existence and some properties of weak solutions for a singular nonlinear parabolic equation

MATHEMATICAL METHODS IN THE APPLIED SCIENCESMath. Meth. Appl. Sci. 2007; 30:1329–1353Published online 1 March 2007 in Wiley InterScience(www.interscience.wiley.com) DOI: 10.1002/mma.851MOS subject classification: 35K 57; 35K 65

Existence and some properties of weak solutions for a singularnonlinear parabolic equation

Liu Qiang1, Yao Zheng’an1 and Zhou Wenshu2,∗,†

1Department of Mathematics, Sun Yat-Sen University, Guangzhou 510275, China2Department of Mathematics, Jilin University, Changchun 130012, China

Communicated by M. Fila

SUMMARY

This paper deals with the initial and boundary value problem for a singular nonlinear parabolic equa-tion. The existence of solutions is established by parabolic regularization. Some properties of solutions,for instance localization and large time behaviour are also discussed. Copyright q 2007 John Wiley &Sons, Ltd.

KEY WORDS: singular parabolic equation; existence; localization; asymptotic behaviour

1. INTRODUCTION

This paper studies a singular nonlinear parabolic equation

�u�t

=�(u) div(|∇u|p−2∇u) in Q = �× (0,∞) (1)

with the Dirichlet boundary condition

u(x, t) = 0 on ��× (0,∞) (2)

and the initial condition

u(x, 0) = u0(x) in � (3)

∗Correspondence to: Zhou Wenshu, Department of Mathematics, Jilin University, Changchun 130012, China.†E-mail: [email protected]

Contract/grant sponsor: NNSFC; contract/grant numbers: 10171113, 10471156, 10626056Contract/grant sponsor: NSFGD; contract/grant number: 4009793

Copyright q 2007 John Wiley & Sons, Ltd. Received 31 March 2006

1330 L. QIANG, Y. ZHENG’AN AND Z. WENSHU

where p>1, � ⊂ RN (N�1) is an open bounded domain with appropriately smooth boundary��, u0�0 and �(u) satisfies the following condition:

(H1) �(u) ∈C1[0,+∞), �(0) = 0 and for any u>0, �′(u)>0

Note that for any �(u) satisfying (H1), 1/�(u) is not integrable near u = 0. A typical example is�(u) = uq with q�1.

The type of equation (with some source term added) arises in biological (see [1]) and astro-physical (see [2, 3]) context. Similar problems can be found in [4–6]. Equation (1) appears alsoin some models describing physical phenomena. For example, Equation (1) is the evolution p-Laplacian when �(u)≡ 1. If �(u)= uq with q ∈ (0, 1), then Equation (1) can be written in di-vergence form after the change of variables (provided m(p − 1) − 1 �= 0): u =mm/[m(p−1)−1]vm ,m = 1/(1 − q), which follows the filtration equation:

�v

�t= div(|∇vm |p−2∇vm)

This equation has been studied extensively, see [7–9] and references therein. We point out that forq�1 and �(u) = uq , Equation (1) does not correspond to any real parabolic equation in divergenceform. From mathematical point of view, it is very necessary and interesting to study it.

Many authors studied Equation (1) with p= 2 and �(u) = uq (q�1) and obtained some resultsin a series of papers (see [1–6, 10–14]). For example, non-uniqueness of solutions is found inde-pendently by Dal Passo and Luckhaus [4] and Ughi [6]. However, there are few papers dealing withEquation (1) with p �= 2. Zhou and Wu considered in [15] the case p>2 and �(u) = uq with q�1and announced some results including existence and localization of solutions. For some technicaldifficulties, however, they did not discuss the existence for the case 1<p<2 and large time be-haviour of solutions. The present paper is devoted to Equation (1) of general form, in particular, thecase 1<p<2. By some similar ideas used in [4, 15], we consider weak solutions since Equation (1)may degenerate at any points where u = 0 or |∇u| = 0 and must overcome some difficulties causedby the nonlinearity and double degeneracies of the equation. As seen below, we present a conceptof weak solutions which seems to be fairly natural. However, weak solutions may not be uniquelydetermined by the initial value (see [4, 6, 15]). We show that the weak solutions of regularization ofproblem (1)–(3) are unique (see Section 3). We remark that although the authors of [4, 6] presenteda uniqueness discussion for p= 2, it seems to be very difficult to do a similar discussion for thegeneral case p �= 2 since the p-Laplacian operator has very strong nonlinearity. Hence, here weuse an entirely different idea to discuss this problem. In the present paper, our main purpose isto show the existence of weak solutions and we also study some properties of weak solutions,for instance, localization property and asymptotic behaviour. Now, we give the concept of weaksolutions.

For T>0, denote �T = �× (0, T ).

Definition 1.1A non-negative function u is called a weak solution of problem (1)–(3), if for any T>0 thefollowing conditions hold:

(a) u ∈ L∞(�T ) ∩ L p(0, T ;W 1,p0 (�)) with

�u�t

∈ L2(�T )

Copyright q 2007 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2007; 30:1329–1353DOI: 10.1002/mma

WEAK SOLUTIONS FOR A SINGULAR NONLINEAR PARABOLIC EQUATION 1331

(b)∫ ∫

�T

(−u

��

�t+ �(u)|∇u|p−2∇u∇� + �′(u)|∇u|p�

)dx dt = 0 ∀� ∈C∞

0 (�T )

(c)∫

�|u(x, t) − u0(x)| dx → 0 (t → 0+)

Remark 1.1To define weak solutions for the problem with arbitrary non-negative function �∈ L∞(�T ) ∩L p(0, T ;W 1,p(�)) as its boundary value, we require u−�∈ L∞(�T ) ∩ L p(0, T ;W 1,p

0 (�)) instead

of u ∈ L∞(�T ) ∩ L p(0, T ;W 1,p0 (�)).

Sometimes, we need to use the concept of weak sub-solutions (sup-solutions) for (1). To definethe weak sub-solutions (sup-solutions), it suffices to replace ‘=’ in (b) by ‘�’ (‘�’) and require� to be non-negative and u ∈ L∞(�T ) ∩ L p(0, T ;W 1,p(�)) with �u/�t ∈ L2(�T ).

Remark 1.2For any weak solution u, the integral identity (b) in Definition 1.1 can be rewritten as

∫ ∫�T

(�u�t

� + �(u)|∇u|p−2∇u∇� + �′(u)|∇u|p�)dx dt = 0 ∀� ∈C∞

0 (�T )

Furthermore, from the denseness of C∞0 (�T ) in L p(0, T ;W 1,p

0 (�)) it follows that the above

integral equality holds for any �∈ L∞(�T ) ∩ L p(0, T ;W 1,p0 (�)).

