Classical and weak solutions of the partial differential equations associated with a class of two-point boundary value problems

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R E S E A R C H

Open Access

Classical and weak solutions of the partial

differential equations associated with a class

of two-point boundary value problems

Ning Ma

1

_{and Zhen Wu}

1*

*_{Correspondence:}

[email protected]

1_{School of Mathematics, Shandong}

University, Jinan, China

Abstract

This paper is concerned with a kind of first-order quasilinear parabolic partial differential equations associated with a class of ordinary differential equations with two-point boundary value problems. We prove that the function given by the solution of an ordinary differential equation is the unique solution of a first-order quasilinear parabolic partial differential equation in both classical and weak senses.

Keywords: Two-point boundary value problems; Ordinary diﬀerential equations; Quasilinear partial diﬀerential equations; Sobolev weak solutions; Viscosity solutions

1 Introduction

In this paper, we study the problem of solving the following ﬁrst-order quasilinear parabolic partial diﬀerential equation (PDE):

⎧ ⎨ ⎩

∂tu(t,x) +xu(t,x)b(t,x,u(t,x)) +f(t,x,u(t,x)) = 0,

u(T,x) =h(x), (t,x)∈[0,T]×Rn_, (1)

wherexu= (∂∂xiu)1≤i≤nis ann-dimensional row vector. We notice that PDE (1) is novel

since its factorbdepends onu(t,x), which is different from the traditional PDE form. It is very complicated and difficult to study the existence and uniqueness of solution of this kind of partial differential equations by traditional methods of the theory partial differ-ential equations; some related studies can be found in [1–4]. However, PDE (1) should be related to a family of coupled ordinary differential equations (ODEs) associated with a kind of two-point boundary problems parameterized by (t,x)∈[0,T]×Rnas follows:

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

˙

Xt,x

s =b(s,Xst,x,Yst,x),

–Y˙_st,x=f(s,Xt,x_s ,Y_st,x),

Xt,x_t =x, YT=h(Xt,xT).

This two-point boundary value problem can be embedded into an optimal control prob-lem when applying the maximum principle; the existence and uniqueness results were

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obtained in [5]. Peng and Pardoux [6] studied the relationship between a system of quasi-linear PDEs and a kind of backward stochastic differential equations. They proved that under different assumptions, the function defined by solution of the backward stochastic differential equation is a classical and viscosity solution of a kind of second-order quasilin-ear PDEs. In the related field, for the stochastic cases, by introducing a family of coupled forward–backward stochastic differential equations (FBSDEs) Wu and Yu [7] gave a prob-abilistic interpretation for a kind of systems of second-order quasilinear parabolic PDEs combined with algebra equations in the viscosity sense. Ouknine and Turpin [8] studied weak solutions of second-order PDEs in Sobolev spaces and gave a probabilistic interpre-tation via the FBSDEs (see also Wei and Wu [9] and Kunita [10]). By some analysis tech-niques of those related references, in this paper, we study PDE (1) in both classical and weak senses, including a Sobolev weak solution and viscosity solution of the two-point boundary value problem.

The paper is organized as follows. In Sect.2, we recall some existence and uniqueness results for the two-point boundary value problem from [5] and give some regularity prop-erties of solutions of the ODEs; Then, in Sect.3, we prove that the function deﬁned by the solution of an ODE is the unique classical solution of PDE (1). Meanwhile, we derive the existence and uniqueness of a solution of PDE (1) in the Sobolev space and in the viscosity sense in Sects.4and5, respectively. Finally, we list some conclusive remarks.

2 Preliminary results of the ODEs

In this paper, we work with a ﬁnite time horizonT> 0. We denote byRnthen-dimensional Euclidean space and byRm×n_{the collection of}_m_×_n_{matrices. For a given Euclidean space,}

we denote by·,·(resp.,| · |) the inner product (resp., norm). The superscriptdenotes the transpose of vectors or matrices.

Let a (terminal) functionh:Rn_→_Rm_{and a couple of coeﬃcients (}_b_,_f_{) : [0,}_T_]_×_Rn_×

Rm_→_Rn+m_{be given. We introduce a family of coupled ODEs parameterized by the initial}

timet∈[0,T] and initial statex∈Rn: ⎧

⎪ ⎪ ⎨ ⎪ ⎪ ⎩

˙

Xt,x

s =b(s,Xst,x,Yst,x),

–Y˙_st,x=f(s,X_st,x,Y_st,x),

X_tt,x=x, Y_Tt,x=h(X_Tt,x).

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Let also anm×nfull-rank matrixGbe given. For each (x,y)∈Rn+m_{, we denote}

=

x y

, A(t,) =

–Gf Gb

(t,).

In this paper, we use the following standard assumptions:

(H1) For each,A(·,) is inL2_(0,_T_);_h_and_A_{are uniformly Lipschitz continuous with}

respect toxand, respectively. HereL2_(0,_T_{) =}_{_ϕ_:T

0 |ϕ(s)|2ds≤ ∞}.

(H2) There exist three nonnegative constantsμ,U1, andU2satisfyingU1+U2> 0 and

μ+U2> 0. Moreover,μ> 0 andU1> 0 (resp.,U2> 0) in the case ofm>n(resp.,n>m),

and for allh= (x,y,z)andh= (x,y),

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and

A(t,) –A(t,),–≤–U1G(x–x) 2

–U2G(y–y) 2

.

Under Assumptions (H1) and (H2), ODEs (2) admit a unique solution

X_st,x,Y_st,x_s_∈_[t,T]∈C(0,T),

(see Theorem 1.1 in [5]). In addition, we have the followingL2-estimates.

Proposition 2.1 Let(b,f,h)and(b,f,h)satisfy(H1)and(H2),and let x,x∈Rn_{and s}_∈

[t,T].Let(Xt,x

s ,Yst,x) (resp., (X t,x s ,Y

t,x

s ))denote the unique solution of ODEs

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

˙

X_st,x=b(t,X_st,x,Y_st,x),

–Y˙t,x

s =f(t,Xt,xs ,Yst,x),

Xtt,x=x, YTt,x=h(X t,x T).

⎛ ⎜ ⎜ ⎜ ⎝resp.,

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

˙

Xt,x

s =b(t,X t,x s ,Y

t,x s ),

–Y˙t,x

s =f(t,X t,x s ,Y

t,x s ),

Xt,x_t =x, Y_Tt,x=h(X_Tt,x) ⎞ ⎟ ⎟ ⎟ ⎠.

