Huashui ZHAN

1 Introduction

In this paper,we consider the initial boundary value problem of the following equation:

where Ω ? R2is a domain with the suitably smooth boundary ?Ω.Equation(1.1)arises in mathematical finance(see[1]),and arises when studying nonlinear physical phenomena such as the combined effects of diffusion and convection of matter(see[2]).Antonelli,Barucci and Mancino[1]introduced a new model for the agent’s decision under risk,in which the utility function is the solution of Equation(1.1),and in particular,0≤u≤1.Under the assumption that f is a uniformly Lipschitz continuous function,Crandall,Ishii and Lions[3],Citti,Pascucci and Polidoro[4],Antonelli and Pascucci[5],step by step,proved that there is a local classical solution of Cauchy problem of Equation(1.1).

Clearly,Equation(1.1)is a strong degenerate parabolic equation since it lacks the secondorder partial derivative term?yyu.There are some different ways to deal with the existence and uniqueness of the global weak solution of the Cauchy problem of Equation(1.1).For example,Equation(1.1)is the special case of the degenerate parabolic equations discussed in[6–7],etc.,and one can refer to[8–12]for the related results.However,Zhan[13]showed that the global weak solution of Equation(1.1)can not be classical generally.In other words,some blow-up phenomena happen in finite time.Based on these facts,we are interested in the initial boundary value problem of Equation(1.1).

It is well known that there are some rules on how to quote the initial boundary value problem of a linear degenerate parabolic equation,for which one can refer to Oleinik’s books[14–15]etc.,and we call these rules as Oleinik rules for simplicity.If Ω =(0,R)× (0,N)? R2,considering the nonnegative solutions,according to Oleinik rules,and using Oleinik’s line method(see[16]),the local classical solution of Equation(1.1)has been discussed in[17].The main aim of this paper is to discuss the initial boundary value problem of Equation(1.1),provided that the spatial variables(x,y)lie in a general domain Ω ? R2,and the boundary ?Ω is suitably smooth.Certainly,the initial value condition is always required,i.e.,

To assure the well-posedness of Equation(1.1),according to Oleinik rules,we should impose

as the homogeneous boundary value condition,where

and={n1,n2,0}is the outer unit normal vector of Σ.We shall investigate the solvability of Equation(1.1)with the initial value(1.2)and the partial boundary value condition(1.3).The most important innovation of the paper lies in how to get a suitable entropy solution of(1.1)–(1.3)to arrive at its well-posedness.We shall use the general parabolic regularization method,i.e.,considering the initial boundary value problem of the following equation:

to prove the existence of the solution.In order to prove the compactness of{uε},we need some estimates on{uε}.Based on the estimates,by Kolmogoroff’s theorem,and using some ideas of[6–7]and[18],the existence of the solution is proved.

Theorem 1.1 Suppose that u0(x)∈ L∞(Ω)is suitably smooth.If fx,fy,ftare bounded functions,and fuis bounded too when u is bounded,then Equation(1.1)with the initial boundary value conditions(1.2)–(1.3)has an entropy solution.

The entropy solution in Theorem 1.1 is in the BV sense,which is defined in the following definition 2.1.Moreover,we shall use Kruzkov’s double variables method(see[19]),to discuss the stability of the solutions.Due to the complicated formula of the entropy solution defined,some special techniques are used.Beyond one’s imagination,if we consider the special domain,such as the half space Ω =R2+,or the unit disc Ω ={(x,y):x2+y2<1},then the stability of the solution may be free from the limitation of the boundary value condition.

Kobayasi K.and Ohwa H.[25]studied the well-posedness of anisotropic degenerate parabolic equations

with the inhomogeneous boundary condition on a bounded rectangle by using the kinetic formulation which was introduced in[26].Li Y.and Wang Q.[27]considered the entropy solutions of the homogeneous Dirichlet boundary value problem of(1.6)in an arbitrary bounded domain.Since the entropy solutions defined in[25,27]are only in the L∞space,the existence of the trace(defined in the traditional way,which was called the strong trace in[27])on the boundary is not guaranteed,the appropriate definition of entropy solutions are quoted,and the trace of the solution on the boundary is defined in an integral formula sense,which was called the weak trace in[27].So,not only definition 2.1 in our paper is different from the definitions of entropy solutions in[25,27],but also the trace of the solution in our paper is in the traditional way.

2 definition of the Solution

Following references[20–21],u∈BV(QT),QT=Ω×(0,T),if and only ifand

where

and K is a positive constant.This is equivalent to that the generalized derivatives of every function in BV(QT)are regular Radon measures on QT.

