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Research article Special Issues

A reversible database watermarking method with low distortion

  • In this paper, a low distortion reversible database watermarking method based on histogram gap is proposed in view of the large gap in histogram of database integer data. By using the method, the tolerance of the attribute column containing all integer data is firstly calculated and the prediction error is obtained according to the tolerance. Then according to the watermark bits to be embedded, the database tuples will be randomly grouped and the histogram can be constructed by using the prediction error. Finally, the histogram correction rule is used to find the histogram peak bin, the number of consecutive non-zero prediction errors on the left and right sides of the peak is obtained, and the histogram shift is performed on the side with a smaller number of non-zero prediction errors, and then the watermark embedding will be realized. The results of the experiments based on the published dataset of FCTD (Forest Cover Type Dataset) show that compared with the existing GAHSW which also considers distortion, the proposed method significantly reduces the number of histogram column shift while embedding the watermarks, greatly reduces the changes to the carrier data, and effectively reduces the database's data distortion caused by watermark embedding.

    Citation: Yan Li, Junwei Wang, Shuangkui Ge, Xiangyang Luo, Bo Wang. A reversible database watermarking method with low distortion[J]. Mathematical Biosciences and Engineering, 2019, 16(5): 4053-4068. doi: 10.3934/mbe.2019200

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  • In this paper, a low distortion reversible database watermarking method based on histogram gap is proposed in view of the large gap in histogram of database integer data. By using the method, the tolerance of the attribute column containing all integer data is firstly calculated and the prediction error is obtained according to the tolerance. Then according to the watermark bits to be embedded, the database tuples will be randomly grouped and the histogram can be constructed by using the prediction error. Finally, the histogram correction rule is used to find the histogram peak bin, the number of consecutive non-zero prediction errors on the left and right sides of the peak is obtained, and the histogram shift is performed on the side with a smaller number of non-zero prediction errors, and then the watermark embedding will be realized. The results of the experiments based on the published dataset of FCTD (Forest Cover Type Dataset) show that compared with the existing GAHSW which also considers distortion, the proposed method significantly reduces the number of histogram column shift while embedding the watermarks, greatly reduces the changes to the carrier data, and effectively reduces the database's data distortion caused by watermark embedding.


    In scientific research and engineering technology, many problems, such as the deformation of engineering building beams and the load of turbine blades under the impact of airflow in fluid mechanics, can be attributed to the existence of solutions of differential equations. And in elastic mechanics and engineering physics, elastic beam is one of the basic components of engineering building. The nonlinear boundary value problem of the fourth order differential equation with different boundary conditions can describes the deflection of elastic beam under external force, the reflects, and the stress. The static beam equation:

    y(4)(x)=f(x,y(x)),0<x<1, (1.1)

    is most studied under the following boundary conditions:

    y(0)=y(1)=y(0)=y(1)=0, (1.2)

    or

    y(0)=y(1)=y(0)=y(1)=0, (1.3)

    where f:[0,1]×[0,)[0,) is continuous. They are often used to describe the equilibrium state of the elastic beams. While the problem (1.1), (1.2) describes the static beam with simple support at both ends, the problem (1.1), (1.3) describes the static beam with simple support at one end and sliding support at the other end. These two types of problems, can be converted into second-order equation by simple transformation. Therefore, some methods of studying ordinary differential equations can be applied to the study of problems (1.1), (1.2) and (1.1), (1.3).

    The problem (1.1), (1.2) has been investigated in many literutures, see [1,2,3,4,5,6]. Among them, Bai [1] used a new maximum principle to give the solutions for the problem. Gabriele [2] used a local minimum theorem to give the existence of at least one non-trivial solution. Li [3] got the existence of positive solutions based on the fixed point index theory. And Yao [6] used the approximation by operators of completely continuous operator sequence and the Guo-Krasnosel'skii fixed point theorem for cone expansion and compression.

    The problem (1.1), (1.3) also has been investigated by many authors, see [7,8,9]. Yao gave the existence of n solutions by choosing suitable cone and using the Krasnosel'skii fixed point theorem in [7]. Yao and Li [8] used the Leray-schauder fixed point theorem to get the existence of positive solutions. Zhao [9] used the fixed point theorem due to Avery-Peterson and Leggett-williams to get the existence of positive solutions. The upper and lower solution method and the fixed-point techniques have been used to study many other problems. We provide reference alone this line for some research on beam equation [10,11].

