New method for the existence and uniqueness of solution of nonlinear parabolic equation

Abstract

There are two contributions in this paper. The first is that the abstract result for the existence of the unique solution of certain nonlinear parabolic equation is obtained by using the properties of H-monotone operators, consequently, the proof is simplified compared to the corresponding discussions in the literature. The second is that the connections between resolvent of H-monotone operators and solutions of nonlinear parabolic equations are shown, and this strengthens the importance of H-monotone operators, which have already attracted the attention of mathematicians because of the connections with practical problems.

Wei et al. Boundary Value Problems (2015) 2015:88 DOI 10.1186/s13661-015-0343-3

R E S E A R C H Open Access

http://crossmark.crossref.org/dialog/?doi=10.1186/s13661-015-0343-3&domain=pdf

Web End = New method for the existence and uniqueness of solution of nonlinear parabolic equation

Li Wei1, Ravi P Agarwal2,3 and Patricia JY Wong4*

*Correspondence: mailto:[email protected]

Web End [email protected]

4School of Electrical and Electronic Engineering, Nanyang Technological University, 50 Nanyang Avenue, Singapore, 639798, SingaporeFull list of author information is available at the end of the article

Abstract

There are two contributions in this paper. The rst is that the abstract result for the existence of the unique solution of certain nonlinear parabolic equation is obtained by using the properties of H-monotone operators, consequently, the proof is simplied compared to the corresponding discussions in the literature. The second is that the connections between resolvent of H-monotone operators and solutions of nonlinear parabolic equations are shown, and this strengthens the importance of H-monotone operators, which have already attracted the attention of mathematicians because of the connections with practical problems.

MSC: 47H05; 47H09Keywords: H-monotone operator; resolvent; subdierential; parabolic equation

1 Introduction and preliminaries1.1 Introduction

Nonlinear boundary value problems involving the generalized p-Laplacian operator arise from many physical phenomena, such as reaction-diusion problems, petroleum extraction, ow through porous media and non-Newtonian uids, just to name a few. Thus, the study of such problems and their generalizations have attracted numerous attention in recent years. In particular, we would mention the books of Lieberman [, ] where in [] the theory of linear and quasilinear parabolic second-order partial dierential equations is elaborated, with emphasis on the Cauchy-Dirichlet problem and the oblique derivative problem in bounded space-time domains; while in [] a detailed qualitative analysis of second-order elliptic boundary value problems that involve oblique derivatives is presented. A sample of other research work that contributes to the literature of parabolic and elliptic problems includes [] listed chronologically as well as the references cited therein. For time-periodic case which is the concern of this paper, we refer the reader to [].

In , Wei and Agarwal [] studied the following nonlinear elliptic boundary value problem involving the generalized p-Laplacian:

div[(C(x) + |u|)pu] + |u|qu + g(x, u(x)) = f (x), a.e. in ,

, (C(x) + |u|)

p

(.)

u x(u(x)), a.e. on ,

2015 Wei et al. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License

(http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro

vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and

indicate if changes were made.

Wei et al. Boundary Value Problems (2015) 2015:88 Page 2 of 18

where C(x) Lp( ), is a non-negative constant and denotes the exterior normal

derivative of . It is shown that (.) has solutions in Ls( ) under some conditions, where

N

N+ < p s < +, q < + if p N, and q

Np

Np if p < N, for N . We observe

that the proof, which uses Theorem . (stated in Section .) as the main tool, is very complicated, since one needs to check that conditions (.) and (.) and the compactness of A + C are satised.

In , Wei et al. [] extended the work on elliptic equation to the following nonlinear parabolic equation involving the generalized p-Laplacian with mixed boundary conditions:

Some new technique has been used to tackle the existence of solutions of (.); specifically, the problem is divided into the following two auxiliary equations: (i) a parabolic equation with Dirichlet boundary conditions (.), and (ii) a parabolic equation with Neumann boundary value conditions (.):

ut div[(C(x, t) + |u|)pu] + |u|pu = f (x, t), (x, t) (, T),

, (C(x, t) + |u|)

p

t div[(|u|p)|u|pu] + |u|ru

+ |u|ru + g(x, u, ut, u) = f (x, t), (x, t) (, T),

, (|u|p)|u|pu x(u(x, t)), (x, t) (, T),

u(x, ) = u(x, T), x ,

R+ is a continuous nonlinear mapping such that pt (t)+(p)(t) > , (t) k, for t , limt+ (t) = k > , here k and k are positive constants.