Let Γube the set of all jump points of u ∈ BV(QT),v be the normal of Γuat X=(x,y,t),and u+(X)and u?(X)be the approximate limits of u at X ∈ Γuwith respect to(v,Y?X)>0 and(v,Y?X)<0 respectively.For the continuous function p(x,y,t,u)and u∈BV(QT),define

which is called the composite mean value of p.For a given t,we denoteas all jump points of u(·,t),the Housdorffmeasure of,the unit normal vector of,and the asymptotic limit of u(·,t)respectively.Moreover,if f(s) ∈ C1(R)and u ∈ BV(QT),then f(u)∈BV(QT)and

Letdτ for small η>0,whereObviously hη(s)∈ C(R)and

where sgn represents the sign function.

definition 2.1 A function u is said to be the entropy solution of(1.1)–(1.3),if

(1)u∈BV(QT)∩L∞(QT),and there exists a function g1∈L2(QT),such that

for any ?(x,y,t)∈ L2(QT).

(2)For any 0≤?∈(QT)any k∈R,and any small η>0,u satisfies

For any k∈R,η>0,here

(3)The trace on the boundary

(4)The initial value condition is true in the sense that

Clearly,by(2.5),we have

Let η→ 0 in this inequality.We have

This is just the entropy solution defined in[23–24].Thus if u is the entropy solution in definition 2.1,then u is an entropy solution defined in general cases.

3 Proof of Theorem 1.1

Lemma 3.1(see[28])Assume that Ω ? RNis an open bounded set and let gk,f ∈ Lq(Ω),as k→∞,gk?f weakly in Lq(Ω),1≤q<∞.Then

We now consider the following regularized problem:

with the initial value(1.2)and the homogeneous boundary value condition

Under the assumptions of Theorem 1.1,it is well known that there is a classical solution uεof the initial boundary value problem of(3.1)with(1.2)and(3.2),and e.g.,one can refer to the chapter 8 of[29].

We need to make some estimates for uεof(3.1).Firstly,since u0(x)∈ L∞(Ω)is suitably smooth,by the maximum principle,we have

Secondly,let’s make the BV estimates of uε.

Lemma 3.2(see[18])Let uεbe the solution of(3.1)with(1.2)and(3.2).If the assumptions of Theorem 2.2 are true,then

with constants ci,i=1,2 independent of ε,where={n1,n2}is the outer normal vector of Ω,and ?uε={?xuε,?yuε}.

Theorem 3.1 Let uεbe the solution of(3.1)with(1.2)and(3.2).If the assumptions of Theorem 1.1 are true,then

wherec is independent of ε,and x1=x,x2=y.

Proof In what follows,we simply denote the solution of(3.1)with(1.2)and(3.2),uε,as u,x1=x,x2=y,x3=t sometimes from the context,and the dual index of i represents the sum from 1 to 2,while the dual index of s or p represents the sum from 1 to 3.Differentiate(3.1)with respect to xs,s=1,2,3,and sum up for s after multiplying the resulting relationThen integrating over Ω yields

where ξs=uxs,and dσ is the surface integrable unit.

By the assumption that ft,fx,fyare bounded,and fuis bounded due to|u|≤c,then

From(3.5)–(3.9),we have

Observe that on Σ = ?Ω ×[0,T),

which implies that

Let the surface integral in(3.10)be

Then,by Lemma 3.2,using(3.11),can be estimated by|gradu|L1(Ω),and one can refer to[18]for details.

Thus,by(3.10),letting η → 0,and noticing that

we have

and by the well known Gronwall lemma,we have

where c is a constant independent of t.By(3.13),using Equation(3.1),it is easy to show that

Now,we denote back that uεis the solution of(3.1).Thus by Kolmogoroff’s theorem,there exists a subsequence{uεn}of uεand a function u ∈ BV(QT)∩L∞(QT)such that uεnis strongly convergent to u,so uεn→ u a.e.on QT.By(3.14),there exist functions g1∈ L2(QT)and a subsequence of{ε},and we can simply denote this subsequence as ε,such that when ε→0,

We now prove that u is a generalized solution of(1.1)–(1.3).Let ? ∈ C20(QT), ? ≥ 0.Multiplying(3.1)by ?Sη(uε? k),and integrating over QT,we obtain

Let’s calculate every term in(3.15)by the part integral method:

From(3.15)–(3.20),we have

By(3.21),we have

By Lemma 3.1,

Let ε→ 0 in(3.22).By(3.23),we get(2.5).At the same time,(2.6)is naturally concealed in the limiting process.

The proof of(2.7)is similar to that in[2,6],so we omit the details here.

4 Double Variables Method

Lemma 4.1(see[6])Let u be a solution of(1.1).Then

which is true in the sense of the Hausdorffmeasure H2(Γu).