The rest of this paper is organized as follows. In Section 2, we show that problem (1)–(3) admitsat least one weak solution. In Section 3, we present a uniqueness discussion. The localization andasymptotic behaviour of weak solutions are studied in Sections 4 and 5, respectively.

In the sequel, the following lemma will be used repeatedly.

Lemma 1.1Assume p>1. Then, there exist positive constants C1, C2 depending only on p, such that for any�, �′ ∈ RN , the following inequalities hold:

(|�|p−2� − |�′|p−2�′)(� − �′) �C1(|�| + |�′|)p−2|� − �′|2 (4)

||�|p−2� − |�′|p−2�′| �C2(|�| + |�′|)p−2|� − �′| (5)

and

||�|p−2� − |�′|p−2�′| �C2(|�| + |�′|)[(m+1/2)p−(m+1)]/(m+1)

× ((|�| + |�′|)(p−2)/2|� − �′|)1/(m+1) (6)

for any positive constant m satisfying (m + 1/2)p − (m + 1)>0.



ProofThe proof of (4) and (5) could be found in [16]. We prove (6). For any positive constant msatisfying (m + 1/2)p − (m + 1)>0, we have

(|�| + |�′|)p−2|� − �′|= ((|�| + |�′|)(m+1/2)(p−2)|� − �′|m)1/(m+1)[(|�| + |�′|)(p−2)/2|� − �′|]1/(m+1)

�(|�| + |�′|)[(m+1/2)p−(m+1)]/(m+1)((|�| + |�′|)(p−2)/2|� − �′|)1/(m+1)

which implies (6) by using (5). �

2. EXISTENCE OF WEAK SOLUTIONS

In this section, we prove the existence of weak solutions of problem (1)–(3) by parabolic regulari-zation. It is obtained by using two limit processes. The main theorem is as follows.

Theorem 2.1Let 0�u0 ∈C(�) ∩W 1,p

0 (�) and (H1) be satisfied. Then, there exists at least one weak solutionof problem (1)–(3) in the sense of Definition 1.1.

To prove Theorem 2.1, we first consider the regularized equation

�uε,�

�t=�(uε,�) div((|∇uε,�|2 + �)(p−2)/2∇uε,�) in �T (7)

with the boundary value

uε,�(x, t) = ε on �� × (0, T ) (8)

and the initial value

uε,�(x, 0) = u0(x) + ε in � (9)

where ε ∈ (0, 1) and � ∈ (0,�(ε)).According to the standard quasi-linear parabolic equation theory, cf. Theorem 4.1 of Chapter VI

in [17], there exists a classical solution uε,� of problem (7)–(9) in the space C2,1(�T ). Moreover,by the comparison theorem, it is easy to see that

uε2,��uε1,� for 1�ε2>ε1>0, ‖u0‖L∞(�) + ε�uε,��ε (10)

Remark 2.1To prove the existence of a classical solution uε,� of the regularized problem (7)–(9), one can use,for instance, Theorem 4.1 of Chapter VI in [17]. We only need to replace the factor �(uε,�) inEquation (7) by �(g(uε,�)), where g satisfies

g(s) ∈ C1(R), 0<C1�g(s)�C2, g′(s) → 0 as s → + ∞

g(s) = s for ε�s�‖u0‖∞ + ε



Then, one can consider the new equation instead of (7)

�uε,�

�t=�(g(uε,�)) div((|∇uε,�|2 + �)(p−2)/2∇uε,�) in �T (7)′

with the boundary value (8) and the initial value (9). The maximum principle implies that thesolution of problem (7)′, (8), and (9) is just the solution of problem (7)–(9). All conditions (a)–(c)in Theorem 4.1 can be verified easily. Since the initial and boundary conditions assumed here donot satisfy condition (d) in Theorem 4.1, a suitable smooth approximation is required.

First, we give some estimates for uε,�.

Lemma 2.1For any �∈ (0, T ), the following estimates hold:∫

�(|∇uε,�(x, �)|2 + �)p/2 dx +

∫ ∫��

(�uε,�

�t

)2

[�(uε,�)]−1 dx dt�C (11)

and ∫ ∫��

�′(uε,�)(|∇uε,�|2 + �)p/2

[�(uε,�)]� dx dt�C if �′(0)>0 (12)

where � ∈ (0,min{1, p/2}), C is a constant independent of �, ε.

ProofMultiplying (7) by �uε,�/�t[�(uε,�)]−1 and integrating over ��, we have

∫ ∫��

(�uε,�

�t

)2

[�(uε,�)]−1 dx dt

=∫ ∫

��

div((|∇uε,�|2 + �)(p−2)/2∇uε,�)�uε,�

�tdx dt

=∫ ∫

��

div

((|∇uε,�|2 + �)(p−2)/2∇uε,�

�uε,�

�t

)dx dt

−∫ ∫

��

(|∇uε,�|2 + �)(p−2)/2∇uε,� · ∇(

�uε,�

�t

)dx dt

=∫ �

0

∫��

(|∇uε,�|2 + �)(p−2)/2 �uε,�

�t�uε,�

��d� dt − 1

p

∫ ∫��

��t

(|∇uε,�|2 + �)p/2 dx dt

=− 1

p

∫�(|∇uε,�(x, �)|2 + �)p/2 dx + 1

p

∫�(|∇u0(x)|2 + �)p/2 dx

where � denotes the outward normal to �� × (0, T ). Then, it shows that (11) is valid.



To prove (12), we multiply (7) by [�(uε,�)]−�, integrate over �T and get

∫ ∫��

�uε,�

�t[�(uε,�)]−� dx dt

=∫ ∫

��

div(|∇uε,�|2 + �)(p−2)/2∇uε,�)[�(uε,�)]1−� dx dt

=∫ ∫

��

div(|∇uε,�|2 + �)(p−2)/2∇uε,�[�(uε,�)]1−�) dx dt

−∫ ∫

��

(|∇uε,�|2 + �)(p−2)/2∇uε,�∇[�(uε,�)]1−� dx dt

=∫ �

0

∫��

[�(uε,�)]1−�(|∇uε,�|2 + �)(p−2)/2 �uε,�

��d� dt

− (1 − �)

∫ ∫��

�′(uε,�)(|∇uε,�|2 + �)(p−2)/2|∇uε,�|2[�(uε,�)]� dx dt

For (10), uε,��ε, �uε,�/��0 on ��× (0, T ). Hence,

∫ ∫��

�′(uε,�)(|∇uε,�|2 + �)(p−2)/2|∇uε,�|2[�(uε,�)]� dx dt� 1

1 − �

∫�

∫ u0(x)+ε

0[�(s)]−� ds dx (13)

Since �′(0)>0, �′(u)>0 on [0,∞), we have M = min{�′(u); u ∈ [0, ‖u0‖L∞ + 1]}>0. On theother hand, for any s ∈ (0, ‖u0‖L∞ + 1] there exists ∈ (0, s) such that �(s) = �′()s�Ms. Sofor any ∈ (0, 1), 1/[�(s)] is integrable on [0, ‖u0‖L∞ + 1], and hence the right side of (13) isuniformly bounded.

