Zheng et al. Journal of Inequalities and Applications (2017) 2017:234 DOI 10.1186/s13660-017-1508-7

RESEARCH

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  of Rd 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 time T. 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), ⎪ ⎪ ⎪ ⎪ ⎨z (x, t) – z(x, t) – c(x, t)y(x, t) – d(x, t)z(x, t) = , (x, t) ∈  × (, T), t ⎪y(x, t) = , ⎪ (x, t) ∈ ∂ × (, T), ⎪ ⎪ ⎪ ⎩ z(x, t) = , (x, t) ∈ ∂ × (, T),

(.)

where  is a bounded, convex domain of Rd (d ≥ ), with a smooth boundary ∂. Let ω ⊂  be a nonempty and open subset of , and T is a positive number. Now we suppose that   a(x, t), b(x, t), c(x, t), d(x, t) ∈ L∞  × (, T)

(.)

 max aL∞ (×(,T)) , bL∞ (×(,T)) , cL∞ (×(,T)) , dL∞ (×(,T)) ≤ M,

(.)

and

where M is 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. © The Author(s) 2017. 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, provided 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.

Zheng et al. Journal of Inequalities and Applications (2017) 2017:234

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Theorem . Let ω be 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







–p        C       × ≤ C exp + C MT + M T y (x) + z (x) dx T 

p

     y(x, T) + z(x, T) dx . ×

(.)

ω

Theorem . Let ω be 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 () C T

C(MT+M T  )

≤ Ce e

y H  () + z H  ()   CM T exp C( + M)Te y L () + z L ()

     y(x, T) + z(x, T) dx.

×

(.)

ω

Remark . The constant C in (.) 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 continuation for parabolic equations was derived in  in the literature []. In [], the authors establish the unique continuation for the parabolic equations with time independent coefficients 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 number λ, we define Gλ (x, t) =

|x–x |  – e (T–t+λ) , d/ (T – t + λ)

(x, t) ∈  × (, T).

(.)

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Then, for each t ∈ [, T], we write

Hλ (t) =

     y(x, t) + z(x, t) Gλ (x, t) dx,

(.)

     ∇y(x, t) + ∇z(x, t) Gλ (x, t) dx,

(.)



Dλ (t) = 

and Nλ (t) =

Dλ (t) , Hλ (t)

(.)

where (y(x, t), z(x, t)) are the solutions of equation (.). The function Nλ (t) was first discussed in [], and it was called frequency function (see also [, ], and []). Throughout this section, we always work under the assumption Hλ (t) = . Next, we will discuss the properties for the functions Gλ (x, t), Hλ (t), Dλ (t) and Nλ (t). Now, we first fix a positive number r and a point x in the subset ω such that Br ⊂ ω. Br denotes the open ball, centered at the point x and of radius r, in Rd . Write m = supx∈ |x – x | . Lemma . is taken from [, ]. Lemma . For each λ > , the function Gλ given in (.) holds the following four identities over Rd × [, T]: ∂t Gλ (x, t) + Gλ (x, t) = ,

(.)

∇Gλ (x, t) =

–(x – x ) Gλ (x, t), (T – t + λ)

(.)

∂i Gλ (x, t) =

– |xi – xi | Gλ (x, t) + Gλ (x, t), (T – t + λ) (T – t + λ)

(.)

and, for i = j, ∂i ∂j Gλ (x, t) =

(xi – xi )(xj – xj ) Gλ (x, t). (T – t + λ)

(.)

Lemma . For each λ > , the following identity holds for t ∈ (, T): d  ln Hλ (t) = –Nλ (t) + dt Hλ (t)



  y(∂t y – y) + z(∂t z – z) Gλ dx. 

Proof By a direct computation, we can obtain Hλ (t)





=



(y∂t y + z∂t z)Gλ dx +









(y∂t y + z∂t z)Gλ dx –

=



=

 y + z Gλ dx





   y + z Gλ dx

(y∂t y + z∂t z)Gλ dx –

 y + z ∂t Gλ dx







(y∂t y + z∂t z)Gλ dx – 

= 





 (∇y) + yy + (∇z) + zz Gλ dx

(.)

