Abstract
In this paper, we consider a time-delayed free boundary problem with time dependent Robin boundary conditions. The special case where is a mathematical model for the growth of a solid nonnecrotic tumor with angiogenesis. In the problem, both the angiogenesis and the time delay are taken into consideration. Tumor cell division takes a certain length of time, thus we assume that the proliferation process leg behind as compared to the process of apoptosis. The angiogenesis is reflected as the time dependent Robin boundary condition in the model. Global existence and uniqueness of the nonnegative solution of the problem is proved. When is sufficiently small, the stability of the steady state solution is studied, where is the ratio of the time scale of diffusion to the tumor doubling time scale. Under some conditions, the results show that the magnitude of the delay does not affect the final dynamic behavior of the solutions. An application of our results to a mathematical model for tumor growth of angiogenesis is given and some numerical simulations are also given.
1. Introduction
In the past a few decades, there are a lot of focus on mathematical models with regard to tumor growth for biological and mathematical interests. Many researchers developed various mathematical models from different aspects to detail the process of tumor growth (see, e.g., [1–8]). The tumor growth process can be classified into two different stages: the stage without a necrotic core (see, e.g., [2, 9–13]) and the stage with a necrotic core (see, e.g., [3, 14–16]). Almost all mathematical models are established by using reaction-diffusion dynamics and mass conservation law for the processes of proliferation and apoptosis.
This paper focus on a time-delayed free boundary problem with the time dependent Robin boundary condition. The model is as follows:
where and are two unknown functions. is a positive constant. and are given functions, and
The special case where is a mathematical model describing the growth of a nonnecrotic tumor with angiogenesis. In particular, when , the biological meaning is as follows: is the nutrient concentration at time and radius . represents the outer radius of tumor at time . represents the ratio between time scale of the diffusion and time scale of the tumor doubling, and is a constant represents the time delay in the process of proliferation, i.e., is the average time required from the beginning of cell division to the completion of division. In order to obtain nutrients, tumors attract blood vessels at a rate proportional to , so that holds on the boundary, where is the nutrients concentration outside the tumor. It should be pointed out that the boundary condition (3) is a time dependent Robin boundary condition since the boundary changes with time. Equation (4) describes the changes of the volume of the tumor. Equations (3), (2), (5), and (6) are boundary and initial conditions. , and are given functions. represents the nutrient consumption rate. It is assumed that the rate of nutrient consumption by tumor cells is an increasing function of nutrient concentration. represents the proliferation rate of tumor cells and represents the apoptosis rate of tumor cells. It is reasonable to assume that the rate of tumor cell proliferation is an increasing function of nutrient concentration and the rate of tumor cell apoptosis is a nonincreasing function of nutrient concentration.
The motivation for studying this model is as follows: Experiments have shown that changes in the proliferation rate modify apoptotic cell loss which does not occur immediately– there exists a time delay for this modification (see [1]), i.e., the proliferation process lags behind as compared to the process of apoptosis. As a result of this research, many researchers have grown interest in the study of mathematical models for tumor growth with time delays (see, e.g., [6, 11, 17–19] and their references). The idea of considering the time delay in the process of proliferation is motivated by the work of Byrne [1], Cui and Xu [11], Foryś and Bodnar [18] and Xu et al. [20] where either linear or constant functions , and are considered in the above mentioned papers. The motivation of considering the nonlinear functions , and is from the work of Cui [10] (where and , i.e., the time delay is not considered). In this paper, we study a more general case, which not only considers both time-delay and nonlinear functions , and , but also takes as any positive integer greater than or equal to 3. The main aim of this paper is to study the time-delayed problem (1)–(6) for Robin boundary conditions and general nonlinear functions , and .
It also should be pointed out that only Dirichlet boundary conditions are considered in [1, 10–12, 20]. In the recent work of Friedman and Lam [21], the authors studied the special case of the problem (1)–(6) where , the functions and are linear and is a constant (but where is a given function of ). The special cases of the model have been extensively studied by many researchers, such as for linear functions
and , where , and are positive constants, Xu et al. [20] have studied the model with Gibbs–Thomson relation, which appears as the Dirichlet boundary condition. In [20], by rigorous mathematical derivation and using theories of functional differential equations, the authors studied the asymptotic behavior of steady state solutions.
