Research article Special Issues

Research on chest radiography recognition model based on deep learning

  • With the development of medical informatization and against the background of the spread of global epidemic, the demand for automated chest X-ray detection by medical personnel and patients continues to increase. Although the rapid development of deep learning technology has made it possible to automatically generate a single conclusive sentence, the results produced by existing methods are not reliable enough due to the complexity of medical images. To solve this problem, this paper proposes an improved RCLN (Recurrent Learning Network) model as a solution. The model can generate high-level conclusive impressions and detailed descriptive findings sentence-by-sentence and realize the imitation of the doctoros standard tone by combining a convolutional neural network (CNN) with a long short-term memory (LSTM) network through a recurrent structure, and adding a multi-head attention mechanism. The proposed algorithm has been experimentally verified on publicly available chest X-ray images from the Open-i image set. The results show that it can effectively solve the problem of automatic generation of colloquial medical reports.

    Citation: Hui Li, Xintang Liu, Dongbao Jia, Yanyan Chen, Pengfei Hou, Haining Li. Research on chest radiography recognition model based on deep learning[J]. Mathematical Biosciences and Engineering, 2022, 19(11): 11768-11781. doi: 10.3934/mbe.2022548

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  • With the development of medical informatization and against the background of the spread of global epidemic, the demand for automated chest X-ray detection by medical personnel and patients continues to increase. Although the rapid development of deep learning technology has made it possible to automatically generate a single conclusive sentence, the results produced by existing methods are not reliable enough due to the complexity of medical images. To solve this problem, this paper proposes an improved RCLN (Recurrent Learning Network) model as a solution. The model can generate high-level conclusive impressions and detailed descriptive findings sentence-by-sentence and realize the imitation of the doctoros standard tone by combining a convolutional neural network (CNN) with a long short-term memory (LSTM) network through a recurrent structure, and adding a multi-head attention mechanism. The proposed algorithm has been experimentally verified on publicly available chest X-ray images from the Open-i image set. The results show that it can effectively solve the problem of automatic generation of colloquial medical reports.



    The nonlinear Schrödinger equations (NLSEs) are extensively used to describe numerous crucial phenomena and dynamic processes in various fields such as fluid dynamics, plasma, chemistry, biology, optical fibers[1,2,3,4], nuclear physics, stochastic mechanics, biomolecule dynamics, dynamics of accelerators, and Bose-Einstein condensates [5,6,7,8,9,10]. The last few decades have seen significant advancements in the field of nonlinear optics [11,12,13]. These equations also appear in other forms of nonlinearities, such as cubic-quintic (CQ), cubic-quintic-septic (CQS), power-law, logarithmic nonlinearities, and various other forms. There is a growing interest in studying soliton pulses that can propagate without changing their shape in optical fibers [14,15].

    Therefore, obtaining the exact soliton solutions of these NLSEs can help us to understand these phenomena better. In recent years, several effective approaches have been developed to construct accurate solutions of these equations. For example, the simple equation method [16], Kudryashov's method [17], the Jacobi elliptic function method [18], the inverse scattering method [19], the extended trial function method [20], the tanh method [21], the F-expansion method [22], generalized extended tanh-function method, the sine–cosine method [23], the generalized Riccati equation method [24], the new ϕ6-model expansion method [25], Hirota bilinear method [26], the exp-function method [27], the Darboux transformation [28], the auxiliary equation method [29], the Binary Bell polynomials [30], the extended hyperbolic function method [31], the homogeneous balance Method [32], the (GG)-expansion method [33,34], the exponential rational function method [35], the Bäcklund transformation [36], the homotopy perturbation method [37], the modified Kudryashov method, the sine-Gordon expansion approach [38], the Riccati-Bernoulli sub-ODE method [39], the modified extended direct algebraic method [40], the truncated Painlevé expansion method [41], the (G/G,1/G)-expansion method [42], the tan(ϕ/2)-expansion method [43,44], the soliton ansatz method [45,46], and so on.

    Researches have been conducted on the study of NLSEs with the polynomial law of nonlinearity. For instance, Seadawy et al. [47] established soliton solutions using the extended simplest equation method, and they also described this model [48], which includes the conservation principles for optical soliton (OS) with polynomial and triple power law nonlinearities, while Aziz et al. [49] discovered chirped soliton solutions by utilizing Jacobi elliptic functions. Furthermore, this model was discussed by Dieu-donne et al. [50], who examined the optical soliton (OS) solutions with two types of nonlinearities, which are triple power and polynomial laws. Sugati et al.[51] also examined this model, obtaining the traveling pulse solutions to the NLSE when it is treated with both spatio-temporal dispersion (STD) and group velocity dispersion (GVD).

    In this paper, our aim is to investigate soliton solutions for two models of the NLSE that carry the polynomial law of nonlinearity (cubic-quintic-septic). We will use the variational principle based on finding Lagrangian, which we will then apply it with different trial functions that have one or two nontrivial variational parameters. Furthermore, we will employ another technique called the amplitude ansatz method to extract new solitary wave solutions.

    The paper is categorized as follows. In Section 2, we discuss model-I of NLSE with polynomial law nonlinearity in terms of formulation of the variational principle and finding solitary wave solutions. In Section 3, we apply the same steps on model-II of NLSE with polynomial law nonlinearity. Finally, the work concludes in Section 4.

    Nonlinear Schrödinger equation (NLSE) with the polynomial nonlinear law is [47]:

    iΓt+aΓxx+b1Γ|Γ|2+b2Γ|Γ|4+b3Γ|Γ|6=0. (2.1)

    Such that Γ is a complex function on the form: Γ(x,t)=Θ(x,t)+iΨ(x,t). Since Θ and Ψ are real functions of x and t, furthermore |Γ|2=(Θ+iΨ)(ΘiΨ). Using the variational approach, we will search for solutions to NLSE with the polynomial nonlinear law. As a result, we derive Γ with respect to x and t to investigate the existence of a Lagrangian and the invariant variational principle for this equation, which are expressed in the following way:

    Let M and N are functionals in Θ and Ψ:

    M(Θ,Ψ)=Θt+a2Ψx2+b1ΨΘ2+b1Ψ3+b2ΨΘ4+2b2Θ2Ψ3+b2Ψ5+b3ΨΘ6+3b3Θ4Ψ3+3b3Θ2Ψ5+b3Ψ7, (2.2)
    N(Θ,Ψ)=Ψt+a2Θx2+b1Θ3+b1ΘΨ2+b2Θ5+2b2Ψ2Θ3+b2ΘΨ4+b3Θ7+3b3Ψ2Θ5+3b3Θ3Ψ4+b3ΘΨ6. (2.3)