Zheng et al. Journal of Inequalities and Applications (2017) 2017:234





  y(∂t y – y) + z(∂t z – z) Gλ dx – 

=

Page 4 of 17



  |∇y| + |∇z| Gλ dx 

  y(∂t y – y) + z(∂t z – z) Gλ dx – Dλ (t).

=

(.)



Therefore, for each t ∈ (, T), we have d H (t)  ln Hλ (t) = λ = –Nλ (t) + dt Hλ (t) Hλ (t)





 y(∂t y – y) + z(∂t z – z) Gλ dx.

(.)





This completes the proof of this lemma.

Lemma . For each λ >  and t ∈ (, T), the functions Hλ and Dλ defined in (.) and (.), respectively, satisfy  Hλ (t)Dλ (t) = –  y ∂t y –

 x – x ∇y + (y – ∂t y) Gλ dx (T – t + λ)  



 x – x ∇z + (z – ∂t z) Gλ dx +  z ∂t z – (T – t + λ)  



+ y(y – ∂t y)Gλ dx + z(z – ∂t z)Gλ dx . 

(.)



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





=



(y∂t y + z∂t z)Gλ dx –



=





  ∇ y + z ∇Gλ dx

(y∂t y + z∂t z)Gλ dx +



 y + z Gλ dx





x – x Gλ dx (T – t + λ)  



x – x x – x =  y ∂t y – ∇y Gλ dx +  z ∂t z – ∇z Gλ dx (T – t + λ) (T – t + λ)  

 x – x =  y ∂t y – ∇y + (y – ∂t y) Gλ dx (T – t + λ)  

 x – x ∇z + (z – ∂t z) Gλ dx +  z ∂t z – (T – t + λ)  



(.) – y(y – ∂t y)Gλ dx – z(z – ∂t z)Gλ dx. (y∂t y + z∂t z)Gλ dx –

=



(y∇y + z∇z)



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

  |∇y| + |∇z| Gλ dx

Dλ (t) =



∇y∇yGλ dx +

=



=

∇z∇zGλ dx 





div(y∇yGλ ) dx – 

y div(∇yGλ ) dx + 

z div(∇zGλ ) dx

– 

div(z∇zGλ ) dx 

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x – x = – yyGλ dx + y∇y Gλ dx – zzGλ dx (T – t + λ)   

x – x Gλ dx + z∇z (T – t + λ) 



 x – x  = – y ∂t y – ∇y + (y – ∂t y) Gλ dx – y(y – ∂t y)Gλ dx (T – t + λ)    

x – x  – z ∂t z – ∇z + (z – ∂t z) Gλ dx (T – t + λ)  

 z(z – ∂t z)Gλ dx. (.) –   This, combining with (.), shows that  Hλ (t)Dλ (t) = –  y ∂t y –

 x – x ∇y + (y – ∂t y) Gλ dx (T – t + λ)  



 x – x +  z ∂t z – ∇z + (z – ∂t z) Gλ dx (T – t + λ)  



+ y(y – ∂t y)Gλ dx + z(z – ∂t z)Gλ dx . 

(.)





This completes the proof of this lemma. Lemma . For each λ >  and t ∈ (, T), then we have D λ (t) = –



|∇y| ∂ν Gλ dσ + 

∂ν y(∇y∇Gλ ) dσ

∂

∂





|∇z| ∂ν Gλ dσ +  

–

∂ν z(∇z∇Gλ ) dσ

∂

∂

∂t y –

  x – x ∇y + (y – ∂t y) Gλ dx (T – t + λ)  



x – x  ∂t z – ∇z + (z – ∂t z) Gλ dx – (T – t + λ)  