Let :

where : R+ {}

R

R is a proper, convex and lower-semicontinuous function with x() = . Let x be the subdierential of x, i.e., x x. Suppose that x() and for each t

R, the function

R is measurable for > . More details of (.) will be presented

u

t div[(C(x, t) + |u|)

u] + |u|pu = f (x, t), (x, t) (, T), , (C(x, t) + |u|)

u (u) h(x, t), (x, t) (, T),

u(x, ) = u(x, T), a.e. x .

p

p

(.)

u

t div[(C(x, t) + |u|)

u] + |u|pu = f (x, t), (x, t) (, T),

u = w, (x, t) (, T),

u(x, ) = u(x, T), a.e. x ,

p

(.)

(.)

By using Theorems . and . (stated in Section .), it is shown that (.) has a unique solution. By employing Theorem ., it is proved that (.) has a unique solution in Lp(, T; W,p( )), which implies that (.) has a unique solution in Lp(, T; W,p( )), where p < +. However, we observe that the inequality (.) is not easy to check during the

discussion.

Motivated by the work of Kawohl et al. [, , , ], Serrin et al. [, , , ] as well as Wei et al. [, ], in this paper we shall consider the following parabolic problem:

u

u (u) h(x, t), (x, t) (, T).

(.)

R be a given function such that, for each x , x = (x, ) :

R

x (I + x)(t)

in Section .

Wei et al. Boundary Value Problems (2015) 2015:88 Page 3 of 18

There are some major dierences between parabolic problems (.) and (.): (i) The main part div[(|u|p)|u|pu] in (.) includes the main part div[(C(x, t) + |u|)

p

u] in (.); (ii) the term g(x, u, ut, u) is considered in (.) but not in (.);

(iii) x(u(x, t)) in (.) is dierent from (u) h(x, t) in (.).

The existence of the unique solution of (.) will be discussed in L(, T; L( )), which

does not change while p is varying from N

N+ to + for N . Hence, the result is dif

ferent from that on (.) in []. Our main tool in this paper will be Theorem . (stated in Section .). Consequently, the proof of our result is dierent from and comparatively simplied with respect to that of [].

Actually, (.) is very general and it includes the following special cases. The related work can be found in [] and the references cited therein.

Example . If we set (t) = + t( + t) , t , then it is obvious that :

R+

{}

R+

is a continuous nonlinear mapping, (t) and limt+ (t) = . Moreover,

pt (t) + (p )(t) = pt

( + t)

+ (p ) + (p ) t

+ t > .

So, if , then (.) becomes the following parabolic capillarity equation:

u

p

t div[( + |

u|

+|u|

p )|u|pu] + |u|r

u + |u|r

u

+ g(x, u, ut, u) = f (x, t), (x, t) (, T),

, ( +

+|u|

p )|u|pu x(u(x, t)), (x, t) (, T),

u(x, ) = u(x, T), x .

(.)

|u|

p

Example . For < p , if we set (t) = (C +t

p )

p

t

pp , t > , where C , then it is ob-

vious that : R+

R+ is a continuous nonlinear mapping, (t) and limt+ (t) = .

Moreover,

pt (t) + (p )(t) =

C + t p

p

t

p

tC( p) + (p )t

C + tp

Ct + (p )tp + > .

If , then (.) becomes the following parabolic equation with generalized p-

Laplacian:

u

t div[(C(x) + |u|)

=

C + t p

p

t

p

p

u] + |u|ru + g(x, u, ut, u)

= f (x, t), (x, t) (, T),

, (C(x) + |u|)

(.)

Example . If, in (.), C(x) , then (.) becomes the following parabolic p-Laplacian

equation:

u

t pu + |u|ru + g(x, u, ut, u) = f (x, t), (x, t) (, T),

, |u|pu x(u(x, t)), (x, t) (, T),

u(x, ) = u(x, T), x .

p

u x(u(x, t)), (x, t) (, T),

u(x, ) = u(x, T), x .

(.)

Wei et al. Boundary Value Problems (2015) 2015:88 Page 4 of 18

mp+p , t > , where m , m + s + =

R+ is a continuous nonlinear mapping, (t) and

+ tp s

m + (p )tp > .

So, if , then (.) becomes the following parabolic curvature equation:

1.2 Preliminaries

Let X be a real Banach space with a strictly convex dual space X. We shall use (, ) to

denote the generalized duality pairing between X and X. We shall use and w-lim

to denote strong and weak convergence, respectively. Let X Y denote the space X

embedded continuously in space Y . For any subset G of X, we denote by int G its interior and G its closure, respectively. For two subsets G and G in X, if G = G and int G =

int G, then we say G is almost equal to G, which is denoted by G G. A mapping

T : X X is said to be hemi-continuous on X [] if w-limt T(x + ty) = Tx for any

x, y X.