    Put Θ=λΘ and Ψ=λΨ,

    10M(λΘ,λΨ)dλ=12Θt+12a2Ψx2+14b1ΨΘ2+14b1Ψ3+16b2ΨΘ4+13b2Θ2Ψ3+16b2Ψ5+18b3ΨΘ6+38b3Θ4Ψ3+38b3Θ2Ψ5+18b3Ψ7,
    10N(λΘ,λΨ)dλ=12Ψt+12a2Θx2+14b1Θ3+14b1ΘΨ2+16b2Θ5+13b2Ψ2Θ3+16b2ΘΨ4+18b3Θ7+38b3Ψ2Θ5+38b3Θ3Ψ4+18b3ΘΨ6.

    The consistency conditions are expressed as follows [52,53], where Θx, Ψx, Θt and Ψt stand for the partial derivatives of Θ and Ψ with respect to variables x and t. If the system of Eqs (2.2) and (2.3) satisfies the the previously mentioned conditions, then a functional integral J(Θ,Ψ) can be written down using the formula given by Tonti [53]:

    J(Θ,Ψ)=ω[14b1Θ4+16b2Θ6+18b3Θ8+12b1Θ2Ψ2+12b2Θ4Ψ2+12b3Θ6Ψ2+14b1Ψ4+12b2Θ2Ψ4+34b3Θ4Ψ4+16b2Ψ6+12b3Θ2Ψ6+18b3Ψ8+12ΨΘt12ΘΨt+12aΘΘxx+12aΨΨxx]dω,

    where dω=dxdt.

    The terms ΘΘxx and ΨΨxx solve integration by parts and choosing the boundary on Θx and Ψx to be such that the boundary terms vanish, we get

    J(Θ,Ψ)=ω[14b1Θ4+16b2Θ6+18b3Θ8+12b1Θ2Ψ2+12b2Θ4Ψ2+12b3Θ6Ψ2+14b1Ψ4+12b2Θ2Ψ4+34b3Θ4Ψ4+16b2Ψ6+12b3Θ2Ψ6+18b3Ψ8+12ΨΘt12ΘΨt12aΘ2x12aΨ2x]dω, (2.4)

    where dω=dxdt.

    We obtain the Lagrangian L as

    L(Θ,Ψ)=14b1Θ4+16b2Θ6+18b3Θ8+12b1Θ2Ψ2+12b2Θ4Ψ2+12b3Θ6Ψ2+14b1Ψ4+12b2Θ2Ψ4+34b3Θ4Ψ4+16b2Ψ6+12b3Θ2Ψ6+18b3Ψ8+12ΨΘt12ΘΨt12aΘ2x12aΨ2x. (2.5)

    We find the value of L in the Euler-Lagrange equations to validate our interpretations:

    LΘt(LΘt)x(LΘx)=0,
    LΨt(LΨt)x(LΨx)=0.

    The resulting derivatives give us the system of Eqs (2.2) and (2.3).

    We demonstrate the simplest example of the application of this technique, by taking the box-shaped initial pulse and an ansatz based on linear Jost functions in a single nontrivial variational parameter in three cases.

    Case 1: We use the following ansatz for Θ(x,t) and Ψ(x,t) functions:

    Θ(x,t)={100exp(μ(t5)(x5)),atx>5,t>5,(t+5)(x+5),at|x|<5,|t|<5,0,atx<5,t<5, (2.6)
    Ψ(x,t)={0,atx>5,t>5,(5t)(5x),at|x|<5,|t|<5,100exp(μ(t+5)(x+5)),atx<5,t<5. (2.7)

    We found the values of the Lagrangian L from substituting Eqs (2.6) and (2.7) into Eq (2.5) (see Figure 1):

    J(Θ,Ψ)=55Ldxdt+5555Ldxdt+55Ldxdt,
    Figure 1.  Show an example of the Lagrangian L(x,t) with Eq (2.5), in the interval 5<x<5 and 5<t<5, by choosing the trial functions (2.6) and (2.7) at a=3,b1=2,b2=12,b3=16.

    where

    55Ldxdt=55Ldxdt=0.

    By using Mathematica program we get the value of J(Θ,Ψ) at interval 5<x<5,5<t<5 as

    J(Θ,Ψ)=10000a3+1850000000b19+304000000000000b2441+1760000000000000000b3567+250003. (2.8)

    Case 2: Assume Θ(x,t) and Ψ(x,t) for the Jost function as the following:

    Θ(x,t)={20exp(μ(t5)(x5)),atx>5,t>5,t+x+10,at|x|<5,|t|<5,0,atx<5,t<5, (2.9)
    Ψ(x,t)={0,atx>5,t>5,10tx,at|x|<5,|t|<5,20exp(μ(t+5)(x+5)),atx<5,t<5. (2.10)

    We found the values of the Lagrangian L from substituting by Eqs (2.9) and (2.10) into Eq (2.5), then we get

    J(Θ,Ψ)=10063(63a+882000b1+145800000b2+28120000000b3+630). (2.11)

    Case 3:

    Θ(x,t)={12(exp(6πtx)exp(2πtx)),atx>π,t>π,sinh(t+x+2π),at|x|<π,|t|<π,0,atx<π,t<π, (2.12)
    ψ(x,t)={0,atx>π,t>π,sinh(2πtx),at|x|<π,|t|<π,12(exp(6π+t+x)exp(2π+t+x)),atx<π,t<π. (2.13)

    We found the values of the Lagrangian L from substituting by Eqs (2.12) and (2.13) into Eq (2.5), then we have

    J(Θ,Ψ)=1.02783×1010a+2.64108×1019b1+1.60864×1029b2+1.39506×1039b3+2.83012×106. (2.14)

    Case 1: We try the following Jost functions:

    Θ(x,t)={12(exp(π2(3+2α+α2)πtπx)exp(π2(12αα2)πtπx)),atx>π,t>π,sinh((π+αt)(π+αx)),at0<x<π,0<t<π,sinh((t+π)(x+π)),atπ<x<0,π<t<0,0,atx<π,t<π, (2.15)
    Ψ(x,t)={0,atx>π,t>π,sinh((πt)(πx)),at0<x<π,0<t<π,sinh((παt)(παx)),atπ<x<0,π<t<0,12(exp(π2(3+2α+α2)+πt+πx)exp(π2(12αα2)+πt+πx)),atx<π,t<π. (2.16)

    This ansatz now contains nontrivial variational parameter α. Substituting Eqs (2.15) and (2.16) into Eq (2.5) (see Figure 2), then we obtain the functional integral at α=1 as follows:

    J(Θ,Ψ)=4.43569×107a+1.13554×1014b1+6.23167×1021b2+4.89178×1029b3. (2.17)
    Figure 2.  Show an example of the Lagrangian L(x,t) with Eq (2.5), in the interval 0<x<π and 0<t<π, by choosing the trial functions (2.15) and (2.16) at the parameter α=1 and a=3,b1=2,b2=12,b3=16.