   Dλ (t). (y – ∂t y) Gλ dx + (z – ∂t z) Gλ dx + +     (T – t + λ) –

Proof By a direct computation, we can derive D λ (t) = 



∇y∂t ∇yGλ dx + 

|∇y| ∂t Gλ dx 





∇z∂t ∇zGλ dx +

+ 



|∇z| ∂t Gλ dx 



∇y∂t ∇yGλ dx –

= 





∇z∂t ∇zGλ dx –

+ 



|∇y| Gλ dx 

|∇z| Gλ dx 





div(∇y∂t yGλ ) dx – 

= 

|∇y| Gλ dx

∂t y div(∇yGλ ) dx – 



(.)

Zheng et al. Journal of Inequalities and Applications (2017) 2017:234





+

div(∇z∂t zGλ ) dx – 



Page 6 of 17





















|∇z| Gλ dx

∂t z∇z∇Gλ dx –

∂t yyGλ dx + 



∂t y∇y 





|∇y| Gλ dx 

∂t zzGλ dx – 

= –





–



∂t y∇y∇Gλ dx –



|∇z| Gλ dx

∂t z div(∇zGλ ) dx – 

∂t yyGλ dx – 

= –



x – x Gλ dx – (T – t + λ)



x – x Gλ dx – –  ∂t zzGλ dx +  ∂t z∇z (T – t + λ)  

|∇y| Gλ dx 



|∇z| Gλ dx.

(.)



However,     |∇y| Gλ = div |∇y| ∇Gλ –  div ∇y(∇y∇Gλ ) + y∇y∇Gλ + 

d 

∇y∂i y∂i ∇Gλ

(.)

i=

and     |∇z| Gλ = div |∇z| ∇Gλ –  div ∇z(∇z∇Gλ ) + z∇z∇Gλ + 

d 

∇z∂i z∂i ∇Gλ .

(.)

i=

Now, we write



 |∇y| ∂ν Gλ dσ –  A= ∂

∂ν y(∇y∇Gλ ) dσ

(.)

∂ν z(∇z∇Gλ ) dσ .

(.)

∂

and



|∇z| ∂ν Gλ dσ – 

B= ∂

∂

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



|∇y| Gλ dx = A +  

y∇y∇Gλ dx +  

d

 i=



∇y∂i y∂i ∇Gλ dx



x – x Gλ dx (T – t + λ) 

d

 |x – x | – Gλ + ∂i y∂i y Gλ dx + (T – t + λ) (T – t + λ) i=  

(xi – xi )(xj – xj ) + ∂j y∂i y Gλ dx (T – t + λ) i =j  y∇y

=A–



x – x Gλ dx + (T – t + λ) 



x – x + ∇y Gλ dx.  (T – t + λ)

=A–





y∇y

|∇y| 

– Gλ dx (T – t + λ) (.)

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In the same way, we get



x – x Gλ dx + (T – t + λ) 



x – x ∇z Gλ dx. +  (T – t + λ)

|∇z| Gλ dx = B –  





z∇z

|∇z| 

– Gλ dx (T – t + λ) (.)

Thus, we can rewrite (.) as





x – x x – x ∇y Gλ dx –  ∇z Gλ dx  (T – t + λ)  (T – t + λ)



x – x x – x ∇yGλ dx +  (z + ∂t z) ∇zGλ dx +  (y + ∂t y) (T – t + λ) (T – t + λ)  



 –  ∂t yyGλ dx –  ∂t zzGλ dx + Dλ (t) (T – t + λ)  



 x – x = –A – B –  ∂t y – ∇y + (y – ∂t y) Gλ dx (T – t + λ)  



 x – x ∇z + (z – ∂t z) Gλ dx ∂t z – – (T – t + λ)  



 (y + ∂t y) Gλ dx –  ∂t yyGλ dx +   



  + (z + ∂t z) Gλ dx –  ∂t zzGλ dx + Dλ (t)  (T – t + λ)  



 x – x ∂t y – = –A – B –  ∇y + (y – ∂t y) Gλ dx (T – t + λ)  



x – x  ∂t z – ∇z + (z – ∂t z) Gλ dx – (T – t + λ)  



   (y – ∂t y) Gλ dx + (z – ∂t z) Gλ dx + (.) Dλ . +     (T – t + λ)

D λ (t) = –A – B – 



This completes the proof of this lemma.