Then the following estimates hold:

sup t≤s≤T

X_st,x2+ sup t≤s≤T

Y_st,x2≤C1 +|x|2, (3)

sup t≤s≤T

X_st,x–Xt,x_s 2+ sup t≤s≤T

Y_st,x–Yt,x_s 2

≤Cx–x2+hXt,x_T–hXt,x_T2

+ T

t

_b

s,Xt,x_s ,Yt,x_s –bs,Xt,x_s ,Yt,x_s 2

+fs,X_st,x,Yt,x_s –fs,Xt,x_s ,Yt,x_s 2. (4)

Proof TheseL2_{-estimates are standard (see [}₇_,₁₁_{]). For reader’s convenience, we only}

prove (4). Let

(Xˆs,Yˆs) =

Xt,x_s –Xt,x_s ,Y_st,x–Yt,x_s ;

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We considerGXˆs,Yˆsand, using Assumptions (H1) and (H2), by diﬀerential equation we

get

μ| ˆXT|2+U1

T

t

GXˆs,GXˆsds+U2

T

t

GYˆs,GYˆs

ds

≤Cλx–x2+C λt≤sups≤T| ˆ

Yt|2+C

T

t

α(s)2+β(s)2ds, (5)

whereλ∈Rcan be any positive number. For| ˆXs|2 and| ˆYs|2, by the Gronwall inequality

we have

sup t≤s≤T| ˆ

Xs|2≤Cx–x 2

+C

T

t

| ˆYs|2+β(s) 2

ds, (6)

sup t≤s≤T| ˆ

Ys|2≤C| ˆXT|2+C

T

t

| ˆXs|2+α(s) 2

ds. (7)

Combining (5) and (7), we have

μ–C

λ

| ˆXT|2+

U1– C

λ

T

t | ˆ

Xs|2ds+U2

T

t | ˆ Ys|2ds

≤Cλx–x2+C

T

t

α(s)2+β(s)2ds. (8)

Case 1:μ> 0,U1> 0: We can chooseλ=λ0such that

μ– C λ0

> 0, U1– C

λ0

> 0.

Combining with (7), we get

sup t≤s≤T| ˆ

Ys|2≤Cx–x2+C

T

t

α(s)2+β(s)2ds.

Now let us prove (4). Case 2:U2> 0:

Combining (6) and (8), we have

sup t≤s≤T| ˆ

Xs|2≤Cλx–x 2

+C λ

| ˆXT|2+

T

t

| ˆXs|2ds

+C

T

t

_α(_s₎2

+β(s)2ds.

Then

sup t≤s≤T

| ˆXs|2≤Cλx–x2+ C

λ(1 +T)t≤sups≤T

| ˆXs|2+C

T

t

α(s)2+β(s)2ds.

We chooseλ=λ1such that

C

λ1

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Then we get

sup t≤s≤T| ˆ

Xs|2≤Cx–x 2

+C

T

t

α(s)2+β(s)2ds. (9)

Combining (7) and (9), we get (4) and conclude the proof.

We deﬁneYtt,xby a functionufrom [0,T]×RntoRm:

u(t,x) :=Y_tt,x. (10)

Remark2.1 Assumption (H2) can be relaxed to ensure the existence and uniqueness of the solution of the ODEs (2); see [5].

Remark2.2 For each deterministic (t,x)∈[0,T]×Rn, from the uniqueness of a solution of ODEs (2) we have

us,Xt,x_s =Ys,Xst,x s =Yst,x.

Proposition 2.2 Let Assumptions(H1)and(H2)hold.The function u deﬁned by(10)is continuous with respect to(t,x).In particular,u is Lipschitz continuous in x.

Proof First, from Proposition2.1we get the Lipschitz continuity ofuinx:

u(t,x+δ) –u(t,x)=Yt,x+δ

t –Ytt,x≤Cx–x.

Next, we prove thatu(t,x) is continuous int. We have

u(t+δ,x) –u(t,x)=u(t,x) –ut+δ,X_t+t,x_δ+ut+δ,X_t+t,x_δ–u(t+δ,x)

≤Y_tt,x–Yt+δ,X t,x t+δ

t+δ +Y

t+δ,Xt,x_t+_δ t+δ –Y

t+δ,x t+δ

≤Y_tt,x–Y_t+t,xδ+CX

t,x t+δ–x

≤

t+δ

t

fr,X_Rt,x,Y_Rt,xdr+CX_t+t,x_δ–x

:=ρ(δ),

whereρ(δ)→0 asδ→0. The continuity intand the Lipschitz continuity inximply the

joint continuity ofuin (t,x).

To improve the smoothness of the solutions of ODEs (2), we add the following assump-tion:

(H3) For any s∈[t,T], the functions b(s,·,·), f(s,·,·), and h(·) are of class C2 _with

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Theorem 2.1 Under Assumptions(H1)–(H3),the function x→(Xt,x_,_Yt,x₎_{is diﬀerentiable} with continuous derivatives given by(∂xiXt,x,∂xiYt,x)1≤i≤nsatisfying the following ODEs:

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

d∂xiXst,x=xb(s,Xt,xs ,Yst,x)∂xiXst,xds+yb(s,Xst,x,Yst,x)∂xiYst,xds,

–d∂xiYst,x=xf(s,Xst,x,Yst,x)∂xiXst,xds+yf(s,Xst,x,Yst,x)∂xiYst,xds,

∂xiXtt,x= (0, 0, . . . , 1i, . . . , 0, 0), ∂xiYTt,x=xh(XTt,x)∂xiXTt,x,s∈[t,T].

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Proof A similar technique can be found in [11] and [12]. Here we give a detailed proof. Lethi=h·ei,hiXt=hi–1(Xst,x+hi–Xst,x), andhiYt=hi–1(Yst,x+hi–Yst,x). Then

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

dhiXst,x=hi–1[b(t,Xst,x+hi,Yst,x+hi) –b(t,Xst,x,Yst,x)]ds,

–dhiYst,x=hi–1[f(t,Xst,x+hi,Yst,x+hi) –f(t,Xst,x,Yst,x)]ds,

hiXtt,x= 1, hiYTt,x=hi–1[h(XTt,x+h) –h(XTt,x)].

