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Review Special Issues

Risk of lung cancer due to external environmental factor and epidemiological data analysis


  • Received: 15 May 2021 Accepted: 30 June 2021 Published: 08 July 2021
  • Lung cancer is a cancer with the fastest growth in the incidence and mortality all over the world, which is an extremely serious threat to human's life and health. Evidences reveal that external environmental factors are the key drivers of lung cancer, such as smoking, radiation exposure and so on. Therefore, it is urgent to explain the mechanism of lung cancer risk due to external environmental factors experimentally and theoretically. However, it is still an open issue regarding how external environment factors affect lung cancer risk. In this paper, we summarize the main mathematical models involved the gene mutations for cancers, and review the application of the models to analyze the mechanism of lung cancer and the risk of lung cancer due to external environmental exposure. In addition, we apply the model described and the epidemiological data to analyze the influence of external environmental factors on lung cancer risk. The result indicates that radiation can cause significantly an increase in the mutation rate of cells, in particular the mutation in stability gene that leads to genomic instability. These studies not only can offer insights into the relationship between external environmental factors and human lung cancer risk, but also can provide theoretical guidance for the prevention and control of lung cancer.

    Citation: Lingling Li, Mengyao Shao, Xingshi He, Shanjing Ren, Tianhai Tian. Risk of lung cancer due to external environmental factor and epidemiological data analysis[J]. Mathematical Biosciences and Engineering, 2021, 18(5): 6079-6094. doi: 10.3934/mbe.2021304

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  • Lung cancer is a cancer with the fastest growth in the incidence and mortality all over the world, which is an extremely serious threat to human's life and health. Evidences reveal that external environmental factors are the key drivers of lung cancer, such as smoking, radiation exposure and so on. Therefore, it is urgent to explain the mechanism of lung cancer risk due to external environmental factors experimentally and theoretically. However, it is still an open issue regarding how external environment factors affect lung cancer risk. In this paper, we summarize the main mathematical models involved the gene mutations for cancers, and review the application of the models to analyze the mechanism of lung cancer and the risk of lung cancer due to external environmental exposure. In addition, we apply the model described and the epidemiological data to analyze the influence of external environmental factors on lung cancer risk. The result indicates that radiation can cause significantly an increase in the mutation rate of cells, in particular the mutation in stability gene that leads to genomic instability. These studies not only can offer insights into the relationship between external environmental factors and human lung cancer risk, but also can provide theoretical guidance for the prevention and control of lung cancer.



    1. Introduction

    Consider the following Euler-Poisson system for the bipolar hydrodynamical model of semi-conductor devices:

    {n1t+j1x=0,j1t+(j21n1+p(n1))x=n1Ej1,n2t+j2x=0,j2t+(j22n2+q(n2))x=n2Ej2,Ex=n1n2D(x), (1)

    in the region Ω=(0,1)×R+. In this paper, n1(x,t), n2(x,t), j1(x,t), j2(x,t) and E(x,t) represent the electron density, the hole density, the electron current density, the hole current density and the electric field, respectively. In this note, we assume that the p and q satisfy the γ-law:p(n1)=n21 and q(n2)=n22 (γ=2), which denote the pressures of the electrons and the holes. The function D(x), called the doping profile, stands for the density of impurities in semiconductor devices.

    For system (1), the initial conditions are

    ni(x,0)=ni0(x)0,ji(x,0)=ji0(x),i=1,2, (2)

    and the boundary conditions at x=0 and x=1 are

    ji(0,t)=ji(1,t)=0,i=1,2,E(0,t)=0. (3)

    So, we can get the compatibility condition

    ji0(0)=ji0(1)=0,i=1,2. (4)

    Moreover, in this paper, we assume the doping profile D(x) satisfies

    D(x)C[0,1] and D=supxD(x)infxD(x)=D. (5)

    Now, the definition of entropy solution to problem (1)(4) is given. We consider the locally bounded measurable functions n1(x,t), j1(x,t), n2(x,t), j2(x,t), E(x,t), where E(x,t) is continuous in x, a.e. in t.