A function is called a proper convex function on X [] if is dened from X to (, +], not identically +, such that (( )x + y) ( ) (x) + (y), whenever

x, y X and .

A function : X (, +] is said to be lower-semicontinuous on X [] if

lim infyx (y) (x), for any x X.

Given a proper convex function on X and a point x X, we denote by (x) the set

of all x X such that (x) (y) + (x y, x), for every y X. Such element x is called

the subgradient of at x, and (x) is called the subdierential of at x [].

Let Jr denote the duality mapping from X into X, which is dened by

Jr(x) =

f X : (x, f ) = x r, f = x r

Example . For s , if we set (t) = ( + t

p, then it is obvious that : R+

p ) s t

limt+ (t) = . Moreover,

pt (t) + (p )(t) = t

mp+ p

u

|u|mu x(u(x, t)), (x, t) (, T),

u(x, ) = u(x, T), x .

(.)

t div[( + |u|) s

|u|mu] + |u|ru + g(x, u, ut, u)

= f (x, t), (x, t) (, T),

, ( + |u|) s

, x X,

where r > is a constant. We use J to denote the usual normalized duality mapping. It is known that, in general, Jr(x) = x rJ(x), for all x = . Since X is strictly convex, J is a

single-valued mapping [].

A multi-valued mapping A : X X is said to be accretive [, ] if (v v, Jr(u u))

, for any ui D(A) and vi Aui, i = , . The accretive mapping A is said to be m-accretive

if R(I + A) = X for some > .

A multi-valued operator B : X X is said to be monotone [] if its graph G(B) is a

monotone subset of X X in the sense that (u u, w w) , for any [ui, wi] G(B),

i = , . Further, B is called strictly monotone if (u u, w w) and the equality holds if and only if u = u. The monotone operator B is said to be maximal monotone if G(B)

is maximal among all monotone subsets of X X in the sense of inclusion. Also, B is

maximal monotone if and only if R(B + J) = X, for any > . The mapping B is said to be

Wei et al. Boundary Value Problems (2015) 2015:88 Page 5 of 18

coercive [] if limn+ (xn, xn)/ xn = + for all [xn, xn] G(B) such that limn+ xn =

+.

Let B : X X be a maximal monotone operator such that [, ] G(B), then the equa

tion J(ut u) + tBut has a unique solution ut D(B) for every u X and t > . The

resolvent JBt and the Yosida approximation Bt of B are dened by []

JBtu = ut,

Btu =

t J(ut u),

for every u X and t > . (Hence, [JBtu, Btu] G(B).)

Denition . ([]) Let C be a closed convex subset of X and let A : C X be a multi-

valued mapping. Then A is said to be a pseudo-monotone operator provided that(i) for each x C, the image Ax is a non-empty closed and convex subset of X;(ii) if {xn} is a sequence in C converging weakly to x C and if fn Axn is such that

lim supn(xn x, fn) , then to each element y C, there corresponds an

f (y) Ax with the property that (x y, f (y)) lim infn(xn x, fn);(iii) for each nite-dimensional subspace K of X, the operator A is continuous from C K to X in the weak topology.

Denition . ([, ]) Let H be a Hilbert space. Let H : H H be a single-valued

mapping and A : H H be a multi-valued mapping. We say that A is H-monotone if A

is monotone and R(H + A)(H) = H, for every > .

Lemma . ([]) If A : X X is a everywhere dened, monotone, and hemi-continuous

mapping, then A is maximal monotone. If, moreover, A is coercive, then R(A) = X.

Lemma . ([]) If : X (, +] is a proper convex and lower-semicontinuous

function, then is maximal monotone from X to X.

Lemma . ([]) If A and A are two maximal monotone operators in X such that (int D(A)) D(A) = , then A + A is maximal monotone.

Theorem . ([]) Let X be a real Banach space with a strictly convex dual space X. Let

J : X X be a duality mapping on X and there exists a function : X [, +) such

that for all u, v X,

Ju Jv (u v). (.)

Let A, C : X X be accretive mappings such that

(i) either both A and C satisfy condition (.), or D(A) D(C) and C satises

condition (.):

for u D(A) and v Au, there exists a constant C(a, f ) such that

v f , J(u a) C(a, f ).

(ii) A + C is m-accretive and boundedly-inversely-compact.

(.)

Wei et al. Boundary Value Problems (2015) 2015:88 Page 6 of 18

Let C : X X be a bounded continuous mapping such that, for any y X, there is a

constant C(y) satisfying (C(u + y), Ju) C(y) for any u X. Then the following results

hold:(a) [R(A) + R(C)] R(A + C + C);

(b) int[R(A) + R(C)] int R(A + C + C).

