Global Existence And Blow-Up Of Generalized Self-Similar Solutions To Nonlinear Degenerate Di¤usion Equation Not In

Global Existence And Blow-Up Of Generalized Self-Similar Solutions To Nonlinear Degenerate Di¤usion Equation Not In

Divergence Form

Bilal Basti

, Nouredine Benhamidouche

Received 11 August 2019

Abstract

This paper investigates the problem of existence and uniqueness of positive solutions under the general self-similar form of the degenerate parabolic partial di¤erential equation which is known as "nonlinear di¤usion equation not in divergence form". By applying the properties of Banach’s …xed point theorems, we establish several results on the existence and uniqueness of the general form of self-similar solutions of this equation.

1 Introduction

Many problems and models in physics, chemistry, biology and economics are modeled by partial di¤erential equations. In this work, we shall give an example of a class of renowned equations, which allow to describe the di¤usion phenomena; that is, a parabolic PDE and known as nonlinear di¤usion equation not in divergence form and is written as:

@u

@t =u^m@²u

@x²; m2(0;1)[(1;1); (1)

whereu=u(x; t)is a nonnegative scalar function of space variablesx2Rand timet >0:

The nonlinear di¤usion equation not in divergence form (1), is often studied by researchers (see for example [1, 6, 15–17]), who gave some results of existence and uniqueness of the global solutions in time and solutions which blow-up in a …nite time, this is applied for a certain class of the function uwhich satis…ed some su¢ cient conditions.

In general, for some PDEs which has the characterization of symmetries, (see for example [11,12,13]), we can determine their exact solutions with certain (…nite or in…nite) transformations. Here, a PDE becomes an ordinary di¤erential equation, in this case the solutions are called self-similar solutions ([3, 7, 9, 16]), which often play a central role in the study of a PDE, since it is equivalent to these solutions to solve locally or globally.

C. Wang et al. [16], tackled in detail the existence and the uniqueness of a shrinking self-similar solution to the equation (1) form 1: The proposed solution was:

u(x; t) = 1

(t+ 1) ! (t+ 1) jxj² ; t >0andx2Rⁿ; where! is a positive function that satis…es some properties and

0; =1 +

m ;

Mathematics Sub ject Classi…cations: 34A05, 35B06, 35B44, 35C06.

yLaboratory for Pure and Applied Mathematics, University of Mohamed Boudiaf M’sila, 28000, Algeria. Prof in Department of Mathematics, University of Ziane Achour Djelfa, 17000, Algeria

zLaboratory for Pure and Applied Mathematics, University of Mohamed Boudiaf M’sila, 28000, Algeria

367

are constants chosen so that the solutions exist.

The equation (1) is a representation of a large class of the nonlinear parabolic equations. Obviously, for m= 0 we recognize the well-known heat equation, which in fact has quite di¤erent properties from the two nonlinear ranges,m >1 (usually called not in divergence form) andm2(0;1):For the case m2(0;1);if we consider the implicit change of variables:

v=p^p11u^1p; withp= 1

1 m; m2(0;1);

the equation (1) can be written in divergence form of the porous medium equation (see [2,5,10,14])

@v

@t = @²

@x²(v^p); p >1: (2)

The porous medium equation (2), is an equation, which admits the properties of similarity. There are several known fundamental families of self-similar solutions, maybe the most important one is formed by the Barenblatt solutionsB( ); discovered independently by Barenblatt in [2] and by Zeldovich and Kompaneets in [18], which are written under the following form:

BC(x; t) = 8<

:

t ^p+11 _2p(p+1)^{p 1} C² x²t ^p+12

1 p 1

; for jxj< Ct^p+11 ; C >0;

0; otherwise.

(3)

Our objective in this work is to study the existence and uniqueness of positive solutions of the degenerate parabolic PDE (1), under the generalized self-similar form which is:

u(x; t) =c(t)f x

a(t) ; x2R; t >0:

The functionsa(t)andc(t)depend on timetand the basic pro…lef >0;are not known in advance and are to be identi…ed.

2 De…nitions and Preliminary Results

To discuss the generalized self-similar solutions:

u(x; t) =c(t)f( ); with = x

a(t); x2R; t >0: (4)

We should …rst deduce the equation satis…ed by the functionf( )in (4) used for the de…nition of self-similar solutions.

2.1 Basic Idea to Compute the Self-Similar Solutions

The functional-di¤erential equation resulting from the substitution of expression (4) in the original PDE (1), should be reduced to the standard bilinear functional equation (see [13]); in this case we obtain the following equation:

_

c(t)f( ) c(t) a_(t)

a²(t)xf⁰( ) = c^m+1(t)

a²(t) f^m( )f⁰⁰ ( ); witha; c2R+: (5) By expressing x from (4) in terms of ; substituting into (5) and divide by c(t); we get the functional equation in two variablestand ;as follows:

_ c(t)

c(t)f( ) a_(t)

a(t) f⁰( ) c^m(t)

a²(t)f^m( )f⁰⁰ ( ) = 0:

This can be rewritten as the standard bilinear functional equation:

'_{1 1}+'_{2 2}+'_{3 3}= 0;

with a direct calculation shows thata=a(t); c=c(t)andf =f( );we have:

'₁= c_

c; '₂= a_

a; '₃= c^m a² and

1=f; ₂= f⁰; ₃=f^mf⁰⁰:

Substituting these expressions into the solution of the three-term functional equation [13], yields the deter- mining system of ordinary di¤erential equations (see also [3,4])

8>

><

>>

:

_ c

c =k₁^c_a^m2;

_ a

a =k₂^c_a^m2;

f^mf⁰⁰ = f+ f⁰:

(6)

Where ; ; k1andk2 are arbitrary constants.