    Also, we can use α=2, then we get

    0π0πLdxdt=π0π0Ldxdt=4.42653×1075a+5.59314×10148b1+5.89626×10224b2+8.85443×10300b32.66975×1036.

    The functional integral at α=2 becomes

    J(Θ,Ψ)=1.77562×1076a+4.12756×10150b1+4.35525×10226b2+6.5433×10302b35.33949×1036. (2.18)

    Case 2: We now try the following Jost function:

    Θ(x,t)={12(exp(4πtx+2πα)exp(tx2πα)),atx>π,t>π,sinh(2π+αt+αx),at0<x<π,0<t<π,sinh(2π+t+x),atπ<t<0,π<x<0,0,atx<π,t<π, (2.19)
    Ψ(x,t)={0,atx>π,t>π,sinh(2πtx),at0<x<π,0<t<π,sinh(2παtαx),atπ<x<0,π<t<0,12(exp(4π+t+x+2πα)exp(t+x2πα)),atx<π,t<π. (2.20)

    This ansatz contains nontrivial variational parameter α. Substituting Eqs (2.19) and (2.20) into Eq (2.5), then we get the functional integral at α=0.5 as follows:

    J(Θ,Ψ)=1.84004×107a+2.29576×1014b1+2.6187×1021b2+4.24151×1028b3+270561. (2.21)

    Also, we can take α=0.2, then we obtain

    0π0πLdxdt=π0π0Ldxdt=13516.2a+2.22124×109b1+7.54018×1013b2+3.79568×1018b3+19035.7.

    The functional integral at α=0.2 becomes

    J(Θ,Ψ)=28484a+4.44353×109b1+1.50805×1014b2+7.59137×1018b3+38071.3. (2.22)

    We assume the Jost function by quadratic polynomials at interval |x|<5,|t|<5:

    Case 1: We use the following Jost functions:

    Θ(x,t)={(100+10000α)exp(μ(t5)(x5)),atx>5,t>5,(t+5)(x+5)+α(5+t)2(5+x)2,at|x|<5,|t|<5,0,atx<5,t<5, (2.23)
    Ψ(x,t)={0,atx>5,t>5,(5t)(5x)+α(5t)2(5x)2,at|x|<5,|t|<5,(100+10000α)exp(μ(t+5)(x+5)),atx<5,t<5. (2.24)

    We found the values of Lagrangian calculation from Eqs (2.23) and (2.24) into Eq (2.5), and we get

    J(Θ,Ψ)=5.55556×106α22.66667×107aα2500000aα+555556α3333.33a+8.65053×1030α8b3+7.81255×1029α7b3+1.97241×1023α6b2+3.11121×1028α6b3+1.38897×1022α5b2+7.14358×1026α5b3+6.1741×1015α4b1+4.13321×1020α4b2+1.03576×1025α4b3+3.12755×1014α3b1+6.67171×1018α3b2+9.72777×1022α3b3+6.14172×1012α2b1+6.18569×1016α2b2+5.79228×1020α2b3+5.61111×1010αb1+3.14172×1014αb2+2.00529×1018αb3+2.05556×108b1+6.89342×1011b2+3.10406×1015b3+8333.33. (2.25)

    We choose a=3, b1=2, b2=12, and b3=16, then the functional integral J(Θ,Ψ) becomes

    J(Θ,Ψ)=1.44175×1030α8+1.30209×1029α7+5.18545×1027α6+1.19067×1026α5+1.72648×1024α4+1.62163×1022α3+9.65689×1019α2+3.34372×1017α+5.17688×1014. (2.26)

    Derive Eq (2.26) with respect to α, then values of α are:

    α=0.0136245,α=0.0129520.00218808i,α=0.012952+0.00218808i,α=0.01109590.00364364i,α=0.0110959+0.00364364i,α=0.00865180.00394467i,α=0.0086518+0.00394467i.

    We substitute the exact roots of α into Eq (2.26) give the following analytical equations:

    J(Θ,Ψ)=4.83907×1010,J(Θ,Ψ)=3.84513×10104.55557×1010i,J(Θ,Ψ)=3.84513×1010+4.55557×1010i,J(Θ,Ψ)=3.14336×1091.06115×1011i,J(Θ,Ψ)=3.14336×109+1.06115×1011i,J(Θ,Ψ)=1.8701×10112.04551×1011i,J(Θ,Ψ)=1.8701×1011+2.04551×1011i. (2.27)

    Case 2: We try the following Jost functions:

    Θ(x,t)={(20+200α)exp(μ(t5)(x5)),atx>5,t>5,(10+t+x)+α(t+5)2+α(x+5)2,at|x|<5,|t|<5,0,atx<5,t<5, (2.28)
    Ψ(x,t)={0,atx>5,t>5,(10tx)+α(5t)2+α(5x)2,at|x|<5,|t|<5,(20+200α)exp(μ(t+5)(x+5)),atx<5,t<5. (2.29)

    We found the values of Lagrangian calculation from Eqs (2.28) and (2.29) into Eq (2.5) (see Figure 3), and we have

    J(Θ,Ψ)=50000α213333.3aα22000aα+15000α100a+9.90698×1017α8b3+9.04639×1017α7b3+5.97273×1013α6b2+3.65565×1017α6b3+4.25597×1013α5b2+8.55415×1016α5b3+4.78095×109α4b1+1.28615×1013α4b2+1.27048×1016α4b3+2.39683×109α3b1+2.11556×1012α3b2+1.22953×1015α3b3+4.62698×108α2b1+2.00368×1011α2b2+7.59375×1013α2b3+4.08889×107αb1+1.03937×1010αb2+2.7454×1012αb3+1.4×106b1+2.31429×108b2+4.46349×1010b3+1000. (2.30)
    Figure 3.  Show an example of the Lagrangian L(x,t) with Eq (2.5), in the interval 5<x<5 and 5<t<5, by choosing the trial functions (2.28) and (2.29) at the parameter α=0.131984 and a=6,b1=3,b2=14,b3=13.

