Lobachevskii Journal of Mathematics

http://ljm.ksu.ru
ISSN 1818-9962

Vol. 24, 2006, 63–72 \begin{matrix}c \\ ◯ \end{matrix} Changchun Liu

Changchun Liu

SELF-SIMILAR SOLUTIONS FOR A NONLINEAR DEGENERATE PARABOLIC EQUATION

(submitted by A. V. Lapin)

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

\frac{∂u}{∂t} = div(|x|^{α}|∇(|u|^{m−1}u)|^{N−1}∇(|u|^{m−1}u)),

in 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 α = 0. They proved the existence of weak solution; see also [6], [7]. After

tions, Existence. (No. 10526022).

2000 Mathematical Subject Classification. 35K15, 35K55, 35K65.

Key words and phrases. Degenerate parabolic equation, Self-similar solu-

This work is supported by the Tianyuan Fund for Mathematics of China

introducing the radial variable r = |x|, we see that the radial symmetric solution satisfies

\frac{∂u}{∂t} = \frac{1}{r^{n−1}} \frac{∂}{∂r} (r^{α+n−1} |\frac{∂}{∂r}(|u|^{m−1}u)|^{N−1} \frac{∂}{∂r}(|u|^{m−1}u)) ,  (1.1)

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

u(0, t) = u_{0}t^{γ}, u(+∞, t) = 0, t > 0,  (1.2)

and the initial value condition

u(r, 0) = 0.  (1.3)

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

u(r, t) = u_{0}t^{γ}y(x), x = crt^{−β}.

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

[x^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′]′ = λx^{n−1}y(x) − x^{n}y′(x), (1.4)

where m > 1, N > 1, β = \frac{γ(mN−1)+1}{N−α+1}, C^{N+1} = u_{0}^{mn−1}β, λ = \frac{γ}{β}, u_{0} > 0, γ > 0 with the following initial value conditions

y(0) = 1, lim_{x→+∞} x^{[−λ]_{+}}y(x) = 0,  (1.5)

where

[−λ]_{+} = \{ \begin{matrix}−λ, & λ < 0, \\ 0, & λ ≥ 0. \end{matrix}

In recent years, the equation (1.4) with α = 0, m = 1 has also caused much of attention [1, 2, 5]. Definition A function y(x) is said to be a solution of the problem (1.4)–(1.5), if the following conditions are satisfied:

1) y(x) ∈ C^{1}[0, +∞),

2) the function x^{α+n−1}|(|y|^{m−1}y)′|^{N−1}(|y|^{m−1}y)′ is continuously differentiable in [0, +∞), and (1.4) a.e. holds in [0, +∞).

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).

Proposition 2.1. Suppose the function y(x) ∈ C^{1}[0, +∞) is a solution of the problem (1.4)-(1.5). Then y(x) ≥ 0, y′(x) ≤ 0, as λ ≥ 0.

Proof. Otherwise, if there exists a point x^{∗} ∈ (0, +∞) such that y(x^{∗}) < 0. By y(0) = 1 and y(+∞) = 0, we know that there would exist a point a_{1} ∈ (0, +∞) and y(a_{1}) is a minimum value. Hence, we have y′(a_{1}) = 0. Again by y(+∞) = 0, y(a_{1}) < 0, we know that there exists a point a_{2} ∈ [a_{1}, +∞), such that y(a_{2}) = \frac{1}{2}y(a_{1}). In addition, the mean value theorem yields,

y(a_{2}) − y(a_{1}) = y′(b)(a_{2} − a_{1}),

where b ∈ [a_{1}, a_{2}]. Since y(a_{2}) − y(a_{1}) > 0 and a_{2} − a_{1} > 0, we obtain y′(b) > 0. Taking [c_{1}, c_{2}] ⊂ [a_{1}, a_{2}] such that y′(c_{1}) = 0 and y(x) < 0, y′(x) > 0 in (c_{1}, c_{2}). Integrating the equation (1.4) with respect to x over (c_{1}, c_{2}), we see that

∫_{c_{1}}^{c_{2}} [x^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′]′ dx

= c_{2}^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′(c_{2})
− c_{1}^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′(c_{1})

= c_{2}^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′(c_{2}) > 0.