The system of ordinary di¤erential equations (6) depends on many unknown parameters. The functions a(t); c(t)andf( )are explicitly determined.

2.2 Statement of the Problem

In this part, we …rst attempt to …nd the equivalent approximate to the following problem of the nonlinear di¤usion equation not in divergence form:

8>

>>

>>

>>

>>

<

>>

>>

>>

>>

>:

@u

@t =u^{m @}_@x^2u2; x 0; m2(0;1)[(1;1);

u(0; t) =c(t)U; U 0; t >0;

@u( a(t);t)

@x = ₍₁ ^c_m)a(t)^m(t) lim

x! a(t)[u(x; t)]^{1 m}; >0;

a(t)R

x

u(x;t) u(s;t)

m 1

ds ₂a(t); 8m >1; 8x2[0; a(t)):

(7)

Under the generalized self-similar form which is:

u(x; t) =c(t)f( ); with = x

a(t) anda; c2R+: (8)

According to the preceding part (the system (6)), we consider this problem:

8>

>>

>>

>>

><

>>

>>

>>

>>

:

f^m( )f⁰⁰( ) = f( ) + f⁰( ); 0; m2(0;1)[(1;1);

f(0) =U; U 0;

f⁰( ) = _{1 m} lim

!

f^{1 m}( ); >0;

R f^{1 m}( )d ₂f^{1 m}( ); 8m >1; 8 2[0; );

(9)

in which ; are arbitrary real constants.

3 Existence and Uniqueness of Basic Pro…le

In this section we study the existence and uniqueness of a class of positive solutions under the generalized self-similar form (8) for the problem of the nonlinear di¤usion equation not in divergence form (7).

We …nd also exact solutions under certain conditions, which depend on the similarity coe¢ cients and :

As in [7] it is necessary to consider the weak solutions of the problem (9).

De…nition 1 (Weak solution) A function f will be called a weak solution with compact support of the problem (9) if and only if

1) f is a continuous function, bounded and nonnegative on[0;1): 2) f has a continuous derivative in a left neighborhood of = : 3) f satis…es the identity:

Z1 0

f⁰( ) +

m 1 f^{1 m}( ) '⁰( )d + +

m 1

Z1 0

f^{1 m}( )'( )d = 0;

for all '2C₀¹(0;1):

We discuss the existence of basic pro…le f of the weak solutions with compact support, then as we shall see later, f is positive in a right neighborhood of = 0; more speci…cally for some " > 0; there exists a number" < <1such that:

f >0; on (0; ) and

f 0; on [ ;1):

Assume thatf is a weak solution of the problem (9) with compact support and let be an arbitrary positive number, we shall be mainly concerned with proving the existence and uniqueness of a positive solution of problem (9) on an interval (0; )which satis…es the boundary conditions:

f(0) =U; f( ) = 0: (10)

Then we shall show that for suitable and there exists a unique positive solution of the equation:

f^m( )f⁰⁰( ) = f( ) + f⁰( ); (11)

in the left neighborhood of = and that this solution can be continued back to = 0:We then ask whether can be chosen so that condition (10) is satis…ed.

Before dealing with the question of existence, we give the necessary conditions of the parameters and for the existence of a nontrivial weak solution with compact support of the problem (9).

Lemma 1 The functionf is a nontrivial weak solution with compact support of the boundary value problem (9)–(10), if and only if

(i) is a negative coe¢ cient for m2(0;1): (ii) is a positive coe¢ cient for m >1:

(iii) = 0and takes strictly positive value.

Proof. Supposef is a nontrivial weak solution of problem (9) with compact support. Then for some" >0;

we have:

f >0 in ( "; )

= 0 in [ ;1) ; >0:

It results thatf is a weak solution of (9) which satis…es (10) in the left neighborhood of = :The integration of (11) starting from to ;where " < < gives:

(1 m)f⁰( ) = f^{1 m}( ) + [ (m 1) + ] Z

f^{1 m}( )d : (12)

The continuity off andf⁰ left of ;ensures the existence of 2( "; )such thatf⁰( )<0:

In fact, if we apply the mean value theorem on the interval( "; );we have:

9 2( "; ) such thatf⁰( ) = f( ) f( ")

( ") = f( ")

" <0:

This implies that the member in the left of (12) at the point = is negative for m2(0;1)and positive form >1:Hence, and (m 1) + cannot both be greater than zero form2(0;1);(resp. less than zero form >1) and = 0implies that >0. Hence the estimate (iii).