Throughout this paper, we suppose that the functions , and satisfy the following conditions: (P1) (P2) for all and (P3) , for all and there exists such that (P4) .
Moreover, we suppose the initial value functions and satisfy the following conditions: for and . and .
The paper is arranged as follows: Section 2 provides proof for the existence and uniqueness of a global solution to problem (1)–(6). Section 3 is devoted to studing asymptotic behavior of the solutions to problem (1)–(6). In the final section, an application of our results to a mathematical model for tumor growth of angiogenesis is given and some numerical simulations are also given.
2. Global Existence and Uniqueness
Lemma 1. Let () be a solution to the problem (1)–(6). The following priori estimates are valid.where and where .
Proof. Obviously, and are upper and lower solutions to the problem (1)–(3), by the maximum principle, we immediately have
From Equation (4), we have
It follows that and
where . Moreover, from the the inequality on the left-hand side of (12), we can get
It infer that when ,
For , by the fact that , one can get
where . Noticing the inequality on the right-hand side of (12) and , we have
where The inequality (11) follows from (10). This completes the proof.
Theorem 1. Suppose the conditions , , and are satisfied. Suppose further that the functions and satisfy the conditions –. Then, there exists a unique solution to (1)–(6) for all .
Proof. By setting , one can change to . Let
Then
Let which will be given later. Consider the problem (1)–(6), by (19), it is equivalent to the following problem:
We define the following metric space : The set consists of vector functions satisfying
(I) , for and
where and
(II) and
Define a metric by
It is obvious that is a complete metric space.
Next, create a mapping as follows. For any , consider the following initial value problem:
Then, one can get
where
By the facts that and one can get that
where we use the monotonicity of the functions and It follows that
Thus, satisfies the condition (I). Taking similar arguments as that in [24], it is not hard to prove is a contractive mapping for is sufficiently small. By the Banach fixed point theorem, we have the local existence and uniqueness of a solution to the problem (1)–(6). To prove global existence and uniqueness, we only need to prove that it is impossible for the local solution to blow up or tend to zero in a finite time. This follows from the priori estimates (see Lemma 1). The proof of Theorem 1 is complete.
3. Asymptotic Stability of Steady State
First, we study the existence of a unique steady state solution of (1)–(6). If is a steady state solution to (1)–(6), it must satisfy the following equations:
Consider the auxiliary boundary problem
where and
Lemma 2 (see Lemma 2.1 [22]). Suppose that the conditions (P1)–(P4) are satisfied. For any the problem (40) and (41) has a unique solution and the following assertions hold:(1)For all and , , . For all and , .(2)For all and , where (3)For any fixed , the function for .(4)For all , , and .
Lemma 3. Assume the conditions (P1)–(P4) are satisfied. LetThen.(1)If , there exists a unique steady state solution to problem (1)–(6), where is a unique solution of and . Moreover, for ; for .(2)If , the problem (1)–(6) has none steady state solution.
Proof. For given , the function satisfies the equations (36)–(38). Substituting it into (39) and letting , one can get.Therefore, the problem (36)–(39) has a solution iff the function has a solution Noticing the facts thatandit follows that (1)If , by intermediate value theorem, it can be inferred that the function has a unique solution (2)If , then for all since . Thus, the problem (1)–(6) has none steady state solution. This completes the proof.
Lemma 4 (see Lemma 3.1 in [23]). Suppose that (P1)–(P4) are satisfied. Let be the solutions of the problem (1)–(6) and letwhere is the unique solution to the following problem:where and Suppose further that for some , and ,and for Then there exists a positive constant and independent of , and (but may dependent on and ) such thatfor all and
Let
where Therefore, could be rewritten in the following form:
Lemma 5. Suppose the conditions (P1)–(P4) are satisfied. Suppose further that for . Consider the following two initial value problemsThen there exists a unique solution to problem (54) and the following assertions hold: If , there exists such that if and the problem (54) has a unique steady state solution , where is a unique solution of . Moreover, the steady state solution is globally asymptotic stable, i.e., for any nonnegative continuous initial value function ,
Proof. Let then (54) takes the form: whereandFrom Lemma 2, we know is continuously differentiable on . Since one can get that are continuous. It is apparent that the initial value problem (54) has one unique solution which exists on , since we may rewrite this problem in the following form:and solve it using the method of steps (see, e.g., [24]) on intervals . Since for Thanks to Lemma 1.1 in [25], we obtain that the nonnegativity of the solution to equation (54) for any nonnegative initial value .