Now, let us estimate the upper bound of I defined by

I =∫ ∫

��

��′(uε,�)(|∇uε,�|2 + �)(p−2)/2[�(uε,�)]−� dx dt

If p�2, using (10) and Young’s inequality and noting uε,��ε, �<�(ε), 0<�<p/2, we obtain

I =∫ ∫

��

��′(uε,�)(|∇uε,�|2 + �)(p−2)/2�(uε,�)−� dx dt

� C�∫ ∫

��

[�′(uε,�)](p−2)/p(|∇uε,�|2 + �)(p−2)/2[�(uε,�)]−� dx dt

= C∫ ∫

��

[�′(uε,�)](p−2)/p(|∇uε,�|2 + �)(p−2)/2

[�(uε,�)]�(p−2)/p

�

[�(uε,�)]2�/pdx dt



� C∫ ∫

��

[�′(uε,�)](p−2)/p(|∇uε,�|2 + �)(p−2)/2

[�(uε,�)]�(p−2)/pdx dt

� 1

2

∫ ∫��

�′(uε,�)(|∇uε,�|2 + �)p/2

[�(uε,�)]� dx dt + C (14)

If 1<p<2, using (10) and noting that uε,��ε, �<�(ε) and 0<�<p/2 yield

I =∫ ∫

��

��′(uε,�)(|∇uε,�|2 + �)(p−2)/2[�(uε,�)]−� dx dt

� �C∫ ∫

��

(|∇uε,�|2 + �)(p−2)/2[�(uε,�)]−� dx dt

=C∫ ∫

��

(�

|∇uε,�|2 + �

)(2−p)/2 �p/2

[�(uε,�)]� dx dt

�C∫ ∫

��

�p/2

[�(uε,�)]� dx dt�C∫ ∫

��

dx dt�C (15)

Combining (13), (14) and (15) yields∫ ∫��

�′(uε,�)(|∇uε,�|2 + �)p/2[�(uε,�)]−� dx dt

=∫ ∫

��

�′(uε,�)(|∇uε,�|2 + �)(p−2)/2|∇uε,�|2[�(uε,�)]−� dx dt

+∫ ∫

��

��′(uε,�)(|∇uε,�|2 + �)(p−2)/2[�(uε,�)]−� dx dt

�1

2

∫ ∫��

�′(uε,�)(|∇uε,�|2 + �)p/2[�(uε,�)]−� dx dt + C

Thus (12) holds and the proof of Lemma 2.1 is completed. �

Equation (10) and Lemma 2.1 imply that for any fixed ε ∈ (0, 1), there exists a subsequence ofuε,�, denoted by uε,�k , and a function uε with uε − ε ∈ L∞(�T ) ∩ L p(0, T ;W 1,p

0 (�)), such thatas � = �k → 0,

uε,� → uε a.e. in �T (16)

∇uε,� ⇀ ∇uε in L p(�T ) (17)

�uε,�

�t⇀

�uε

�tin L2(�T ) (18)

where ⇀ denotes weak convergence.



Lemma 2.2As � = �k → 0, there hold

∫ ∫�T

(|∇uε,�| + |∇uε|)p−2|∇uε,� − ∇uε|2 dx dt → 0 (19)

∫ ∫�T

||∇uε,�|p − |∇uε|p| dx dt → 0 (20)

∫ ∫�T

|�′(uε,�)(|∇uε,�|2 + �)(p−2)/2|∇uε,�|2 − �′(uε)|∇uε|p| dx dt → 0 (21)

∫ ∫�T

|�(uε,�)(|∇uε,�|2 + �)(p−2)/2∇uε,� − �(uε)|∇uε|p−2∇uε| dx dt → 0 (22)

ProofFirst, we prove (19). For (uε,�−uε)/(�(uε,�))∈L p(0, T ;W 1,p

0 (�)), multiplying (7) by (uε,�−uε)/

(�(uε,�)), integrating both sides of the equality over �T and integrating by parts, we have

∫ ∫�T

(�uε,�

�tuε,� − uε

�(uε,�)+ (|∇uε,�|2 + �)(p−2)/2∇uε,�∇(uε,� − uε)

)dx dt = 0

So

∫ ∫�T

�uε,�

�tuε,� − uε

�(uε,�)dx dt

+∫ ∫

�T

((|∇uε,�|2 + �)(p−2)/2∇uε,� − |∇uε,�|p−2∇uε,�)∇(uε,� − uε) dx dt

+∫ ∫

�T

(|∇uε,�|p−2∇uε,� − |∇uε|p−2∇uε)∇(uε,� − uε) dx dt

=−∫ ∫

�T

|∇uε|p−2∇uε∇(uε,� − uε) dx dt (23)

Obviously, (17) implies that

∫ ∫�T

|∇uε|p−2∇uε∇(uε,� − uε) dx dt → 0 (�= �k → 0) (24)



By virtue of Holder’s inequality, Lemma 2.1 and noting uε,��ε, we obtain

∣∣∣∣∫ ∫

�T

�uε,�

�tuε,� − uε

�(uε,�)dx dt

∣∣∣∣

�(∫ ∫

�T

(�uε,�

�t

)2

[�(uε,�)]−1 dx dt

)1/2 (∫ ∫�T

(uε,� − uε)2

�(uε,�)dx dt

)1/2

�C(ε)

(∫ ∫�T

(uε,� − uε)2 dx dt

)1/2

→ 0 (�= �k → 0) (25)

Next, we define J by

J =∫ ∫

�T

((|∇uε,�|2 + �)(p−2)/2∇uε,� − |∇uε,�|p−2∇uε,�)∇(uε,� − uε) dx dt

If p�2, by the inequality

|ar − br |�|a − b|r ∀a, b�0, r ∈ [0, 1] (26)

and Lemma 2.1 we get

|J | �∫ ∫

�T

|(|∇uε,�|2 + �)(p−2)/2∇uε,� − |∇uε,�|p−2∇uε,�||∇(uε,� − uε)| dx dt

=∫ ∫

�T

|(|∇uε,�|2 + �)(p−2)/2 − |∇uε,�|p−2||∇uε,�||∇(uε,� − uε)| dx dt

�∫ ∫

�T

|(|∇uε,�|2 + �)(p−1)/2 − |∇uε,�|p−1||∇(uε,� − uε)| dx dt

� (p − 1)∫ ∫

�T

(|∇uε,�|2 + �)(p−2)/2|(|∇uε,�|2 + �)1/2 − |∇uε,�|||∇(uε,� − uε)| dx dt

� (p − 1)�1/2∫ ∫

�T

(|∇uε,�|2 + �)(p−2)/2|∇(uε,� − uε)| dx dt

→ 0 (� = �k → 0)