    We put a=6, b1=3, b2=14, and b3=13, then the functional integral J(Θ,Ψ) becomes

    J(Θ,Ψ)=3.30233×1017α8+3.01546×1017α7+1.2187×1017α6+2.85245×1016α5+4.23816×1015α4+4.10379×1014α3+2.5364×1013α2+9.17853×1011α+1.49404×1010. (2.31)

    Derive Eq (2.31) with respect to α, then values of α are:

    α=0.131984,α=0.1291840.0176052i,α=0.129184+0.0176052i,α=0.1176820.0366915i,α=0.117682+0.0366915i,α=0.08663870.051817i,α=0.0866387+0.051817i.

    We substitute the exact roots of α into Eq (2.31) give the following analytical equations:

    J(Θ,Ψ)=207360,J(Θ,Ψ)=102453293854i,J(Θ,Ψ)=102453+293854i,J(Θ,Ψ)=1.38256×106+542198i,J(Θ,Ψ)=1.38256×106542198i,J(Θ,Ψ)=1.31177×107+1.75734×107i,J(Θ,Ψ)=1.31177×1071.75734×107i. (2.32)

    We inspect here the exact solutions of the nonlinear Schrödinger equation with the polynomial nonlinear law. This equation is defined as Eq (2.1).

    Case 1: We suppose the ansatz function of the NLSE with the polynomial nonlinear law is in the form of a bright solitary wave solution

    h1(x,t)=Asech(w(xtv)),Γ(x,t)=Asech(w(xtv))ei(kxωt), (2.33)

    where A,w and v are the amplitude, the pulse width, and velocity of soliton in normalized unites. Substituting from Eq (2.33) in Eq (2.1), and separating the real and imaginary parts, we obtain

    ak2v+avw2+vω+(A2b1v2avw2)sech2(w(xtv))+A4b2vsech4(w(xtv))+A6b3vsech6(w(xtv))=0, (2.34)
    (w2akvw)tanh(w(xtv))=0. (2.35)

    Equating the coefficients of the linearly independent terms to zero, we obtain the dynamical system in A,w,v,k,ω,a,b1,b2,b3 by solving this system we get:

    Family I:

    A=±wb1kv,a=12kv,ω=k2w22kv. (2.36)

    The sufficient conditions for solitary wave solution stability are

    b1kv>0,kv0. (2.37)

    Family II:

    A=±k2vωb1v,a=12kv,w=±k(k2vω), (2.38)

    provided that

    k2vωb1v>0,kv0,k(k2vω)>0. (2.39)

    Family III:

    A=±w2ωb1k2b1w2,a=ωk2w2,v=k2w22kω, (2.40)

    whenever

    ωb1(k2w2)>0,kω0,k2w20. (2.41)

    Family IV:

    A=±1b1(v4ω2+v2w2v2ω),a=v4ω2+v2w2v2ω2v2w2,k=v4ω2+v2w2v+vω, (2.42)

    provided that

    v4ω2+v2w2>0,1b1(v4ω2+v2w2v2ω)>0,v2w20. (2.43)

    Then, the solutions of the NLSE with the polynomial nonlinear law as bright solitary wave solutions are (see Figures 4 and 5):

    Γ11(x,t)=±wb1kvsech(w(xtv))ei(kxωt), (2.44)
    Γ12(x,t)=±k2vωb1vsech(w(xtv))ei(kxωt), (2.45)
    Γ13(x,t)=±w2ωb1k2b1w2sech(w(xtv))ei(kxωt), (2.46)
    Γ14(x,t)=±1b1(v4ω2+v2w2v2ω)sech(w(xtv))ei(kxωt). (2.47)
    Figure 4.  Representation of solitary wave solution Γ11. These figures are obtained by b1=0.5,k=0.25,v=0.7,w=1,ω=0.3, (4a) is plotted in 3D while (4b) is plotted in 2D at different positions, and (4c) plotted as contour.
    Figure 5.  Representation of solitary wave solution Γ13. These figures are obtained by b1=0.25,k=0.3,v=0.2,w=1,ω=0.7, (5a) is plotted in 3D while (5b) is plotted in 2D at different positions, and (5c) plotted as contour.

    Case 2: Another choice of the dark solitary wave solution of the NLSE with the polynomial nonlinear law is

    h2(x,t)=A+Btanh(w(xtv)),Γ(x,t)=(A+Btanh(w(xtv)))ei(kxωt). (2.48)

    By replacement from Eq (2.48) in Eq (2.1) and separating the real and imaginary parts

    aAk2+A7b3+21A5b3B2+A5b2+35A3b3B4+10A3b2B2+A3b1+7Ab3B6+5Ab2B4+3Ab1B2+Aω+(21A5b3B270A3b3B410A3b2B221Ab3B610Ab2B43Ab1B2)×sech2(w(xtv))+(35A3b3B4+21Ab3B6+5Ab2B4)×sech4(w(xtv))7Ab3B6sech6(w(xtv))+(aBk2+7A6b3B+35A4b3B3+5A4b2B+21A2b3B5+10A2b2B3+3A2b1B+b3B7+b2B5+b1B3+Bω)×tanh(w(xtv))+(2aBw235A4b3B342A2b3B510A2b2B33b3B72b2B5b1B3)×tanh(w(xtv))sech2(w(xtv))+(21A2b3B5+3b3B7+b2B5)tanh(w(xtv))sech4(w(xtv))B7b3tanh(w(xtv))sech6(w(xtv))=0, (2.49)
    (2aBkwBwv)sech2(w(xtv))=0 (2.50)

    Equating the coefficients of the linearly independent terms to zero, we deduce the coefficients A,B,w,v,k,ω,a,b1,b2,b3 in the form:

    Family I:

    A=0,B=±ib23b3,b1=27b23ω+2b3227ab23k29b2b3,w=±13b327b23ωb3227ab23k26a,v=12ak. (2.51)