    The typical static elastic beam equation with fixed ends can be described by Eq (1.1) with boundary condition

    y(0)=y(1)=y(0)=y(1)=0. (1.4)

    Compared with the above two type problems, this problem can not be solved by directly converting it into some second-order problems. But due to its wide application, many authors set out to find a positive solution to problem (1.1), (1.4). In 1984, Agarwal [12] first considered the problem, and used the compression mapping principle and numerical iteration method to study the existence of solutions. Later, Wu and Ma [13,14] used the Krasnoselskii's fixed point theorem to get the existence results under the following conditions:

    (1)¯liml0min{f(x,v)|(x,v)[14,34]×[l24,l]}/l>A;(2)¯limlmin14x34f(x,l)/l>A24.

    Caballero [15] applied a fixed point theorem in partially ordered metric spaces to get a unique symmetric positive solution for problem (1.1), (1.4), where f(x,y) is a nondecreasing function with respect to y for each x[0,1], and satisfy the Lipschitz type condition. Obviously the function f(x,y) has strong constraints in above literatures.

    This paper study the existence of positive solutions for the following fourth-order two-point boundary value problem:

    y(4)(x)=f(x,y(x)),0<x<1, (1.5)
    y(0)=y(1)=y(0)=y(1)=0, (1.6)

    where f:[0,1]×[0,)[0,) is continuous. In section 2, the Green's function is given by the use of the Laplace transform, and some preliminary lemmas also be obtained. In section 3, the eigenvalue of the linear equation corresponding to problem (1.5), (1.6) is given by Ritz method in calculus of variations with the help of the mathematical tool software. The fixed point index value is obtained by using the theory of the self-adjoint operator, and the existence of positive solutions is given by the use of the fixed point index theorem. In section 4, an examples is given to illustrate the main results.

    Compared with other literature, in this paper, the way to get the eigenvalue of problem considered is new. We use Laplace transform to get the Green function of this problem, this way is more convenient and easier. And the constraint of f(t,u) in this paper is more weaker than which get in other paper.

    For convenience, we introduce some symbols,

    ˉf0=¯limy0+maxx[0,1]f(x,y)y,f_0=lim_y0+minx[0,1]f(x,y)y,
    ˉf=¯limy+maxx[0,1]f(x,y)y,f_=lim_y+minx[0,1]f(x,y)y.

    Lemma 2.1. Given hC[0,1], the unique solution of

    y(4)(x)h(x)=0,0<x<1, (2.1)
    y(0)=y(1)=y(0)=y(1)=0 (2.2)

    is

    y(x)=10G(x,s)h(s)ds, (2.3)

    where

    G(x,s)={s2(1x)2[(xs)+2(1s)x]6,0sx1;x2(1s)2[(sx)+2(1x)s]6,0xs1. (2.4)

    Proof. By the use of Laplace transform on equation (2.1), one has

    s4Y(s)s3y(0)s2y(0)sy(0)y(0)=H(s),

    where Y(s)=£[y(x)],H(s)=£[h(x)]. Thus,

    Y(s)=1s4H(s)+1s4y(0)++1s3y(0)+1s2y(0)+1sy(0).

    By the use of the inverse Laplace transform and the boundary condition (2.2), one has

    y(x)=£1[Y(s)]=16x3h(x)+16x3y(0)+12x2y(0)=16x0(xs)3h(s)ds+16x3y(0)+12x2y(0), (2.5)

    then

    y(x)=x012(xs)2h(s)ds+12x2y(0)+xy(0). (2.6)

    Let x=1 in (2.5) and (2.6), then we can get

    y(0)=10s(1s)2h(s)ds,
    y(0)=10(1+2s)(1s)2h(s)ds.

    Put them into Eq (2.5), we get

    y(x)=x016(xs)3h(s)ds1016x3(1+2s)(1s)2h(s)ds+1012x2s(1s)2h(s)ds=x016s2(1x)2[(xs)+2(1s)x]h(s)ds+1x16x2(1s)2[(sx)+2(1x)s]h(s)ds=10G(x,s)h(s)ds.