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Hence we treat this equation as a linear one:

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

dhiXst,x=φ(s,hiXst,x,hiYst,x)ds,

–dhiYst,x=ψ(s,hiXst,x,hiYst,x)ds,

hiXtt,x= 1, hiYTt,x=hi–1[h(XTt,x+hi) –h(X t,x T )],

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whereφandψ are deﬁned by

φh(s,x,y) =Ah(s)x+Bh(s)y,

ψhi(s,x,y) =Chi(s)x+Dhi(s)y,

and

Ahi(s) =

⎧ ⎨ ⎩

b(s,Xst,x+hi,Yst,x+hi)–b(s,Xt,xs ,Yst,x+hi)

Xst,x+hi–Xst,x

ifXst,x+hi=Xt,xs , xb(s,Xt,xs ,Y

t,x+hi

s ) otherwise,

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Bhi(s) =

⎧ ⎨ ⎩

b(s,Xt,xs ,Yst,x+hi)–b(s,Xt,xs ,Yst,x) Yst,x+h–Yst,x

ifYt,x+hi s =Yst,x, yb(s,Xst,x,Yst,x) otherwise,

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Chi(s) =

⎧ ⎨ ⎩

f(s,Xst,x+hi,Yst,x+hi)–f(s,Xst,x,Yst,x+hi)

Xt,x+s hi–Xst,x

ifXst,x+hi=Xst,x, xf(s,Xt,xs ,Y

t,x+hi

s ) otherwise,

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Dhi(s) =

⎧ ⎨ ⎩

f(s,Xst,x,Yst,x+hi)–f(s,Xst,x,Yst,x)

Yst,x+hi–Yst,x

ifYst,x+hi=Yst,x, yf(s,Xst,x,Yst,x) otherwise,

(17)

forh= 0. Let

φ0(s,x,y) =xb

s,X_st,x,Y_st,xx+yb

s,X_st,x,Y_st,xy,

ψ0(s,x,y) =xf

s,X_st,x,Y_st,xx+yf

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From Proposition 2.1 we know that (Xt,x+hi_,_Yt,x+hi_{) converges to (}_Xt,x_,_Yt,x_{). We have}

to prove that (hiX,hiY) converges to (∂xiX,∂xiY), the solution of ODEs (11). To use

the same convergence argument, we have to show thatφhi(s,∂xiXt,x,∂xiYt,x) converges to

φ0(s,∂xiXt,x,∂xiYt,x) andψhi(s,∂xiXt,x,∂xiYt,x) converges toψ0(s,∂xiXt,x,∂xiYt,x) ashigoes

to 0. Notice that

Ahi(s) =

1

0 xb

s,X_st,x+λXt,x+hi s –Xst,x

,Yt,x+hi

s

dλ.

Thus T

0

Ahi(s) –xb

s,X_st,x,Y_st,x+hi2∂xiXst,x

2 ds

≤ T

0

1

0

xb

s,X_st,x+λX_st,x+hi–X_st,x,Y_st,x+hi

–xb

s,X_st,x,Y_st,x+hi2∂xiXst,x

2 dλds.

We split this integral into two terms on the sets{|Xt,x+his –Xst,x| ≤η}and{|X t,x+hi

s –Xst,x|>η}.

By Assumption (H3),xb(s,x,y) is uniformly continuous and bounded (by a constantK).

It follows that, for eachε> 0, there existsη> 0 such that Ahi(s) –xb

s,Xt,x_s ,Y_st,x+hi∂xiXst,x

≤ε2∂xiXst,x+K 2

T

0

1_{|

Xst,x+hi–Xst,x|>η}

∂xiXst,x 2

ds.

We split the term into two parts corresponding to the set{|∂xiXst,x| ≤M}and its

comple-ment. Then we have T

0

1_{|

Xt,x+s hi–Xt,xs |>η}

∂xiXst,x 2

ds≤M 2

η2X t,x+hi s –X

t,x s

2

+ T

0

1_{|_∂

xiXst,x|>M}∂xiX t,x s

2 ds.

By the Lebesgue theorem, since∂xiXt,xis square integrable,

T

0 1{|∂_xiYt,x_|_>M_}|∂_xiXt,x

s |2ds

con-verges to 0 asM→ ∞. ChoosingMsuﬃciently large and using the convergence ofXst,x+hi

toX_st,x, it follows that

lim hi→0

Ahi(s) –xb

s,X_st,x,Y_st,x+hi∂xiXst,x= 0.

By the same method we get

lim hi→0

_xbs,X_st,x,Y_st,x+hi∂xiXst,x

–xb

s,X_st,x,Y_st,x)∂xiXst,x= 0.

Hence it follows that

lim hi→0

Ahi(s) –xb

s,X_st,x,Y_st,x∂xiXst,x= 0.

Similar arguments give that

lim hi→0

Bhi(s) –yb

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lim hi→0

Chi(s) –xf

s,X_st,x,Y_st,x∂xiXst,x= 0,

lim

hi→0Dhi(s) –yf

s,X_st,x,Y_st,x∂xiYst,x= 0.

It is easy to verify that the functions φ and ψ are continuous with respect to pa-rameter hi at point 0. By Proposition2.1, as hi→0, φhi(s,∂xiXst,x,∂xiYst,x) converges to

φ0(s,∂xiXst,x,∂xiYst,x), andψh(s,∂xiXst,x,∂xiYst,x) converges toψ0(s,∂xiXst,x,∂xiYst,x). By

Propo-sition2.1, (hiX,hiY) converges to (∂xiX,∂xiY).

We also have the following property.

Proposition 2.3 Under Assumptions(H1)–(H3),there exists a constant C> 0,depending

only on L,μ,and T,such that

sup 0≤t≤T

_xX_st,x2+ sup 0≤t≤T

_xY_st,x2≤C.

Proof From Proposition2.1and Theorem2.1we get

sup 0≤t≤T

_xX_st,x2+ sup 0≤t≤T

_xY_st,x2

= sup 0≤t≤T

lim h→0

_hXt,x_s +hYst,x 2

= sup 0≤t≤T

lim h→0|h|

–1_Xt,x+h s –Xst,x

2

+Y_st,x+h–Y_st,x2

≤C.

3 Classical solution to the PDE

We now relate the functionudeﬁned by (10) to the parabolic partial diﬀerential equa-tions (1).

Theorem 3.1 Let Assumptions(H1)–(H3)hold.Then the function u deﬁned by(10)is of class C1,1_([0,_T_]_×_Rn₎_{and solves PDE}₍₁_)._{In particular}_,_u₍_t_,_x₎_{is the unique solution of} PDE(1).

Proof By Theorem2.1,u∈C0,1_([0,_T_]_×_Rn_{). Let}_h_{> 0 be such that}_t₊_h_≤_T_{. Clearly,} Y_t+ht,x =Yt+h,X

t,x t+h t+h . Hence

u(t+h,x) –u(t,x) =ut+h,X_t+ht,x–ut+h,X_t+ht,x+u(t+h,x) –u(t,x). (18)

Diﬀerentiatingu(t,Xt,x

s ), we get

Dut,Xt,x_s =xu

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HereDu(t,x) = (du1(t,x),du2(t,x), . . . ,dum(t,x))is anm-dimensional rank victor.