In what follows, because the functionf is positive and tends to zero if ! ; the mean value theorem enables us to prove that there exists " ₀< ;such thatf⁰( )<0;8 2( ₀; ):We divide the proof into two cases:

(i) Form2(0;1);let us consider >0;for all 2( ₀; );this implies that (m 1) + <0:It results from (12) that:

(1 m)f⁰( )f^{m 1}( ) >[ (m 1) + ] ( ): (13) In fact, we will use:

8 2[ 0; ); 8 2( ; ); we havef( )< f( ); which implies for anym2(0;1);that:

f^{1 m}( )< f^{1 m}( ) andf^{1 m}( )f^{m 1}( )<1;

also Z

f^{1 m}( )f^{m 1}( )d < :

If ! in (13), the left part is negative and the right part tends towards zero, contradiction, which implies that <0form2(0;1):

(ii) For m >1;we consider <0; for all 2( ₀; ); this implies that (m 1) + >0; consequently

>0:It results from (12) that:

(1 m)f⁰( ) f^{1 m}( )< (m 1) Z

f^{1 m}( )d : (14)

If ! in (14), the left part is positive and the right part tends towards zero, contradiction.

Thus, we have already proved that the coe¢ cients

>0form >1; or <0form2(0;1); or = 0and >0;

are the only cases for which the nontrivial weak solution with compact support of the boundary value problem (9)–(10) exists.

New explicit solutions for = 0andm2(0;1)[(1;2)

Let be an arbitrary positive number. It is clear from the proof of Lemma 1 that a necessary condition for the existence of a positive solution of problem (9)–(10) in a left neighborhood of = is either >0 form >1; <0form2(0;1) or = 0and >0:

The objective of this part is to show that this condition is also su¢ cient.

We begin by assuming that = 0 and >0; we can then solve problem (9)–(10) uniquely. After an elementary computation that the function:

f( ; ) = m²

2 (2 m)( )²

1 m

; m2(0;1)[(1;2); 0 < ; (15)

is the unique solution of the problem (9)–(10).

In fact, starting from (11), we put = 0and >0;we get f^m( )f⁰⁰( ) = f( ); also

f⁰⁰( )f⁰( ) = f^{1 m}( )f⁰( ): (16) Form2(0;1)[(1;2);we have from (9) that f⁰( ) = 0:Then, the integration of (16) starting from to gives:

1

2[f⁰( )]²=

2 mf^{2 m}( ):

Also, after a simple integration from to ; we have the functionf( ; )which is presented in (15).

Because f(0; )is a continuous, monotonically increasing function of ;such that f(0; 0) = 0andf(0;1) =1;

the equationf(0; ) =U is uniquely solvable for everyU >0:

Let (U) be its solution, then f = f( ; (U)) is the unique solution of the problem (9)–(10), with compact support de…ne for all0 < and satis…es:

f(0; ) =U andf( ; ) = 0;

where

(U) =

r2 (2 m)

m² U^m: In addition,f is a continuous and monotonous decreasing function.

Next, we discuss the case 6= 0and we give some elementary lemmas for the two following cases <0 form2(0;1)and >0form >1:

Lemma 2 Suppose that0 1< and f is a positive solution of the boundary value problem (9)–(10) on [ 1; ): Then

1. f⁰( )<0 on [ ₁; );if and only if >_{1 m}:

2. If <_{1 m} andf⁰( ₀) = 0;for some ₀2[ 1; ): Thenf has a maximum at ₀ such that:

0

[ (m 1) + ]

(m 1) ; for the case m2(0;1) and

0

[ (m 1) + ]

(m 1) ; for the case m2(1;1):

3. Suppose thatf is a positive solution of (9) on [0; ). Then:

f⁰(0)<0; for >

1 m: 4. If

= 1 m; we have:

f⁰(0) = 0 andf⁰( )<0; 8 2(0; ); i.e. f( ) f(0) =U; 8 2[0; ]: (17) In this case, we …nd a new explicit solution of the problem (9)–(10) as follows:

f( ; ) = m

2 (m 1)

2 2

1 m

; m2(0;1)[(1;1); j j< : (18) Proof. For 2[ 1; ); the integral equation (12) gives:

(1 m)f⁰( ) = f^{1 m}( ) + [ (m 1) + ] Z

f^{1 m}( )d : (19)

1. Because <0form2(0;1);the right member in (19) is always negative i.e. f⁰( )<0if and only if (m 1) + takes a negative value. In the same way, we have >0form2(1;1);then the left member in (19) is always positive i.e. f⁰( )<0if and only if (m 1) + takes a positive value, therefore

f⁰( )<0on [ ₁; ) when >

1 m: 2. If < _{1 m} then <0 for anym2(0;1)[(1;1):According to (11);

f⁰⁰( ₀)<0 whenf⁰( ₀) = 0:

Sof has a maximum at = ₀and strictly decreasing on( ₀; );i.e. f⁰( )<0 on( ₀; ):We put = ₀ in (19), we get:

0 = ₀f^{1 m}( ₀) + [ (m 1) + ] Z

0

f^{1 m}( )d :

Form2(0;1);we have <0and (m 1) + >0;then:

0f^{1 m}( ₀) + [ (m 1) + ] ( ₀)f^{1 m}( ₀) 0:

So

0

[ (m 1) + ]

(m 1) :

In the same way, form2(1;1); we have >0 and (m 1) + <0;then we …nd:

0

[ (m 1) + ]

(m 1) :

3. With = 0;(19) becomes:

(1 m)f⁰(0) = [ (m 1) + ] Z

0

f^{1 m}( )d : (20)

The sign off⁰(0)results immediately from (20).