In this section,we shall prove how to use the double variables method to consider the stability of the solutions.Let u,v be two entropy solutions of(1.1)with initial values

and with the homogeneous boundary value u(x,y,t)=v(x,y,t)=0 when(x,y,t)∈ Σ3.For simplicity,we denote the spatial variables(x,y)as(x1,x2)or(y1,y2)in what follows,and correspondingly,dx=dx1dx2,dy=dy1dy2.

By definition 2.1,we have

Letand

We choose k=v(y,τ),l=u(x,t), ? = ψ(x,t,y,τ)in(4.3)–(4.4),integrate over QT,add them together,and then we get

Clearly,

Noticing that

as η → 0,we have

and as h→0,we have

For the third term and the forth term in the bracket of(4.7),we have

Noticing that

and by the properties of the BV function(the equality(2.2)and Lemma 4.1)

We have

Noticing thatwe have

Combining(4.7)–(4.13),and letting η → 0,h → 0,we get

5 Stability of the Solutions

In the last section of this paper,we shall discuss the stability of the solutions.

Theorem 5.1 Let 0≤u≤1,0≤v≤1 be two solutions of Equation(1.1)with the homogeneous boundary value γu|Σ3= γv|Σ3=0,and with the different initial values u0(x1,x2),v0(x1,x2) ∈ L∞(Ω)respectively.Suppose that|fu(·,u)|≤ c,and that the distance function d(x)=dist(x,?Ω)satisfies

so then for any t∈(0,T),

where

Proof Let δεbe the mollifier as usual.In detail,for the known function

where

and for any given ε>0, δε(s)is defined as

Now,we can choose φ in(4.14)by

where η(t) ∈(Ω)is defined as follows.For any given small enough 0< λ,0≤ ωλ≤ 1,ω|?Ω=0 and

When 0≤d(x)≤λ,

and when d<0,we let ωλ(d)=0.Then

Now,

Using the condition(5.1),and|dx1x1|≤ c,and with the fact that|dxi|≤ |?d|=1,i=1,2,0≤fu(·,u)≤ c,0 ≤ u,v ≤ 1,and from(4.14),we have

Here Ωλ={x:d(x)=dist(x,?Ω)< λ}.By(5.3),

As λ → 0,according to the definition of the trace,by γu|Σ3= γv|Σ3=0,we have

Let 0

Here α?(t)is the kernel of the mollifier with α?(t)=0 for t/∈(??,?).Let?→0.Then

By the Gronwall lemma,the desired result follows by letting s→0,i.e.,

So we have the conclusion.

Theorem 5.2 Let u,v be two solutions of Equation(1.1)with the different initial values u0(x1,x2),v0(x1,x2)∈ L∞(Ω)respectively.0≤ u≤ 1,0≤ v≤ 1.Suppose that the domain Ω is just the unit discand suppose|fu(·,u)|≤ c.Then

where

Proof We can choose φ = ?(x)η(t)in(4.14),where η(t) ∈(0,T).Then we have

Due to that 0≤u≤1,0≤v≤1,|x2|≤1,

We have

and as the proof of Theorem 5.1,we have

Theorem 5.3 It is supposed that the domaintwo solutions of Equation(1.1)with the different initial values u0(x1,x2),v0(x1,x2)∈ L∞(Ω)∩L1(Ω)respectively.If|fu(·,u)|≤ c,then

where

and d(x)=dist(x,?Ω)=x2.

Remark 5.1 The condition of the initial values u0(x1,x2),v0(x1,x2)∈ L∞(Ω)∩L1(Ω)in the theorem is stronger than the solutions obtained in Theorem 1.1.At the same time,due to Ω=R2+={(x1,x2):x2>0},it implies that

is an empty set.It means that the solution of the equation is free from the limitation of the boundary value in this case.

Proof We can choose φ in(4.14)by

whereand then

From(4.14),we have

whereand by(5.10),we have

As the proof of Theorem 5.1,we have

The proof of Theorem 5.3 is complete.

At the end of this paper,let us consider a special domain

Thenand let

For small enough λ,we set

Theorem 5.4 It is supposed that the domain Ω = ΩR.Let u,v be two solutions of Equation(1.1)with the homogeneous boundary value

and with the different initial values u0(x1,x2),v0(x1,x2)∈ L∞(Ω)∈ L∞(Ω)respectively.Then

Proof Through a limit process,we can choose φ in(4.14)by

where

When 0

When

Certainly,when πλ ≤ x2≤ 1? πλ,

At the same time,it is clear that

From(4.14),we have

Here

implies

implies

Due to

we have

Letλ→0 in(5.15).As the proof of Theorem 5.1,we have the conclusion.

AcknowledgementThe author expresses his sincere thanks to the reviewers for their kindly amendments to the paper.

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