Com-bining (18) with (19), we get

u(t+h,x) –u(t,x) = – t+h

t xu

t+h,X_st,xbs,X_st,x,Y_st,xds

– t+h

t

fs,Xt,x_s ,Y_st,xds. (20)

Lett=t0<t1<· · ·<tn=T. We have

u(T,x) –u(t,x) = –

n–1

i=0

ti+1

ti xu

ti+1,Xsti,x

bs,X_sti,x,Y_sti,xds

–

n–1

i=0

ti+1

ti

fs,Xti,x_s ,Y_sti,xds. (21)

It follows from Theorem2.1that if we take a sequence of meshest=t0<t1<· · ·<tn=T

such thatlimn→∞supi≤n–1(tni+1–tin) = 0, then we obtain:

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

t

xu(t,x)b

t,x,u(t,x)+ft,x,u(t,x).

Henceu∈C1,1_([0,_T_]_×_Rn_{) and satisﬁes PDE (}₁_).

Now we consider the uniqueness of the solution. It suﬃces to show that (X_st,x,u(s,X_st,x);

t≤s≤T) solves ODEs (2). From the uniqueness of the solution of ODEs (2) we get the uniqueness of the solution of the PDE (1). Suppose that a functionv(t,x)∈C1,1_([0,_T_]_×_Rn₎

is a solution of PDE (1). Set

DX_st,x=bs,X_st,x,vs,X_st,xds.

Lettingt=t0<t1<t2<· · ·<tn=T, we have n–1

i=0

vti,Xtit,x

–vti+1,Xti+1t,x

=

n–1

i=0

vti,Xtit,x

–vti,Xti+1t,x

+

n–1

i=0

vti+1,Xtit,x

–vti+1,Xti+1t,x

= –

n–1

i=0

ti+1

ti xv

ti,Xst,x

bs,X_st,x,vs,X_st,xds

+

n–1

i=0

ti+1

ti

xv

s,X_tt,x_i+1bs,X_tt,x_i+1,vs,X_tt,x_i+1+fs,X_tt,x_i+1,us,X_tt,x_i+1ds.

Here we apply the diﬀerential equation to u(ti,·) and calculate v(ti,Xtit,x) –v(ti,Xt,xti+1).

Then we compute v(ti+1,Xtit,x) –v(ti+1,Xti+1t,x) from the PDE (1). Finally, by the fact that v∈C1,1_([0,_T_]_×_Rn_{) and by the assumption we let the mesh size go to zero to obtain}

vs,X_st,x–hX_Tt,x= T

s

(10)

where (Xt,x

s ,v(s,Xt,xs );t≤s≤T) solves ODEs (2). By the uniqueness of solution of ODEs,

(Xt,x

s ,v(s,Xst,x)) = (Xst,x,Yst,x). In particular,v(t,x) =Y t,x

t .

4 Weak solutions in Sobolev space

In this section, we prove that the functionu(t,x) deﬁned by (10) is the unique weak solu-tion of PDE (1) in the Sobolev space under some usual assumptions. First, we recall the deﬁnition of a Sobolev weak solution for PDE (1) from [8] and [9].

Deﬁnition 4.1 A function u is called a Sobolev weak solution(solution in L2

ρ(Rn;Rm))of

PDE(1)if u∈L2_([0,_T_];_L2

ρ(Rn;Rm))and for an arbitraryϕ∈Cc1,∞([0,T]×Rn;Rm),

T

t

Rn

u(s,x)∂sϕ(s,x)dx ds+

Rn

u(t,x)ϕ(t,x)dx–

Rn

u(T,x)ϕ(T,x)dx

+ T

t

Rnx

bs,x,u(s,x)ϕ(s,x)u(s,x)dx ds

= T

t

Rn

fs,x,u(s,x)ϕ(s,x)dx ds, (22)

whereρis the weight function deﬁned asρ(x) := (1 +|x|2₎q_,_q_≤_–2,_{and L}2

ρis the Hilbert

space with the inner product

u1,u2L2ρ=

Rnu1(x)u2(x)ρ(x)dx,

where u1(x)u2(x)is the inner product of the Euclidean space.

We make the following assumption:

(H4) For anys∈[t,T], the functionb(s,·,·) is inC1,1₍_Rn_×_Rm_;_Rn_{) with bounded}

deriva-tives.

Theorem 4.1 Under Assumptions(H1), (H2),and(H4),Xt,x

s is the solution deﬁned in the forward equation in ODEs(2).Then the map Xt,·

s :Rn→Rnis a homeomorphism.This means the map Xt,·

s is one-to-one and onto,so that its inverse map exists.Moreover,the inverse map,denoted byXˆ_st,·:Rn→Rn,is also continuous.

Proof The one-to-one property of the mapX_st,xfollows from Proposition2.1. The rest of proof is similar to [10] (pp. 225–227), and hence we omit it.

Lemma 4.1(Norm equivalence principle) Assume Assumptions(H1), (H2),and(H4).Let

X_st,xbe the solution of forward equation in ODEs(2),letρbe a weight function.Then there exist constants c,C> 0such that,for any s∈[t,T]andϕ∈L1

ρ(Rn,Rm),

c

Rn

ϕ(x)ρ(x)dx≤

Rn

ϕXt,x_s ρ(x)dx≤C

Rn

(11)

and,for any∈L1

ρ([t,T]⊗Rn;Rm),

c

T

t

Rn

(s,x)ρ(x)dx≤

T

t

Rn

s,X_st,xρ(x)dx

≤C

T

t

Rn

(s,x)ρ(x)dx, (24)

where c and C depend on T,L,ρ,and the bounds of the ﬁrst derivatives of b,f,h,but do not depend on the initial value x.

Proof First, we takeρ(x) := (1 +|x|2₎q_,_q_∈_R_{. We claim that there exists constants}_c_,_C_{> 0}

such that

c≤J(Xˆ t,y s )ρ(Xˆt,ys )

ρ(x) ≤C, ∀y∈R

n_,_t_≤_s_≤_T_. ₍₂₅₎

HereXˆst,yis the inverse ﬂow ofXst,y,J(Xˆst,y) :=detyXˆst,yis the determinant of the Jacobian

matrix ofXˆst,y. The existence ofxˆt,ys is given by Theorem4.1. Now we prove (25). Assume

thatT–h≤t≤Tfor some smallh> 0. We substitutex=Xˆst,yinto ODEs (2) withXt, ˆ Xst,y

s =

Xt,· s ◦ ˆX

t,y

s =y. Then

⎧ ⎨ ⎩

ˆ

Xst,y=y–

s t b(r,Xˆ

r,y s ,Yt,

ˆ Xst,y r )dr,

Yt,Xˆ t,y s

s =h(Xt, ˆ Xst,y T ) +

T s f(r,X

t,Xˆst,y r ,Yt,

ˆ Xst,y r )dr.