If 1<p<2, then 0<(p − 1)/2<1. It follows from the inequality (26) that

|J | �∫ ∫

�T

|(|∇uε,�|2 + �)(p−2)/2∇uε,� − |∇uε,�|p−2∇uε,�||∇(uε,� − uε)| dx dt

=∫ ∫

�T

|(|∇uε,�|2 + �)(p−2)/2|∇uε,�|2−p − 1|||∇uε,�|p−2∇uε,�||∇(uε,� − uε)| dx dt



=∫ ∫

�T

|(|∇uε,�|2 + �)(p−2)/2|∇uε,�| − |∇uε,�|p−1||∇(uε,� − uε)| dx dt

�∫ ∫

�T

|(|∇uε,�|2 + �)(p−1)/2 − |∇uε,�|p−1||∇(uε,� − uε)| dx dt

+∫ ∫

�T

|(|∇uε,�|2 + �)1/2 − |∇uε,�||(|∇uε,�|2 + �)(p−2)/2|∇(uε,� − uε)| dx dt

� �(p−1)/2∫ ∫

�T

|∇(uε,� − uε)| dx dt

+ �1/2∫ ∫

�T

(|∇uε,�|2 + �)(p−2)/2|∇(uε,� − uε)| dx dt

= �(p−1)/2∫ ∫

�T


+ �(p−1)/2∫ ∫

�T

(�

|∇uε,�|2 + �

)(2−p)/2


� 2�(p−1)/2∫ ∫

�T


→ 0 (�= �k → 0)

Thus for any p>1 there holds

J → 0 (�= �k → 0) (27)

Combining (23)–(25) with (27) yields

lim sup�=�k→0

∫ ∫�T

(|∇uε,�|p−2∇uε,� − |∇uε|p−2∇uε)∇(uε,� − uε) dx dt�0

From (4), we obtain (19).Now, we prove (20). From Holder’s inequality, Lemma 2.1 and (19), we derive∫ ∫

�T

||∇uε,�|p − |∇uε|p| dx dt

�p∫ ∫

�T

(|∇uε,�| + |∇uε|)p−1|∇uε,� − ∇uε| dx dt

= p∫ ∫

�T

(|∇uε,�| + |∇uε|)p/2(|∇uε,�| + |∇uε|)(p−2)/2|∇uε,� − ∇uε| dx dt

�p

(∫ ∫�T

(|∇uε,�|+|∇uε|)p dx dt)1/2(∫ ∫

�T

(|∇uε,�|+|∇uε|)p−2|∇uε,�−∇uε|2 dx dt)1/2



�C

(∫ ∫�T

(|∇uε,�| + |∇uε|)p−2|∇uε,� − ∇uε|2 dx dt)1/2

→ 0 (� = �k → 0)

To prove (21), we note∫ ∫�T

|�′(uε,�)(|∇uε,�|2 + �)(p−2)/2|∇uε,�|2 − �′(uε)|∇uε|p| dx dt

�∫ ∫

�T

�′(uε,�)|(|∇uε,�|2 + �)(p−2)/2|∇uε,�|2 − |∇uε,�|p| dx dt

+∫ ∫

�T

�′(uε,�)||∇uε,�|p − |∇uε|p| dx dt

+∫ ∫

�T

|�′(uε,�) − �′(uε)||∇uε|p dx dt

= J1 + J2 + J3

Obviously, (16) implies that

J3 → 0 (�= �k → 0)

From (20), it follows that

J2�C∫ ∫

�T

||∇uε,�|p − |∇uε|p| dx dt → 0 (� = �k → 0)

Let us now estimate J1. If p�2, it follows from (10) that

J1 � C∫ ∫

�T

|(|∇uε,�|2 + �)(p−2)/2|∇uε,�|2 − |∇uε,�|p| dx dt

� C∫ ∫

�T

|(|∇uε,�|2 + �)p/2 − |∇uε,�|p| dx dt

� �pC

2

∫ ∫�T

(|∇uε,�|2 + �)(p−2)/2 dx dt

→ 0 (� = �k → 0)

If 1<p<2, then 0<p/2<1. Using inequality (26) yields

J1 �C∫ ∫

�T

|(|∇uε,�|2 + �)(p−2)/2|∇uε,�|2 − |∇uε,�|p| dx dt

�C∫ ∫

�T

|(|∇uε,�|2 + �)p/2 − |∇uε,�|p| dx dt



+C�∫ ∫

�T

(|∇uε,�|2 + �)(p−2)/2 dx dt

�C�p/2∫ ∫

�T

dx dt + C�p/2∫ ∫

�T

(�

|∇uε,�|2 + �

)(2−p)/2

dx dt

� 2C�p/2∫ ∫

�T

dx dt → 0 (� = �k → 0)

Summing up the above arguments, we conclude that for any p>1,

J1 → 0 (�= �k → 0)

and thus (21) holds.Finally, we prove (22). We note that∫ ∫

�T

|�(uε,�)(|∇uε,�|2 + �)(p−2)/2∇uε,� − �(uε)|∇uε|p−2∇uε| dx dt

�∫ ∫

�T

�(uε,�)|(|∇uε,�|2 + �)(p−2)/2∇uε,� − |∇uε,�|p−2∇uε,�| dx dt

+∫ ∫

�T

�(uε,�)||∇uε,�|p−2∇uε,� − |∇uε|p−2∇uε| dx dt

+∫ ∫

�T

|�(uε,�) − �(uε)||∇uε|p−1 dx dt

= H1 + H2 + H3

Firstly, (16) implies that

H3 → 0 (� = �k → 0)

Secondly, we estimate H1. If p�2, by Lemma 2.1 and inequality (26), we have

H1 � C∫ ∫

�T

|(|∇uε,�|2 + �)(p−2)/2∇uε,� − |∇uε,�|p−2∇uε,�| dx dt

� C∫ ∫

�T

|(|∇uε,�|2 + �)(p−1)/2 − |∇uε,�|p−1| dx dt

� C(p − 1)∫ ∫

�T

(|∇uε,�|2 + �)(p−2)/2|(|∇uε,�|2 + �)1/2 − |∇uε,�|| dx dt

� C(p − 1)�1/2∫ ∫

�T

(|∇uε,�|2 + �)(p−2)/2 dx dt

→ 0 (� = �k → 0)



