prime }\left(x_0\right)=0$. As $x_0=+\infty$, by (3.6), we have $y^{\prime }\left(x\right)=-v^{1∕N}\left(t\right)∕\left[\left(N-n+\alpha -1\right)∕\left(N-1\right)W^{1∕N}\right]$. Again by $v\left(0\right)=0$ and $W\left(0\right)=+\infty$, we know that $y^{\prime }\left(+\infty \right)=0$. Substituting $t=y\left(x\right)$ in (1.4), (1.5), it is easily seen that the function $y\left(x\right)\in C^1\left[0,+\infty \right)$ is the solution of the problem (1.4), (1.5). $\square$

By the Theorem 3.3, we have the following conclusion.

References

Changchun Liu, Department of Mathematics, Jilin University,
Changchun 130012, P. R. China

E-mail address: lcc@email.jlu.edu.cn

Received June 8, 2006