    Definition 1.1. The vector function (n1,n2,j1,j2,E) is a weak solution of problem (1)(4), if it satisfies the equation (1) in the distributional sense, verifies the restriction (2) and (3). Furthermore, a weak solution of system (1)(4) is called an entropy solution if it satisfies the entropy inequality

    ηet+qex+j21n1+j22n2j1E+j2E0, (6)

    in the sense of distribution. And the (ηe,qe) are mechanical entropy-entropy flux pair which satisfy

    {ηe(n1,n2,j1,j2)=j212n1+n21+j222n2+n22,qe(n1,n2,j1,j2)=j312n21+2n1j1+j322n22+2n2j2. (7)

    For bipolar hydrodynamic model, the studies on the existence of solutions and the large time behavior as well as relaxation-time limit have been extensively carried out, for example, see [1][2][3][4][5][6] etc. Now, we make it into a semilinear ODE about the potential and the pressures with the exponent γ=2. We can get the existence, uniqueness and some bounded estimates of the steady solution. Then, using a technical energy method and a entropy dissipation estimate, we present a framework for the large time behavior of bounded weak entropy solutions with vacuum. It is shown that the weak solutions converge to the stationary solutions in L2 norm with exponential decay rate.

    The organization of this paper is as follows. In Section 2, the existence, uniqueness and some bounded estimates of stationary solutions are given. we present a framework for the large time behavior of bounded weak entropy solutions with vacuum in Section 3.


    2. Steady solutions

    In this part, we will prove the existence and uniqueness of steady solution to problem (1)(4). Moreover, we can obtain some important estimates on the steady solution (N1,N2,E).

    The steady equation of (1)(4) is as following

    {J1=J2=0,2N1N1x=N1E,2N2N2x=N2E,Ex=N1N2D(x), (8)

    and the boundary condition

    E(0)=0. (9)

    We only concern the classical solutions in the region where the density

    infxN1>0   and   infxN2>0. (10)

    hold.

    Now, we introduce a new variation Φ(x), and make Φ(x): = E(x). To eliminate the additive constants, we set 10Φ(x)dx=0. Then (2.1) turns into

    {2N1x=Φx,2N2x=Φx,Φxx=N1N2D(x). (11)

    Obviously, (11)1 and (11)2 indicate

    {N1(x)=12Φ(x)+C1,N2(x)=12Φ(x)+C2,Φxx(x)=12Φ(x)+C1+12Φ(x)C2D(x). (12)

    where C1 and C2 are two unknown positive constants. To calculate these two constants, we suppose*

    *Using the conservation of the total charge: integrating (1)1 and (1)3 from 0 to 1

    (10nidx)t=10jixdx=0,  for  i=1,2,

    we see this assumption is right.

    10(ni(x,0)Ni(x))dx=0  for  i=1,2, (13)

    then

    ˉn1:=10n1(x,0)dx=10N1(x)dx=10(Φ(x)2+C1)dx=C1,ˉn2:=10n2(x,0)dx=10N2(x)dx=10(Φ(x)2+C2)dx=C2. (14)

    Substituting (14) into (12)3, we have

    Φxx=Φ(x)+ˉn1ˉn2D(x). (15)

    Clearly, we can prove the existence and uniqueness of solutions to (15) with the Neumann boundary condition

    Φx(0)=Φx(1)=0. (16)

    Integrate(15) from x=0 to x=1, we get

    ˉn1ˉn2=10D(x)dx. (17)

    Suppose Φ(x) attains its maximum in x0[0,1], then we get Φxx(x0)0 and

    If x0(0,1), then Φx(x0)=0, Φxx(x0)0 clearly. If x0=0 or x0=1, the Taylor expansion

    Φ(x)=Φ(x0)+Φ(x0)(xx0)+Φ(x0)2(xx0)2+o(xx0)2,

    the boundary condition (16) indicates Φ(x0)0.

    Φ(x0)+ˉn1ˉn2D(x0)0.