    The proof is completed.

    Remark 2.1. The Green's function (2.4) have been obtained before in some literatures, but to my best knowledge, the way to get it by Laplace transform hadn't been mentioned yet. Obviously, this way is more convenient and easier.

    Lemma 2.2. [13] Let a(s):[0,1][0,1] define as

    G(a(s),s)=max0x1G(x,s),

    then a(s)=132s, for 0s12;a(s)=21+2s, for 12s1.

    Lemma 2.3. [13] The function G(x,s) defined by (2.4) satisfies:

    (i) G(x,s)=G(s,x)>0, for x,s[0,1];

    (ii) G(x,s)G(a(s),s)q(x), where q(x)=min{23x2,23(1x)2}, for x,s[0,1];

    (iii) G(x,s)124G(a(s),s), for x[14,34],s[0,1].

    Let C+[0,1] be the cone of all nonnegative functions in C[0,1] which is a Banach space with y=maxx[0,1]|y(x)|. And the cone

    D={yC+[0,1]|y(x)124y,x[14,34]}C+[0,1].

    Then denote

    m=min14s34G(12,s)=34,

    and define T:C+[0,1]C+[0,1] by

    (Ty)(x):=10G(x,s)f(s,y(s))ds. (2.7)

    Lemma 2.4. [13] By Lemma 2.3 and formula (2.7), there holds

    (Ty)(x)=10G(x,s)f(s,y(s))dsq(x)Ty,x[0,1].

    Lemma 2.5. Let f:[0,1]×[0,)[0,) be continuous, then

    (1) the operator T:DD is completely continuous,

    (2) the solution of boundary value problem (1.5), (1.6) y(x) satisfies

    y(x)q(x)y.

    Proof. The positive solution of problem (1.5), (1.6) is equivalent to the nonzero fixed point of T. On the one hand,

    minx[14,34](Ty)(x)=minx[14,34]10G(x,s)f(s,y(s))ds124Ty.

    By the arbitrariness of y and G(x,s)0, we obtain T(D)D. Then by the continuity of f and Arzela-Ascoli Theorem, we get (1) holds. On the other hand, by the definition of the operator T and the function a(s), we have

    Ty10G(a(s),s)f(s,y(s))ds.

    Suppose yD is a solution of problem (1.5), (1.6), then

    y(x)=(Ty)(x)=10G(x,s)f(s,y(s))dsq(x)Ty=q(x)y.

    The proof is completed.

    Given b>0, let

    Db={yD|y(x)<b}andDb={yD|y(x)=b}.

    Lemma 2.6. [16] Suppose T:DD is completely continuous and

    (i)infyDbTy>0,

    (ii)αTyy for yDb and α1.

    Then i(T,Db,D)=0.

    Lemma 2.7. [16] Suppose T:DD is completely continuous. If αTyy for yDb and 0<α1, then i(T,Db,D)=1.

    Lemma 2.8. Assume that G(x,s) is defined as formula (2.4). Then the operator H:C[0,1]C[0,1],

    (Hz)(x)=10G(x,s)z(s)ds,

    is a self-adjoint operator.

    Proof. By Lemma 2.1, it is clear that G(x,s) is a real symmetric function, that is to say that, G(x,s)=G(s,x) for 0s,x1. Thus, for z1=z1(s),z2=z2(s)C[0,1], there holds

    (Hz1,z2)=10(Hz1)(x)z2(x)dx=1010G(x,s)z1(s)z2(x)dsdx=10z1(s)10G(x,s)z2(x)dxds=10z1(s)10G(s,x)z2(x)dxds=10z1(s)(Hz2)(s)ds=(z1,Hz2).

    So, the operator H is a self-adjoint operator.

    Lemma 3.1. The first eigenvalue λ1 of the problem

    y(4)(x)λy(x)=0, (3.1)
    y(0)=y(1)=y(0)=y(1)=0, (3.2)

    is λ1500.564.

    Proof. Define the linear differential operator L:{yC4[0,1]y(0)=y(1)=y(0)=y(1)=0}C[0,1] as

    (Ly)(x)=y(4)(x). (3.3)

    With Ritz method, the problem which to get the first eigenvalue λ1 can be transformed to get the extreme value of the functional

    J[y]=10yLydx=10y(x)[y(4)(x)]dx, (3.4)

    under the normalization condition

    10y2(x)dx=1. (3.5)

    Combing the definition of the functional J and the boundary condition (3.2), one has

    J[y]=10y(x)[y(4)(x)]dx=10[y(x)]2dx.