    The sufficient conditions for dark solitary wave solution stability are

    b2b3<0,b2b30,ak0,27b23ωb3227ab23k2a>0. (2.52)

    Family II:

    A=0,B=±ib23b3,w=±13b33b1b2b3b322a,v=12ak,ω=27ab23k22b32+9b1b3b227b23, (2.53)

    provided that

    b2b3<0,3b1b2b3b32a>0,ak0,b30. (2.54)

    Family III:

    A=0,B=±6k22kvω+2w22b3kv,a=12kv,b1=3b3(3k26kvω+4w2)34k2v2(k22kvω+2w2),b2=33b23(k22kvω+2w2)2kv, (2.55)

    such that

    k22kvω+2w2b3kv>0,kv0,4k2v2(k22kvω+2w2)0. (2.56)

    Family IV:

    A=0,B=±6ak2+2aw2ωb3,b1=(3ak2+4aw23ω)3b3ak2+2aw2ω,b2=33b23(ak2+2aw2ω),v=k2+2w22k(ak2+2aw2), (2.57)

    whenever

    ak2+2aw2ωb3>0,ak2+2aw2ω0,k(ak2+2aw2)0. (2.58)

    Then, the dark soliton solutions of the NLSE with the polynomial nonlinear law Eq (2.1) are (see Figure 6):

    Γ21(x,t)=±ib23b3tanh(w(xtv))ei(kxωt), (2.59)
    Γ23(x,t)=±6k22kvω+2w22b3kvtanh(w(xtv))ei(kxωt), (2.60)
    Γ24(x,t)=±6ak2+2aw2ωb3tanh(w(xtv))ei(kxωt). (2.61)
    Figure 6.  Representation of solitary wave solution Γ24. These figures are obtained by a=0.7,b3=0.25,k=1,v=0.5,w=0.9,ω=0.2, (6a) is plotted in 3D while (6b) is plotted in 2D at different positions, and (6c) plotted as contour.

    The model II of the NLSE with polynomial nonlinearity is given by [49]:

    iΨt+aΨxx+bΨxt+(k1|Ψ|2+k2|Ψ|4+k3|Ψ|6)Ψ=0, (3.1)

    where Ψ(x,t) denotes the complex valued function. The coefficients a and b indicate group velocity dispersion (GVD) and spatio-temporal dispersion (STD), respectively. The first term represents the linear evolution of pulses in nonlinear optical fibers. The final term represents the nonlinearity of the non-Kerr law. Since Ψ(x,t) is a complex function on the form Ψ(x,t)=u(x,t)+iv(x,t) such that u and v are real functions of x and t, also |Ψ|2=(u+iv)(uiv). We will use the variational technique to find solutions for the Eq (3.1), where we derive Ψ with respect to x and t to obtain the Lagrangian of this equation, which is expressed in the following way:

    Let M and N are functionals in u and v,

    M(u,v)=ut+a2vx2+b2vxt+k1vu2+k1v3+k2vu4+2k2u2v3+k2v5+k3vu6+3k3v3u4+3k3u2v5+k3v7, (3.2)
    N(u,v)=vt+a2ux2+b2uxt+k1u3+k1uv2+k2u5+2k2v2u3+k2uv4+k3u7+3k3v2u5+3k3u3v4+k3uv6. (3.3)

    The system of Eqs (3.2) and (3.3) satisfies the conditions mentioned in [52,53], then a functional integral J(u,v) can be written down using the formula given by Tonti [53]:

    J(u,v)=ω[12auuxx+12avvxx+12buuxt+12bvvxt+12k3v2u6+34k3u4v4+12k2v2u4+12k3u2v6+12k2u2v4+12k1u2v2+18k3u8+16k2u6+14k1u4+18k3v8+16k2v6+14k1v4+12vut12uvt]dω, (3.4)

    where dω=dxdt.

    Then the Lagrangian L is given by

    L(u,v)=12auuxx+12avvxx+12buuxt+12bvvxt+12k3v2u6+34k3u4v4+12k2v2u4+12k3u2v6+12k2u2v4+12k1u2v2+18k3u8+16k2u6+14k1u4+18k3v8+16k2v6+14k1v4+12vut12uvt. (3.5)

    As a necessary check to our calculations, we use the value of L in the Euler-Lagrange equations:

    Lut(Lut)+2x2(Luxx)+2xt(Luxt)=0,
    Lvt(Lvt)+2x2(Lvxx)+2xt(Lvxt)=0,

    which yields the system of Eqs (3.2) and (3.3).

    We provide an example of applying this technique by using a box-shaped initial pulse and a single nontrivial variational parameter based on linear Jost functions in three cases.

    Case 1: We use the following ansatz for u(x,t) and v(x,t) functions:

    u(x,t)={100exp(μ(t5)(x5)),atx>5,t>5,(t+5)(x+5),at|x|<5,|t|<5,0,atx<5,t<5, (3.6)
    v(x,t)={0,atx>5,t>5,(5t)(5x),at|x|<5,|t|<5,100exp(μ(t+5)(x+5)),atx<5,t<5. (3.7)

    We found the values of the Lagrangian L from substituting Eqs (3.6) and (3.7) into Eq (3.5), and by using Mathematica program we get the value of J(u,v) at interval 5<x<5,5<t<5 as

    J(u,v)=2500b+2.05556×108k1+6.89342×1011k2+3.10406×1015k3+8333.33. (3.8)

    Case 2: Assume u(x,t) and v(x,t) for the Jost function as the following:

    u(x,t)={20exp(μ(t5)(x5)),atx>5,t>5,t+x+10,at|x|<5,|t|<5,0,atx<5,t<5, (3.9)
    v(x,t)={0,atx>5,t>5,10tx,at|x|<5,|t|<5,20exp(μ(t+5)(x+5)),atx<5,t<5. (3.10)

    We found the values of the Lagrangian L from substituting by Eqs (3.9) and (3.10) into Eq (3.5) (see Figure 7), then we have

    J(u,v)=100063(88200k1+14580000k2+2812000000k3+63). (3.11)
    Figure 7.  Show an example of the Lagrangian L(x,t) with Eq (3.5), in the interval 5<x<5 and 5<t<5, by choosing the trial functions (3.9) and (3.10) at a=2,b=6,k1=2,k2=3,k3=4.

