Quantitative unique continuation for the linear coupled heat equations

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R E S E A R C H

Open Access

Quantitative unique continuation for the

linear coupled heat equations

Guojie Zheng, Keqiang Li

*

_{and Jun Li}

*_{Correspondence:}

[email protected] College of Mathematics and Information Science, Henan Normal University, Xinxiang, 453007, P.R. China

Abstract

In this paper, we established a quantitative unique continuation results for a coupled heat equations, with the homogeneous Dirichlet boundary condition, on a bounded convex domain

ofRd_{with smooth boundary}

_∂

_{. Our result shows that the value} of the solutions can be determined uniquely by its value on an arbitrary open subset

ω

of

at any given positive timeT.

MSC: 35K05; 93D15

Keywords: coupled heat equations; unique continuation; frequency functions

1 Introduction

In this paper, we consider an unique continuation of the following linear coupled heat equations:

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

yt(x,t) –y(x,t) –a(x,t)y(x,t) –b(x,t)z(x,t) = , (x,t)∈×(,T),

zt(x,t) –z(x,t) –c(x,t)y(x,t) –d(x,t)z(x,t) = , (x,t)∈×(,T),

y(x,t) = , (x,t)∈∂×(,T),

z(x,t) = , (x,t)∈∂×(,T),

(.)

whereis a bounded, convex domain ofRd ₍_d_≥_{), with a smooth boundary}_{∂. Let}

ω⊂be a nonempty and open subset of, andTis a positive number. Now we suppose that

a(x,t),b(x,t),c(x,t),d(x,t)∈L∞×(,T) (.)

and

maxaL∞(×(,T)),bL∞(×(,T)),cL∞(×(,T)),dL∞(×(,T)) ≤M, (.)

whereMis a positive number.

In this paper, we are concerned with the unique continuation for the solution of Eq. (.). The main results obtained in this work can be stated as follows.

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Theorem . Letωbe a nonempty open subset of.Then there are two positive numbers p=p(,ω)and C=C(,ω)such that for each T> and for each potential a,b,c,d∈ L∞(×(,T))with condition(.),any solution(y,z)with y(·, )≡y(·)∈L(),z(·, )≡

z(·)∈L(),to equation(.)has the following estimate:

_y₍_x_,_T₎

+z(x,T)dx

≤Cexp

C T +C

MT+MT×

y(x)+z(x) 

dx

–p

×

ω

y(x,T)+z(x,T)dx

p

. (.)

Theorem . Letωbe a nonempty open subset of.If the initial data(y,z)≡(, ),and

y,z∈H(),then there exists a positive number C=C(,ω)such that for each T> and

for each potential a,b,c,d∈L∞(×(,T))with condition(.),any solution(y,z)with y(·, )≡y(·),z(·, )≡z(·),to equation(.)has the following estimate:

y_L₍₎+z_L₍₎

≤CeCT_eC(MT+MT)_exp

C( + M)TeCMT

y_H ()

+z_H ()

y_L₍₎+z_L₍₎

×

ω

y(x,T)+z(x,T)dx. (.)

Remark . The constantCin (.) or (.) stands for a positive constant only dependent on,ω. This constant varies in diﬀerent contexts.

The study of unique continuation for the solutions for PDEs began at the early of the last centaury. Besides its own interesting in PDEs theory, it also plays a key role in both inverse problem and control theory. The first quantitative result of strong unique contin-uation for parabolic eqcontin-uations was derived in  in the literature []. In [], the authors establish the unique continuation for the parabolic equations with time independent co-efficients in terms of the eigenfunctions of the corresponding elliptic operator, and their results did not apply to parabolic equations with time dependent coefficients. After , there were more results of unique continuation for parabolic equations, and we refer the reader to [–], and the rich work cited therein. In [], the author discusses the unique continuation for stochastic counterpart. In our paper, we mainly study this property for the coupled heat equations. To the best of our knowledge, this topic has not been studied in past publications. With the aid of frequency function methods, we can establish these quantitative estimates (see [, ]).

The paper is organized as follows. In Section , some preliminary results are presented. The proofs of Theorem . and Theorem . will be given in Section .