Theorem . ([]) Let T : X X be a bounded and pseudo-monotone operator, K be

a closed and convex subset of X. Suppose that is a lower-semicontinuous and convex function dened on K which is not always + such that (v) (, +] for all v K.

Suppose there exists v K such that (v) < + and

(v v, Tv) + (v)

v

Then there exists u K such that

(u v, Tu) (v) (u), v K.

Theorem . ([]) Let X be a real reexive Banach space with both X and its dual X being convex spaces. Let S : D(S) X X be a linear maximal monotone operator and

T : X X be a pseudo-monotone and coercive operator. Then, for each f X, there exists

an u D(S) such that, in the weak sense, Su + Tu = f .

Theorem . ([]) Let X be a real reexive Banach space with both X and its dual X being strictly convex. Let J be the normalized duality mapping. Let A and B be two maximal monotone operators in X. Suppose there exist k < and C, C > such that

a, J(Btv)

k Btv C Btv C (.)

for v D(A), a Av and t > , where Bt is the Yosida approximation of B. Then R(A) +

R(B) R(A + B).

Theorem . ([]) Let A : H H be a maximal monotone operator and H : H H be

a bounded, coercive, hemi-contiunuous, and monotone mapping. Then A is H-monotone.

2 Main results

In this paper, unless otherwise stated, we shall assume that

N ,

r +

, as v , v K.

N

N + < p < +, ri

min

p, p

, i = , ,

r = .

In (.), is a bounded conical domain of a Euclidean space RN with its boundary C

[], T is a positive constant, , and are non-negative constants, and denotes the exterior normal derivative of . We shall assume that Greens formula is available.

p +

p = ,

r = ,

r +

Wei et al. Boundary Value Problems (2015) 2015:88 Page 7 of 18

Suppose that g :

RN+

R is a given function satisfying the following conditions:

(a) Carathodorys conditions

x g(x, r) is measurable on , for all r

RN+;

r g(x, r) is continuous on

RN+, for almost all x .

(b) Growth condition

g(x, s, . . . , sN+) h(x) + k|s|min{p/p ,},

where (s, s, . . . , sN+)

RN+, h(x) L( ) Lp ( ) and k is a positive constant.(c) Monotone condition g is monotone with respect to r, i.e.,

g(x, s, . . . , sN+) g(x, t, . . . , tN+) (s t)

for all x and (s, . . . , sN+), (t, . . . , tN+)

RN+.

(d) Coercive condition

g(x, s, . . . , sN+)s ks,

where k is a xed positive constant.

Now, we present our discussion in the sequel.

Lemma . ([]) Let X denote the closed subspace of all constant functions in W,p( ). Let X be the quotient space W,p( )/X. For u W,p( ), dene the mapping P : W,p( )

X by

Pu =

meas( )

u dx.

Then there is a constant k > such that for all u W,p( ),

u Pu p k u (L

p( ))N .

Lemma . Dene the mapping B : Lp(, T; W,p( )) Lp (, T; (W,p( ))) by

(w, Bu) =

T

|u|p

|u|pu, w dx dt +

T

|u|ruw dx dt

+

T

|u|ruw dx dt

for any u, w Lp(, T; W,p( )). Then B is strictly monotone, pseudo-monotone, and coer

cive.

(Here, , and | | denote the Euclidean inner-product and Euclidean norm in

RN, re-

spectively.)

Wei et al. Boundary Value Problems (2015) 2015:88 Page 8 of 18

Proof Step . B is everywhere dened. For u, w Lp(, T; W,p( )), we nd

(w,

Bu)

T

k|u|p|w| dx dt +

T

|u|r|w| dx dt

+

T

|u|r|w| dx dt

k u p/p

Lp(,T;W,p( )) w L

p(,T;W,p( )) + w L

r (,T;Lr ( )) u r/r

Lr (,T;Lr ( ))

+ w L

r (,T;Lr ( )) u r/r

Lr (,T;Lr ( )).

Since W,p( ) Lp( ) Lr

( ) and W,p( ) Lp( ) Lr

( ), for v W,p( ),

we have v L

r ( ) k v W

,p( ), v L

r ( ) k v W

,p( ), where k and k are positive con-

stants. Hence,

(w,

Bu)

k u p/p

Lp(,T;W,p( )) w L

p(,T;W,p( ))

+ k u r/r

Lp(,T;W,p( )) w L

p(,T;W,p( ))

+ k u r/r

Lp(,T;W,p( )) w L

p(,T;W,p( )),

which implies that B is everywhere dened.Step . B is strictly monotone.