    So we get

    Φ(x0)D+ˉn2ˉn1. (18)

    Similarly, if Φ attains its minimum in x1[0,1], we obtain

    Φ(x1)D+ˉn2ˉn1. (19)

    Moreover, from (12),(14),(15),(18), and (19), we have

    D+ˉn2+ˉn12N1(x)D+ˉn2+ˉn12,D+ˉn2+ˉn12N2(x)D+ˉn2+ˉn12, (20)
    D(N1N2)(x)D  for any  x[0,1]. (21)

    Above that, the theorem of existence and uniqueness of steady equation is given.

    Theorem 2.1. Assume that (5) holds, then problem (8), (9) has an unique solution (N1,N2,E), such that for any x[0,1]

    nN1(x)n,  nN2(x)n, (22)

    and

    D(N1N2)(x)D, (23)

    satisfy, where

    n:=max{D+ˉn2+ˉn12,D+ˉn2+ˉn12},n:=min{D+ˉn2+ˉn12,D+ˉn2+ˉn12}, (24)

    ˉn1, ˉn2 are defined in (14).


    3. Large time behavior

    Now, our aim is to prove the weak-entropy solution of (1)(4) convergences to corresponding stationary solution in L2 norm with exponential decay rate. For this purpose, we introduce the relative entropy-entropy flux pair:

    η(x,t)=2i=1(j2i2ni+n2iN2i2Ni(niNi))(x,t)=(ηe2i=1Qi)(x,t)0, (25)
    q(x,t)=2i=1(j3i2n2i+2niji2Niji)(x,t)=(qe2i=1Pi)(x,t), (26)

    where

    Qi=N2i+2Ni(niNi),Pi=2Niji,

    ηe and qe are the entropy-entropy flux pair defined in (1.7).

    The following theorem is our main result in section 3.

    Theorem 3.1(Large time behavior) Suppose (n1,n2,j1,j2,E)(x,t) be any weak entropy solution of problem (1.1)(1.4) satisfying

    2(2Dˉn1ˉn2)<(n1n2)(x,t)<2(2D+ˉn1+ˉn2), (27)

    for a.e. x[0,1] and t>0. (N1,N2,E)(x) is its stationary solution obtained in Theorem 2.1. If

    10η(x,0)dx<,  10(ni(s,0)Ni(s))ds=0, (28)

    then for any t>0, we have

    10[j21+j22+(EE)2+(n1N1)2+(n2N2)2](x,t)dxC0e˜C0t10η(x,0)dx. (29)

    holds for some positive constant C0 and ˜C0 .

    Proof. We set

    yi(x,t)=x0(ni(s,t)Ni(s))ds,    i=1,2, x[0,1], t>0. (30)

    Clearly, yi(i=1,2) is absolutely continuous in x for a.e. t>0. And

    yix=(niNi),yit=ji,y2y1=EE,yi(0,t)=yi(1,t)=0, (31)

    following (1.1), (2.1), and (2.1). From (1.1)2 and (2.1)2, we get y1 satisfies the equation

    y1tt+(y21tn1)xy1xx+y1t=n1EN1E. (32)

    Multiplying y1 with (32) and integrating over (0,1), we have

    For weak solutions, (1) satisfies in the sense of distribution. We choose test function φn(x,t)C0((0,1)×[0,T)) and let φn(x,t)yi(x,t) as n+ for i=1,2.

    ddt10(y1y1t+12y21) dx10(y21tn1)y1x dx10(n21N21)y1xdx10y21t dx=10(N1(y2y1)y1+Ex2y21)dx. (33)

    In above calculation, we have used the integration by part. Similarly, from (1.1)4 and (2.1)3, we get

    ddt10(y2y2t+12y22) dx10(y22tn2)y2x dx10(n22N22)y2x dx10y22t dx=10(N2(y2y1)y2+Ex2y22) dx. (34)

