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.
Consider the following Euler-Poisson system for the bipolar hydrodynamical model of semi-conductor devices:
{n1t+j1x=0,j1t+(j21n1+p(n1))x=n1E−j1,n2t+j2x=0,j2t+(j22n2+q(n2))x=−n2E−j2,Ex=n1−n2−D(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+j22n2−j1E+j2E≤0, | (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.
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=N1−N2−D(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=N1−N2−D(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)−C2−D(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)+ˉn1−ˉn2−D(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
ˉn1−ˉn2=∫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)(x−x0)+Φ″(x0)2(x−x0)2+o(x−x0)2, |
the boundary condition (16) indicates Φ″(x0)≤0.
Φ(x0)+ˉn1−ˉn2−D(x0)≤0. |
So we get
Φ(x0)≤D∗+ˉn2−ˉn1. | (18) |
Similarly, if Φ attains its minimum in x1∈[0,1], we obtain
Φ(x1)≥D∗+ˉn2−ˉn1. | (19) |
Moreover, from (12),(14),(15),(18), and (19), we have
D∗+ˉn2+ˉn12≤N1(x)≤D∗+ˉn2+ˉn12,−D∗+ˉn2+ˉn12≤N2(x)≤−D∗+ˉn2+ˉn12, | (20) |
D∗≤(N1−N2)(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]
n∗≤N1(x)≤n∗, n∗≤N2(x)≤n∗, | (22) |
and
D∗≤(N1−N2)(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).
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)=2∑i=1(j2i2ni+n2i−N2i−2Ni(ni−Ni))(x,t)=(ηe−2∑i=1Qi)(x,t)≥0, | (25) |
q∗(x,t)=2∑i=1(j3i2n2i+2niji−2Niji)(x,t)=(qe−2∑i=1Pi)(x,t), | (26) |
where
Qi=N2i+2Ni(ni−Ni),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(2D∗−ˉn1−ˉn2)<(n1−n2)(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+(E−E)2+(n1−N1)2+(n2−N2)2](x,t)dx≤C0e−˜C0t∫10η∗(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=−(ni−Ni),yit=ji,y2−y1=E−E,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)x−y1xx+y1t=n1E−N1E. | (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)∈C∞0((0,1)×[0,T)) and let φn(x,t)→yi(x,t) as n→+∞ for i=1,2.
ddt∫10(y1y1t+12y21) dx−∫10(y21tn1)y1x dx−∫10(n21−N21)y1xdx−∫10y21t dx=∫10(N1(y2−y1)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
ddt∫10(y2y2t+12y22) dx−∫10(y22tn2)y2x dx−∫10(n22−N22)y2x dx−∫10y22t dx=−∫10(N2(y2−y1)y2+Ex2y22) dx. | (34) |
Add (33) and (34), we have
ddt∫10(y1y1t+12y21+y2y2t+12y22) dx−∫10(n21−N21)y1xdx−∫10(n22−N22)y2x dx=∫10((y21tn1)y1x +(y22tn2)y2x) dx+∫10(y21t+y22t) dx+∫10(N1(y2−y1)y1+Ex2y21−N2(y2−y1)y2−Ex2y22) dx. | (35) |
Since
∫10(N1(y2−y1)y1+Ex2y21−N2(y2−y1)y2−Ex2y22) dx=∫10n1−N1−n2+N2−D(x)2y21dx+∫10n2−N2−n1+N1+D(x)2y22dx−∫10N1+N22(y1−y2)2dx, | (36) |
then, from (31)1 and (36) we get
ddt∫10(y1y1t+12y21+y2y2t+12y22) dx+∫10(N1+n1)y21x+∫10(N2+n2)y22xdx+∫10N1+N22(y1−y2)2dx=∫10((y21tn1)y1x+(y22tn2)y2x) dx+∫10(y21t+y22t) dx+∫10(n1−N1−n2+N2−D(x)2y21+n2−N2−n1+N1+D(x)2y22)dx. | (37) |
Moreover, since
|yi(x)|=|∫x0yis(s)ds|≤x12(∫x0y2isds)12≤x12(∫10y2isds)12,x∈[0,1], | (38) |
we can obtain
‖yi‖2L2=∫10|yi|2dx≤12‖yix‖2L2, | (39) |
verifies for i=1,2. If the weak solutions n1(x,t) and n2(x,t) satisfy (27) then
infx{N1+n1}>supx{n1−N1−n2+N2−D(x)4}, | (40) |
and
infx{N2+n2}>supx{n2−N2−n1+N1+D(x)4}, | (41) |
hold, where we have used the assumption (5) and the estimate (23).
Following (39), (40) and (41), we have
∫10n1−N1−n2+N2−D(x)2y21dx<∫10(N1+n1)y21xdx, | (42) |
and
∫10n2−N2−n1+N1+D(x)2y22dx<∫10(N2+n2)y22xdx. | (43) |
Thus (36), (42), and (43) indicate there is a positive constant β>0, such that
ddt∫10(y1y1t+12y21+y2y2t+12y22) dx+β∫10(y21x+y22x)dx+∫10N1+N22(y1−y2)2dx≤∫10((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+j22n2−j1E+j2E=η∗t+2∑i=1Qit+q∗x+2∑i=1Pix+j21n1+j22n2−j1E+j2E=η∗t+q∗x+j21n1+j22n2−j1E+j2E+j1E−j2E≤0. | (45) |
Since
−j1E+j2E+j1E−j2E=(E−E)(j2−j1)=(y2−y1)(y2t−y1t), | (46) |
then (44) turns into
η∗t+q∗x+y21tn1+y22tn2+(y2−y1)(y2t−y1t)≤0. | (47) |
We use the theory of divergence-measure fields, then
ddt∫10(η∗+12(y2−y1)2)dx+∫10(y21tn1+y22tn2) dx ≤0, | (48) |
where we use the fact
∫10q∗x dx =0. | (49) |
Let λ>2+2n∗>0. Then, we multiply (48) by λ and add the result to (44) to get
ddt∫10(λη∗+λ2(y2−y1)2+y1y1t+12y21+y2y2t+12y22)dx+β∫10(y21x+y22x)dx+∫10N1+N22(y1−y2)2dx+∫10((λ−N1)y21tn1+(λ−N2)y22tn2)dx≤0. | (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.
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).
The author declare no conflicts of interest in this paper.
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