The steady state solution of (54) satisfies the equationBy (P3), we know that , then we haveNoticing is strictly monotone decreasing (see the proof of Lemma 3) and when ,Therefore, one can get that there exists such that if and the problem (54) has a unique steady state solution , where is a unique solution of .
Since for , for and for By Lemma 3.2 in [11], we can get (24) hold. This completes the proof.
Lemma 6. Suppose (P1)–(P4) are satisfied and for . Let be the solutions of the problem (1)–(6). If and for some , there exists a constant depending on such thatfor all and , where and are as before.
Proof. By (11), we can get that
for all If (53) is not true for some . Then there exists such that
for and either or .
If , then By (10) in Lemma 1, we obtain
Noticing for by Lemma 4, one can get that there exists positive constants and independent of and (but may dependent on and ) such that
for all and Denote and . By using the differential mean value theorem, we obtain
for , and Therefore,
Then for
where for has been used. Choosing small such that , one can get . Then there exists (sufficiently small), for , there holds which is a contraction to the fact that .
If , by similar analysis, one can also show the contradiction. This completes the proof.
Remark 1. When , . Thus, if for , Lemma 6 above extends Lemma 3.2 in [23] from the case to the case The assumption that could be satisfied for some special cases. For example, in [21], when , and , where are two constants, the existence, uniqueness, and stability of steady state solutions are proved. For the above special case, in the last section, we will prove for and apply our results to prove the existence, uniqueness and stability of steady state solutions when .
Lemma 7. Assume that (P1)–(P4) are satisfied and for . Let be the solutions of the problem (1)–(6). If , assume that there exists such that
for . Then there exists positive constants and independent of and , for any and , where is a given constant, whenfor and for the following estimateshold for and .
Proof. For the convenience of notation expression, in the following of the paper we use to represents various constants independent of and . By Lemma 2 and (85), one can getfor Thenfor Specially,Noticing that for by Lemma 4 we know that there exists positive constant and such thatfor and , where is independent of and .
Sinceby (78), we havefor where we have used the facts that .
Consider the auxiliary initial value problemBy Lemma 5, there exists unique solutions denoted by to problem (81). Moreover, if , there exists such that if and the problem (81) has unique steady state solutions , where is a unique solution of . The steady state solutions are globally asymptotic stable, i.e.,for any nonnegative initial value function .
By the comparison principle (see Lemma 3.1 in [11]), we obtain thatfor all Since is decreasing, and , we can getFor both stationary solutions , using the linearization theorem, one can get that the characteristic equations are equal towhereandSince , and (see Lemma 2(3)) for noticing that for , one can get that which infers that all complex roots of Equations (85) have negative real parts. Then, there exists positive constant , , and such that for any where It follows thatBy Lemma 2(2) and (72), using the differential mean value theorem, we obtainfor Thenfor Specially, for Noting for all by Lemma 4, there exists a positive constant independent and such thatfor arbitrary and SetThen for By the differential mean value theorem and (72), we obtain that for Then by the equation and the inequality (72), we have By (93), we haveThis completes the proof of Lemma 7.
Theorem 2. Suppose that the conditions (P1)–(P4) are satisfied and for . Let be the solution to the problem (1)–(6). If , then for any , if , there exist positive constants and C such that if , we have the following estimates: for all
Proof. First, we prove that there exist positive constants and such that if , (99) holds. Choosing sufficiently small such that , by Lemma 6 we know there exists a positive constant such that.for all and , where and are as before. Thenfor all By Lemma 1 and Equation (2.4), we obtain that for all ,Obviously holds for all Therefore, the conditions of Lemma 7 are satisfied for Then by Lemma 7, one can gethold for all For any given satifying , we define by By induction, we obtainhold for all
Then, determine by using the following formula:and for given , there exists an integer satisfying It follows thatBy similar arguments, one can get for all
Next, we prove when , (109) is also valid. From (48) and (48), we know thatis the unique solution to (1)–(3). Substituting (110) into (4), we havewhere is defined in (52). Noting , by Lemma 3, we have: If , there exists a unique steady state solution to problem (1)–(6), where is a unique solution of and . Moreover, for ; for . Since is strictly monotone increasing in , thanks to Lemma 3 [11], it follows that
By using the linearization method, linearizing the equation (111) at the steady state solution , one can get the characteristic equation of the linearized equationwhereandBy the facts , and (see Lemma 2(3)) for noticing that , one can get that which infer that all complex roots of equation (112) have negative real parts. Therefore, there exits positive constant , and such that such that for any From (115), one can get when is sufficiently large, for . Notice that is bounded, and when is sufficiently large, there is a positive lower bound of for , and notice thatandusing the differential mean value theorem, one can get for all The proof of Theorem 2 is complete.