For u, v Lp(, T; W,p( )), we have

(u v, Bu Bv)

T

|u|p

|u|p

|v|p

|v|p |u| |v| dx dt

+

T

|u|r |v|r |u| |v| dx dt

+

T

|u|r |v|r |u| |v| dx dt.

If we set f (s) = s

p (s), s > , then in view of the assumption of , we have

f (s) =

sp > ,

which implies that f is strictly monotone. Hence, B is strictly monotone.

Step . B is hemi-continuous.

It suces to show that for any u, v, w Lp(, T; W,p( )) and t [, ], (w, B(u + tv)

Bu) as t . Since

w,

B(u + tv) Bu

p

(s) + s (s)

T

|u

+ tv|p

|u + tv|p(u + tv)

|u|p

|u|pu

|w| dx dt

Wei et al. Boundary Value Problems (2015) 2015:88 Page 9 of 18

+

T

|u

+ tv|r(u + tv) |u|ru

|w|

dx dt

+

T

|u

+ tv|r(u + tv) |u|ru

|w|

dx dt,

by Lebesques dominated convergence theorem and noting that is continuous, we nd

lim

t

w, B(u + tv) Bu = .

Hence, B is hemi-continuous.Step . B is coercive.

We shall rst show that for u Lp(, T; W,p( )),

u L

p(,T;W,p( )) k

|u|p dx dt p+ k, (.)

where k and k are positive constants.In fact, using Lemma ., we know that, for u Lp(, T; W,p( )),

u

T

meas( )

u dx

Lp( ) k |u|p dx p.

Thus,

u

meas( )

u dx p W,p( )

=

u

meas( )

u dx pLp( ) +

u meas( )

u dx p (Lp( ))N

kp + |u|p dx.

Since

u

meas( )

u dx W,p( ) u W,p( )

u dx W,p( ),

we have

u W

,p( )

u

meas( )

u dx

W,p( ) + Const.

Therefore,

u L

p(,T;W,p( ))

u

meas( )

u dx Lp(,T;W,p( ))+ k

kp + p

T

|u|p dx dt p+ k.

If we set k = (kp + )

p , then (.) is true.

Wei et al. Boundary Value Problems (2015) 2015:88 Page 10 of 18

Since limt+ (t) = k > , there exists suciently large K > such that (t) > l whenever t > K. Now, for u Lp(, T; W,p( )), let u L

p(,T;W,p( )) +. Using (.), we nd

(u, Bu)

u L

p(,T;W,p( ))

T

(|u|p)|u|p dx dt u L

T

|u|r dx dt

u L

T

|u|r dx dt

u L

p(,T;W,p( ))

=

p(,T;W,p( )) +

p(,T;W,p( )) +

> u L

p(,T;W,p( ))

l

T

|u|p dx dt +

T

|u|r dx dt

+

T

|u|r dx dt

> u L

p(,T;W,p( ))

l

T

|u|p dx dt +.

This completes the proof.

Lemma . The mapping : Lp(, T; W,p( ))

R dened by

(u) =

T

x

u| (x, t)

d (x) dt,

for any u Lp(, T; W,p( )), is proper, convex, and lower-semicontinuous on Lp(, T;

W,p( )). Moreover, the subdierential of is maximal monotone in view of Lemma ..

Proof The proof is similar to that of Lemma . in [].

Lemma . ([]) Dene S : D(S) Lp (, T; (W,p( ))) by

Su(x, t) =

u

t ,

where

D(S) =

u

t Lp

u Lp

, T; W,p( )

, T;

W,p( )

, u(x, ) = u(x, T)

.

The mapping S is linear maximal monotone.

Denition . Dene a mapping A : L(, T; L( )) L(,T;L( )) by

Au =

w(x) L

, T; L( )

|w(x) Bu + (u) + Su

for u D(A) = {u L(, T; L( ))| there exists a w(x) L(, T; L( )) such that w(x)

Bu + (u) + Su}.

Lemma . Dene the mapping F : Lp(, T; W,p( )) Lp (, T; (W,p( ))) by

(v, Fu) =

T

g x, u,ut , u

v(x, t) dx dt

for u(x, t), v(x, t) Lp(, T; W,p( )). Then F is everywhere dened.

Wei et al. Boundary Value Problems (2015) 2015:88 Page 11 of 18

Proof Step . For u(x, t) Lp(, T; W,p( )), x g(x, u, ut, u) is measurable on .

From the fact that u(x, t), uxi Lp( ), i = , , . . . , N, we see that x (u, ux , . . . , uxN ) is

measurable on . Combining with the fact that g satises Carathodorys conditions, we know that x g(x, u, ut, u) is measurable on .

Step . F is everywhere dened.