2 Preliminary results

Given a positive numberλ, we deﬁne

Gλ(x,t) =

 (T–t+λ)d/e

–_(|x_T–_–x_t₊|_λ₎

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Then, for eacht∈[,T], we write

Hλ(t) =

y(x,t)+z(x,t)Gλ(x,t)dx, (.)

Dλ(t) =

_∇y(x,t)+∇z(x,t)Gλ(x,t)dx, (.)

and

Nλ(t) =

Dλ(t) Hλ(t)

, (.)

where (y(x,t),z(x,t)) are the solutions of equation (.). The functionNλ(t) was ﬁrst

dis-cussed in [], and it was called frequency function (see also [, ], and []). Throughout this section, we always work under the assumptionHλ(t)= . Next, we will discuss the

properties for the functionsGλ(x,t),Hλ(t),Dλ(t) andNλ(t). Now, we ﬁrst ﬁx a positive

numberrand a pointxin the subsetωsuch thatBr⊂ω.Brdenotes the open ball,

cen-tered at the pointxand of radiusr, inRd. Writem=supx∈|x–x|. Lemma . is taken from [, ].

Lemma . For eachλ> ,the function Gλgiven in(.)holds the following four identities overRd_×_[,_T_]:

∂tGλ(x,t) +Gλ(x,t) = , (.)

∇Gλ(x,t) =

–(x–x)

(T–t+λ)Gλ(x,t), (.)

∂_iGλ(x,t) =

–

(T–t+λ)Gλ(x,t) +

|xi–xi|

(T–t+λ)Gλ(x,t), (.)

and,for i=j,

∂i∂jGλ(x,t) =

(xi–xi)(xj–xj)

(T–t+λ) Gλ(x,t). (.)

Lemma . For eachλ> ,the following identity holds for t∈(,T):

d

dtlnHλ(t) = –Nλ(t) +



Hλ(t)

y(∂ty–y) +z(∂tz–z)

Gλdx. (.)

Proof By a direct computation, we can obtain

Hλ(t) = 

(y∂ty+z∂tz)Gλdx+

y+z∂tGλdx

= 

(y∂ty+z∂tz)Gλdx–

y+zGλdx

= 

y+zGλdx

= 

(y∂ty+z∂tz)Gλdx– 

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= 

Gλdx– 

|∇y|+|∇z|Gλdx

= 

Gλdx– Dλ(t). (.)

Therefore, for eacht∈(,T), we have

d

dtlnHλ(t) = Hλ(t) Hλ(t)

= –Nλ(t) +



Hλ(t)

Gλdx. (.)

This completes the proof of this lemma.

Lemma . For eachλ> and t∈(,T),the functions Hλand Dλdeﬁned in(.)and

(.),respectively,satisfy

H_λ(t)Dλ(t) = –



y

∂ty–

x–x (T–t+λ)∇y+



(y–∂ty)

Gλdx

+ 

z

∂tz–

x–x (T–t+λ)∇z+



(z–∂tz)

Gλdx



+

y(y–∂ty)Gλdx+

z(z–∂tz)Gλdx



. (.)

Proof By (.), (.), we can rewriteH_λ(t) as follows:

Hλ(t) = 

y+zGλdx

= 

(y∂ty+z∂tz)Gλdx+

∇y+z∇Gλdx

= 

(y∇y+ z∇z) x–x

(T–t+λ)Gλdx

= 

y

∂ty–

x–x (T–t+λ)∇y

Gλdx+ 

z

∂tz–

x–x (T–t+λ)∇z

Gλdx

= 

y

∂ty–

x–x (T–t+λ)∇y+



(y–∂ty)

Gλdx

+ 

z

∂tz–

x–x (T–t+λ)∇z+



(z–∂tz)

Gλdx

–

y(y–∂ty)Gλdx–

z(z–∂tz)Gλdx. (.)

It follows from (.) that for eacht∈(,T)

Dλ(t) =

|∇y|+|∇z|Gλdx

=

∇y∇yGλdx+

∇z∇zGλdx

=

div(y∇yGλ)dx–

ydiv(∇yGλ)dx+

div(z∇zGλ)dx

–

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= –

yyGλdx+

y∇y x–x

(T–t+λ)Gλdx–

zzGλdx

+

z∇z x–x

= –

y

∂ty–

x–x (T–t+λ)∇y+



(y–∂ty)

Gλdx–

 

y(y–∂ty)Gλdx

–

z

∂tz–

x–x (T–t+λ)∇z+



(z–∂tz)

Gλdx

– 

z(z–∂tz)Gλdx. (.)