    Add (33) and (34), we have

    ddt10(y1y1t+12y21+y2y2t+12y22) dx10(n21N21)y1xdx10(n22N22)y2x dx=10((y21tn1)y1x +(y22tn2)y2x) dx+10(y21t+y22t) dx+10(N1(y2y1)y1+Ex2y21N2(y2y1)y2Ex2y22) dx. (35)

    Since

    10(N1(y2y1)y1+Ex2y21N2(y2y1)y2Ex2y22) dx=10n1N1n2+N2D(x)2y21dx+10n2N2n1+N1+D(x)2y22dx10N1+N22(y1y2)2dx, (36)

    then, from (31)1 and (36) we get

    ddt10(y1y1t+12y21+y2y2t+12y22) dx+10(N1+n1)y21x+10(N2+n2)y22xdx+10N1+N22(y1y2)2dx=10((y21tn1)y1x+(y22tn2)y2x) dx+10(y21t+y22t) dx+10(n1N1n2+N2D(x)2y21+n2N2n1+N1+D(x)2y22)dx. (37)

    Moreover, since

    |yi(x)|=|x0yis(s)ds|x12(x0y2isds)12x12(10y2isds)12,x[0,1], (38)

    we can obtain

    yi2L2=10|yi|2dx12yix2L2, (39)

    verifies for i=1,2. If the weak solutions n1(x,t) and n2(x,t) satisfy (27) then

    infx{N1+n1}>supx{n1N1n2+N2D(x)4}, (40)

    and

    infx{N2+n2}>supx{n2N2n1+N1+D(x)4}, (41)

    hold, where we have used the assumption (5) and the estimate (23).

    Following (39), (40) and (41), we have

    10n1N1n2+N2D(x)2y21dx<10(N1+n1)y21xdx, (42)

    and

    10n2N2n1+N1+D(x)2y22dx<10(N2+n2)y22xdx. (43)

    Thus (36), (42), and (43) indicate there is a positive constant β>0, such that

    ddt10(y1y1t+12y21+y2y2t+12y22) dx+β10(y21x+y22x)dx+10N1+N22(y1y2)2dx10((y21tn1)y1x+(y22tn2)y2x) dx+10(y21t+y22t) dx=10(N1y21tn1+N2y22tn2) dx. (44)

    In view of the entropy inequality (6), and the definition of η and q in (25) and (26), the following inequality holds in the sense of distribution.

    ηet+qex+j21n1+j22n2j1E+j2E=ηt+2i=1Qit+qx+2i=1Pix+j21n1+j22n2j1E+j2E=ηt+qx+j21n1+j22n2j1E+j2E+j1Ej2E0. (45)

    Since

    j1E+j2E+j1Ej2E=(EE)(j2j1)=(y2y1)(y2ty1t), (46)

    then (44) turns into

    ηt+qx+y21tn1+y22tn2+(y2y1)(y2ty1t)0. (47)

    We use the theory of divergence-measure fields, then

    ddt10(η+12(y2y1)2)dx+10(y21tn1+y22tn2) dx 0, (48)

    where we use the fact

    10qx dx =0. (49)

    Let λ>2+2n>0. Then, we multiply (48) by λ and add the result to (44) to get

    ddt10(λη+λ2(y2y1)2+y1y1t+12y21+y2y2t+12y22)dx+β10(y21x+y22x)dx+10N1+N22(y1y2)2dx+10((λN1)y21tn1+(λN2)y22tn2)dx0. (50)

    Using the estimate (22) in Theorem 2.1. and the Poincaˊre inequality (39), we have

    {d\over{dt}}\int_0^1 (\lambda \eta^*+{\lambda\over 2}(y_2-y_1)^2 + y_1y_{1t}+\frac12y_1^2+y_2y_{2t}+\frac12y_2^2) dx+{\beta\over 2}\int_0^1(y_{1x}^2+y_{2x}^2)dx\\\\ \;\;\;\; +{\beta\over 2}\int_0^1(y_{1}^2+y_{2}^2)dx+n_*\int_0^1(y_1-y_2)^2dx+\int_0^1\bigg{(} \frac{y_{1t}^2}{n_1} +\frac{y_{2t}^2}{n_2}\bigg{)} dx\leq 0. (51)