    Case 3:

    u(x,t)={12(exp(6πtx)exp(2πtx)),atx>π,t>π,sinh(t+x+2π),at|x|<π,|t|<π,0,atx<π,t<π, (3.12)
    v(x,t)={0,atx>π,t>π,sinh(2πtx),at|x|<π,|t|<π,12(exp(6π+t+x)exp(2π+t+x)),atx<π,t<π. (3.13)

    We found the values of the Lagrangian L from substituting by Eqs (3.12) and (3.13) into Eq (3.5), then we get

    J(u,v)=1.02783×1010a+1.02783×1010b+2.64108×1019k1+1.60864×1029k2+1.39506×1039k3+2.83012×106. (3.14)

    Case 1: We try the following Jost functions:

    u(x,t)={12(exp(π2(3+2α+α2)πtπx)exp(π2(12αα2)πtπx)),atx>π,t>π,sinh((π+αt)(π+αx)),at0<x<π,0<t<π,sinh((t+π)(x+π)),atπ<x<0,π<t<0,0,atx<π,t<π, (3.15)
    v(x,t)={0,atx>π,t>π,sinh((πt)(πx)),at0<x<π,0<t<π,sinh((παt)(παx)),atπ<x<0,π<t<0,12(exp(π2(3+2α+α2)+πt+πx)exp(π2(12αα2)+πt+πx)),atx<π,t<π. (3.16)

    This ansatz now contains nontrivial variational parameter α. Substituting Eqs (3.15) and (3.16) into Eq (3.5), then we obtain the functional integral at α=1 as follows:

    J(u,v)=4.43568×107a+4.92254×107b+1.13554×1014k1+6.23167×1021k2+4.89178×1029k3. (3.17)

    Also, we can use α=2, then we get

    0π0πLdxdt=π0π0Ldxdt=4.42653×1075a+4.47679×1075b+5.59314×10148k1+5.89626×10224k2+8.85443×10300k32.66975×1036.

    The functional integral at α=2 becomes

    J(u,v)=1.77562×1076a+1.78568×1076b+4.12756×10150k1+4.35525×10226k2+6.5433×10302k35.33949×1036. (3.18)

    Case 2: We now try the following Jost function:

    u(x,t)={12(exp(4πtx+2πα)exp(tx2πα)),atx>π,t>π,sinh(2π+αt+αx),at0<x<π,0<t<π,sinh(2π+t+x),atπ<t<0,π<x<0,0,atx<π,t<π, (3.19)
    v(x,t)={0,atx>π,t>π,sinh(2πtx),at0<x<π, 0<t<π,sinh(2παtαx),atπ<x<0,π<t<0,12(exp(4π+t+x+2πα)exp(t+x2πα)),atx<π,t<π. (3.20)

    This ansatz contains nontrivial variational parameter α. Substituting Eqs (3.19) and (3.20) into Eq (3.5) (see Figure 8), then we get the functional integral at α=0.2 as follows:

    J(u,v)=28473.8a+28473.8b+4.44353×109k1+1.50805×1014k2+7.59137×1018k3+38071.3. (3.21)
    Figure 8.  Show an example of the Lagrangian L(x,t) with Eq (3.5), in the interval 0<x<π and 0<t<π, by choosing the trial functions (3.19) and (3.20) at the parameter α=0.2 and a=2,b=6,k1=2,k2=3,k3=4.

    Also, we can take α=0.5 then we obtain

    0π0πLdxdt=π0π0Ldxdt=4.40169×106a+4.40169×106b+9.17623×1013k1+1.04745×1021k2+1.6966×1028k3+135281.

    The functional integral at α=0.5 becomes

    J(u,v)=1.84004×107a+1.84004×107b+2.29576×1014k1+2.6187×1021k2+4.24151×1028k3+270561. (3.22)

    We assume the Jost function by quadratic polynomials at interval |x|<5,|t|<5:

    Case 1: We use the following Jost functions:

    u(x,t)={(100+10000α)exp(μ(t5)(x5)),atx>5,t>5,(t+5)(x+5)+α(5+t)2(5+x)2,at|x|<5,|t|<5,0,atx<5,t<5, (3.23)
    v(x,t)={0,atx>5,t>5,(5t)(5x)+α(5t)2(5x)2,at|x|<5,|t|<5,(100+10000α)exp(μ(t+5)(x+5)),atx<5,t<5. (3.24)

    We found the values of Lagrangian calculation from Eqs (3.23) and (3.24) into Eq (3.5) (see Figure 9), and we get

    J(u,v)=5.55556×106α2+1.33333×107aα2+250000aα+555556α+2.5×107α2b+555556αb+2500b+8.65053×1030α8k3+7.81255×1029α7k3+1.97241×1023α6k2+3.11121×1028α6k3+1.38897×1022α5k2+7.14358×1026α5k3+6.1741×1015α4k1+4.13321×1020α4k2+1.03576×1025α4k3+3.12755×1014α3k1+6.67171×1018α3k2+9.72777×1022α3k3+6.14172×1012α2k1+6.18569×1016α2k2+5.79228×1020α2k3+5.61111×1010αk1+3.14172×1014αk2+2.00529×1018αk3+2.05556×108k1+6.89342×1011k2+3.10406×1015k3+8333.33. (3.25)
    Figure 9.  Show an example of the Lagrangian L(x,t) with Eq (3.5), in the interval 5<x<5 and 5<t<5, by choosing the trial functions (3.28) and (3.29) at the parameter α=0.131961 and a=3,b=2,k1=2,k2=4,k3=6.

    We put a=2, b=6, k1=2, k2=3, and k3=4, then the functional integral J(u,v) becomes

    J(u,v)=3.46021×1031α8+3.12502×1030α7+1.24449×1029α6+2.85747×1027α5+4.14318×1025α4+3.89131×1023α3+2.3171×1021α2+8.02211×1018α+1.24183×1016. (3.26)

    Derive Eq (3.26) with respect to α, then values of α are:

    α=0.0136235,α=0.01295140.00218597i,α=0.0129514+0.00218597i,α=0.01109620.00364064i,α=0.0110962+0.00364064i,α=0.008652630.00394263i,α=0.00865263+0.00394263i.