This, combining with (.), shows that

H_λ(t)Dλ(t) = –



y

∂ty–

x–x (T–t+λ)∇y+



(y–∂ty)

Gλdx

+ 

z

∂tz–

x–x (T–t+λ)∇z+



(z–∂tz)

Gλdx



+



. (.)

Lemma . For eachλ> and t∈(,T),then we have

D_λ(t) = –

∂

|∇y|∂νGλdσ+ 

∂

∂νy(∇y∇Gλ)dσ

–

∂

|∇z|∂νGλdσ+ 

∂

∂νz(∇z∇Gλ)dσ

– 

∂ty–

x–x (T–t+λ)∇y+



(y–∂ty) 

Gλdx

– 

∂tz–

x–x (T–t+λ)∇z+



(z–∂tz) 

Gλdx

+ 

(y–∂ty)Gλdx+

 

(z–∂tz)Gλdx+



(T–t+λ)Dλ(t). (.)

Proof By a direct computation, we can derive

Dλ(t) = 

∇y∂t∇yGλdx+

|∇y|∂tGλdx

+ 

∇z∂t∇zGλdx+

|∇z|∂tGλdx

= 

∇y∂t∇yGλdx–

|∇y|Gλdx

+ 

∇z∂t∇zGλdx–

|∇z|Gλdx

= 

div(∇y∂tyGλ)dx– 

∂tydiv(∇yGλ)dx–

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+ 

div(∇z∂tzGλ)dx– 

∂tzdiv(∇zGλ)dx–

|∇z|Gλdx

= –

∂tyyGλdx– 

∂ty∇y∇Gλdx–

|∇y|Gλdx

– 

∂tzzGλdx– 

∂tz∇z∇Gλdx–

|∇z|Gλdx

= –

∂tyyGλdx+ 

∂ty∇y

x–x

|∇y|Gλdx

– 

∂tzzGλdx+ 

∂tz∇z

x–x

|∇z|Gλdx. (.)

However,

|∇y|Gλ=div

|∇y|∇Gλ

– div∇y(∇y∇Gλ)

+ y∇y∇Gλ+ 

d

i=

∇y∂iy∂i∇Gλ (.)

and

|∇z|Gλ=div

|∇z|∇Gλ

– div∇z(∇z∇Gλ)

+ z∇z∇Gλ+ 

d

i=

∇z∂iz∂i∇Gλ. (.)

Now, we write

A=

∂

|∇y|∂νGλdσ– 

∂

∂νy(∇y∇Gλ)dσ (.)

and

B=

∂

|∇z|∂νGλdσ– 

∂

∂νz(∇z∇Gλ)dσ. (.)

Then, by (.), (.), (.), (.), we obtain

|∇y|Gλdx=A+ 

y∇y∇Gλdx+ 

d

i=

∇y∂iy∂i∇Gλdx

=A– 

y∇y x–x

+ 

d

i=

∂iy∂iy

–

(T–t+λ)Gλ+

|x–x| (T–t+λ)Gλ

dx

+ 

i=j

∂jy∂iy

(xi–xi)(xj–xj)

(T–t+λ) Gλdx

=A– 

y∇y x–x

(T–t+λ)Gλdx+

|∇y|

– (T–t+λ)

Gλdx

+ 



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In the same way, we get

|∇z|Gλdx=B– 

z∇z x–x

(T–t+λ)Gλdx+

|∇z|

– (T–t+λ)

Gλdx

+ 



Gλdx. (.)

Thus, we can rewrite (.) as

Dλ(t) = –A–B– 



Gλdx– 



Gλdx

+ 

(y+∂ty)

x–x

(T–t+λ)∇yGλdx+ 

(z+∂tz)

x–x

(T–t+λ)∇zGλdx

– 

∂tyyGλdx– 

∂tzzGλdx+



(T–t+λ)Dλ(t)

= –A–B– 

∂ty–

x–x (T–t+λ)∇y+



(y–∂ty) 

Gλdx

– 

∂tz–

x–x (T–t+λ)∇z+



(z–∂tz) 

Gλdx

+ 



(y+∂ty) _G

λdx– 

∂tyyGλdx

+ 



(z+∂tz) _G

λdx– 

∂tzzGλdx+



(T–t+λ)Dλ(t)

= –A–B– 

∂ty–

x–x (T–t+λ)∇y+



(y–∂ty) 

Gλdx

– 

∂tz–

x–x (T–t+λ)∇z+



(z–∂tz) 

Gλdx

+ 

 

(z–∂tz)Gλdx+



(T–t+λ)Dλ. (.)