    Now, we consider \eta^* in (25). Clearly

    n_i^2-N_i^2-2N_i(n_i-N_i), (52)

    is the quadratic remainder of the Taylor expansion of the function n_i^{2} around N_i>n_*>0 for i = 1, 2. And then, there exist two positive constants C_1 and C_2 such that

    C_1y_{ix}^2 \le n_i^2-N_i^2-2N_i(n_i-N_i) \le C_2y_{ix}^2. (53)

    Making C_3 = \min\{C_1, {{1\over 2}}\} and C_4 = \max\{C_2, {{1\over 2}\}}, then we get

    C_3({{y_{1t}^2}\over {n_1}}+{y_{2t}^2\over {n_2}}+y_{1x}^2+y_{2x}^2) \leq \eta^* \leq C_4({{y_{1t}^2}\over {n_1}}+{y_{2t}^2\over {n_2}}+y_{1x}^2+y_{2x}^2). (54)

    Let

    F(x, t) = \lambda \eta^*+{\lambda\over 2}(y_2-y_1)^2 + y_1y_{1t}+\frac12y_1^2+y_2y_{2t}+\frac12y_2^2,

    then there exist positive constants C_5, C_6, and C_7, depending on \lambda, n_*, \beta, such that

    \int_0^1F(x, t)dx = \int_0^1[\lambda \eta^*+{\lambda\over 2}(y_2-y_1)^2 + y_1y_{1t}+\frac12y_1^2+y_2y_{2t}+\frac12y_2^2]dx \\\\ \leq C_5\int_0^1[({y_{1t}^2\over {n_1}} +{y_{2t}^2\over {n_2}})+ n_*(y_2-y_1)^2+ {\beta\over 2}(y_{1x}^2 +y_{2x}^2)~ + {\beta\over 2}(y_{1}^2 +y_{2}^2)]dx\\\\ \le C_6 \int_0^1\eta^*dx, (55)

    and

    0<C_7\int_0^1[({y_{1t}^2\over {n_1}} +{y_{2t}^2\over {n_2}})+ n_*(y_2-y_1)^2+ {\beta\over 2}(y_{1x}^2 +y_{2x}^2)~ + {\beta\over 2}(y_{1}^2 +y_{2}^2)]dx\\\\ \leq \int_0^1[\lambda \eta^*+{\lambda\over 2}(y_2-y_1)^2 + y_1y_{1t}+\frac12y_1^2+y_2y_{2t}+\frac12y_2^2]dx = \int_0^1F(x, t)dx. (56)

    Then

    {d\over{dt}}\int_0^1 F(x, t) ~dx + {1\over {C_5}}\int_0^1 F(x, t)dx \leq 0, (57)

    and

    \int_0^1[({y_{1t}^2\over {n_1}} +{y_{2t}^2\over {n_2}})+ n_*(y_2-y_1)^2+ {\beta\over 2}(y_{1x}^2 +y_{2x}^2)~ + {\beta\over 2}(y_{1}^2 +y_{2}^2)]dx\\ \le{1\over {C_7}}\int_0^1F(x, t)dx \le {1\over {C_7}}e^{-{{t}\over {C_5}}}\int_0^1F(x, 0)dx\\ \le C_8e^{-{t\over {C_5}}}\int_0^1\eta^*(x, 0)dx. (58)

    are given, following the Growall inequality and the estimates (55) and (56). Up to now, we finish the proof of Theorem 3.1.


    Acknowledgments

    In the process of the selected topic and write a paper, I get the guidance from my tutor: Huimin Yu. In the teaching process, my tutor helps me develop thinking carefully. The spirit of meticulous and the rigorous attitude of my tutor gives me a lot of help. Gratitude to my tutor is unable to express in words. And this paper supported in part by Shandong Provincial Natural Science Foundation (Grant No. ZR2015AM001).


    Conflict of interest

    The author declare no conflicts of interest in this paper.




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