If 1<p<2, then 0<(p − 1)/2<1. Using inequality (26) yields

H1 �C∫ ∫

�T

|(|∇uε,�|2 + �)(p−2)/2∇uε,� − |∇uε,�|p−2∇uε,�| dx dt

=C∫ ∫

�T

|(|∇uε,�|2 + �)(p−2)/2|∇uε,�|2−p − 1|||∇uε,�|p−2∇uε,�| dx dt

=C∫ ∫

�T

|(|∇uε,�|2 + �)(p−2)/2|∇uε,�| − |∇uε,�|p−1| dx dt

�C∫ ∫

�T

|(|∇uε,�|2 + �)(p−1)/2 − |∇uε,�|p−1| dx dt

+C∫ ∫

�T

|(|∇uε,�|2 + �)1/2 − |∇uε,�||(|∇uε,�|2 + �)(p−2)/2 dx dt

�C�(p−1)/2∫ ∫

�T

dx dt + C�1/2∫ ∫

�T

(|∇uε,�|2 + �)(p−2)/2 dx dt

=C�(p−1)/2∫ ∫

�T

dx dt + C�(p−1)/2∫ ∫

�T

(�

|∇uε,�|2 + �

)(2−p)/2

dx dt

� 2C�(p−1)/2∫ ∫

�T

dx dt → 0 (� = �k → 0)

So for any p>1, there holds

H1 → 0 (� = �k → 0)

Thirdly, we estimate H2. For any fixed p>1, there exists a positive constant m =m(p) such that

0<� = (2m + 1)p − 2(m + 1)

m + 1<p

Hence, it follows from Lemma 2.1 and Holder’s inequality that∫ ∫�T

(|∇uε,�| + |∇uε|)� dx dt�C

where C is a constant independent of �. By this estimate, Holder’s inequality, Lemmas 1.1 and2.1 we conclude that

H2 � C∫ ∫

�T

||∇uε,�|p−2∇uε,� − |∇uε|p−2∇uε| dx dt

� C∫ ∫

�T

(|∇uε,�| + |∇uε|)[(m+1/2)p−(m+1)]/(m+1)

× [(|∇uε,�| + |∇uε|)(p−2)/2|∇uε,� − ∇uε|]1/(m+1) dx dt



� C

(∫ ∫�T

(|∇uε,�| + |∇uε|)� dx dt)1/2

×(∫ ∫

�T

[(|∇uε,�| + |∇uε|)p−2|∇uε,� − ∇uε|2]1/(m+1) dx dt

)1/2

� C

(∫ ∫�T

[(|∇uε,�| + |∇uε|)p−2|∇uε,� − ∇uε|2]1/(m+1) dx dt

)1/2

� C

(∫ ∫�T

(|∇uε,�| + |∇uε|)p−2|∇uε,� − ∇uε|2 dx dt)1/[2(m+1)]

→ 0 (�= �k → 0) (28)

To sum up, we prove (22) and so Lemma 2.2. �

Proposition 2.1Let uε be denoted in (16)–(18). Then, uε is a weak solution of the problem

ut = �(u) div(|∇u|p−2∇u) in �T

u(x, t) = ε on ��× (0, T )

u(x, 0) = u0(x) + ε in �

(29)

Moreover,‖u0‖L∞ + ε�uε�ε, uε2�uε1 for ε2>ε1∫ ∫

�t

(�uε

��

)2

dx d� +∫

�|∇uε(x, t)|p dx�C

(30)

where t ∈ (0, T ], and C is a constant independent of ε and t .

ProofObviously, for all ε ∈ (0, 1), uε − ε ∈ L p(0, T ;W 1,p

0 (�)). Equation (30) is obtained from (10),(16)–(18) and Lemma 2.1. To prove that uε satisfies the integral equality (b) in Definition 1.1,we multiply (7) by �∈C∞

0 (�T ), integrate both sides of the equality over �T and integrate byparts. Hence, ∫ ∫

�T

(−uε,��t + �(uε,�)(|∇uε,�|2 + �)(p−2)/2∇uε,�∇�

+�′(uε,�)(|∇uε,�|2 + �)(p−2)/2|∇uε,�|2�) dx dt = 0

Letting � = �k → 0 and using (16) and Lemma 2.2, we derive∫ ∫�T

(−uε�t + �(uε)|∇uε|p−2∇uε∇� + �′(uε)|∇uε|p�) dx dt = 0



Finally, applying Lemma 2.1, we conclude that∫�

|uε(x, t) − u0(x) − ε| dx�Ct1/2

where C is independent of ε, � and t . The proof is complete. �

By (30), we conclude that there exists a function u ∈ L∞(�T ), such that as ε → 0,

uε → u a.e. in �T (31)

∇uε ⇀ ∇u in L p(�T ) (32)

�uε

�t⇀

�u�t

in L2(�T ) (33)

Denote �(s) : [0,∞) → [0,∞) by

�(s) =∫ s

0[�(y)]1/(p−1) dy

Lemma 2.3As ε → 0, there hold:∫ ∫

�T

(|∇�(uε)| + |∇�(u)|)p−2|∇�(uε) − ∇�(u)|2 dx dt → 0 (34)

∫ ∫�T

||∇�(uε)|p − |∇�(u)|p| dx dt → 0 (35)

∫ ∫Qεc

||∇uε|p − |∇u|p| dx dt → 0 (36)

∫ ∫Qc

||∇uε|p − |∇u|p| dx dt → 0 (37)

where Qεc ={(x, t) ∈ �T ; uε�c>0} and Qc ={(x, t) ∈ �T ; u�c>0}.

ProofLet �=�(uε) − �(ε) − �(u). Then, �∈ L p(0, T ;W 1,p

0 (�)). Substituting it into the integralequality (b) of Definition 1.1 yields

∫ ∫�T

�uε

�t(�(uε) − �(ε) − �(u)) dx dt

=−∫ ∫

�T

�(uε)|∇uε|p−2∇uε∇(�(uε) − �(ε) − �(u)) dx dt

−∫ ∫

�T

�′(uε)|∇uε|p(�(uε) − �(ε) − �(u)) dx dt



=−∫ ∫

�T

|∇�(uε)|p−2∇�(uε)∇(�(uε) − �(u)) dx dt

−∫ ∫

�T

�′(uε)|∇uε|p(�(uε) − �(ε) − �(u)) dx dt

Note that uε�u and �(s) is non-decreasing. We derive∫ ∫�T

�uε

�t(�(uε) − �(ε) − �(u)) dx dt +

∫ ∫�T

|∇�(uε)|p−2∇�(uε)∇(�(uε) − �(u)) dx dt

��(ε)