4. If

(m 1) + = 0for any m2(0;1)[(1;1); we have from (19):

(1 m)f⁰( ) = f^{1 m}( ): (21)

Which implies easily (17).

The integration of (21) starting from to gives:

1

mf^m( ) =

2 (1 m)

2 2 ;

which implies the functionf( ; )presents in (18) by:

f( ; ) = m

2 (m 1)

2 2

1 m

; m2(0;1)[(1;1); j j< :

Let (U)be its solution.

Because f(0; (U))is a continuous, monotonically increasing function of (U);such that:

f(0; 0) = 0andf(0;1) =1:

Then the equationf(0; (U)) =U is uniquely solvable for everyU 0and f =f( ; (U))is the unique solution of problem (9)–(10), with compact support de…nite for allj j< and it satis…es:

f(0; ) =U andf( ; ) = 0;

where

(U) = s

2 (m 1) m U^m: In addition,f is a continuous and monotonous decreasing function.

We now turn to the question of existence.

Lemma 3 For any m2(0;1)[(1;1)and any >0; the problem (9) with the boundary conditions (10), has a unique positive solution in the left neighborhood of = :

The proof of this lemma concerns the analysis of the of Banach’s …xed point theorem, which is:

Theorem 4 (Banach’s …xed point [8]) Let X be a non-empty closed subset of a Banach space E; then any contraction mappingM ofX into itself has a unique …xed point.

Now, we prove the Lemma3.

Proof. Assume thatf is a positive solution in the left neighborhood of = :By Lemma2,f⁰( )<0for 2( "; )for some" >0:

1. For m2(0;1);we have <0:In this case, the writing of (19) is given as follows:

f⁰( ) = f^{1 m}( ) + [ (1 m) ]

Z f⁰( )

f^m( )d : (22)

WithG(f) = we have

dG

df = 1

Gf^{1 m} [ (1 m) ]

Rf 0

G(') '^m d'

: (23)

By integrating the di¤er-integral equation (23) from0 tof;we obtain

G(f) =

Zf 0

d'

G'^{1 m} [ (1 m) ]

R' 0

G(g) g^m dg

: (24)

We putH(f) = 1 ¹G(f); then the equation (24) becomes H(f) = 1

2

Zf 0

d'

H'^{1 m} _{1 m}'^{1 m} [ (1 m) ]

R' 0

H(g) g^m dg

: (25)

Using the principle of contraction of Banach Theorem, we prove now that the equation (25) admits a unique positive solution to the right off = 0:

LetX be the set of bounded functionsH(f)on[0; h]; h >0;satisfying:

0 H(f) = m

j j(1 m) +j (1 m) j:

Letk kbe the norm sup de…ned onX;thenX is a complete metric space. OnX we de…ne the operator:

M(H) (f) = 1

2

Zf 0

d'

H'^{1 m} _{1 m}'^{1 m} [ (1 m) ]

R' 0

H(g) g^m dg

: (26)

First, we prove thatM is a mapping from X toX. LetH 2X:It is clear that

H(')'^{1 m}

1 m'^{1 m} [ (1 m) ] Z' 0

H(g) g^m dg

1 m'^{1 m} j jH(')'^{1 m} j (1 m) j kHk Z' 0

g ^mdg

1 m'^{1 m}: (27)

Therefore, starting from (26), we have:

M(H) (f) 1

2

Zf 0

d'

1 m'^{1 m}

(m 1)h^m

2m : (28)

Thus,M(H)is well de…ned onX andM(H) : [0; h]!Ris nonnegative and continuous. The right side of (28) shows that we can …ndh₀>0;such thath h₀ and

kM(H)k ; for allH 2X:

SoM is a mapping fromX toX for some h h₀:

In the next step, we show that M is a contracting mapping on X. Let H₁; H₂ 2 X and h h₀; be a positive real number, we put:

K(H) = H(')'^{1 m}

1 m'^{1 m} [ (1 m) ] Z' 0

H(g) g^m dg:

Then

kK(H2) K(H1)k j j kH2(') H1(')k'^{1 m} +j (1 m) j

1 m kH2(') H1(')k'^{1 m} j j+j (1 m) j

1 m kH2(') H1(')k'^{1 m}: So

kM(H1) M(H2)k = 1

2

Zf 0

d' K(H₁)

1

2

Zf 0

d' K(H₂)

= 1

2

Zf 0

K(H₂) K(H₁) K(H1)K(H2) d'

(1 m)²

2 2

Zf 0

kK(H2) K(H1)k

'^{2(1 m)} d'

(1 m) [j j(1 m) +j (1 m) j]

mh ^{m 2 2} kH2(') H1(')k:

Hence, there exists a h₁ 2 (0; h₀] such that if h h₁; M is a contraction on X: After the principle of contraction of Banach [8], M has a unique …xed point in X; consequently, there exists a unique positive solution for the problem (9) in the interval( "; )for some " >0:

2. After using some techniques, we use the same steps for the case m2(1;1);to prove the existence of a single positive solution for the problem (9).