(26)

We diﬀerentiate (26) with respect toyto get

yXˆst,y=I–

s

t

b_xr,Xt,Xˆt,ys r ,Yt,

ˆ Xt,ys r

yXt,Xˆ t,y s r dr–

s

t

b_yr,Xt,Xˆt,ys r ,Yt,

ˆ Xst,y r

yYt,Xˆ t,y s r dr

=:I+J_st(y). (27)

We deﬁne ρ(x) :=eF(X)_{, so} _F₍_x_{) is} _C2

l,b(Rn). Using the diﬀerential equation for F(Xˆ t,y s )

(see[10], pp. 262–263), we get

FXˆt,y_s –F(y) = s

t

br,Xˆ_rt,y,Yt,Xˆst,y r

xF

_ˆ

X_rt,ydr.

It follows that

ρ(Xˆst,y)

ρ(y) =exp

FXˆ_st,y–F(y)=exp

s

t

br,Xˆ_rt,y,Yt,Xˆst,y r

xF

_ˆ

X_rt,ydr

:=N_ts(y).

Since the ﬁrst derivatives ofFare bounded, it is easy to verify thatNs

t(x) is bounded by

Assumptions (H1) and (H4). Then there exist two constantsr> 0 andR> 0 such that

r≤N_ts(y)≤R, ∀y∈Rn. (28)

SinceJ(Xˆst,y) :=detyXˆst,y, from(27) we obtain

(12)

We consider (27), apply the diﬀerential equation, and use a similar method in Proposi-tion2.3. Then there exists a constantc0> 0, depending only onL,μ,T, and the bounds of

the ﬁrst-order derivatives ofb,f, andh, such that

sup t≤r≤T

_yXt,Xˆst,y

r

2

+yYt,Xˆ t,y s

r

2_≤ c0.

So

J_st(y)2≤c0(s–t). (29)

By (28) and (29) the upper and lower bounds can be estimated as

r1 –c0(s–t)

≤J(Xˆ

t,y s )ρ(Xˆst,y)

ρ(y) ≤R

1 +c0(s–t)

.

Ifs–tis small enough, the lower boundr(1 –√c0(s–t)) > 0. Therefore, we can takeh

small enough such that (25) holds forT–h≤s≤T. Note thatcandCdo not depend on the initial valuey. So we use the ﬂow propertyXˆst,y=Xˆrt,·◦ ˆX

r,y

s ,∀t≤r≤s≤T (from

Remark2.1) to drop the restrictionT–h≤t≤Tand extend inequality (25) to the whole interval of [t,T].

Finally, we prove (23). Using the change of variabley=Xt,x s , we get

Rn

ϕX_st,xρ(x)dx=

Rn

ϕ(y)ρ(y)J(Xˆ

t,y s )ρ(Xˆst,y)

ρ(y) dy.

By (25) we get (23). Moreover, for a function (s,x)→(s,x), we considerx→(s,x) in the same way as before. We integrate with respect tos∈[t,T] to get (24). The lemma is

proved.

Let the molliﬁerKdbe deﬁned asKd(x) :=Cdexp(_1––1_|_x_|2) for|x|< 1 andKd(x) = 0

other-wise, whereCdis chosen such that

RnKd(x)dx= 1. DenoteKdm(x) :=mdKd(mx). Suppose

thatφ:Rn_→_R_{is a Hölder-continuous function with exponent}_γ _∈_{(0, 1) and deﬁne}

φm(x) :=

RnK m

d(x–y)φ(y)dy

form> 0.

By [13]φmis aC∞function and is Hölder-continuous with respect to exponentγ, More-over,φm→φuniformly onRasm→ ∞. Similarly, we deﬁne

hm(x) =

RnK m d

x–xhxdx,

bm(r,x,y) =

Rn_×_Rm

K_dmx–xK_kmy–ybr,x,ydy,

fm(r,x,y) =

Rn_×_RmK m d

x–xK_kmy–yfr,x,ydy.

It is easy to see that (hm₍_·_),_bm₍_t_,_·_,_·_),_fm₍_t_,_·_,_·₎₎

(13)

deﬁnition we can easily check that, when mis large enough,hm_,_bm_,_fm _{also satisfy}

As-sumptions (H1)–(H3) independently ofm. By [5] and Proposition2.1the smootherized ODEs

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

˙

Xt,x

s,m=bm(t,Xs,mt,x,Ys,mt,x),

–Y˙t,x

s,m=fm(t,Xs,mt,x,Ys,mt,x),

X_t,mt,x =x, YT=h(XT,mt,x ),

(30)

have a unique solution (Xt,·

·,m,Y·t,,m·)∈L2([0,T];L2ρ(Rn;Rn))⊗L2([0,T];L2ρ(Rn;Rm)). We deﬁneY_t,mt,x by a functionumfrom [0,T]×RntoRm:

um(t,x) :=Yt,mt,x.

Lemma 4.2 Under Assumptions (H1) and (H2), (Xt,·

·,m,Y·t,,m·)→ (X·t,·,Y·t,·) in L2([0,T]; L2

ρ(Rn;Rn))⊗L2([0,T];L2ρ(Rn;Rm))as m→ ∞.

Proof We setX˜t,x

s =Xt,xs,m–Xst,xandY˜st,x=Ys,mt,x–Yst,x. Applying the diﬀerential equation tor

GX˜t,x

s ,Y˜st,xand a similar method as in Proposition2.1, we get GXˆ_Tt,x,hm(XT) –h(XT)

+U1

T

t

GXˆs,GXˆsds+U2

T

t

GYˆs,GYˆs

ds

≤– T

t

GXˆs,fm

s,X_st,x,Y_st,x–fs,X_st,x,Y_st,xds

+ T

t

GXˆs,bm

s,X_st,x,Y_st,x–b(s,Xt,x_s ,Y_st,xds

→0 asm→ ∞.

Eventually, by (3) and the deﬁnition ofρ, similarly to the discussion in Proposition2.1, we have (Xt,·

·,m,Y·t,,m·)→(X·t,·,Y·t,·) inL2([0,T];L2ρ(Rn;Rn))⊗L2([0,T];L2ρ(Rn;Rm)) asm→ ∞.

Theorem 4.2 Under Assumptions(H1), (H2),and(H4),the function u(t,x)deﬁned by(10)

is the unique Sobolev weak solution of PDE(1)with u(T,x) =h(x).

Proof Existence. By Lemma4.1and Proposition2.1we have T

0

Rn

u(s,x)2ρ–1(x)dx ds≤C

T

0

Rn

us,Xt,x_s 2ρ–1(x)dx ds

=C

T

0

Rn

Y_st,x2ρ–1(x)dx ds<∞.

Sou(s,x)∈L2([0,T];L2ρ(Rn;Rm)).