For u, v Lp(, T; W,p( )), we have

(v,

Fu)

T

h(x) v(x,

t)

dx dt + k

T

u(x,

t)

p/p

v(x,

t)

dx dt

T p

h(x)

Lp ( ) + k u p/p

Lp(,T;W,p( ))

v L p(,T;W,p( )),

which implies that F is everywhere dened.This completes the proof.

Denition . Dene the mapping H : L(, T; L( )) L(, T; L( )) by

Hu(x) = v(x) L

, T; L( )

|v(x) = Fu(x)

for u D(H) = {u(x) L(, T; L( ))| there exists v(x) L(, T; L( )) such that v(x) =

Fu(x)}, where F is the same as in Lemma ..

Lemma . The mapping H : L(, T; L( )) L(, T; L( )) dened in Denition .

is bounded, coercive, hemi-continuous, and monotone.

Proof Step . H is bounded.From condition (b) of g, we know that

Hu L(,T;L( )) =

T

g

x, u,ut , u

dx dt

k u L(,T;L( )) + k

h(x)

L( ),

where k and k are positive constants. This implies that H is bounded.

Step . H is coercive.

From condition (d) of g, we know that

(u, Hu) =

T

g x, u,ut , u

u dx dt k

T

|u| dx dt

= k u L(,T;L( )) +,

as u L

(,T;L( )) +. Hence, H is coercive.

Step . H is hemi-continuous.

Since g satises condition (a), we have, for any w(x, t) L(, T; L( )),

w, H(u + tv) Hu

=

g

x, u + tv,ut + tvt ,(u + tv)

g

x, u,ut ,u w dx dt ,

as t , which implies that H is hemi-continuous.

Wei et al. Boundary Value Problems (2015) 2015:88 Page 12 of 18

Step . H is monotone.

In view of condition (c) of g, we have

(u v, Hu Hv)

=

T

x, u,ut , u

g

x, v,vt ,v u(x, t) v(x, t)

dx dt ,

g

which implies that H is monotone.This completes the proof.

Lemma . For all u, v Lp(, T; W,p( )), we have

v, (u) =

T

x

u| (x, t)

v| (x, t) d (x) dt.

Moreover, ().

Proof The idea of the proof mainly comes from Proposition .(ii) in []. For completeness, we give the outline of the proof as follows.

Dene the mapping G : Lp(, T; Lp( )) Lp (, T; Lp ( )) by Gu = x(u), for any u

Lp(, T; Lp( )). Also, dene the mapping K : Lp(, T; W,p( )) Lp(, T; Lp( )) by K(v) =

v| , for any v Lp(, T; W,p( )). Then KGK = , where is the same as in Lemma ..

In fact, it is obvious that G is continuous. For u(x, t), v(x, t) Lp(, T; Lp( )), we have

(u v, Gu Gv) =

T

(x(u) x(v))(u v) d (x) dt , since x is monotone. Thus, G is monotone. In view of Lemma ., G : Lp(, T; Lp( )) Lp (, T; Lp ( )) is maximal

monotone.Dene : Lp(, T; Lp( ))

R by (u) =

x(u) d (x) dt. It is easy to see that is a proper, convex, and lower-semicontinuous function on Lp(, T; Lp( )), which implies that : Lp(, T; Lp( )) Lp (, T; Lp ( )) is maximal monotone in view of Lemma ..

Since

(u) (v) =

T

T

d (x) dt

x(u) x(v)

T

x(v)(u v) d (x) dt = (Gv, u v)for all u(x, t), v(x, t) Lp(, T; Lp( )), we have Gv (v). So G = .

Now, it is clear that KGK : Lp(, T; W,p( )) Lp (, T; (W,p( ))) is maximal mono

tone since both K and G are continuous. Finally, for any u, v Lp(, T; W,p( )), we have

(v) (u) = (Kv) (Ku)

=

T

v| (x, t)

x

u| (x, t)

d (x) dt

x

T

x

u| (x, t)

v|

(x, t) u| (x, t)

d (x) dt

= (GKu, Kv Ku) =

KGKu, v u

.

Hence, we get KGK and so KGK = .

Wei et al. Boundary Value Problems (2015) 2015:88 Page 13 of 18

It now follows that for all u, v Lp(, T; W,p( )),

v, (u) =

T

x

v| (x, t) d (x) dt.

Moreover, () since x(). This completes the proof.

Lemma . The mapping A : L(, T; L( )) L(, T; L( )) dened in Denition . is

maximal monotone.

Proof Noting Lemmas .-., we can easily get the result that A is monotone.

Next, we shall show that R(I + A) = L(, T; L( )), which ensures that A is maximal monotone.