We now continue f backwards as a function of : By the standard theory, this can be done uniquely so long asf remains positive and bounded. There are now three possibilities:

(A) f( )! 1 as ! 1 for some ₁2[0; ); (B) f( )can be continued back to = 0;

(C) f( )!0as ! 2for some ₂2[0; ): We begin by ruling out possibility (A).

Lemma 5 Let 2[0; )be a positive real number. If f is a positive solution of the problem (9) on( ; ); thenf is bounded on ( ; )and

sup

< <

f( )

2m

2jm 1jmaxfj (m 1) + 2 j;j jg

1 m

; for any m2(0;1)[(1;1):

Proof. Assume thatf is a positive solution of (9) on ( ; ). We prove this lemma for the following two cases: (i) (m 1) + <0;(ii) (m 1) + 0:

1. For the case m2(0;1) ; <0:

(i) If (m 1) + <0, in this case f⁰( )<0; 8 2 ( ; )by Lemma 2, i.e f( )< f( ); 8 2( ; ) and from (19),

(1 m)f⁰( ) f^{1 m}( ) + [ (m 1) + ]f^{1 m}( ) ( ); 8 2( ; ):

So, for any < < ;we have:

(1 m)f^{m 1}( )f⁰( ) + (m 1) ( ) + ( )

+ (m 1) ( ): (29)

The integration of (29) from to gives:

1 m

m f^m( ) +1

2 (m 1) ( ) ( )

f^m( ) m

2 (1 m)[ (1 m) ( ) 2 ] ( ): So

sup

< <

f^m( )

2m

2 (1 m)[ (m 1) + 2 ]: (30)

(ii) If (m 1) + 0, by equation (19), we have for any < < ;

(1 m)f^{m 1}( )f⁰( ) : (31)

The integration of (31) from to gives:

1 m

m f^m( ) 2

2 2 :

This implies that:

sup

< <

f^m( )

2m

2 (1 m): (32)

2. The casem2(1;1) ; >0.

(i) If (m 1) + <0, by equation (19), we have for any < < ;

(1 m)f^{m 1}( )f⁰( ) : (33)

After the integration of (33) from to ;we have:

sup

< <

f^m( )

2m

2 (m 1): (34)

(ii) If (m 1) + 0,8 2( ; );we have from (19):

(1 m)f⁰( ) = f^{1 m}( ) + [ (m 1) + ] Z

f^{1 m}( )d ;

f^{1 m}( ) +

2[ (m 1) + ]f^{1 m}( ): Then

(1 m)f^{m 1}( )f⁰( ) +

2[ (m 1) + ]: (35)

After the integration (35) from to ;we have:

sup

< <

f^m( )

2m

2 (m 1)[ (m 1) + 2 ]: (36)

Note that the bounds of (30), (32), (34) and (36) are independent of and consequently,f( )can not be unlimited at ! :

The following lemmas distinguish between the possibilities (B) and (C).

First, we give an elementary lemma for the casem2(0;1) ; <0:

Lemma 6 Assume thatf is a positive solution of problem (9)–(10) in the left neighborhood of = :Then (i) If (m 1) + 2 <0; thenf( )>0on [0; ):

(ii) If (m 1) + 2 = 0; thenf( )>0on (0; ) andf(0) = 0:

(iii) If (m 1) + 2 >0; there exists an 2(0; )such that f( )>0 on ( ; )andf( ) = 0:

Proof. For the casem2(0;1) ; <0;assume thatf is a positive solution of the problem (9)–(10) on[0; ): The integration of (19) from to gives:

f( ) =

m 1

Z d

f^{m 1}( )+ (m 1) + 2

m 1

Z

f^{m 1}( )d : (37)

Hence the lemma.

Now, we give an elementary lemma for the case m2(1;1) ; >0:

Lemma 7 Assume thatf is a positive solution of the problem (9)–(10) in the left neighborhood of = : Then

(i) If (m 1) + 0;thenf( )>0 on[0; ):

(ii) If (m 1) + 0 and (m 1) + 2 >0;then f( )>0 on [0; ):

(iii) If (m 1) + 2 0; then there exists an 2(0; )such that f( )>0 on( ; )andf( ) = 0:

Proof. For the case m 2(1;1) ; > 0; assume that f is a positive solution of the problem (9)–(10) on [0; ):

(i) If (m 1) + 0;then (m 1) + 2 >0;also we have from (19) that:

(1 m)f⁰( ) f^{1 m}( ) (38)

and the integration of (38) from to gives:

f^m( ) m

2 (m 1)

2 2 : (39)

It is easy to see from (39) thatf( )is always positive on[0; ):

Now we prove the estimate (ii) and (iii). If (m 1) + 0;we have from (19) that:

(1 m)f⁰( ) f^{1 m}( ) +

2[ (m 1) + ]f^{1 m}( ): Which implies after an integration from to ;that

f^m( ) m

2 (m 1)( + [ (m 1) + 2 ]) ( ): We observe in this case the validity of estimate (ii) and (iii). We can …x as follows:

[ (m 1) + 2 ]

; for the case < (m 1) + 2 0:

Hence the lemma.