Lemma . For eachλ> and t∈(,T),it follows that

d dt

(T–t+λ)Nλ(t)

≤M(T+λ). (.)

Proof Firstly, we computeN_λ(t) ast∈(,T). By (.) and (.), we derive that

N_λ(t) = 



Hλ(t)



D_λ(t)Hλ(t) –Dλ(t)Hλ(t)

= 

Hλ(t)

–A–B+ 

(T–t+λ)Dλ(t)

– 

∂ty–

x–x (T–t+λ)∇y+



(y–∂ty) 

Gλdx+

 

(y–∂ty)Gλdx

– 

∂tz–

x–x (T–t+λ)∇z+



(z–∂tz) 

Gλdx+

 

(z–∂tz)Gλdx

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–



Hλ(t)

 –



y

∂ty–

x–x (T–t+λ)∇y+



(y–∂ty)

Gλdx

+ 

z

∂tz–

x–x (T–t+λ)∇z+



(z–∂tz)

Gλdx



+



= 

(T–t+λ)Nλ– 

A+B Hλ(t)

+



Hλ(t)



y

∂ty–

x–x (T–t+λ)∇y+



(y–∂ty)

Gλdx

+

z

∂tz–

x–x (T–t+λ)∇z+



(z–∂tz)

Gλdx



–

∂ty–

x–x (T–t+λ)∇y+



(y–∂ty) 

Gλdx

y+zGλdx

–

∂tz–

x–x (T–t+λ)∇z+



(z–∂tz) 

Gλdx

y+zGλdx

–



Hλ(t)



+ 

Hλ(t)

. (.)

Let

α=∂ty–

x–x (T–t+λ)∇y+



(y–∂ty), β=∂tz–

x–x (T–t+λ)∇z+



(z–∂tz). It follows from the Cauchy-Schwarz inequality that

αGλdx

y+zGλdx+

βGλdx

y+zGλdx

=

α+βGλdx

y+zGλdx

≥

yαGλdx+

zβGλdx

 .

It, together with (.), shows that

N_λ(t) – 

(T–t+λ)Nλ(t) + 

A+B Hλ(t)

– 

Hλ(t)

≤. (.)

In what follows, we will discuss the properties ofAandB. Sincey=z=  on∂, we have

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According to (.) and (.), then

A=

∂

|∇y|∂νGλdσ– 

∂

∂νy(∇y∇Gλ)dσ

= – 

(T–t+λ)

∂

|∇y|(x–x)·ν

Gλdσ+

 (T–t+λ)

×

∂

∂νy

(x–x)∇y

Gλdσ

= – 

(T–t+λ)

∂

|∇y|(x–x)·ν

Gλdσ+

 (T–t+λ)

×

∂ |∂νy|

(x–x)·ν

Gλdσ

= 

(T–t+λ)×

∂ |∂νy|

(x–x)·ν

Gλdσ≥.

In the same way, we get

B= 

(T–t+λ)×

∂ |∂νz|

(x–x)·ν

Gλdσ≥.

Combining with (.), we get

(T–t+λ)Nλ(t) –Nλ(t)

–(T–t+λ)

Hλ(t)

≤. (.)

Using equations (.) and (.) can be written as

(T–t+λ)N_λ(t) –Nλ(t)≤

(T–t+λ)

Hλ(t)

(ay+bz)Gλdx+

(cy+dz)Gλdx

≤M(T–t+λ)

Hλ(t)

|y|+|z|Gλdx

≤M(T+λ).

Thus,

d dt

(T–t+λ)Nλ(t)

≤M(T+λ), ∀t∈(,T).

Let

KM,y,z,T= ln

(|y(x, )|₊_|_z₍_x_{, )}_|₎_dx

(|y(x,T)|+|z(x,T)|)dx

+m

T + MT+ M

_T₊d .