∫ ∫�T

�′(uε)|∇uε|p dx dt

So ∫ ∫�T

�uε

�t(�(uε) − �(ε) − �(u)) dx dt

+∫ ∫

�T

(|∇�(uε)|p−2∇�(uε) − |∇�(u)|p−2∇�(u))∇(�(uε) − �(u)) dx dt

��(ε)

∫ ∫�T

�′(uε)|∇uε|p dx dt

−∫ ∫

�T

|∇�(u)|p−2∇�(u)∇(�(uε) − �(u)) dx dt (38)

On the other hand, from (30) to (33) and �(0)= 0, it follows that as ε → 0

∇�(uε) → ∇�(u) in L p(�T )∫ ∫�T

�uε

�t(�(uε) − �(ε) − �(u)) dx dt → 0

�(ε)

∫ ∫�T

�′(uε)|∇uε|p dx dt → 0

(39)

Combining (38) and (39) yields

lim supε→0

∫ ∫�T

(|∇�(uε)|p−2∇�(uε) − |∇�(u)|p−2∇�(u))∇(�(uε) − �(u)) dx dt�0

Recalling Lemma 1.1, we obtain (34). Equation (35) can be proved by using (34) and the samearguments as in Lemma 2.2.

Let us now prove (36) and (37). From the equality

[�(uε)]p/(p−1)(|∇uε|p − |∇u|p)= (|∇�(uε)|p − |∇�(u)|p) − |∇u|p{|�(uε)|p/(p−1) − |�(u)|p/(p−1)}



we conclude that ∫ ∫�T

[�(uε)]p/(p−1)||∇uε|p − |∇u|p| dx dt

�∫ ∫

�T

||∇�(uε)|p − |∇�(u)|p| dx dt

+∫ ∫

�T

||�(uε)|p/(p−1) − |�(u)|p/(p−1)||∇u|p dx dt

Equations (31) and (35) imply that the right side tends to zero as ε → 0. Since uε�c in Qεc and �

is non-decreasing, (36) is proved. Equation (37) is an immediate consequence of (36). Thus, theproof of Lemma 2.3 is complete. �

Lemma 2.4Suppose that �′(0)>0. Then, for all ε ∈ (0, 1), there holds∫ ∫

�T

�′(uε)|∇uε|p[�(uε)]� dx dt�C

where � ∈ (0,min{1, p/2}), C is a constant independent of ε.

ProofFrom Lemmas 2.1 and 2.2, it is easy to see that∫ ∫

�T

�′(uε,�)|∇uε,�|p[�(uε,�)]� dx dt � C (40)

∫ ∫�T

||∇uε,�|p − |∇uε|p| dx dt → 0 (� = �k → 0) (41)

where C is a constant independent of � and ε. Using uε,�, uε�ε, (41) and Proposition 2.1, we have

∫ ∫�T

∣∣∣∣�′(uε,�)|∇uε,�|p[�(uε,�)]� − �′(uε)|∇uε|p

[�(uε)]�∣∣∣∣ dx dt

�∫ ∫

�T

�′(uε,�)

[�(uε,�)]� ||∇uε,�|p − |∇uε|p| dx dt

+∫ ∫

�T

|∇uε|p∣∣∣∣ �′(uε,�)

[�(uε,�)]� − �′(uε)

[�(uε)]�∣∣∣∣ dx dt

�C1

∫ ∫�T

∣∣|∇uε,�|p − |∇uε|p∣∣ dx dt

+C2

∫ ∫�T

|∇uε|p|�′(uε,�)[�(uε)]� − �′(uε)[�(uε,�)]�| dx dt

→ 0 (� = �k → 0) (42)



where C1 and C2 are constants independent of ε, �. Then, the proof of Lemma 2.4 is complete bycombining (40) and (42). �

Lemma 2.5As ε → 0, there hold ∫ ∫

�T

|�′(uε)|∇uε|p − �′(u)|∇u|p| dx dt → 0 (43)

∫ ∫�T

|�(uε)|∇uε|p−2∇uε − �(u)|∇u|p−2∇u| dx dt → 0 (44)

ProofNote that∫ ∫

�T

|�′(uε)|∇uε|p − �′(u)|∇u|p| dx dt

�∫ ∫

�T

�′(uε)||∇uε|p − |∇u|p| dx dt +∫ ∫

�T

|�′(uε) − �′(u)||∇u|p dx dt

= I1 + I2

Obviously, by (31), we have

I2 → 0 (ε → 0) (45)

Next, we estimate I1. Let � and �(ε) be the characteristic functions of {(x, t) ∈ �T ; u(x, t)< }

and {(x, t) ∈ �T ; uε(x, t)< }, respectively. Then,

I1 =∫ ∫

�T

�′(uε)||∇uε|p − |∇u|p|�(ε) dx dt

+∫ ∫

�T

|�′(uε)||∇uε|p − |∇u|p|(1 − �(ε) ) dx dt

= K1 + K2

If �′(0)= 0, then we have

K1� maxs∈[0, ]

�′(s)∫ ∫

�T

||∇uε|p − |∇u|p| dx dt → 0 ( → 0+)

If �′(0)>0, taking � = 12 min{1, p/2} in Lemma 2.4 and noting uε�u and �(0)= 0, we have

K1 �∫ ∫

�T

�′(uε)|∇uε|p�(ε) dx dt +

∫ ∫�T

�′(uε)|∇u|p�(ε) dx dt

� [�( )]�∫ ∫

�T

�′(uε)|∇uε|p[�(uε)]� dx dt + C

∫ ∫�T

|∇u|p�(ε) dx dt



� C[�( )]� + C∫ ∫

�T

|∇u|p� dx dt

→ 0 ( → 0+)

Summing up, K1 → 0 ( → 0+), i.e. for any �>0, letting sufficiently small, we obtain

K1<�/2

For fixed >0, it follows from Lemma 2.3 that

K2�C∫ ∫

�T

||∇uε|p − |∇u|p|(1 − �(ε) ) dx dt → 0 (ε → 0)

where C is independent of ε. Hence, there exists ε0 ∈ (0, 1) such that K2<�/2 as ε ∈ (0, ε0).It follows that

I1 = K1 + K2<� ∀ε<ε0

Combining this estimate with (45) implies (43).Finally, we prove (44). First, we have

F =∫ ∫

�T

|�(uε)|∇uε|p−2∇uε − �(u)|∇u|p−2∇u| dx dt

=∫ ∫

�T

||∇�(uε)|p−2∇�(uε) − |∇�(u)|p−2∇�(u)| dx dt

Then using (43) and by the same arguments as in (28), we derive that F → 0 (ε → 0). So theproof of Lemma 2.5 is complete. �

By means of (31) and Lemma 2.5, it is easy to check that u satisfies the integral identity (b) inDefinition 1.1. Finally, from (30), it follows that∫

�|u(x, t) − u0(x)| dx�Ct1/2

and hence condition (c) is satisfied. Thus, Theorem 2.1 is completely proved.