New explicit solutions for (m 1) + 2 = 0and m2(0;1)[(1;2) Let be an arbitrary positive number.

Starting from (37)form2(0;1); if we put (m 1) + 2 = 0;we have:

f( ) =

m 1

Z d

f^{m 1}( ): (40)

The change of variableg( ) =R

f^{1 m}( )d ;givesg( ) = 0andg⁰( ) = f^{1 m}( )<0;8 2(0; ):Then the equation (40) becomes:

f( ) = [ g⁰( )]

1 1 m =

m 1 g( ): (41)

The last equation can be resolved for anym2(0;1)[(1;2):

In fact, from the proof of Lemma1 that a necessary condition for the existence of a positive solution of problem (9)–(10) in a left neighborhood of = ; is either >0 form >1; <0for m2(0;1) or = 0 and >0:

Then the value _{m 1} is always positive. In this case the equation (41) is equivalent to g⁰( )g^{m 1}( ) =

m 1

1 m

1 m: (42)

The integration of (42) starting from to gives:

1

mg^m( ) = 1

2 m m 1

1 m

2 m 2 m :

From (41), we have after an elementary computation that:

f( ; ) = m

(m 1) (2 m)

m 2 m 2 m

1 m

; 0 < ; (43)

is a continuous function, for any choice of >0;with compact support and satis…es:

f(0; ) =f( ; ) = 0:

Theorem 8 The following statements (i) and (ii) hold.

(i) Assume thatU= 0:Then for every >0there exists a solutionf( ; )of the boundary value problem (9)–(10) which is positive in (0; ) if and only if (m 1) + 2 = 0 and <0 form2(0;1); or if

(m 1) + 2 = 0and >0 form >1:

(ii) Assume that U > 0: Then the boundary value problem (9)–(10) admits a unique solution and there exists a unique (U)>0 such that f( ; (U))is positive on (0; )if and only if (m 1) + 2 <0 and 0form2(0;1); or if (m 1) + 2 >0 and 0 form >1:

Proof. By Lemma1 a necessary condition for the existence of such a solution is that 0for m2(0;1) or 0 form >1:

For = 0; >0;we already gave the solution (15) of (19) form2(0;1)[(1;2)which is:

f( ; ) = m²

2 (2 m)( )²

1 m

; 0 < :

This solution is continuous and satis…es for everyU >0;that f(0; ) =U andf( ; ) = 0;where (U) =

r2 (2 m)

m² U^m:

In the following, for the case 6= 0; we put <0 form2(0;1)and >0 form >1:

We already proved in Lemma3, the local existence of positive solution in the left neighborhood of = of problem (9)–(10). This local solution is unique and can be continued back to = 0as a positive solution withf(0)>0if and only if <0and (m 1)+2 <0form2(0;1);also if >0and (m 1)+2 >0 form >1(by Lemmas6 and7).

Now, the boundary condition (10) at = 0is satis…ed if we can …nd a (U)such that

f(0; ) =U: (44)

If only one such a exists, the solution is unique. We distinguish two cases:

(i)U = 0: By Lemmas6and 7, the equation (44) can only be satis…ed if (m 1) + 2 = 0:Moreover (44) is then satis…ed for any >0:Then there exists a nontrivial weak solutionf with compact support of the problem (9)–(10) with the property:

f( )>0 on (0; ); f( ) = 0 on f0g [[ ;1):

(ii)U >0: It follows from Lemmas6and7 that now a necessary condition for (44) to have a solution is that <0and (m 1) + 2 <0form2(0;1)or >0and (m 1) + 2 >0form >1:

With this intention, let us suppose that f( ; )is a solution of the problem (9)–(10) on(0; );then for any >0 the function ^m2f( ; )is a solution of (9)–(10) on(0; ):If = ¹;then:

2

mf(0; 1) =U: (45)

Because f(0; 1) >0 in the cases (m 1) + 2 <0 for m2 (0;1) or (m 1) + 2 >0 form > 1: We obtain a unique solution = (U)of (45).

Thus, the function f( ; (U))is the unique solution of (9)–(10), with the property:

f( )>0 on [0; ); f( ) = 0 on [ ;1): Hence the theorem.

We denote by(z)₊ the positive part of z;which isz ifz >0and else is zero.

Now, we give the principal theorem of this work.

Theorem 9 (Global existence and blow-up of self-similar solutions) Let a(t); c(t) be positive real functions oft;which satisfy:

a(0) =c(0) = 1; a_(0) = ; c_(0) = : (46) Then, forf 2C([0; ];R⁺);the problem (7) admits an exact solution in the form:

u(x; t) =c(t)f( ); with = x

a(t); x2R; t >0; (47)

if the basic pro…lef is a solution of the problem (9) on [0; ]and satis…es in each point:

f^m( )f⁰⁰( ) = f( ) + f⁰( ): Thereupon, we separate the following cases:

(1) If m+ 2 <0; the functionsa(t); c(t)are given by:

8<

:

a(t) = (1 ( m+ 2 )t) ^m+2 ;

c(t) = (1 ( m+ 2 )t) ^{m+2 ;} 8t >0: (48) (2) If m+ 2 = 0; the functionsa(t); c(t)in this case are given by:

( a(t) =e ^t;

c(t) =e ^t; 8t >0: (49)

In each case (1) and (2). The problem (7) admits a global solution in time under the generalized self-similar form, this solution de…ned for all t >0. Moreover, if <0; we have

t!lim+1u(x; t) = 0 for allx2R:

(3) If m+ 2 >0; the functionsa(t); c(t)are given by:

8<

:

a(t) = (1 ( m+ 2 )t)₊^m+2 ; c(t) = (1 ( m+ 2 )t)₊ ^m+2 ;

0< t < T: (50)

The momentT = _m+2¹ represents the maximal existence value of the functionsa(t); c(t):Moreover;

if >0;the problem (7) admits a solution under the generalized self-similar form, which blows up in a …nite time. The solution is de…ned for all t 2 (0; T); where T represents the blow-up time of the solution such that:

for allx2R; lim

t!T u(x; t) = +1; with T = 1

m+ 2 >0:

Proof. We have already proved in Theorem 8 the existence of solution f of (9), which is with compact support[0; ]if and only if

1. (m 1) + 2 = 0and[ <0form2(0;1) or >0form >1]; 2. (m 1) + 2 <0and 0form2(0;1);

3. (m 1) + 2 >0and 0form >1:

Now, to determine the functionsa(t); c(t);just solve the system (6) which is ( _{c_}

c =k1c^m a²;

_ a

a =k2c^m a²:

(51) The conditions (46), implyk1= andk2= ;then the system (51), can be resolved as follows

( _{c_}

c = ^c_a^m2;

_ a

a = ^c_a^m2: ) a_

a= c_ c; then we deduce after an integration from0 totthat:

a(t) =c (t): (52)

If we replace (52) in (51), we obtain

c ^{m+2 1}dc= dt: (53)

If m+ 2 6= 0;we obtain easily the solution of (53) as follows c(t) = (1 ( m+ 2 )t)₊ ^m+2 in the same way we …nd:

a(t) = (1 ( m+ 2 )t)₊^m+2 :

We deduce that the functionsa(t); c(t) are globally de…ned if m+ 2 <0;and a(t); c(t)are maximal functions if m+ 2 >0;and well de…ned if and only if

0< t < T = 1 m+ 2 :

If m+ 2 = 0;the functionsa(t); c(t)are de…ned globally, we obtain in this case:

( a(t) =e ^t;

c(t) =e ^t; 8t >0:

We notice from this theorem that we have two time behaviors of functionsa(t); c(t);their behaviors depend on parameters of similarity ; :

In (1) and (2) i.e m+ 2 0;the functionsa(t); c(t)are de…ned globally in time. Now, the pro…lef is a bounded positive function for allx2R: If <0 we have:

t!lim+1c(t) = 0:

In each case (1) and (2). Thus:

t!lim+1u(x; t) = lim

t!+1c(t)f x

a(t) = 0:

(3) In the case m+ 2 >0;we have a(t); c(t)given in (50), is well de…ned if and only if:

0< t < T = 1 m+ 2 :

We recall that the solution blows up in …nite time if there exists a timeT <+1;which we call it the blow-up time, such that the solution is well de…ned for all0< t < T;while

sup

x2Rju(x; t)j !+1; whent!T : If >0;the valueT represents the blow-up time of the solution, thus lim

t!T c(t) = +1and lim

t!T u(x; t) = lim

t!T c(t)f x

a(t) = +1:

4 Examples

Example 1 (Blow-up solution) For = 0and >0;we have from Theorem 9that m+ 2 = m >0 and

a(t) = 1; c(t) = (1 mt)

1 m

+ ; where0< t < T = 1 m:

In this case, the self-similar form is a solution of separation of variables u(x; t) =c(t)f(x); where f is given in (15), by:

f(x) = m²

2 (2 m)( x)²

1 m

; m2(0;1)[(1;2); 0 x < :

Then, the solution of (1) in the generalized self-similar form (47) for = 0; >0 is given by:

u(x; t) = 8<

:

m²

2(2 m)(1 mt)( x)²

1

m; 0 x < ; m2(0;1)[(1;2):

0; otherwise.

This solution blows up when t! ¹m: In this case

(U) =

r2 (2 m)

m² U^m; for everyU >0:

Example 2 (Global and blow-up solutions) Now, we present the new explicit solutions on the general- ized self-similar form of the equation (1). Because the pro…le f is bounded with compact support, then it is an integrable function on R; where

Z

R

f( )d =M; for someM >0:

Therefore, form2(0;1)[(1;1)and Z

R

u(s; t)ds=c^m(t): We can …nd explicitly a new exact solution of (1). Indeed

Z

R

u(s; t)ds= Z

R

c(t)f s

a(t) ds=a(t)c(t) Z

R

f( )d =c^m(t);

this implies that

c^{m 1}(t)

a(t) =M: (54)

According to the formulasa(t)andc(t)in (48) and (49), the equality (54) implies that:

= (1 m): Because <0 form2(0;1) and >0 form >1;then <0:

For (m 1) + = 0;we already gave the solution (18) of (9) for m2(0;1)[(1;1)which is:

f( ; ) = m

2 (m 1)

2 2

1 m

; 0 < :