    We substitute the exact roots of α into Eq (3.26) give the following analytical equations:

    J(u,v)=1.15381×1012,J(u,v)=9.16876×10111.08685×1012i,J(u,v)=9.16876×1011+1.08685×1012i,J(u,v)=7.64569×10102.53327×1012i,J(u,v)=7.64569×1010+2.53327×1012i,J(u,v)=4.47645×10124.88381×1012i,J(u,v)=4.47645×1012+4.88381×1012i. (3.27)

    Case 2: We try the following Jost functions:

    u(x,t)={(20+200α)exp(μ(t5)(x5)),atx>5,t>5,(10+t+x)+α(t+5)2+α(x+5)2,at|x|<5,|t|<5,0,atx<5,t<5, (3.28)
    v(x,t)={0,atx>5,t>5,(10tx)+α(5t)2+α(5x)2,at|x|<5,|t|<5,(20+200α)exp(μ(t+5)(x+5)),atx<5,t<5. (3.29)

    We found the values of Lagrangian calculation from Eqs (3.28) and (3.29) into Eq (3.5), and we have

    J(u,v)=50000α2+13333.3aα2+2000aα+15000α+9.90698×1017α8k3+9.04639×1017α7k3+5.97273×1013α6k2+3.65565×1017α6k3+4.25597×1013α5k2+8.55415×1016α5k3+4.78095×109α4k1+1.28615×1013α4k2+1.27048×1016α4k3+2.39683×109α3k1+2.11556×1012α3k2+1.22953×1015α3k3+4.62698×108α2k1+2.00368×1011α2k2+7.59375×1013α2k3+4.08889×107αk1+1.03937×1010αk2+2.7454×1012αk3+1.4×106k1+2.31429×108k2+4.46349×1010k3+1000. (3.30)

    We put a=3, b=2, k1=2, k2=4, and k3=6, then the functional integral J(u,v) becomes

    J(u,v)=5.94419×1018α8+5.42784×1018α7+2.19363×1018α6+5.13419×1017α5+7.62802×1016α4+7.38564×1015α3+4.56427×1014α2+1.6514×1013α+2.68738×1011. (3.31)

    Derive Eq (3.31) with respect to α, then values of α are:

    α=0.131961,α=0.1291420.0174743i,α=0.129142+0.0174743i,α=0.1176820.0366515i,α=0.117682+0.0366515i,α=0.08669090.0518096i,α=0.0866909+0.0518096i.

    We substitute the exact roots of α into Eq (3.31) give the following analytical equations:

    J(u,v)=3.53554×106,J(u,v)=1.84827×1065.13825×106i,J(u,v)=1.84827×106+5.13825×106i,J(u,v)=2.46359×107+1.0125×107i,J(u,v)=2.46359×1071.0125×107i,J(u,v)=2.31965×108+3.16423×108i,J(u,v)=2.31965×1083.16423×108i. (3.32)

    We survey here the exact solutions of the nonlinear Schrödinger equation with polynomial law nonlinearity Eq (3.1).

    Case 1: We suppose the ansatz function of the NLSE with polynomial law nonlinearity is in the form of a bright solitary wave solution.

    h1(x,t)=Asech(w(xtv)),Ψ(x,t)=Asech(w(xtv))ei(μxωt), (3.33)

    where A,w and v are the amplitude, the pulse width, and velocity of soliton in normalized unites. Substituting from Eq (3.33) in Eq (3.1), and separating the real and imaginary parts, we obtain

    avw2+bμvωaμ2vbw2+vω+(2avw2+A2k1v+2bw2)×sech2(w(xtv))+A4k2vsech4(w(xtv))+A6k3vsech6(w(xtv))=0, (3.34)
    (2aμvw+bvwω+bμw+w)tanh(w(xtv))=0. (3.35)

    Equating the coefficients of the linearly independent terms to zero, we obtain the dynamical system in A,w,v,μ,ω,a,b,k1,k2,k3 by solving this system we get:

    Family I:

    A=±2(μw2vw2ω)k1μ2v+k1vw2,a=v2ω2w2v(μ2+w2)(vωμ),b=2μvω+w2μ2(μ2+w2)(μvω). (3.36)

    The sufficient conditions for solitary wave solution stability are:

    μw2vw2ωk1μ2v+k1vw2>0,v(μ2+w2)(vωμ)0,(μ2+w2)(μvω)0. (3.37)

    Family II:

    A=±bμ2bμvω+μ2vωk1v,a=bμ+bvω+12μv,w=±bμ3bμ2vω+μ22μvωbvω+1bμ, (3.38)

    provided that

    bμ2bμvω+μ2vωk1v>0,bμ3bμ2vω+μ22μvωbvω+1bμ>0,μv0. (3.39)

    Family III:

    A=±w2ωbk1μ3+bk1μw2+k1μ2k1w2,a=ω(b2μ2+b2w2+2bμ+1)bμ3+bμw2+μ2w2,v=bμ3+bμw2+μ2w2ω(bμ2+bw2+2μ), (3.40)

    whenever

    ωbk1μ3+bk1μw2+k1μ2k1w2>0,bμ3+bμw2+μ2w20,ω(bμ2+bw2+2μ)0. (3.41)

    Then, the solutions of the NLSE with polynomial law nonlinearity as bright solitary wave solutions are (see Figures 10 and 11):

    Ψ11(x,t)=±2(μw2vw2ω)k1μ2v+k1vw2sech(w(xtv))ei(μxωt), (3.42)
    Ψ12(x,t)=±bμ2bμvω+μ2vωk1vsech(w(xtv))ei(μxωt), (3.43)
    Ψ13(x,t)=±w2ωbk1μ3+bk1μw2+k1μ2k1w2sech(w(xtv))ei(μxωt). (3.44)
    Figure 10.  Representation of solitary wave solution Ψ11. These figures are obtained by k1=0.3,μ=0.5,v=0.2,w=1,ω=0.25, (10a) and (10b) are plotted in 3D while (10c) is plotted in 2D at different positions, and (10d) plotted as contour.
    Figure 11.  Representation of solitary wave solution Ψ12. These figures are obtained by b=0.5,k1=0.3,μ=0.9,v=0.25,w=1,ω=0.1, (11a) is plotted in 3D while (11b) is plotted in 2D at different positions, and (11c) plotted as contour.