(10)

Lemma . For eachλ> ,we have

λNλ(T) + d

 ≤

λ

T + 

KM,y,z,T. (.)

Proof Integrating (.) over (t,T), we have

λNλ(T)≤(T–t+λ)Nλ(t) + M(T+λ)(T–t) ≤(T+λ)Nλ(t) + MT(T+λ).

Integrating the above over (,T_), we obtain

T

λNλ(T)≤(T+λ) T





Nλ(t)dt+ MT(T+λ). (.)

By integrating (.) over (,T_), we get T





Nλ(t)dt= –

T





H_λ(t)

Hλ(t) dt+

T



 

Hλ(t)

Gλdx dt

= –lnHλ(

T

)

Hλ()

+ T



 

Hλ(t)

–y(ay+bz) –z(cy+dz)Gλdx dt

≤ln Hλ()

Hλ(T_)

+ T



 

Hλ(t)

My+zGλdx dt

≤ln Hλ()

Hλ(T_)

+ MT.

This, alone with (.), shows that

T

λNλ(T)≤(T+λ)

lnHλ()

Hλ(T_)

+ MT+ MT

.

We have

Hλ() Hλ(T_)

=

(|y(x, )|

₊_|_z₍_x_{, )}_|₎₍_T₊_λ)–_d_·_e–|_(x–_Tx₊_λ|₎

dx

(|y(x,

T

)|+|z(x,

T

)|)(

T

 +λ) –d_ _·_e–

|x–x_| (T_{ +}λ)_dx

≤

(|y(x, )|+|z(x, )|)dx·(

T

 +λ)

d



(|y(x,

T

)|+|z(x,

T

)|)·e –|x–x|

(T_{ +}λ)

dx·(T+λ)d

≤

(|y(x, )|

₊_|_z₍_x_{, )}_|₎_dx_·₍T

 +λ)

d



(|y(x,

T

)|+|z(x,

T

)|)dx·(T+λ)

d



·e m

(T_{ +}λ)

≤emT

(|y(x, )|

₊_|_z₍_x_{, )}_|₎_dx

(|y(x,

T

)|+|z(x,

T

(11)

Therefore,

T

λNλ(T)≤(T+λ)

m

T +ln

(|y(x, )|+|z(x, )|)dx

(|y(x,

T

)|+|z(x,

T

)|)dx

+ MT+ MT

.

Using the energy estimate

(|y(x,T)|

₊_|_z₍_x_,_T₎_|₎_dx

(|y(x,

T

)|+|z(x,

T

)|)dx

≤eMT,

we obtain

λNλ(T)≤

λ

T + 

ln (|y(x, )|

₊_|_z₍_x_{, )}_|₎_dx

(|y(x,T)|+|z(x,T)|)dx

+m

T + MT+ M

_T

=

λ

T + 

KM,y,z,T–

d



.

Thus,

λNλ(T) + d

 ≤

λ

T + 

KM,y,z,T–

λ

T · d



≤

λ

T + 

KM,y,z,T.

This is (.).

While most of the proof of the following lemma is similar to Lemma  of [], we would rather give the proof in detail for the sake of completeness.

Lemma . For each T> and y∈L(),z∈L(),the solution y and z,with y(·, ) =

y(·),z(·, ) =z(·),to(.)holds the estimate:

 –λ

r

λ

T + 

KM,y,z,T

|x–x|y(x,T) 

+z(x,T)e–|x–x|

 λ _dx

≤λ

λ

T + 

KM,y,z,T

Br

y(x,T)+z(x,T)e–|x–x|

 λ _dx_.

Proof We ﬁrst point out the following fact: for anyf ∈H

() and for eachλ> , we have ≤|∇(f(x)exp(–|x–x|

λ ))|

_dx_{. By computing the right-hand term, we get}

|x–x| λ f(x)



e–|x–x|



λ dx≤λ

_∇f(x)e–|x–x|

 λ dx

+d 

f(x)e–|x–x|



λ _dx_. _(.)