Observing that λx^{n−1}y(x) ≤ 0 and x^{n}y′(x) > 0, we have λx^{n−1}y(x) − x^{n}y′(x) < 0. Hence

∫_{c_{1}}^{c_{2}} λx^{n−1}y(x) − x^{n}y′(x)dx ≤ 0.

The contradiction shows y ≥ 0.

Similarly, we can prove y′(x) ≤ 0. Otherwise, if there exists x^{∗} ∈ (0, +∞) such that y′(x^{∗}) > 0. By y(0) = 1, y(+∞) = 0, we know that there would exist a x_{1} ∈ (0, +∞) with y(x_{1}) being a maximum value. Hence, we have y′(x_{1}) = 0. By y(x_{1}) > 0 which is a maximum value, there exists a point x_{2} ∈ [x_{1}, +∞), satisfying y(x_{2}) = \frac{1}{2}y(x_{1}). The mean value theorem yields,

y(x_{2}) − y(x_{1}) = y′(b)(x_{2} − x_{1}),

where b ∈ [x_{1}, x_{2}]. By y(x_{2}) − y(x_{1}) < 0 and x_{2} − x_{1} > 0, therefore we have y′(b) < 0. we take [c_{1}, c_{2}] ⊂ [a_{1}, a_{2}] such that y′(c_{1}) = 0 and y(x) > 0, y′(x) < 0 in (c_{1}, c_{2}). Integrating the equation (1.4) with respect to x over (c_{1}, c_{2}), we see that

∫_{c_{1}}^{c_{2}} [x^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′]′ dx

= c_{2}^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′(c_{2})
− c_{1}^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′(c_{1})

= c_{2}^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′(c_{2}) < 0.

Observing that λx^{n−1}y(x) > 0 and −x^{n}y′(x) > 0, then λx^{n−1}y(x) − x^{n}y′(x) > 0. Thus we have

∫_{c_{1}}^{c_{2}} λx^{n−1}y(x) − x^{n}y′(x)dx ≥ 0.

The contradiction shows y′ ≤ 0. □

Proposition 2.2. If the function y(x) ∈ C^{1}[0, +∞) is a solution of the problem (1.4)-(1.5). Then y(x) ≥ 0, y′(x) ≤ 0 as −n ≤ λ < 0.

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

Lemma 2.1. If there exists a x_{0} satisfies y′(x_{0}) = 0, y(x_{0}) > 0. Then, we have y(x) ≥ 0, y′(x) ≤ 0 in [x_{0}, +∞), for −n ≤ λ < 0.

Proof. We first prove y(x) ≥ 0. Otherwise, if there exists a point a_{1} such that y(a_{1}) < 0. By y(+∞) = 0, we know that there would exist a point x_{1} and y(x_{1}) < 0 is a minimum value. Hence, we have y′(x_{1}) = 0. Integrating the equation (1.4) with respect to x over (x_{1}, x_{2}), and using y′(x_{0}) = 0, y′(x_{1}) = 0 and y(x_{0}) > 0, y(x_{1}) < 0, we see that

∫_{x_{0}}^{x_{1}} [x^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′]′ dx

= x_{1}^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′(x_{1})

− x_{0}^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′(x_{0}) = 0, and

∫_{x_{0}}^{x_{1}} λx^{n−1}y(x) − x^{n}y′(x)dx

= ∫_{x_{0}}^{x_{1}} λx^{n−1}y(x)dx + ∫_{x_{0}}^{x_{1}} nx^{n−1}y(x)dx − x^{n}y|_{x_{0}}^{x_{1}}

> (λ + n)\frac{1}{n}y(x_{1})(x_{1}^{n} − x_{0}^{n}) + x_{0}^{n}y(x_{0}) − x_{1}^{n}y(x_{1})

= (\frac{λ}{n} + 1)y(x_{1})x_{1}^{n} − (\frac{λ}{n} + 1)y(x_{1})x_{0}^{n} + x_{0}^{n}y(x_{0}) − x_{1}^{n}y(x_{1})

= \frac{λ}{n}y(x_{1})x_{1}^{n} − (\frac{λ}{n} + 1)y(x_{1})x_{0}^{n} + x_{0}^{n}y(x_{0}) > 0.