3. SOME DISCUSSIONS ABOUT UNIQUENESS

The purpose of this section is to prove the following theorem.

Theorem 3.1Let function u be as in (31)–(33). Then, function u is maximal among all weak solutions ofproblem (1)–(3).

Before proceeding the proof of Theorem 3.1, we first establish the following comparison theorem.



Lemma 3.1Let p>1. Assume that u and v are weak sup-solution and sub-solution of (1), respectively, and u�Ma.e. in �T for some M>0. If u(x, 0)�v(x, 0) a.e. in � and u(x, t)�v(x, t) a.e. on ��× (0, T ),then u�v a.e. in �T .

ProofFirst, for all t ∈ (0, T ) and any �∈ L∞(�t ) ∩ L p(0, t;W 1,p

0 (�)) with ��0, we have∫ ∫�t

(u�� + �(u)|∇u|p−2∇u∇� + �′(u)|∇w|p�) dx d� � 0

∫ ∫�t

(v�� + �(v)|∇v|p−2∇v∇� + �′(v)|∇v|p�

)dx d� � 0

(46)

Since u, v ∈ L∞(�T ) ∩ L p(0, T ;W 1,p(�)) and u�v on �� ×(0, T ), we have (v−u)+ ∈ L∞(�T ) ∩L p(0, T ;W 1,p

0 (�)) and hence sgn�((v − u)+) ∈ L∞(�T ) ∩ L p(0, T ;W 1,p0 (�)), where sgn�(z) =

sgn(z) inf{|z|/�, 1} (�>0). While u�M>0 a.e. in �T , we get �u = sgn�((v − u)+)/�(u)∈L∞

(�T ) ∩ L p(0, T ;W 1,p0 (�)). Define �v=sgn�((v−u)+)/�(v) whenever v>M , �v=0 whenever

v�M . Then, �v ∈ L∞(�T ) ∩ L p(0, T ;W 1,p0 (�)). So �u and �v can be chosen in (46) as test

functions. Therefore,

∫ ∫�t

[u��(u)−1sgn�((v − u)+) + |∇u|p−2∇u∇(v − u)+ sgn′�((v − u)+)] dx d� = 0

∫ ∫�t

[v��(v)−1sgn�((v − u)+) + |∇v|p−2∇v∇(v − u)+ sgn′�((v − u)+)] dx d� = 0

Therefore,

∫ ∫�t

(h(v) − h(u))� sgn�((v − u)+) dx d�

+∫ ∫

�t

(|∇v|p−2∇v − |∇u|p−2∇u)∇(v − u)+ sgn′�((v − u)+) dx d� = 0

where h(s) : (0,+∞) → R is defined by h(s) = ∫ s� 1/(�(u)) du. Then, it follows from Lemma 1.1

that ∫ ∫�t

(h(v) − h(u))� sgn�((v − u)+) dx d��0

Passing to the limit as � → 0+ yields∫ ∫�t

(h(v) − h(u))� sgn((v − u)+) dx d��0



Since sgn((v − u)+) = sgn((h(v) − h(u))+), we have∫ ∫�t

(h(v) − h(u))� sgn((h(v) − h(u))+) dx d��0

It follows from v(x, 0)�u(x, 0) a.e. in � that∫�(h(v) − h(u))+(x, t) dx�0

which implies that (h(v) − h(u))+(x, t) = 0 a.e. in �T , u�v a.e. in �T . The proof iscomplete. �

Proof of Theorem 3.1Let v be any weak solution of problem (1)–(3) and uε be as in (7)–(9). Applying Lemma 3.1 touε and v (note uε�ε) yields uε�v a.e. in �T . Then, letting ε → 0, we derive u�v a.e. in �T . Theproof is complete. �

Remark 3.1Theorem 2.1 shows that we can get a weak solution u of problem (1)–(3) by virtue of parabolic reg-ularization. If we will call u a weak solution of regularization of problem (1)–(3), then Theorem 3.1implies that the weak solutions of regularization of problem (1)–(3) are unique.

4. LOCALIZATION OF WEAK SOLUTIONS

For a non-negative function w : �→ R+ ∪ {0} that may be discontinuous, we define its supportas follows:

suppw ={x ∈G; lim

→ 0+�(G ∩ B (x))

�(B (x))>0

}

where G ={x ∈ �;w(x)>0}, B (x)={y ∈ �; |x − y|< } and �(E) is the Lebesgue measure ofset E in RN . It is easy to see that if w ∈C(�), then suppw =G.

By the similar idea used in [4, 15], we can show the following theorem.

Theorem 4.1Let 0�u0 ∈C(�), u0 = 0 on ��, u0 /≡ 0 and supp u0 ⊆ �. If u is a weak solution of problem(1)–(3), then

supp u(·, t) ⊆ supp u0 a.e. in (0, +∞)

ProofFirst for any t>0 and �∈ L∞(�t ) ∩ L p(0, t;W 1,p

0 (�)), we have∫ ∫�t

(u�� + �(u)|∇u|p−2∇u∇� + �′(u)|∇u|p�) dx d�= 0 (47)

Let �=�� = inf{d(x)/�, 1} where 0<�<1, d(x)= dist(x, supp u0 ∪ ��). It is well known thatthe distance function d(x) is Lipschitz with constant 1, and then Rademacher’s Theorem implies



that it is differentiable almost everywhere (see, for instance [18, pp. 49–51]). Hence, for any ε>0,�=�/(�(u) + ε) can be chosen as a test function. Substituting it into (47) yields

∫ ∫�t

[u�

�

�(u) + ε+ �(u)|∇u|p−2∇u∇ �

�(u) + ε+ �′(u)|∇u|p �

�(u) + ε

]dx d�= 0

and hence

∫ ∫�t

u��

�(u) + εdx dt� −

∫ ∫�t

�(u)|∇u|p−2∇u∇�

�(u) + εdx d��C

where C is independent of ε. For � · u0 = 0, it follows that

∫�

�{supp�}

(∫ u(x,t)

0

1

�(s) + εds

)� dx�C

So for � ∈ (0, 1) and t ∈ (0, T ) a.e. we derive

∫�

�{{x;u(x,t)>�}∩{x;�=1}}∫ �

0

1

�(s) + εds dx�C

And hence

�({x ∈ {x;�= 1}; u(x, t)>�})�C

[∫ �

0

ds

�(s) + ε

]−1

(48)

where C is independent of ε. We further claim that for any �>0,

�({x ∈ {x;� = 1}; u(x, t)>�}) = 0 a.e. in (0, +∞) (49)

Indeed, it is not difficult to show that there exists a positive constant M� such that �(s)�M�s on[0, �]. Therefore,

∫ �

0

ds

�(s) + ε�∫ �

0

ds

M�s + ε� 1

M�ln

(M��

ε+ 1

)→ + ∞ (ε→ 0+)

From (48) it follows (49). Since � is arbitrary, we can infer from (49) that

�({(x, t) ∈ {�= 1} × {t}; u>0}) = 0 a.e. in (0, +∞)

So by the arbitrariness of � ∈ (0, 1) Theorem 4.1 follows. �



5. LARGE TIME BEHAVIOUR OF WEAK SOLUTIONS

First, we give an explicit weak solution for (1). For t�0, x ∈ RN , let

U (x, t; �, L) = [L − ��(x)]+(t + �)1/(p−1)

where L , �>0, �=[(p − 1)N ]−1/(p−1) and �(x)= (p − 1)p−1|x |p/(p−1).

