Mathematical model of hormone sensitive prostate cancer treatment using leuprolide: A small step towards personalization

In this paper we present a new version of a mathematical model of Elishmereni et al. describing androgen deprivation therapy (ADT) for hormone sensitive prostate cancer patients (HSPC). We first focus on the detail description of the model, and then we present mathematical analysis of the proposed model, starting from the simplified model without resistance and ending on the full model with two resistance mechanisms present. We make a step towards personalization proposing an underlying tumor growth law base on a cohort of patients from Mayo hospital. We conclude that the model is able to reflect reality, that is in clinical scenarios the level of testosterone in HSPC patients inevitably rises leading to the failure of ADT.

Using the method of mathematical induction we easily show that L j (t) = α k j 1 t j j! e −k1t , j = 1, . . . n.
Clearly, if L j (t) is described by (22) then which proves that Eq. (22) is valid for all j ≤ n, as we have shown that it is valid for j = 1 previously. Recall, that we assumed k 3 = 0, which means that we obtain the following system of linear non-homogeneous equations of the forṁ L c = α k n+1 1 t n n! e −k1t +αk 2 e −k1t +QL p − (Q + d L )L c , Note that this system could be solved, but as we are not able to recognize the flow between the central and peripheral compartments, we skip this division and consider only one compartment, which yieldṡ where L is the amount of the drug in the organizm, and solving this equation we obtain The formula for L we calculated above is rather complex, so we decided to simplify it even more, assuming k 1 << d L and approximating B Appendix. In this Appendix we present some results related to the analysis of System (8) and simplified System (10). Let us focus on Eqs (8). Clearly, for any smooth decreasing positive function h 1 there exists unique solution of Eqs (8) which is positive for positive initial data. Below we show that solutions are bounded and there exists invariant subset of (R + ) 3 for this system.
As h 1 is positive decreasing, we have 0 < h 1 (z) ≤ h 1 (0). This implieṡ Using this inequality we obtaiṅ Having all the variables bounded from above we are in a position to show boundedness from below:ẋ and z m = p3ym d3 . We obtain the following conclusion.
Now, we present the proof of Proposition 1.
Proof. Looking for stability of SS, we calculate a Jacobian matrix of System 8 obtaining the following characteristic equation According to the Routh-Hurwitz Criterion, stability is guaranteed by the inequality For h 1 of the form (9), we have where d = d 1 d 2 d 3 and p = p 1 p 2 p 3 . Hence, Ineq (23) is equivalent to Note that (24) and proves stability of SS.
On the other hand, if P is large enough then the change of stability is possible. It is obvious that keeping all the parameters d i on the fixed level and increasing the value of P p 2 p 3 we are able to violate (23). One of the possible options is to choose h 1 with a Hill coefficient γ greater than 1. Then for some threshold value γ th of the bifurcation coefficient γ we obtain a pair of purely imaginary eigenvalues λ = ±iω and crossing this threshold γ th we observe a change from stable to oscillatory dynamics via a Hopf bifurcation.
Properties of simplified System (10) are similar to those of System (8). Using the same line of reasoning we are able to find an invariant subset of (R + ) 2 .
Corollary 11. System (10) is positively invariant with respect to the set d3 . Now, we present the proof Proposition 3 for Eqs (10).
Proof. It is easy to check that we have one SS (x,z) satisfyinḡ This state is locally stable, because its Jacobian matrix has negative trace −(d 1 + d 3 ) and positive determinant d 1 d 3 + p 3 |h 1 (z)|. Global stability could be easily proved using the Dulac-Bendixson Criterion. Clearly, it is enough to take a divergence of the right-hand side of (10): This shows that there is no periodic orbits of (10). All solutions are bounded and there is only one steady state which is locally stable, so it must be globally stable.
C Appendix. Now, we focus on the properties of System (12) for constant drug amount L(t) ≡ L ≥ 0, comparing its dynamics for L = 0 with the dynamics of System (10).
Proposition 12. System (12) is positively invariant with respect to the set Proof. The proof is exactly the same as for System (8).
Note that for L = 0 and the functions h 1 , h 3 described by Eqs (9) and (11), respectively, the values x m0 , z m0 , z M 0 tends with b 3 → 0 to the respective values for System (10). Moreover, it is asy to see that z M 0 < z M , x m0 > x m , while the relation between z m and z m0 depends on the model parameters. Now, we prove Proposition 4.
Proof. Any steady state SS (x L ,z L ) of (12) satisfies and we see that the equation forz L has one positive solution as its right-hand side is increasing from 0 to ∞, while the left-hand side is positive decreasing. Global stability could be shown exactly in the same way as for System (10), using the divergence of the system in the Dulac-Bendixson Criterion.
Note that for the functions (9) (14). Because this properties do not depend on the subscript i, we omit it to shorten the notation and consider r˙= βL 1 − r l .
We are interested in the dynamics of solutions of Eq (25) mainly for constant amount of the drug. However, after a single dose of the drug the amount of it in the organism decreases, such that we can assume L ≤ 0.
Proposition 13. For any continuous non-increasing positive function L there exists unique solution of Eq (25) which lies in the interval (0, l] for initial data r(0) ∈ [0, l].
Proof. Existence of unique solution is obvious due to the properties of the right-hand side of Eq (25). Note that if r(0) = 0 then r˙(0) = βL and this value is positive after the drug application. This means that the solution starts to increase and it increases until r(t) < l. Assume that there exists first time-point t > 0 such that r(t) = l and the solution exceeds l at this point, which means that r˙(t) > 0 or r˙(t) = 0 and t is an inflection point. Clearly, we have meaning that t is not a point of inflection and the solution is not able to exceed the threshold l. It is also obvious that for r(0) ≥ 0 we have r(t) > 0.
Note that for any continuous function L we are able to solve Eq (25). In fact, it is enough to take integrable function L. We calculate Hence, for constant amount of the drug, L(t) ≡ L we obtain r(t) = l 1 − e − βL l t , while for any non-increasing function L, r(t) ≤ l 1 − e − βL (0) l