    Choose the primary function as

    yk(x)=(1x)2x2k,(k=1,2,3,),

    then, the n-th order approximate solution of the functional extremum function is

    yn(x)=x2(1x)2(a1+a2x++anxn1). (3.6)

    By (3.4), (3.5), there is

    J[yn(x)]=10[yn(x)]2dx, (3.7)
    10y2n(x)dx=1. (3.8)

    Then the Lagrangian function

    F(a1,a2,,an,τ)=J[yn(x)]+τ[10y2n(x)dx1]=10[yn(x)]2dx+τ[10y2n(x)dx1].

    Now, we solve the following equation system by the use of the mathematical tool software Mathematica,

    Fai=0(i=1,2,3,,n),andFτ=0.

    Firstly, we get ai(i=1,2,3,,n) and τ, then, plug them into (3.6) to get yn(x). By Eq (3.7) we get the corresponding eigenvalue λ1=J[yn(x)]500.564 for n6. The proof is completed.

    Remark 3.1. Although some authors mentioned the eigenvalue before, but to my best knowledge, the eigenvalue of problem (3.1), (3.2) hadn't been obtained yet. However, we use the Ritz-Method to get the concrete eigenvalue λ1 here.

    Theorem 3.1. Suppose one of the following relation is true:

    (i) ˉf0<λ1, f_>λ1;

    (ii) f_0>λ1,ˉf<λ1.

    Then boundary value problem (1.5), (1.6) has at least one positive solution.

    Proof. (i) By ˉf0<λ1, for ϵ(0,λ1), there exists b0>0 such that

    f(x,y)(λ1ϵ)y,x[0,1],y[0,b0]. (3.9)

    Set b(0,b0). Assume there exist y0Db and α0(0,1] such that α0Ty0=y0. Then y0 satisfies

    y(4)0(x)=α0f(x,y0(x)),0x1. (3.10)

    Taking into account (3.9) and (3.10), there is

    y(4)0(x)=α0f(x,y0(x))α0(λ1ϵ)y0(x)(λ1ϵ)y0(x).

    Due to the fact that the inverse operator of a self-adjoint operator is self-adjoint and Lemma 2.8, the operator L defined by (3.3) is a self-adjoint operator. Suppose ˆyD is a eigenfunction of L with respect to λ1 such that 10ˆy(x)dx=1, that is to say that

    Lˆy=λ1ˆy.

    Then

    (Ly0,ˆy)=(y0,Lˆy)=(y0,λ1ˆy)=λ1(y0,ˆy).

    So we get

    λ1(y0,ˆy)(λ1ϵ)(y0,ˆy).

    obviously this is a contradictory. Hence Lemma 2.7 yields that

    i(T,Db,D)=1. (3.11)

    Since f_>λ1, for given ϵ>0, there exists K>0 such that

    f(x,y)(λ1+ϵ)y,x[0,1],yK, (3.12)

    suppose C=max0x1,0yK|f(x,y)(λ1+ϵ)y|+1, then

    f(x,y)(λ1+ϵ)yC,x[0,1],y[0,+).

    Let r>r0:=max{24K,b0}. For yDr, there is y(s)124y>K for s[14,34], then, by (3.12)

    Ty(Ty)(12)=10G(12,s)f(s,y(s))ds3414m(λ1+ϵ)y(s)ds1243414m(λ1+ϵ)yds148m(λ1+ϵ)y (3.13)

    So infyDrTy>0.

    Suppose there exist y0Dr and μ01 such that μ0Ty0=y0, then

    (Ly0)(x)=μ0f(x,y0(x))(λ1+ϵ)y0(x)C.

    Thus,

    λ1(y0,ˆy)=(y0,λ1ˆy)=(y0,Lˆy)=(Ly0,ˆy)(λ1+ϵ)(y0,ˆy)C.

    So,

    (y0,ˆy)Cϵ.

    Similar to the proof of Lemma 2.5, by μ0Ty0=y0, one has

    y0(x)(Ty0)(x)q(x)y0.