Proposition 5.1

(i) For all L , �>0, there holds

Ut =U div(|∇U |p−2∇U ) in B (L) × (0,+∞)

where (L) =[pL/(�(p − 1))](p−1)/p.(ii) Let L>0 satisfy � ⊇ B (L). Then for all �>0, U (x, t; �, L) is a weak solution of problem

(1)–(3) with �(s)= s and the initial value U0(x; �, L) =[L − ��(x)]+/�1/(p−1). Moreover,

suppU (·, t; �, L) = suppU0(·; �, L) in (0, +∞)

ProofBy simple calculations, we have

∇�(0)= 0, ∇�= x |x |(2−p)/(p−1) for any x �= 0

Hence

|∇�|p−2∇�= x ∀x ∈ RN

and

div(|∇�|p−2∇�) = N

By virtue of the fact, it is not difficult to prove (i) and (ii). �

The main result about the asymptotic behaviour of weak solutions is stated in the followingtheorem.

Theorem 5.1Let 0�u0 ∈ L∞(�) and assume that � satisfies (H1) and �′(0)>0. If u is a weak solution ofproblem (1)–(3), then there exist positive constants L and KL depending p, N , � and ‖u0‖∞ suchthat

u(x, t)� L

(KLt + 1)�a.e. in Q, � = 1

p − 1

ProofChoose L>0 sufficiently large such that

L − ��(x)�‖u0‖∞ + 1 on �



Since �′(s)>0 for all s ∈ [0, L], there exists a positive constant KL such that

�(s)�KLs ∀s ∈ [0, L]Let U (x, t) =U (x, t; K−1

L , L) and V (x, t) = K−1/(p−1)L U (x, t). Then, we have

V (x, 0) � ‖u0‖∞ + 1�u0(x) a.e. in �

L � V (x, t)�0 on � × (0,∞)

and for any T>0, V (x, t)�C(T )>0 in �T .Recalling Proposition 5.1, we conclude that

Vt − �(V ) div(|∇V |p−2∇V ) � Vt − KLV div(|∇V |p−2∇V )

= K−1/(p−1)L [Ut −U div(|∇U |p−2∇U )]

= 0 in Q

It follows that for any T>0, V is a weak sup-solution of (1) on �T . By Lemma 3.1, we obtain

V (x, t)�u(x, t) a.e. in Q = RN × (0,∞)

i.e.

u(x, t)�K−1/(p−1)L

L − ��(x)

(t + K−1L )1/(p−1)

a.e. in Q

This ends our proof. �

ACKNOWLEDGEMENTS

The authors would love to thank the referee for his important comments which improve this paper.

REFERENCES

1. Allen LJS. Persistence and extinction in single species reaction–diffusion models. Bulletin of MathematicalBiology 1983; 45:209–277.

2. Low BC. Resistive diffusion of force-free magnetic fields in a passive medium. The Astrophysical Journal 1973;181:209–226.

3. Low BC. Resistive diffusion of force-free magnetic fields in a passive medium II. The Astrophysical Journal1973; 181:917–929.

4. Dal Passo R, Luckhaus S. A degenerate diffusion problem not in divergence form. Journal of DifferentialEquations 1987; 69(1):1–14.

5. Wang S, Wang MX, Xie CH. A nonlinear degenerate diffusion equation not in divergence form. Zeitschrift furAngewandte Mathematik und Physik 2000; 51:149–159.

6. Ughi M. A degenerate parabolic equation modelling the spread of an epidemic. Annali di Matematica Pura edApplicata 1986; 143:385–400.

7. Esteban JR, Vazquez JL. On the equation of turbulent filtration in one dimensional porous medium. NonlinearAnalysis–Theory Methods and Applications 1986; 10(11):1303–1325.



8. Esteban JR, Vazquez JL. Homogeneous diffusion in R with power-like nonlinear diffusivity. Archive for RationalMechanics and Analysis 1988; 103(1):39–80.

9. Wu ZQ, Zhao JN, Yin JX, Li HL. Nonlinear Diffusion Equations. World Scientific: Singapore, 2001.10. Bertsch M, Dal Passo R, Ughi M. Discontinuous viscosity solutions of a degenerate parabolic equation.

Transactions of the American Mathematical Society 1990; 320(2):779–798.11. Bertsch M, Dal Passo R, Ughi M. Nonuniqueness of solutions of a degenerate parabolic equation. Annali di

Matematica Pura ed Applicata 1992; 161:57–81.12. Bertsch M, Ughi M. Positivity property of viscosity solutions of a degenerate parabolic equation. Nonlinear

Analysis—Theory Methods and Applications 1990; 14(7):571–592.13. Friedman A, McLeod B. Blow-up of solutions of nonlinear degenerate parabolic equations. Archive for Rational

Mechanics and Analysis 1986; 96(1):55–80.14. Tsutsumi T, Ishiwata M. Regional blow-up of solutions to the initial boundary value problem for �u/�t=u�(�u+u).

Proceedings of the Royal Society of Edinburgh, Section A—Mathematics 1997; 127(4):871–887.15. Zhou WS, Wu ZQ. Some results on a class of nonlinear degenerate parabolic equations not in divergence form.

Northeastern Mathematical Journal 2003; 4:291–294.16. Damascelli L. Comparison theorems for some quasilinear degenerate elliptic operations and applications to

symmetry and monotonicity results. Annales de l’Institut Henri Poincare, Analyse Non-lineaire 1998; 4(15):493–516.

17. Ladyzenskaja OA, Solonnikov VA, Ural’ceva NN. Linear and Quasilinear Equations of Parabolic Type. AmericanMathematical Society: Providence, RI, 1968.

18. Ziemer WP. Weakly Differentiable Functions. Springer: Beijing, 1991.


existence and some properties of weak solutions for a singular nonlinear parabolic equation

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