The contradiction shows y ≥ 0. Similarly, we can prove y′(x) ≤ 0. □

Lemma 2.2. If there exists a x_{0} satisfies y′(x_{0}) = 0, y(x_{0}) < 0, then, we have y(x) ≤ 0, y′(x) ≥ 0 in [x_{0}, +∞), for −n ≤ λ < 0.

Proof. For simplicity, we only prove the first inequality y(x) ≤ 0, since the other can be shown similarly. Otherwise, if there exists a_{1} such that y(a_{1}) > 0, by y(+∞) = 0, we know that there would exist a x_{1} and y(x_{1}) < 0 is a maximum value. Hence, we have y′(x_{1}) = 0. By y′(x_{0}) = 0, y′(x_{1}) = 0 and y(x_{0}) > 0, y(x_{1}) < 0. Integrating the equation (1.4) with respect to x over (x_{1}, x_{2}), we see that

∫_{x_{0}}^{x_{1}} [x^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′]′ dx

= x_{1}^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′(x_{1})
− x_{0}^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′(x_{0}) = 0

and

∫_{x_{0}}^{x_{1}} λx^{n−1}y(x) − x^{n}y′(x)dx

= ∫_{x_{0}}^{x_{1}} λx^{n−1}y(x)dx + ∫_{x_{0}}^{x_{1}} nx^{n−1}y(x)dx − x^{n}y|_{x_{0}}^{x_{1}}

< (λ + n)\frac{1}{n}y(x_{1})(x_{1}^{n} − x_{0}^{n}) + x_{0}^{n}y(x_{0}) − x_{1}^{n}y(x_{1})

= (\frac{λ}{n} + 1)y(x_{1})x_{1}^{n} − (\frac{λ}{n} + 1)y(x_{1})x_{0}^{n} + x_{0}^{n}y(x_{0}) − x_{1}^{n}y(x_{1})

= \frac{λ}{n}y(x_{1})x_{1}^{n} − (\frac{λ}{n} + 1)y(x_{1})x_{0}^{n} + x_{0}^{n}y(x_{0}) < 0

The contradiction shows y ≤ 0. □

Proof of Proposition 2.2. We first prove that y(x) ≥ 0. Otherwise, if there exists x^{∗} ∈ (0, +∞) such that y(x^{∗}) < 0, by y(0) = 1, y(+∞) = 0, we know that there would exist a a_{1} ∈ (0, +∞) and y(a_{1}) is a minimum value. Therefore, we have y′(a_{1}) = 0. By lemma 2.2, we have y′(x) ≥ 0, y(x) ≤ 0 in [a_{1}, +∞) . Multiplying both sides of the equation (1.4) by −x^{−λ−1}, we have

−x^{−λ−n} [x^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′]′ = [x^{−λ}y(x)]′  (2.1)

Then, integrating the resulting relation with respect to x over (a_{1}, M), we have

∫_{a_{1}}^{M} −x^{−λ−n} [x^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′]′ dx

= −x^{−λ−n}x^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′|_{x_{2}}^{M}
− (λ + n) ∫_{a_{1}}^{M} x^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′x^{−λ−n−1}dx
= M^{−λ+α−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′(M)

− (λ + n) ∫_{a_{1}}^{M} x^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′x^{−λ−n−1}dx ≤ 0

However, the right hand side

∫_{a_{1}}^{M}(x^{−λ}y)′dx = M^{−λ}y(M) − a_{1}^{−λ}y(a_{1}).

By lim x^{−λ}y(x) = 0, we have the right hand side > 0, as M → +∞. The contradiction shows y ≤ 0.

Similarly, we can prove y′(x) ≥ 0. □

Proposition 2.3. Suppose the function y(x) ∈ C^{1}[0, +∞) is a solution of the problem (1.4)-(1.5). Then a.e. y′(x) < 0, as y(x) > 0, where λ ≠ 0.