Case . p . We dene F : Lp(, T; W,p( )) Lp (, T; (W,p( ))) by

Fu = u, (v, Fu)Lp(,T;W,p( ))Lp (,T;(W

,p( ))) = (v, u)L(,T;L( )),

where (, )L

u| (x, t)

(,T;L( )) denotes the inner-product of L(, T; L( )). Then F is everywhere dened, monotone and hemi-continuous, which implies that F is maximal monotone in view of Lemma .. Combining with the facts of Lemmas ., .-., we have R(B + + S + F) = Lp (, T; (W,p( ))).

For f L(, T; L( )) Lp (, T; (W,p( ))), there exists u Lp(, T; W,p( ))

L(, T; L( )) such that

f = Bu + (u) + Su + Fu = Au + u,

which implies that R(I + A) = L(, T; L( )).Case . NN+ < p < , then p . Similar to Lemma ., we dene

B : Lp (, T; W,p( ))

Lp(, T; (W,p( ))) by

(w,

Bu) =

T

|u|p

|u|pu, w dx dt +

T

|u|ruw dx dt

+

T

|u|ruw dx dt

for any u, w Lp (, T; W,p( )). Then

B is maximal monotone and coercive. Similar to Lemma ., dene the mapping

: Lp (, T; W,p( ))

R by

(u) =

T

x

u| (x, t)

d (x) dt,

for any u Lp (, T; W,p( )), then

is maximal monotone. Similar to Lemma .,

dene

S) = {u Lp (, T; W,p( ))|ut Lp(, T; (W,p( ))), u(x, ) = u(x, T)} Lp(, T; (W,p( ))) by

Su(x, t) =

u

t .

S : D(

Wei et al. Boundary Value Problems (2015) 2015:88 Page 14 of 18

Then

then we have R(

u

t div[(|u|p)|u|pu] + |u|ru + |u|ru + g(x, u, ut, u) = f (x, t), a.e.

(x, t) (, T);

(b) , (|u|p)|u|pu x(u(x, t)), a.e. x (, T);

(c) u(x, ) = u(x, T), x .

Proof We split our proof into two steps.Step . There exists a unique u(x, t) which satises Hu + Au = f , where f (x, t)

L(, T; L( )) is a given function.

From Theorem ., Lemmas . and ., we know that A is H-monotone. Thus, R(H + A) = L(, T; L( )). Then, for f (x, t) L(, T; L( )) in (.), there exists u(x, t)

L(, T; L( )) such that Hu(x, t) + Au(x, t) = f (x, t). Next, we shall prove that u(x, t) is unique.

Suppose that u(x, t) and v(x, t) satisfy Hu + Au = f and Hv + Av = f , respectively. Then (u v, Au Av) = (u v, Hu Hv) , which ensures that

= (u v, Au Av) = (u v, Bu Bv) + u v, (u) (v)

+ (u v, Su Sv).

Using Lemmas ., ., and ., we have (u v, Bu Bv) = , which implies that u(x, t) = v(x, t), since B is strictly monotone.

Step . If u(x, t) L(, T; L( )) satises f = Hu + Au, then u(x, t) is the solution of (.).

Since (u + ) = (u) for any C( (, T)), we have (, (u)) = . Then, for

C( (, T)), we have

(, f Hu) = (, Bu) + , (u) + (, Su) = (, Bu) + (, Su).

So

dx dt +

ut dx dt

S is linear maximal monotone. Similar to Case , dene F : Lp (, T; W,p( )) Lp(, T; (W,p( ))) by

Fu = u, (v, Fu)Lp (,T;W,p( ))L

p(,T;(W,p( ))) = (v, u)L(,T;L( )),

S + F) = Lp(, T; (W,p( ))). So, for f L(, T; L( )) Lp(, T; (W,p( ))), there exists u Lp (, T; W,p( )) L(, T; L( )) such that

f =

Bu +

(u) +

Su + Fu = Au + u,

which implies that R(I + A) = L(, T; L( )).

Theorem . For f (x, t) L(, T; L( )), the nonlinear parabolic equation (.) has a

unique solution u(x, t) in L(, T; L( )), i.e.,(a)

B +

+

T

f g

T

dx dt

x, u,ut , u

=

|u|p

|u|ru dx dt +

T

|u|pu,

T

|u|ru dx dt

+

T

Wei et al. Boundary Value Problems (2015) 2015:88 Page 15 of 18

T

= div

|u|p

|u|pu

dx dt +

T

|u|ru dx dt

+

T

|u|ru dx dt +

T

ut dx dt,

which implies that the equationu

t

div

|u|p

|u|pu

+ |u|ru

+ |u|ru + g

x, u,

u

t ,

u

= f (x, t), a.e. x (, T), (.)

is true.