From the structure of smootherized ODEs (30) and Theorem3.1we get thatum(t,x) := Y_t,mt,x is the unique classical solution of the following PDFs:

⎧ ⎨ ⎩

∂tum(t,x) +xum(t,x)bm(t,x,um(t,x)) +fm(t,x,um(t,x)) = 0,

(14)

Moreover, by the integration-by-parts formula it is easy to verify thatum(t,x) also satisﬁes

the following weak formulation: for any smooth test functionϕ∈C1,∞

c ([0,T]×Rn;Rm),

T

t

Rnum(s,x)∂sϕ(s,x)dx ds+

Rnum(t,x)ϕ(t,x)dx–

Rnum(T,x)ϕ(T,x)dx

+ T

t

Rnx

bs,x,um(s,x)

ϕ(s,x)um(s,x)dx ds

= T

t

Rnf

s,x,um(s,x)

ϕ(s,x)dx ds. (31)

Now we verify thatu(t,x) satisﬁes (22) withu(T,x) =h(x) by passing the limit forumin L2

ρ in (31). We only show the convergence of the last term. By the Lipschitz assumption and the fact thatfm→f in theL2ρsense asm→ ∞, for any∈C1,c∞([0,T]×Rn), we

have T

t

Rnf

m_s_,_x_,_um₍_s_,_x₎₍_s_,_x₎_{dx ds}_–

T

t

Rnf

s,x,u(s,x)(s,x)dx ds 2

≤Cp

T

t

Rn

fms,x,um(s,x)–fms,x,u(s,x)2(s,x)dx ds

+ T

t

Rn

fms,x,u(s,x)–fs,x,u(s,x)2(s,x)dx ds

≤Cp,L

T

t

Rn

um(s,x) –u(s,x))2(s,x)dx ds

+ T

t

Rn

fms,x,u(s,x)–fs,x,u(s,x)2(s,x)dx ds

→0 asm→ ∞.

Thereforeu(t,x) satisﬁes (22) and is a Sobolev weak solution of (1) withu(T,x) =h(x). Uniqueness. Letvbe another solution of PDE (1). By Deﬁnition4.1, we get:

T

t

Rnv(s,x)∂sϕ(s,x)dx ds+

Rnv(t,x)ϕ(t,x)dx–

Rnv(T,x)ϕ(T,x)dx

+ T

t

Rnv(s,x)x

bs,x,u(s,x)ϕ(s,x)dx ds

= T

t

Rnf

s,x,u(s,x)ϕ(t,x)dx ds. (32)

By Lemma 4.3 in [8], for the test functionψt(s) =ϕ(Xˆst)J(Xˆst), we obtain

T

s

Rnv(s,x)∂sϕ(s,x)dx ds= –

T

t

Rnv(s,x)x

bs,x,u(s,x)ϕ(s,x)dx ds. (33)

Taking (33) into (32), we get

Rnv(t,x)ϕ(t,x)dx–

Rnv(T,x)ϕ(T,x)dx=

T

t

Rnf

(15)

Let us make the change of variabley=Xˆt,x

s in each term of (34). Then (34) becomes

Rnv

t,X_st,yϕ(y)dy–

Rnh

T,X_Tt,yϕ(y)dy= T

t

Rnf

s,X_st,y,us,Xt,y_s ϕ(y)dy ds.

Sinceϕis arbitrary, we can prove that, for almost everyy,

vt,X_st,y=hT,X_Tt,y+ T

t

fs,X_st,y,us,X_st,yds.

Then we get thatv(s,X_st,x) is a solution of ODEs (2), and we obtain the uniqueness result by the uniqueness of the solution of ODEs (2), and the proof is completed.

5 Viscosity solution to the PDE

In this section, we prove that the functionu(t,x) defined by (10) is the unique viscosity solution of PDE (1) under Assumptions (H1) and (H2). We first recall the definition of a viscosity solution for (1) from [14] and [15].

Deﬁnition 5.1 Let u be a continuous function on[0,T]×Rn_→_Rm_{satisfying u} i(T,x) = hi(x),x∈Rn, 1≤i≤m.It is called a viscosity subsolution(resp.,supersolution)of PDE(1)if for any1≤i≤m, (t,x)∈[0,T)×Rn_,_and_ϕ_∈_C1,1_([0,_T_]_×_Rn_;_R₎_{such that}_(ϕ_–_u

i)attains a local minimum(resp.,maximum)at(t,x),ϕ(t,x) –ui(t,x) = 0,such that

∂tϕ(t,x) +xϕ(t,x)b

t,x,u(t,x)+fi

t,x,u(t,x)≥0,

resp.∂tϕ(t,x) +xϕ(t,x)b

t,x,u(t,x)+fi

t,x,u(t,x)≤0.

A function u is called a viscosity solution of PDE(1)if it is both a viscosity subsolution and a viscosity supersolution.

Theorem 5.1 Let Assumptions(H1)and(H2)hold.The function u(t,x)deﬁned by(10)is continuous and is a viscosity solution of PDE(1).

Proof The continuity ofufollows from Proposition2.2. Next, we only show thatuis a viscosity subsolution of PDE (1). A similar argument would show thatuis also a viscosity supersolution.

Let (t,x)∈[0,T)×Rn,ϕ∈C1,1([0,T)×Rn;R), and let (ϕ–ui) attain a local minimum

at (t,x),ϕ(t,x) –ui(t,x) = 0. Assuming that

∂tϕ(t,x) +xϕ(t,x)b

t,x,u(t,x)+fi

t,x,u(t,x)< 0,

we will obtain a contradiction.

It follows from above that there exists 0 <α<T–tsuch that, for all (s,y)∈[t,T]×Rn

satisfyingt≤s≤t+α,

ui(s,y)≤φ(s,y)

and

∂tϕ(s,y) +xϕ(t,x),b

s,y,u(s,y)+fi

s,y,ui(s,y)

(16)

We denote

t_:=_inf_s_>_t_:_|_X

s–x| ≥α

∧(t+α). From ODEs (2) we have

Y_st,x=YT+

T

s

fr,X_rt,x,ur,X_rt,xdr,

Y_tt,x =YT+

T

t f

r,X_rt,x,ur,X_rt,xdr.

Then we get

Y_st,x=Y_tt,x +

t

s∧tf(r,X

t,x r ,u

r,X_rt,xdr.

SinceY_tt,x =u(t,Xt,x_t), we have

Y_st,x=ut,X_tt,x

+

t

s∧tf(r,X

t,x r ,u

r,X_rt,xdr,

Y_st,x,i=ui

t,X_tt,x

+

t

s∧tfi(r,X

t,x r ,u

r,X_rt,xdr.