    So,

    (y0,ˆy)y010q(x)ˆy(x)dx.

    Thus,

    y0Cϵ[10q(x)ˆy(x)dx]1:=ˉr. (3.14)

    Now set r>max{ˉr,b0}, then there is μTyy for yDr and μ1. Consequently, two conditions of Lemma 2.6 all hold, thus

    i(T,Dr,D)=0. (3.15)

    By (3.11) and (3.15) there is

    i(T,Dr¯Db,D)=i(T,Dr,D)i(T,Db,D)=1. (3.16)

    Hence T has a fixed point in Dr¯Db.

    (ii) Since f_0>λ1, for ϵ>0, there exists h0>0 such that

    f(x,y)(λ1+ϵ)y,x[0,1],0yh0. (3.17)

    Set h(0,h0), similar to (3.13), one has

    Ty148m(λ1+ϵ)y,yDh,

    so infyDbTy>0.

    Assume there exist y1Dh and α11 such that α1Ty1=y1, then

    λ1(y1,ˆy)=(y1,λ1ˆy)=(y1,Lˆy)=(Ly1,ˆy)(λ1+ϵ)(y1,ˆy).

    Obviously this is a contradiction, because (y1,ˆy)>0. Therefore, Lemma 2.5 yields that

    i(T,Dh,D)=0. (3.18)

    On the other hand, since ˉf<λ1, for ϵ(0,λ1), there exists K2>0 such that

    f(x,y)(λ1ϵ)y,x[0,1],yK2.

    Let C=max0x1,0yK2|f(x,y)(λ1ϵ)y|+1, then

    f(x,y)(λ1ϵ)y+C,x[0,1],y[0,+). (3.19)

    Assume there exist y2D and α2(0,1] such that α2Ty2=y2, then

    λ1(y2,ˆy)=(y2,λ1ˆy)=(y1,Lˆy)=(Ly1,ˆy)(λ1ϵ)(y2,ˆy)+C.

    Similar to the previous proof, one has y2ˉr. Set r>max{ˉr,h0}, we get αTyy for yDr and α(0,1]. Hence by Lemma 2.7,

    i(T,Dr,D)=1, (3.20)

    then

    i(T,Dr¯Db,D)=i(T,Dr,D)i(T,Db,D)=1.

    In conclusion, the boundary value problem (1.5), (1.6) has at least one positive solution.

    Remark 3.2. Compared with literatures [14,15] and so on, the constraint of f(x,y) in our results is weaker.

    Example 4.1. Consider the problem:

    y(4)(x)=[y(x)]θ+(300x)y(x),0<x<1,θ1, (4.1)
    y(0)=y(1)=y(0)=y(1)=0. (4.2)

    It's obviously that f(x,y)y=yθ1+300x.

    Case (1): θ>1, there are

    (i)ˉf0=¯limy0+maxx[0,1]f(x,y)y=300<500.564

    Case (2) : \theta < 1 , there are

    \begin{eqnarray*} &(i)& \underline{f}_0 = \varliminf\limits_{y \to 0+} \min\limits_{x\in[0, 1]} \frac{f(x, y)}{y} = +\infty \gt 500.564\approxeq \lambda_1; \\ &(ii)& \bar{f}_{\infty} = \varlimsup\limits_{y \to +\infty} \max\limits_{x\in[0, 1]} \frac{f(x, y)}{y} = 300 \lt 500.564\approxeq \lambda_1, \end{eqnarray*}

    By Theorem 3.1, for \theta\neq1 , Problem (4.1), (4.2) has at least one positive solution.

    The fourth order differential equations have been applied in different aspects of applied mathematics and physics, especially in the theory of elastic beams and stability. In this paper, the fourth-order two-point boundary value problem is studied, which can describe some changes when the elastic beam deforms or rotates when it is subjected to external forces, and provides an important theoretical basis for solving the problem. And compare with other literature, our method in this paper makes the equation more widely used and describes the equilibrium state of the elastic beam better.

    This work is supported by Natural Science Foundation of China (Grant No.11571207), the Taishan Scholar project, and SDUST graduate innovation project SDKDYC190237.

    The authors declare no conflict of interest in this paper.



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