By using (.) and Greens formula, we have

T

|u|p

|u|pu

v| d (x) dt

,

=

T

div

|u|p

|u|pu

v dx dt +

T

|u|p

|u|pu, v

dx dt

=

v,ut +|u|ru + |u|ru + g x, u,ut , u

f

+ v, Bu |u|ru |u|ru

= (v, Su + Bu + Hu f ) =

v, (u)

T

= x

u| (x)

v| (x) d (x) dt. (.)

Then

,

|u|p

|u|pu

, a.e. on (, T). (.)

From the denition of S, we can easily obtain u(x, ) = u(x, T) for all x . Combining

with (.) and (.) we see that u is the unique solution of (.).

This completes the proof.

Lemma . Dene

B : Lp(, T; W,p( )) Lp (, T; (W,p( ))) by

Bu Bu f (x, t), for

u Lp(, T; W,p( )). Then

x

u(x, t)

B is maximal monotone.

Proof Similar to the proof of Lemma ., we know that

B is everywhere dened, mono-

tone, and hemi-continuous. It follows that

B is maximal monotone.

Denition . Dene a mapping

A : L(, T; L( )) L(,T;L( )) by

Au =

w(x) L

, T; L( )

|w(x)

Bu + (u) + Su

A) = {u L(, T; L( ))|there exists w(x) L(, T; L( )) such that w(x)

Bu + (u) + Su}.

Denition . Let H be a Hilbert space and A be a H-monotone operator. The resolvent

operator of A, RHA, : H H, is dened by

RHA,(u) = (H + A)u, u H.

for u D(

Wei et al. Boundary Value Problems (2015) 2015:88 Page 16 of 18

Theorem . u(x, t) = RH A,

() if and only if u(x, t) L(, T; L( )) is the solution

of (.).

Proof Let u(x, t) be the solution of (.). Then, using Greens formula and Lemma ., we have

v, (H +

A)u

=

T

|u|p

|u|pu, v

dx dt +

T

|u|ruv dx dt

+

T

|u|ruv dx dt

T

f (x, t)v(x, t) dx dt

+

T

g x, u,ut , u

v(x, t) dx dt +

v, (u) +

T

ut v dx dt

T

= div

|u|p

|u|pu

v dx dt

T

,

|u|p

|u|pu

v| d (x) dt

+

T

|u|ruv dx dt +

T

|u|ruv dx dt

T

+ f (x, t)v(x, t) dx dt +

T

g x, u,ut , u

v(x, t) dx dt

+

T

x(u| )v| d (x) dt +

T

ut v dx dt

=

T

,

|u|p

|u|pu

v| d (x) dt +

T

x(u| )v| d (x) dt

T

= x(u| )v| d (x) dt +

T

x(u| )v| d (x) dt = .

Thus, u(x, t) = RH A,

().

If u(x, t) = RH A,

(), then noting Lemma ., we have for C( (, T)),

=

T

ut dx dt +

T

|u|p

|u|pu,

dx dt

+

T

|u|ru dx dt +

T

|u|ru dx dt

T

f dx dt +

T

g x, u,ut , u

dx dt,

which implies that the equation

u

t

div

|u|p

|u|pu

+ |u|ru

+ |u|ru + g

x, u,

u

t ,

u

= f (x, t), a.e. (x, t) (, T),

is true.

Wei et al. Boundary Value Problems (2015) 2015:88 Page 17 of 18

Similar to the last part of Theorem ., we know that , (|u|p)|u|pu

x(u(x, t)). From the denition of S, we know that u(x, ) = u(x, T) for all x , which

implies that u(x, t) is the solution of (.). This completes the proof.

Competing interests

The authors declare that they have no competing interests.

Authors contributions

All authors contributed equally to the writing of this paper. All authors read and approved the nal manuscript.

Author details

1School of Mathematics and Statistics, Hebei University of Economics and Business, Shijiazhuang, 050061, China.

2Department of Mathematics, Texas A&M University - Kingsville, Kingsville, TX 78363, USA. 3Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah, 21589, Saudi Arabia. 4School of Electrical and Electronic Engineering, Nanyang Technological University, 50 Nanyang Avenue, Singapore, 639798, Singapore.

Acknowledgements

Li Wei is supported by the National Natural Science Foundation of China (11071053), Natural Science Foundation of Hebei Province (No. A2014207010), Key Project of Science and Research of Hebei Educational Department (ZH2012080) and Key Project of Science and Research of Hebei University of Economics and Business (2013KYZ01).

Received: 17 March 2015 Accepted: 1 May 2015

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