(35)

LetYˆs=ϕ(s,Xst,x). By the diﬀerential equation, dYˆs=∂tϕ

s,X_st,xds+xϕ

s,X_st,xbs,X_st,x,us,X_st,xds,

ˆ

Ys=ϕ

t_,_Xt,x t

–

t

s∧t

∂tϕ

r,X_rt,xds+xϕ

r,X_rt,xbr,X_rt,x,ur,X_st,xdr.

(36)

We consider the diﬀerence between (35) and (36):

ˆ

Ys–Yst,x,i=ϕ

t_,_Xt,x t

–ui

t_,_Xt,x t

– t

s∧t

∂tϕ

r,X_rt,xds+xϕ

r,X_rt,xbr,X_rt,x,ur,X_st,x

+fi

r,X_rt,x,ur,X_rt,xdr.

From the deﬁnition oft_{we have}

ϕt_,_Xt,x t

–ui

t_,_Xt,x t

≥0 and

–β(r) =∂tϕ

r,X_rt,x+xϕ

r,X_rt,xbr,X_rt,x,ur,X_st,x+fi

r,X_rt,x,ur,X_rt,x< 0.

Then

ˆ

Ys–Yst,x,i=ϕ

t_,_Xt,x t

–ui

t_,_Xt,x t

+

t

β(r)dr> 0,

(17)

Remark5.1 It is obvious that if a functionuis Lipschitz continuous inxand continuous int, thenusatisﬁes the linear growth assumption, that is, there exists a constantC> 0 such that

_u₍_t_,_x₎_≤_C

1 +|x|, ∀(t,x)∈[0,T]×Rn. (37) Lemma 5.1 Let Assumptions(H1)and(H2)hold,let u,v∈C([0,T]×Rn₎_{be Lipschitz} continuous in x,and let u be a viscosity subsolution and v be a viscosity supersolution of PDE(1).Then the functionω:=u–v is a viscosity subsolution of the following PDE:

⎧ ⎨ ⎩

∂tω(t,x) +C(1 +|x|)|xω(t,x)|+C|ω(t,x)|= 0,

ω(T,x) = 0, (38)

where C is a constant only depending on the Lipschitz constants and the linear constants of b,f.

Proof Letϕ∈C1,1_([0,_T_]_×_Rn_{), and let (}_t

0,x0)∈[0,t]×Rnbe a global maximum point of

ωi–ϕfor some 1≤i≤m.

We introduce the function

ψα,β(t,x,s,y) =ui(t,x) –vi(s,y) –| x–y|2

α –

(t–s)2

β –ϕ(t,x),

whereα,βare positive parameters devoted to zero.

Let (t¯,x¯,s¯,y¯) be a global maximum point ofψα,βin ([0,T]× ¯BR)2, whereBRis a ball with

large radiusR. We drop the dependence oft¯,x¯,¯s,y¯onαandβfor simplicity of notations. Noting that

ψα,β(¯t,x¯,s¯,y¯)≥ max (t,x)∈[0,T]× ¯BR

ψα,β(t,x,t,x),

we have

ui(¯t,x¯) –vi(¯s,y¯) –|¯ x–y¯|2

α –

|¯t–¯s|2

β –ϕ(t¯,x¯)

≥ max [0,T]× ¯BR

ui(t,x) –vi(t,x) –ϕ(t,x)

:=M. (39)

We set

N= max

(t,x,s,y)∈([0,T]× ¯BR)2

ui(t,x) –vi(s,y) –ϕ(t,x)

.

Then we have

|¯x–y¯|2

α +

|¯t–s¯|2

β ≤–M+N.

Since [0,T]× ¯BRis compact, we can ﬁnd (x0,y0)∈[0,T]× ¯BRandαk,βk> 0 such that (t¯,x¯)

(18)

Relation (39) again implies

0≤lim inf

α,β→0

|¯x–y¯|2

α +

|¯t–¯s|2

β

≤lim sup

α,β→0

_|¯

x–y¯|2

α +

|¯t–¯s|2

β

≤lim sup

α,β→0

ui(t¯,x¯) –vi(¯s,y¯) –ϕ(¯t,x¯)

–M

= 0.

We have

lim

α→0

|¯x–¯y|2

α =βlim→0

|¯t–¯s|2

β = 0.

Lettingαandβtend to 0, we get that (t0,x0) is a global maximum point ofui–vi–ϕin

[0,T]× ¯BR.

Now we are going to use the deﬁnition of the viscosity solution. By the deﬁnition of (¯t,x¯,¯s,y¯) the function

(t,x)→ui(t,x) –

vi(s,y) +| x–y|2

α +

(t–s)2

β +ϕ(t,x)

attains a global maximum point at (¯t,x¯) in [0,T]× ¯BR. Hence we have

2(¯t–¯s)

β +∂tϕ(¯t,x¯) +

2(x¯–y¯)

α +xϕ(¯t,x¯)

bt¯,x¯,u(¯t,x¯)+fi

_¯

t,x¯,u(¯t,x¯)≥0.

Similarly, the function

(s,y)→vi(s,y) –

ui(t,x) –| x–y|2

α –

(t–s)2

β –ϕ(t,x)

attains a global minimum point at (¯s,y¯) in [0,T]× ¯BR. Thus

2(¯t–¯s)

β +

2(x¯–y¯)

α b

¯

s,¯y,v(s¯,y¯)+fi

¯

s,y¯,v(s¯,y¯)≥0.

Considering the diﬀerence between the last inequalities, we obtain

0≤∂tϕ(¯t,x¯) +

2(x¯–y¯) α

b¯t,x¯,u(t¯,x¯)–b¯s,y¯,v(¯s,y¯) +Dϕ(t¯,x¯)bt¯,x¯,u(¯t,x¯)+fi

_¯

t,x¯,u(t¯,x¯)–fi

¯

s,y¯,v(s¯,y¯)

≤∂tϕ(¯t,x¯) +

2|¯x–¯y|

α b _¯

t,x¯,u(¯t,x¯)–b¯t,y¯,u(¯t,x¯)

+bt¯,y¯,u(¯t,x¯)–bt¯,y¯,v(¯t,x¯)+b¯t,¯y,v(¯t,x¯)–b¯t,y¯,v(¯t,y¯) +Dϕ(¯t,x¯)b¯t,x¯,u(t¯,x¯)+fi

_¯

t,x¯,u(t¯,x¯)–fi

_¯

t,¯y,u(t¯,x¯) +fi

_¯

t,y¯,u(t¯,x¯)–fi

_¯

t,y¯,v(¯t,x¯)+fi

_¯

t,y¯,v(t¯,x¯)–fi

_¯

(19)

By the Lipschitz continuity and the linear growth property ofbandf, lettingβ tend to zero, the above inequality becomes

0≤∂tϕ(¯t,x¯) +

2|¯x–¯y|

α ×C

|¯x–y¯|+u(¯t,x¯) –v(¯t,x¯)+v(t¯,x¯) –v(¯t,y¯) +C1 +|¯x|+|u(¯t,x¯)xϕ(t¯,x¯)

+C|¯x–y¯|+u(t¯,x¯) –v(t¯,x¯)+v(¯t,x¯) –v(¯t,y¯). Lettingα→0, we have2|¯x–_αy¯|→0. We get

∂tϕ(t0,x0) +C

1 +|x0|xϕ(t0,x0)+Cωi(t0,x0)≥0.