Proof. If there exists a point x_{0} ∈ [0, +∞), such that y(x_{0}) > 0 and y′(x_{0}) = 0. Without loss of generality, we assume there exists a strictly monotone sequence {x_{j}}_{j=1}^{∞} such that lim_{j→+∞} x_{j} = x_{0}, y(x_{j}) > 0, y′(x_{j}) = 0. Set f(x_{j}) = x_{j}^{α+n−1} |(|y|^{m−1}y(x_{j}))′|^{N−1} (|y|^{m−1}y(x_{j}))′. Hence

0 = \frac{f(x_{j+1}) − f(x_{j})}{x_{j+1} − x_{j}} = [x^{α+n−1} |(|y|^{m−1}y)′|^{N−1} (|y|^{m−1}y)′]′ |_{x=ξ_{j}}

= λξ_{j}^{n−1}y(ξ_{j}) − ξ_{j}^{n}y′(ξ_{j}), where ξ_{j} ∈ [x_{j}, x_{j+1}]. Letting j → +∞, we have y(x_{0}) = 0. The contradiction shows y′ < 0. □

3. Self-similar solution

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

v(t) = mt^{m−1}W^{α+n−1}|W′|^{−N},

and consider the following problem

\frac{dv}{dt} = −λtW^{n−1}W′ + W^{n},  (3.1)

\frac{dW^{(N−n+α+1)/N}}{dt} = −\frac{N − n + α + 1}{N}m^{1/N}t^{(m−1)/N}v^{−1/N}(t),  (3.2)
v(0) = 0, W (1) = 0.  (3.3)

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

v(0) = h > 0, W (1) = 0.  (3.4)

Theorem 3.1. For fixed h > 0, the problem (3.1), (3.2), (3.4) admits a solution v(t, h).

Proof. If v(t) ∈ C[0, 1] satisfies

v(t) = λm^{1/N} ∫_{0}^{t} τ^{(N+m−1)/N}v^{−1/N}(τ)

(\frac{N − n + α + 1}{N}m^{1/N} ∫_{τ}^{1} s^{(m−1)/N}v^{−1/N}(s)ds)^{(Nn−N+α+n−1)/(N−1)} dτ

+ ∫_{0}^{t} (\frac{N − n + α + 1}{N}m^{1/N} ∫_{τ}^{1} s^{(m−1)/N}v^{−1/N}(s)ds)^{Nn/(N−1)} dτ + h (3.5)

and let

W (t) = (\frac{N − n + α + 1}{N}m^{1/N} ∫_{t}^{1} s^{(m−1)/N}v^{−1/N}(s)ds)^{N/(N−1)} . (3.6)

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

On the other hand, clearly, for any a ∈ [0, 1], we have

v(t) = λm^{1/N} ∫_{a}^{t} τ^{(N+m−1)/N}v^{−1/N}(τ)

(\frac{N − n + α + 1}{N}m^{1/N} ∫_{τ}^{1} s^{(m−1)/N}v^{−1/N}(s)ds)^{(Nn−N+α+n−1)/(N−1)} dτ

+∫_{a}^{t} (\frac{N − n + α + 1}{N}m^{1/N} ∫_{τ}^{1} s^{(m−1)/N}v^{−1/N}(s)ds)^{Nn/(N−1)} dτ +v(a). (3.7)

Define the map φ : Ω → Ω,

φv = λm^{1/N} ∫_{0}^{t} τ^{(N+m−1)/N}v^{−1/N}(τ)

(\frac{N − n + α + 1}{N}m^{1/N} ∫_{τ}^{1} s^{(m−1)/N}v^{−1/N}(s)ds)^{(Nn−N+α+n−1)/(N−1)} dτ

+ ∫_{0}^{t} (\frac{N − n + α + 1}{N}m^{1/N} ∫_{τ}^{1} s^{(m−1)/N}v^{−1/N}(s)ds)^{Nn/(N−1)} dτ + h,

where Ω = {v(t) ∈ C[0, 1]; h ≤ v(t) ≤ (φh)(t)}. It is seen that the operator φ is Ω to Ω continuous and compact. By Leray–Schauder principle of fixed point, the operator φ has a fixed point v(t, h) in Ω, which is the desired solution of the (3.5). Hence it is the solutions of the problem (3.1), (3.2), (3.4). □

Lemma 3.1. If h_{1} > h_{2} > 0, then

0 ≤ v(t, h_{1}) − v(t, h_{2}) ≤ h_{1} − h_{2}, on [0, 1].