Theorem 3. Suppose that the conditions (P1)–(P4) are satisfied and for . Let be the solution to the problem (1)–(6). If , then for any and initial value function , ,
Proof. From Lemma 2 (1) and Equation (4), we obtain.where . It follows thatandLet , then (121) is reduced to the following equation:Consider the following auxiliary linear initial value problemSince , by a well known result of functional differential equations, one can get Let . Then is strictly monotone increasing in and for all . By using Lemma 2.1 [11], one can get Then follows from (120) and On account of , we have This completes the proof.
4. An Application
In this section, for the special case of the problem (1)–(6) where , and , we will apply our results to prove the existence, uniqueness and stability of steady state solutions when . In this section we assume , , and .
First, it is obvious that , and satisfy the conditions (P1) and (P2). Since , for and there exists such that the functions , and satisfy the condition (P3). Therefore, by Theorem 2, if the initial value functions and satisfy the conditions –, then, problem (1)–(6) has a unique solution for all .
For any and , by Theorem 3.1 in [21], we know that there exists a unique steady state solution denoted by of (1)–(6) which is determined by
and
where , and
The solution to problem (40) and (41) is
By (49) and a direct computation, one can get that
where . Thus,
Next we prove . From [11], we know for Therefore,
where we used for (see Lemma 2.1 in [21]), it follows that for
Since the condition (P4) and
then all conditions of Theorem 2 are satisfied. By Theorem 2, let be the solution of the system (1)–(6). For any satisfying , there exist corresponding positive constants and such that if such that
for all
Next, using Matlab R2016a, we will do some numerical simulation of the tumor growth model discussed above. First, we take the following parameter values:
The steady state solution is determined by (124). Let
where and are as before. In Figure 1, we plot the curve of (the blue curve). As can be seen from Figure 1, noting the red curve is the curve of , where there is only one positive steady state solution, which can be solved by Matlab R2016a and . Figure 2 shows the dynamic change of tumor radius with parameters taken as (126). From the above analysis, as well as notice
we know all conditions of Theorem 2 are satisfied. As can be seen from Figure 2, whether the initial value is taken or , all the solutions eventually tend to the unique steady state solution . This verifies the results of Theorem 2.
Next, if we take the parameter values as follows:
where
one can get all conditions of Theorem 3 satisfied. As can be seen from Figure 3, whether the initial value is taken or , all the solutions eventually tend to zero, which verifies the results of Theorem 3.
It can be seen from Theorems 2 and 3 that time delay does not affect the final tendency of tumor growth to the steady state or to disappear. In the following, by using the Figures 4–6, we show that the time delay has an effect on the speed of tumor growth towards to the steady state solution or toward extinction. In Figures 4–6, except for the size of time delay, the other parameters take the same value (please refer to captions of Figures 4–6). In Figures 4 and 6, the top curve of three curves corresponds to the larger where , the bottom curve of the three curves corresponds to a smaller where , the remaining curve corresponds In Figure 5, the top curve of three curves corresponds to the smaller where , the bottom curve of the three curves corresponds to a larger where , the remaining curve corresponds From Figures 4–6, we see that when other conditions remain unchanged, the larger the time delay, the slower the tumor tends to the steady state solution or tends to disappear.
Data Availability
No empirical data were used for this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
The authors would like to thank the editor and the referees for their very helpful suggestions on modification of the original manuscript. This work of the first author is supported by NNSF of China (11301474), NSF of Guangdong Province (2018A030313536). This work of the second author is supported by Shanghai Pujiang Program (2019PJC062).