It follows from (.) and (.) that

|x–x|y(x,T) 

+z(x,T)e–|x–x|

 λ _dx

≤λ

λ

_∇y(x,T)+∇z(x,T)e–|x–x|

(12)

+d 

y(x,T)+z(x,T)e–|x–x|

 λ _dx

≤λ

λNλ(T) + d



_y₍_x_,_T₎

+z(x,T)e–|x–x|

 λ _dx

≤λ

λNλ(T) + d



Br

y(x,T)+z(x,T)e–|x–x|

 λ dx

+ 

r

\Br

|x–x|y(x,T) 

+z(x,T)e–|x–x|

 λ _dx

. (.)

It, combining with (.), shows that

|x–x|y(x,T) 

+z(x,T)e–|x–x|

 λ dx

≤λ

λ

T + 

KM,y,z,T

Br

y(x,T)+z(x,T)e–|x–x|

 λ _dx

+ 

r

|x–x|y(x,T) 

+z(x,T)e–|x–x|

 λ _dx

. (.)

We get

 –λ

r

λ

T + 

KM,y,z,T

|x–x|y(x,T) 

+z(x,T)e–|x–x|

 λ _dx

≤λ

λ

T + 

KM,y,z,T

Br

_y₍_x_,_T₎

+z(x,T)e–|x–x|



λ _dx_. _(.)

This completes the proof of the lemma.

3 Results and discussion 3.1 The proof of Theorem 1.1

Proof We start with proving (.). We takeλ>  in the estimate of Lemma . to be such that

λ

r

λ

T + 

KM,y,z,T=



. (.)

By direct computation, we have

λ= 

–T+

T₊ Tr KM,y,z,T

.

Sincem_T ≤KM,y,z,T, it follows that

 λ = 

T+

T₊ Tr KM,y,z,T

Tr KM,y,z,T

= 

T+

T₊ Tr  KM,y,z,T



(13)

≤

T+

Tr KM,y,z,T



TrKM,y,z,T

≤

 +√r

m



rKM,y,z,T. Therefore, we have

emλ ≤_e(m+r √

m)

rKM,y,z,T

≤e(m+r

√

m)

r d

_e(m+r √

m)

r( m

T+MT+MT) × (|y(x, )|

₊_|_z₍_x_{, )}_|₎_dx

(|y(x,T)|+|z(x,T)|)dx

(m+r√m)/r

. (.)

By Lemma ., we get

|x–x|y(x,T) 

+z(x,T)e–|x–x|

 λ dx

≤r

Br

_y₍_x_,_T₎

+z(x,T)e–|x–x|

 λ _dx_.

Then we derive that

y(x,T)+z(x,T)e–mλ_dx

≤

y(x,T)+z(x,T)e–|x–x|

 λ dx

≤

∩|x–x|≥r

_y₍_x_,_T₎

+z(x,T)e–|x–x|

 λ _dx

+

Br

_y₍_x_,_T₎

+z(x,T)e–|x–x|

 λ _dx

≤ 

r

|x–x|y(x,T) 

+z(x,T)e–|x–x|

 λ _dx

+

Br

_y₍_x_,_T₎

+z(x,T)e–|x–x|

 λ _dx

≤

Br

_y₍_x_,_T₎

+z(x,T)e–|x–x|

 λ _dx

≤

Br

_y₍_x_,_T₎

+z(x,T)dx.

Thus,

_y₍_x_,_T₎

+z(x,T)dx

≤emλ

Br

≤e(m+r

√

m)

rd_e(m+r √

m)

(14)

× (|y(x, )|

₊_|_z₍_x_{, )}_|₎_dx

(|y(x,T)|+|z(x,T)|)dx

(m+r√m)/r

×

Br

y(x,T)+z(x,T)dx. (.)

Then we have

≤CerC_e C Tr_e

C r(MT+M

_T₎

(|y(x, )|

₊_|_z₍_x_{, )}_|₎_dx

(|y(x,T)|+|z(x,T)|)dx

C/r

×

Br

_y(x,T)+z(x,T)dx,

which is equivalent to the following inequality:

≤CeC/TeC(MT+MT)

y(x, )+z(x, )dx

C r+C

×

Br

_y(x,T)+z(x,T)dx

r r+C

≤Cexp

C T +C

MT+MT

y(x) 

+z(x) 

dx

C r+C

×

ω

_y₍_x_,_T₎

+z(x,T)dx

r r+C

.