Proof. We first show the left inequality. If this were not true, then there will be a point t_{0} ∈ [0, 1], such that

v(t_{0}, h_{1}) − v(t_{0}, h_{2}) < 0.

By v(0, h_{1}) − v(0, h_{2}) = h_{1} − h_{2} > 0, hence t_{0} ≠ 0.

Since t_{0} ≠ 0, then there exists a interval (a, t_{0}], where 0 < a < t_{0} ≤ 1, such that

v(t, h_{1}) − v(t, h_{2}) < 0, v(a, h_{1}) − v(a, h_{2}) = 0 in(a, t_{0}].

Using (3.7), we obtain v(t_{0}, h_{1})−v(t_{0}, h_{2}) = 0, this yields a contradiction. We can obtain the right inequality, by the left inequality and (3.5). □

By the lemma 3.1, we have the following conclusion.

Lemma 3.2. The problem (3.1), (3.2), (3.4) has a unique solution.

Theorem 3.2. The problem (3.1), (3.2), (3.3) has a unique solution v(t, h).

Proof. By the lemma 3.1, we know that

lim_{h→0} v(t, h) = v(t) uniformly for t ∈ [0, 1].

Substituting v(t, h) into (3.5), then letting h → 0, using lemma 3.2, we have the problem (3.1)-(3.3) admits one and only one nonnegative continuous solution v(t). □

Theorem 3.3. The problem (1.4), (1.5) has a unique solution.

Proof. Now, we construct a solution of the problem (1.4)-(1.5) by the solution of the problem (3.1)-(3.3). Suppose (v(t), W (t)) is a solution of the problem (3.1)-(3.3), then W (t) is a strictly decreasing function in [0, 1]. Hence the inverse function t = y(x) of x = W (t) exists in [0, W (0)). If W (0) < +∞, we define the y(x) = 0, as x ∈ [W (0), +∞). Hence the y(x) is a nonnegative function with y(0) = 1, lim_{x→+∞} x^{[−λ]_{+}}y(x) = 0, in [0, +∞). We set x_{0} = W (0). Observing that x = W (t) in [0, x_{0}) and y′(x) = \frac{1}{W′(t)} < 0 a.e. in (0, x_{0}). As x_{0} < +∞, we have y′_{−}(x_{0}) = \frac{1}{W′_{+}(0)} = 0. Again as x ∈ [x_{0}, +∞), y(x) = 0, hence, we know that y′(x) continuous at x_{0} and y′(x_{0}) = 0. As x_{0} = +∞, by (3.6), we have y′(x) = −v^{1/N}(t)/[(N − n + α − 1)/(N − 1)W^{1/N}]. Again by v(0) = 0 and W (0) = +∞, we know that y′(+∞) = 0. Substituting t = y(x) in (1.4), (1.5), it is easily seen that the function y(x) ∈ C^{1}[0, +∞) is the solution of the problem (1.4), (1.5). □

By the Theorem 3.3, we have the following conclusion.

Theorem 3.4. If m > 1, N > 1, γ > 0, u_{0} > 0, \frac{γ}{β} ≥ −n. Then the problem (1.1)–(1.3) admits one and only one nonnegative self-similar solution

u(r, t) = u_{0}t^{α}y(crt^{−β}),

where y(x) ∈ C^{1}[0, +∞) is a unique nonnegative solution of the problem (1.4), (1.5), c, β are defined in Section 1.

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Changchun Liu, Department of Mathematics, Jilin University, Changchun 130012, P. R. China E-mail address: lcc@email.jlu.edu.cn

Received June 8, 2006