Since (t0,x0) is a maximum point ofωi–ϕ, by Deﬁnition5.1the functionωis a subsolution

of PDE (35), and we conclude the proof.

Lemma 5.2 Let Assumptions(H1)and(H2)hold.For any A> 0,there exists C1> 0such that the function

χ(t,x) =expC1(T–t) +A

ψ(x),

where

ψ(x) =log|x|2+ 1122_, satisﬁes

∂tχ(t,x) +C

1 +|x|xχ(t,x)+Cχ(x) < 0 in[t1,T]×Rn.Here t1=T–_C1A.

Proof By the deﬁnition ofχandψwe have

_xψ(x)≤ 2[ψ(x)]

1 2

(1 +|x|2₎12

.

Then we have

|xχ(t,x)≤

C1(T–t) +A

χ(t,x)xψ(x)

≤Cχ(t,x) [ψ(x)]

1 2

(1 +|x|2₎12

.

Because of the choice oft1, these estimates do not depend onC1. Easy computations yield

∂tχ(t,x) +C

1 +|x|xχ(t,x)+Cχ(x)

≤

–C1ψ(x) +C

1 +|x| [ψ(x)] 1 2

(1 +|x|2₎12

+C

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Sinceψ(x) > 1, it is clear that whenC1 is large enough, the quantity in the brackets is

negative, and the proof is complete.

Theorem 5.2 Let Assumptions(H1)and(H2)hold.Then there exists at most one viscosity solution of PDE(1)in the class of continuous functions that are Lipschitz continuous in the spatial variable x.

In particular,the function u(t,x) =Y_tt,xis the unique viscosity solution of (1)in the class of continuous functions that are Lipschitz continuous in the spatial variable x.

Proof First, we will show thatω=u–vsatisﬁes

ω(t,x)≤αχ(t,x), [0,T]×Rn,

for anyα> 0. Then, we letαtend to zero.

By (37), to prove this inequality, we ﬁrst remark that

lim

|x|→∞

ω(t,x)e–A[log((|x|2₊₁₎12)]2

= 0

uniformly fort∈[0,T] and for someA> 0. This implies, in particular, thatω(t,x) –αχ(t,x) is bounded from above in [t1,T]×Rnfor anyα> 0, and we have that

M:= max

1≤i≤m[t1,T]max×Rn

|ωi|–αχ

(t,x)e–L(T–t)

is achieved at some point (t0,x0)∈[t1,T]×Rn. Since| · |is the supremum norm inRm,

we have

M:= max

[t1,T]×Rn

|ω|–αχ(t,x)e–L(T–t)

and|ωi0(t0,x0)|=|ω(t0,x0)|. We may assume w.l.o.g. that|ωi0(t0,x0)| = 0; otherwise, we are

done.

Then we have|ωi0(t0,x0)|> 0.

From the maximum point property we deduce that

ωi0(t,x) –αχ(t,x)≤

ωi0(t0,x0) –αχ(t0,x0)

e–L(t–t0), (t,x)∈[t1,T]×Rn.

We deﬁne

ϕ(t,x) =αχ(t,x) +ωi0(t0,x0) –αχ(t0,x0)

e–L(t–t0)

and get

(ωi0–ϕ)(t,x)≤(ωi0–ϕ)(t0,x0), (t,x)∈[t1,T]×Rn.

Sinceϕ(t0,x0) =ωi0(t0,x0) > 0 and Lemma5.1, we have

∂tϕ(t0,x0) +C

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for allt0∈[t1,T). By the deﬁnition ofϕwe can rewrite this inequality as

∂tχ(t0,x0) +C

1 +|x0|xχ(t0,x0)+Cχ(t0,x0)≥0.

This is a contradiction with Lemma5.1. Thust0=T. Since|ω(T,x)|= 0, we have

u(t,x) –v(t,x)≤αχ(t,x), (t,x)∈[t1,T]×Rn.

Lettingαtend to zero, we obtain

u(t,x) =v(t,x), (t,x)∈[t1,T]×Rn.

Applying successively the same argument on the intervals [t2,t1], wheret2= (t1–A/C1)+,

and then, ift2> 0, then on [t3,t2], wheret3= (t2–A/C1)+, and so on, we ﬁnally obtain that

u(t,x) =v(t,x), (t,x)∈[0,T]×Rn.

The proof is complete.

6 Conclusion

In this paper, to our best knowledge, we are the first to study this kind of PDE systems as-sociated with the two-point boundary value problems. The distinguishing feature is that we consider the coefficientbof PDE (1) dependent onu(t,x). We give three kinds of so-lutions of PDE (1). The first one is the classical solution, which needs the coefficients of ODEs (2) be twice continuously differentiable with bounded derivatives besides the usual assumptions in [5]. If the coefficientbof ODEs (2) is only once continuously differentiable with bounded derivatives and iffsatisfies the usual Lipschitz condition, then we can prove that the associated PDE has a unique weak solution in the Sobolev space. In addition, we also prove that the function defined by the solution of ODEs (2) is the unique viscosity solution of PDE (1) if the coefficients of ODEs (2) only satisfy the usual assumptions and Lipschitz condition. This kind of two-point boundary problems is quite important in the ordinary differential equations and has meaningful applications in optimal control theory. By virtue of the solution of PDE (1) we give a method to solve the numerical solution of the two-point boundary value problem and provide a powerful tool to solve the related optimal control problems.

Acknowledgements

The authors are grateful to the anonymous reviewers for carefully reading this paper and for their comments and suggestions which have improved the paper.

Funding

This work was supported by the Natural Science Foundation of China (61573217), the National High-level personnel of special support program and the Chang Jiang Scholar Program of Chinese Education Ministry.

Abbreviations

PDE, partial differential equation; ODEs, ordinary differential equations; PDEs, partial differential equations; FBSDEs, forward–backward stochastic differential equations.

Availability of data and materials

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Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

Bothl authors have contributed equally in this paper. Both authors read and approved the ﬁnal manuscript.

Authors’ information

Not applicable.

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Received: 26 April 2018 Accepted: 18 July 2018 References

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