Lemma . For each λ >  and t ∈ (, T), it follows that  d (T – t + λ)Nλ (t) ≤ M (T + λ). dt

(.)

Proof Firstly, we compute Nλ (t) as t ∈ (, T). By (.) and (.), we derive that Nλ (t)

    Dλ (t)Hλ (t) – Dλ (t)Hλ (t) = Hλ (t)    = –A – B + Dλ (t) Hλ (t) (T – t + λ)



 x – x  ∇y + (y – ∂t y) Gλ dx + (y – ∂t y) Gλ dx ∂t y – – (T – t + λ)    

 



 x – x  – ∇z + (z – ∂t z) Gλ dx + ∂t z – (z – ∂t z) Gλ dx (T – t + λ)    

Zheng et al. Journal of Inequalities and Applications (2017) 2017:234

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  

 x – x  –  y ∂t y – ∇y + (y – ∂t y) Gλ dx Hλ (t) (T – t + λ)  



 x – x ∇z + (z – ∂t z) Gλ dx +  z ∂t z – (T – t + λ)   

 

+ y(y – ∂t y)Gλ dx + z(z – ∂t z)Gλ dx

–





 A+B Nλ –  (T – t + λ) Hλ (t)

 

 x – x  ∇y + (y – ∂t y) Gλ dx y ∂t y – + Hλ (t) (T – t + λ)  



 x – x ∇z + (z – ∂t z) Gλ dx + z ∂t z – (T – t + λ)  





    x – x ∂t y – y + z Gλ dx ∇y + (y – ∂t y) Gλ dx – (T – t + λ)   

 

   x – x  ∂t z – ∇z + (z – ∂t z) Gλ dx – y + z Gλ dx (T – t + λ)   

 



 y(y – ∂t y)Gλ dx + z(z – ∂t z)Gλ dx – Hλ (t)   



 + (y – ∂t y) Gλ dx + (z – ∂t z) Gλ dx . (.) Hλ (t)  

=

Let α = ∂t y –

 x – x ∇y + (y – ∂t y), (T – t + λ) 

β = ∂t z –

 x – x ∇z + (z – ∂t z). (T – t + λ) 

It follows from the Cauchy-Schwarz inequality that





α  Gλ dx 

 y + z Gλ dx +







=

 α  + β  Gλ dx





≥

β  Gλ dx







 y + z Gλ dx



 y + z Gλ dx



 zβGλ dx .

yαGλ dx +









It, together with (.), shows that  A+B Nλ (t) +  (T – t + λ) Hλ (t) 



 (y – ∂t y) Gλ dx + (z – ∂t z) Gλ dx ≤ . – Hλ (t)  

Nλ (t) –

(.)

In what follows, we will discuss the properties of A and B. Since y = z =  on ∂, we have ∇y = ∂ν yν, ∇z = ∂ν zν on ∂. If the domain  is convex, we can derive that ((x – x ) · ν) ≥ .

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According to (.) and (.), then



|∇y| ∂ν Gλ dσ –  

A= ∂

∂ν y(∇y∇Gλ ) dσ ∂

    |∇y| (x – x ) · ν Gλ dσ + (T – t + λ) ∂ (T – t + λ)

  × ∂ν y (x – x )∇y Gλ dσ

=–

∂

    =– |∇y| (x – x ) · ν Gλ dσ + (T – t + λ) ∂ (T – t + λ)

  |∂ν y| (x – x ) · ν Gλ dσ × ∂

=

 × (T – t + λ)



  |∂ν y| (x – x ) · ν Gλ dσ ≥ .