Letp=_rr₊_Cand then the above inequality can be written as

_y₍_x_,_T₎

+z(x,T)dx

≤Cexp

C T +C

MT+MT

y(x)+z(x) 

dx

–p

×

ω

_y₍_x_,_T₎

+z(x,T)dx

p

. (.)

Thus, we can obtain (.). This completes the proof of this theorem.

3.2 The proof of Theorem 1.2

Proof In order to give (.), we should ﬁrst prove the following backward uniqueness es-timate:

y(x, )_L+z(x, )_L

y(x,T)

L+z(x,T)_L

≤exp

C( + M)TeCMT

y(x, )

H_+z(x, )



H_

y(x, )

L+z(x, )_L

(15)

Fory,z∈H(), the solutions of equation (.)y(x,t),z(x,t)∈L(,T;H()∩H()). Then we can deﬁne a function(t) as follows:

(t) =

y(x,t)

H_+z(x,t)



H_

y(x,t)_L+z(x,t)_L

≥C.

Letf =ay+bz, andg=cy+dz. By direct computation, we obtain

d dt

 y



L

= –y

H–ay+bz,y, (.)

d dt

 z



L

= –z_H

–cy+dz,z, (.)

d dt

 y



H_

= –y_L+

y(ay+bz)dx,

d dt

 z



H 

= –z_L+

z(cy+dz)dx.

Then

d dt(t) =

(y_H 

+z_H 

)(y_L+z_L) – (y_H 

+z_H 

)(y_L+z_L) (y_L+z_L)

= 

(y_L+z_L)

×

–y_L–z_L+

fy dx+

gz dxy_L+z_L

+

y_H +z



H_+

fy dx+

gz dxy_H +z



H_

= 

(y_L+z_L)

×

–y_L–z_L+

fy dx+

+

y

H+z 

H+  

fy dx+

gz dx

 –



fy dx+

gz dx

 .

Integrating by parts and using Cauchy-Schwarz inequality, we have

d dt(t) =

 (y

L+z_L)

×

–y_L–z_L+

fy dx+

+

–y+f



y dx+

–z+g



z dx

 –



fy dx+

gz dx



≤ 

(y_L+z_L)

×

–y_L–z_L+

fy dx+

(16)

+–y+f 

L

yL+ –z+g



L

zL 

–  

fy dx+

gz dx



≤ 

(y

L+z_L)

×

–y_L–z_L+

fy dx+

+–y+f 



L

+–z+g  

L

y_L+z_L

.

Thus,

d dt(t)≤

 (y_L+z_L)

_f

L +g

 

L

y_L+z_L

= f 

L+g_L (y_L+z_L)

=ay+bz 

L+cy+dz_L (y_L+z_L)

≤CM

y

H_+z



H_

y

L+z_L

=CM(t).

With the aid of Gronwall’s inequality, we obtain, fort∈(,T),

(t)≤eCMT(). (.)

This, combining with (.) and (.), indicates that

 =  

d dt

y_L+z_L

+y_H +z



H+ay+bz,y+cy+dz,z

≤ 



d dt

y_L+z_L

+(t)y_L+z_L

+ M(t)y_L+z_L

≤ 



d dt

y_L+z_L

+ ( + M)eCMT()y_L+z_L

. (.)

Integrating (.) on (,T), we get the desired estimate

y(x, )_L+z(x, ) 

L≤exp

( + M)TeCMT()y(x,T)_L+z(x,T) 

L

.

Thus,

y(x, )_L+z(x, )_L

y(x,T)

L+z(x,T)_L

≤exp

( + M)TeCMT

y(x, )

H+z(x, ) 

H

y(x, )

L+z(x, )_L

.

This is the proof of (.), and we have completed the proof of Theorem ..

4 Conclusions

(17)

other words, if the solution of equation (.) satisﬁesy(·,T) =z(·,T) =  onω, theny=z=  on×(,T). This is the quantitative version of the unique continuation property for equation (.).

Acknowledgements

The authors would like to thank professor Xin Yu for his valuable suggestions on this article.

Funding

This work was partially supported by the Natural Science Foundation of China (11501178), the Natural Science Foundation of Henan Province (No. 162300410176).

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

GZ provided the question. GZ, KL and JL gave the proof for the main result together. All authors read and approved the ﬁnal manuscript.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations. Received: 15 May 2017 Accepted: 8 September 2017

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