    Case 2: Another choice of the dark solitary wave solution of the NLSE with polynomial law nonlinearity is

    Ψ(x,t)=(A+Btanh(w(xtv)))ei(μxωt). (3.45)

    By replacement from Eq (3.45) in Eq (3.1) and separating the real and imaginary parts

    A7k3aAμ2+21A5B2k3+A5k2+35A3B4k3+10A3B2k2+A3k1+Abμω+7AB6k3+5AB4k2+3AB2k1+Aω+(21A5B2k370A3B4k310A3B2k221AB6k310AB4k23AB2k1)×sech2(w(xtv))+(35A3B4k3+21AB6k3+5AB4k2)×sech4(w(xtv))7AB6k3sech6(w(xtv))+(aBμ2+7A6Bk3+35A4B3k3+5A4Bk2+21A2B5k3+10A2B3k2+3A2Bk1+bBμω+B7k3+B5k2+B3k1+Bω)tanh(w(xtv))+(2aBw235A4B3k342A2B5k310A2B3k2+2bBw2v3B7k32B5k2B3k1)tanh(w(xtv))sech2(w(xtv))+(21A2B5k3+3B7k3+B5k2)tanh(w(xtv))sech4(w(xtv))B7k3tanh(w(xtv))sech6(w(xtv))=0, (3.46)
    (2aBμwbBμwvbBwωBwv)sech2(w(xtv))=0. (3.47)

    Equating the coefficients of the linearly independent terms to zero, we deduce the coefficients A,B,w,v,μ,ω,a,b,k1,k2,k3 in the form:

    Family I:

    A=0,B=±ik23k3,a=k32v2ω+k32μv+54k23w227k23v2ω227k23v(μ22w2)(vωμ),b=54k23μvω27k23μ22k32μv54k23w227k23(μ22w2)(μvω),k1=3k32μ2v54k23vw2ω4k32vw2+54k23μw29k2k3v(μ22w2). (3.48)

    The sufficient conditions for dark solitary wave solution stability are

    k2k3<0,k23v(μ22w2)(vωμ)0,k2k3v(μ22w2)0. (3.49)

    Family II:

    A=0,B=±ik23k3,a=27b2k23μ254b2k23w2+54bk23μ+bk32v+27k2327k23v(bμ22bw2+2μ),k1=3bk32μ2v4bk32vw2+6k32μv+54k23w29k2k3v(bμ22bw2+2μ),ω=27bk23μ354bk23μw2+27k23μ2+2k32μv+54k23w227k23v(bμ22bw2+2μ), (3.50)

    provided that

    k2k3<0,k2k3v(bμ22bw2+2μ)0,k23v(bμ22bw2+2μ)0. (3.51)

    Family III:

    A=0,B=±ik23k3,a=bμ+bvω+12μv,k1=27bk23μvω27bk23μ227k23μ+54k23vω+4k32v18k2k3v,w=μ(27bk23μvω27k23μ+54k23vω2k32v27bk23μ2)54bk23vω+54k2354bk23μ, (3.52)

    whenever

    k2k3<0,μv0,k2k3v0,μ(27bk23μvω27k23μ+54k23vω2k32v27bk23μ2)54bk23vω+54k2354bk23μ>0. (3.53)

    Family IV:

    B=±6aμ2v2ω2av2w2ωaμ3v+2aμvw2+v2ω22w2v(2w2μ2)(vωμ)(k3(μ+vω)(2w2μ2)(μvω)),A=0,b=2aμv1μ+vω,k1=1v2/3(μ+vω)32w2(av(vωμ)+1)v(aμ2(vωμ)+vω2)×(3(k3(μ+vω)(2w2μ2)(μvω))(2w2μ2)(vωμ)×(3v(aμ2(vωμ)+vω2)4w2(av(vωμ)+1))),k2=3k332w2(av(vωμ)+1)v(aμ2(vωμ)+vω2)3v(2w2μ2)(vωμ)(k3(μ+vω)(2w2μ2)(μvω)), (3.54)

    such that

    aμ2v2ω2av2w2ωaμ3v+2aμvw2+v2ω22w2v(2w2μ2)(vωμ)(k3(μ+vω)(2w2μ2)(μvω))>0,μ+vω0,v2/3(μ+vω)32w2(av(vωμ)+1)v(aμ2(vωμ)+vω2)0,3v(2w2μ2)(vωμ)(k3(μ+vω)(2w2μ2)(μvω))0. (3.55)

    Then, the dark soliton solutions of the NLSE with polynomial law nonlinearity Eq (3.1) are (see Figure 12)

    Ψ21(x,t)=±ik23k3tanh(w(xtv))ei(μxωt), (3.56)
    Ψ24(x,t)=±6aμ2v2ω2av2w2ωaμ3v+2aμvw2+v2ω22w2v(2w2μ2)(vωμ)(k3(μ+vω)(2w2μ2)(μvω))×tanh(w(xtv))ei(μxωt). (3.57)
    Figure 12.  Representation of solitary wave solution Ψ24. These figures are obtained by a=0.5,k3=0.25,μ=0.9,v=0.7,w=0.3,ω=1, (12a) and (12b) are plotted in 3D while (12c) and (12d) are plotted in 2D at different positions, and (12e) plotted as contour.

    In this paper, we discussed the two models of the nonlinear Schrödinger equation (NLSE) with polynomial law nonlinearity by effective and understandable techniques, such as the variational principle method based on the Lagrangian and the amplitude ansatz method. We found the functional integral and the Lagrangian of these models. Meanwhile, the solutions of the proposed equations were extracted by choice of different ansatz functions based on the Jost function, and they are continuous at all intervals. Firstly, the Jost function was approximated by piecewise linear ansatz function with a single nontrivial variational parameter in three cases from a region of a rectangular box. Then, the Jost function was approximated by the piecewise ansatz function containing the hyperbolic function in two cases of the two-box potential and was also approximated by quadratic polynomials with two free parameters rather than a piecewise linear ansatz function. Finally, this trial function had been approximated by the tanh function. Besides, we applied the amplitude ansatz method to obtain the new soliton solutions of the offered equations that contain bright soliton, dark soliton, bright-dark solitary wave solutions, rational dark-bright solutions, and periodic solitary wave solutions. The conditions for the stability of solutions were conducted. Graphical models, such as 2D, 3D, and contour plots, were induced using appropriate parameter values. These solutions have vital applications in applied sciences and might provide valuable support for investigators and physicists to solve more complex physical phenomena.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors extend their appreciation to the Deputyship for Research and Innovation, Ministry of Education in Saudi Arabia for funding this research work through project number 445-9-693. Furthermore, the authors would like to extend their appreciation to Taibah University for its supervision support.

    The authors declare that they have no competing interests.



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