∂

In the same way, we get B=

 × (T – t + λ)



  |∂ν z| (x – x ) · ν Gλ dσ ≥ .

∂

Combining with (.), we get (T – t + λ)Nλ (t) – Nλ (t)  

(T – t + λ) – (y – ∂t y) Gλ dx + (z – ∂t z) Gλ dx ≤ . Hλ (t)  

(.)

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

+ λ)Nλ (t) – Nλ (t)

(T – t + λ) ≤ Hλ (t) ≤









(ay + bz) Gλ dx + 

M (T – t + λ) Hλ (t)

(cy + dz) Gλ dx 



   |y| + |z| Gλ dx 

≤ M (T + λ). Thus,  d (T – t + λ)Nλ (t) ≤ M (T + λ), dt

∀t ∈ (, T).

This completes the proof of this lemma. Let

 (|y(x, )| + |z(x, )| ) dx m d + + MT + M T  + . KM,y,z,T =  ln    + |z(x, T)| ) dx T  (|y(x, T)|  Then we have the following lemma.



Zheng et al. Journal of Inequalities and Applications (2017) 2017:234

Page 10 of 17

Lemma . For each λ > , we have λNλ (T) +

d ≤ 



λ +  KM,y,z,T . 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 





T

   Hλ (t)  y(∂t y – y) + z(∂t z – z) Gλ dx dt dt + Hλ (t) Hλ (t)    T



   Hλ ( T )  –y(ay + bz) – z(cy + dz) Gλ dx dt = – ln + Hλ () H (t) λ  

T    Hλ ()  ≤ ln M y + z Gλ dx dt + T Hλ (t)  Hλ (  ) 

Nλ (t) dt = –

≤ ln

T 

Hλ () + MT. Hλ ( T )

This, alone with (.), shows that   Hλ () T   . λNλ (T) ≤ (T + λ) ln T + MT + M  Hλ ( T ) We have |x–x |  – – d   · e (T+λ) dx Hλ ()  (|y(x, )| + |z(x, )| )(T + λ) = |x–x | Hλ ( T )  – T T  T  T (  +λ) – d (|y(x, )| + |z(x, )| )( + λ) · e dx      d T     (|y(x, )| + |z(x, )| ) dx · (  + λ) ≤  |x–x | – T  d T  T  (  +λ) (|y(x, )| + |z(x, )| ) · e dx · (T + λ)      d m (|y(x, )| + |z(x, )| ) dx · ( T + λ)  ( T  +λ) ≤  · e d T  T    (|y(x,  )| + |z(x,  )| ) dx · (T + λ)  (|y(x, )| + |z(x, )| ) dx m ≤ e T   . T  T   (|y(x,  )| + |z(x,  )| ) dx

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Therefore,   

  T m    (|y(x, )| + |z(x, )| ) dx λNλ (T) ≤ (T + λ) + ln  + MT + M T . T  T   T  (|y(x,  )| + |z(x,  )| ) dx Using the energy estimate  

(|y(x, T)| + |z(x, T)| ) dx

T  T   (|y(x,  )| + |z(x,  )| ) dx

≤ eMT ,

we obtain

   (|y(x, )| + |z(x, )| ) dx m   + T  ln   + MT + M   T  (|y(x, T)| + |z(x, T)| ) dx



λ d = +  KM,y,z,T – . T 

λNλ (T) ≤

λ + T

Thus,

λ d λ +  KM,y,z,T – · T T 

λ +  KM,y,z,T . ≤ T

d λNλ (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: 



    |x–x |  λ λ +  K |x – x | y(x, T) + z(x, T) e– λ dx M,y,z,T  r T 

  |x–x |    λ y(x, T) + z(x, T) e– λ dx. ≤ λ +  KM,y,z,T T Br

–

Proof We first point out the following fact: for any f ∈ H () and for each λ > , we have   |  ≤  |∇(f (x) exp(– |x–x ))| dx. By computing the right-hand term, we get λ



 |x–x | |x – x |  f (x) e– λ dx ≤ λ λ +



  |x–x | ∇f (x) e– λ dx



d 



  |x–x | f (x) e– λ dx. 