Now, we determine the functions a(t)andc(t):We have from Theorem9 m+ 2 = m+ 2 (1 m) = (2 m):

If

(2 m)6= 0; i:e: m6= 2;

the functions a(t)andc(t)are given by (

a(t) = (1 (2 m)t)

m 1 m 2

+ ;

c(t) = (1 (2 m)t)

1 m 2

+ ;

m6= 2; 0< t < T:

WhereT represents the maximal existence value

T = ₍₂¹_m); if m >2;

T = +1; if m <2:

We obtain the exact solution of (1), form2(0;1)[(1;2)[(2;1)as follows

u(x; t) = 8>

<

>:

c(t) _2(m^m₁₎ ² x²(1 (2 m)t)

2(1 m) m 2

+

1 m

; jxj< r1;

0; otherwise,

(55)

where

r1= (1 (2 m)t)

m 1 m 2

+ :

This solution is de…ned globally in time ifm2(0;1)[(1;2); and blows up when

t! 1

(2 m); form2(2;1): In this case

(U) = s

2 (m 1)

m U^m; for everyU >0:

If m= 2; the functionsa(t)andc(t) are written as (49), in this case we obtain = and

u(x; t) = (

e ^t

q 2 x²e^{2 t} ; jxj< e ^t;

0; otherwise.

This solution is de…ned globally in time. In this case (U) = U

p ; for everyU >0:

For the functions a(t)andc(t)that satisfy the parameters of the classical self-similar form, i.e.

c(t) =t ; a(t) =t : The self-similar solution (8) of the equation (1) is written as:

u(x; t) =t f( ); with =xt ;

where and are exponents which satisfy the similarity condition ([11], [12], [13])

m+ 2 + 1 = 0; (56)

and the functionf is the self similar pro…le which to be determined by the solution of the following di¤ erential equation:

f^m( )f⁰⁰( ) = f( ) + f( ):

If = (1 m);the similarity condition (56) is equivalent to:

m+ 2 + 1 = (2 m) + 1 = 0;

this implies that:

= 1

m 2 and =1 m

m 2:

We get form2(0;1) that ; <0: So the solution (55) forc(t) =t ; a(t) =t ;is written as:

U(x; t) =t^m12 m 2 (2 m)

2 x²t^{2(1m m)2}

1 m

; for jxj< t^{mm 12}; m2(0;1): Also

U(x; t) = 8>

<

>:

1 1 m

1

m t^{1mm2 m(1}_{2(2 m)}^{m) 2} x²t^{2(1m m)2}

1 m m

1 1 m

; jxj< t^{mm 12};

0; otherwise.

(57)

If we put form2(0;1)the parameterp=₁¹_m >1;then the solution (57) of the equation (1) in the classical self-similar form is written as:

U(x; t) =p^pp1B^p(x; t); p >1:

WhereB( ) is the Barenblatt solutions of the porous medium equation (2), which are given in (3) under the following form:

B (x; t) = 8<

:

t ^p+11 _2p(p+1)^{p 1 2} x²t ^p+12

1 p 1

; for jxj< t^p+11 ; p >1;

0; otherwise.

Example 3 (Global solution) We present the second new explicit solutions on the generalized self-similar form of the equation (1). For0< m6= 1and

Z

R

u(s; t)ds=c^m(t) a(t) ; we can …nd explicitly a new exact solution of (1). In fact

Z

R

u(s; t)ds= Z

R

c(t)f s

a(t) ds=a(t)c(t) Z

R

f( )d =c^m(t) a(t) ; this implies that

c^{m 1}(t)

a²(t) =M: (58)

According to the formulasa(t)andc(t)in (48) and (49), the equality (58) implies that:

= (1 m)

2 ; then <0:

For (m 1) + 2 = 0;we gave the solution (43) of (9) form2(0;1)[(1;2) which is:

f( ; ) = m

(m 1) (2 m)

m 2 m 2 m

1 m

; 0 < :

Now, we determine the functions a(t)andc(t):We have from Theorem9 m+ 2 = m+ (1 m) = <0:

The functionsa(t)andc(t) are given by:

(

a(t) = (1 t)^12m;

c(t) = ₁¹ _t; 8t >0:

We obtain the exact solution of (1), form2(0;1)[(1;2)as follows:

u(x; t) = 8<

:

1 1 t

mx^m (m 1)(2 m)a^m(t)

2 m x^{2 m}c^{(1 m)(22 m)}(t)

1

m; x2r₂;

0; otherwise,

where

r₂= 0; (1 t)^12m :

This solution is de…ned globally in time for anym2(0;1)[(1;2): In this case >0:

5 Conclusion

In this paper we have discussed the existence and uniqueness of positive solutions for a class of nonlinear di¤usion equation not in divergence form, under the generalized self-similar form. Also we have found new solutions; the behavior of these solutions depends on some parameters (that satisfy some conditions), which make their existence global or local, and we generalized the families of self-similar solutions of the porous medium equation, which is formed by the Barenblatt solutions.

Acknowledgments. The authors are deeply grateful to the reviewers and editors for their insightful comments that have helped to improve the quality of this research work that is supported by the General Direction of Scienti…c Research and Technological Development (DGRSTD)- Algeria.

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