It follows from (.) and (.) that

    |x–x |  |x – x | y(x, T) + z(x, T) e– λ dx 



    |x–x |  ∇y(x, T) + ∇z(x, T) e– λ dx ≤ λ λ 

(.)

Zheng et al. Journal of Inequalities and Applications (2017) 2017:234



    – |x–x | d     λ y(x, T) + z(x, T) e + dx  

  |x–x |    d y(x, T) + z(x, T) e– λ dx ≤ λ λNλ (T) +  



    |x–x |  d y(x, T) + z(x, T) e– λ dx ≤ λ λNλ (T) +  Br 

    – |x–x |        λ +  |x – x | y(x, T) + z(x, T) e dx . r \Br

Page 12 of 17

(.)

It, combining with (.), shows that

    |x–x |  |x – x | y(x, T) + z(x, T) e– λ dx







    |x–x |  λ y(x, T) + z(x, T) e– λ dx ≤ λ +  KM,y,z,T T Br 

    – |x–x |        λ |x – x | y(x, T) + z(x, T) e dx . +  r 

(.)

We get 



    |x–x |  λ λ +  K |x – x | y(x, T) + z(x, T) e– λ dx M,y,z,T  r T 



    |x–x |  λ y(x, T) + z(x, T) e– λ dx. +  KM,y,z,T ≤ λ T Br

–

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 take λ >  in the estimate of Lemma . to be such that

λ λ  +  KM,y,z,T = . r T 

(.)

By direct computation, we have 

 Tr λ= –T + T  + .  KM,y,z,T Since

m 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

 KM,y,z,T Tr

Zheng et al. Journal of Inequalities and Applications (2017) 2017:234

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 ≤  T +

 Tr KM,y,z,T KM,y,z,T Tr

r  ≤  + √ KM,y,z,T . m r Therefore, we have √ (m+r m)  KM,y,z,T r

m

e λ ≤ e

√ √ (m+r m)  d (m+r m)  ( m +MT+M T  ) r r T e

≤e

√ 

 (|y(x, )| + |z(x, )| ) dx (m+r m)/r  ×  .    (|y(x, T)| + |z(x, T)| ) dx

By Lemma ., we get

    |x–x |  |x – x | y(x, T) + z(x, T) e– λ dx





    |x–x |  y(x, T) + z(x, T) e– λ dx.

≤ r Br

Then we derive that

    m  y(x, T) + z(x, T) e– λ dx





    |x–x |  y(x, T) + z(x, T) e– λ dx

≤

≤



∩|x–x |≥r



    |x–x |  y(x, T) + z(x, T) e– λ dx

    |x–x |  y(x, T) + z(x, T) e– λ dx

+ Br

≤

 r



    |x–x |  |x – x | y(x, T) + z(x, T) e– λ dx



    |x–x |  y(x, T) + z(x, T) e– λ dx

+

Br

    |x–x |  y(x, T) + z(x, T) e– λ dx

≤

Br

     y(x, T) + z(x, T) dx.

≤ Br

Thus,

     y(x, T) + z(x, T) dx



≤ e

m λ



     y(x, T) + z(x, T) dx

Br √ √ (m+r m)  d (m+r m)  ( m +MT+M T  ) r r T ≤ e e

(.)

Zheng et al. Journal of Inequalities and Applications (2017) 2017:234

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(m+r√m)/r     (|y(x, )| + |z(x, )| ) dx  ×    (|y(x, T)| + |z(x, T)| ) dx

     y(x, T) + z(x, T) dx. ×

(.)

Br

Then we have

     y(x, T) + z(x, T) dx

 C r

≤ Ce e

C Tr

e

C (MT+M T  ) r

  (|y(x, )| + |z(x, )| ) dx C/r      (|y(x, T)| + |z(x, T)| ) dx

     y(x, T) + z(x, T) dx,

× Br

which is equivalent to the following inequality:

     y(x, T) + z(x, T) dx

 C/T C(MT+M T  )

≤ Ce



e

     y(x, ) + z(x, ) dx

C r +C





     y(x, T) + z(x, T) dx

×

r r +C

Br



C        r +C C       y (x) + z (x) dx + C MT + M T ≤ C exp T 

     y(x, T) + z(x, T) dx

×

r r +C

.

ω r r +C

Let p =

and then the above inequality can be written as

     y(x, T) + z(x, T) dx







–p        C       ≤ C exp y (x) + z (x) dx + C MT + M T T 

p          × y(x, T) + z(x, T) dx .

(.)

ω

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 estimate: y(x, )L + z(x, )L y(x, T)L + z(x, T)L

y(x, )H  + z(x, )H    CM T ≤ exp C( + M)Te . y(x, )L + z(x, )L

(.)

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Page 15 of 17

For y , 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.

Let f = ay + bz, and g = cy + dz. By direct computation, we obtain

d  yL = –yH  – ay + bz, y,  dt 

d  zL = –zH  – cy + dz, z,  dt 



d  yH  = –yL + y(ay + bz) dx,  dt  

d  zH  = –zL + z(cy + dz) dx.  dt  

(.) (.)

Then (yH  + zH  ) (yL + zL ) – (yH  + zH  )(yL + zL ) d     (t) = dt (yL + zL ) =

(yL ×

 + zL )





  –yL – zL + f y dx + gz dx yL + zL 









      + yH  + zH  + fy dx + gz dx yH  + zH  

=

(yL ×











 + zL )





  –yL – zL + f y dx + gz dx yL + zL 









 



  . + yH  + zH  + fy dx + gz dx – fy dx + gz dx         Integrating by parts and using Cauchy-Schwarz inequality, we have  d (t) =  dt (yL + zL ) 



    × –yL – zL + f y dx + gz dx yL + zL 







 



f g  –y + –z + + y dx + z dx – fy dx + gz dx        ≤

 (yL + zL ) 



  × –yL – zL + f y dx + gz dx yL + zL 



Zheng et al. Journal of Inequalities and Applications (2017) 2017:234

  +  –y + ≤

Page 16 of 17

   f  yL + –z +    L





 

g  zL –  fy dx + gz dx  L   

 (yL + zL ) 



    × –yL – zL + f y dx + gz dx yL + zL   +  –y +

   f  + –z +    L





     g   yL + zL .   L

Thus,         f    d   + g  (t) ≤ yL + zL        dt  L  L (yL + zL ) =

f L + gL (yL + zL )

≤ CM



=

yH  + zH  



yL + zL

ay + bzL + cy + dzL (yL + zL ) = CM (t).

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

(t) ≤ eCM T ().

(.)

This, combining with (.) and (.), indicates that   d  yL + zL + yH  + zH  + ay + bz, y + cy + dz, z    dt        d ≤ yL + zL + (t) yL + zL + M(t) yL + zL  dt     d   yL + zL + ( + M)eCM T () yL + zL . ≤  dt

=

(.)

Integrating (.) on (, T), we get the desired estimate           y(x, ) + z(x, ) ≤ exp ( + M)TeCM T () y(x, T) + z(x, T) . L L L L Thus, y(x, )L + z(x, )L y(x, T)L + z(x, T)L

y(x, )H  + z(x, )H    CM T . ≤ exp ( + M)Te y(x, )L + z(x, )L

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



4 Conclusions In this work, we discuss the unique continuation for the linear coupled heat equations. Our results demonstrate that the value of the solution equation (.) can be determined uniquely by its value on an arbitrary open subset ω of  at any given positive time T. In

Zheng et al. Journal of Inequalities and Applications (2017) 2017:234

Page 17 of 17

other words, if the solution of equation (.) satisfies y(·, T) = z(·, T) =  on ω, then y = 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.

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