Models of Cancer Growth Revisited

In the present paper we study models of cancer growth, initiated in Jens Chr. Larsen: Models of cancer growth [1]. We consider a cancer model in variables C cancer cells, growth factors , 1, , i GF i p =  , (oncogene, tumor suppressor gene or carcinogen) and growth inhibitor , 1, , i GI i q =  (cells of the immune system or chemo or immune therapy). For 1 q = this says, that cancer grows if (1) below holds and is eliminated if the reverse inequality holds. We shall prove formulas analogous to (1) below for arbitrary , , p q p q ∈ ≥  . In the present paper, we propose to apply personalized treatment using the simple model presented in the introduction.


Introduction
Cancer grows if 0 g = and  (1) and is eliminated if the reverse inequality holds. Here  [1]. So if you have many (few) growth inhibitors compared to growth factors, cancer is eliminated (cancer grows).
In [1] we proved, that Formula (1) when 1 p = implied that cancer grows and is eliminated if the reverse inequality holds. In the present paper we prove, that cancer grows if 0, g p q = ≥ and Here ( ) (  )   T  T  3 , , , , , C F I g g g g y C GF GI = = ∈  , where T denotes a transpose. If you fit my model to measurements, you will get some information about the particular cancer. γ ∈  is the cancer agressiveness parameter. If this parameter is high cancer initially proliferates rapidly. α + ∈  is the carcinogen severity. β − ∈  is the fitness of the immune system, its response to cancer. , F I µ µ − ∈  are decay rates. g is a vector of birth rates. δ ∈  gives the growth factor response to cancer and σ ∈  gives the growth inhibitor response to cancer. So fitting my model may have prognostic and diagnostic value. If we have a toxicology constraint for chemo therapy or immune therapy with a suitable safety margin GI P + ≤ ∈ (4) then we can keep the system at the toxicology limit by requiring ( ) 1 I I P C P g σ µ = + + + (5) which is equivalent to If , 0 I σ µ < , then we can give chemo therapy at this rate. Then we get the induced system 2 2 : We shall prove that this treatment benefits the patient in section 2. To get the system to the toxicology limit P assume that we have ] [ , 0,1 GI P η η = ∈ (9) Then looking at the third coordinate of T we see that we shall require ( ) 1 I I P C P g σ µ η = + + + (10) which implies that k GF k C GI aC k H c c k k GF k k C GI k k GI (12) ( ) 3 , , c C GF GI = ∈  . Iterating this map will give an approximation to the flow.
Then 64 k is the rate at which you give chemo therapy. If we have the constraint GI P ≤ (13) then looking at the third coordinate of H we see that to keep the system at the toxicology limit with a suitable safety margin, we must have ( ) 43 64 46 P P k C P k k P = + − ⋅ + −  (14) Solving for 64 k we get Since the ij k are positive we can give the chemo therapy at this rate. To get this system to the toxicology limit we shall require I felt I had to suggest this. If you want to try this you may want to do it stepwise.
In Figure 1, I have plotted a fit of T to three Gompertz functions From a paper from 1964 [2] we know that solid tumors grow like Gompertz functions. That is the cancer burden is approximately a Gompertz function. Define the error functions The result is I have also fitted S to two Gompertz functions In Maple, there is a command QPSolve that minimizes a quadratic error function with constraints on the signs of the parameters estimated. There are several important monographs relevant to the present paper, see [3]- [8]. There are several

The Routh Hurwitz Criterion for Maps
We shall derive a well known criterion for stability of a fixed point of a map. To this end define the Möbius transformation which maps the left hand plane −  to the interior  of the unit disc. This is because when z x iy = + lies in the interior of  . Also This shows, that g is a bijective map with inverse denote the characteristic polynomial of the two by two matrix A in (7). Note, If this polynomial is a Routh Hurwitz polynomial, i. e. the roots lie in −  , then the roots of But this implies, by adding these two inequalities, that However, then ( ) γ µ γµ αδ γ µ γµ αδ by the Routh Hurwitz criterion. We shall find the fixed points of S, with 0 P = .
From the second coordinate find But the denominator is positive if fix c is stable, by the above. Hence the treatment benefits the patient, because we assume that 0 F µ < , and (88) and we are lowering C g by 0 P β < .

Now suppose
The assumption 1 We claim that fix c is stable, when 1 b < . But then 1 are the distinct eigenvalues of A. So there is a change of basis matrix D such that have unique fixed points fix c  and fix c and we clearly have We need the following definition.
But now stability follows from the estimate and this implies that fix c is stable, because S and T  are conjugate: as n → +∞ . So fix c  is unstable and since T  and S are conjugate, fix c is unstable.

Models of Cancer Growth
Consider the mapping ( ) The matrix here is denoted A. T maps to itself. g is a vector of birth rates and 1 1 , , , , , after the last column to obtain, assuming the formula Then the characteristic polynomial of A is So the eigenvalues are 1 µ is a factor of ( ) For the moment assume 0, p q ∆ > ≥ . Define the matrix of eigenvectors of A by We shall find formulas for the complements Proof. Suppose 2, , . We are deleting row r. So in column which is what we wanted to prove.
Decompose after rows 1, , p  , to get ( ) ( ) ( ) which gives The proposition follows, because Proof. We have Initially let 1 q = . Now The proposition follows. The aim of our computations is to show that there exists an affine vector field X on Then we shall find a formula for Now define when , ,1 0 where we denote the last vector ( ) d t . This is readily shown by differentiating We shall require ( ) because then the time one map of Y is ( ) Then the flows of X and Y are related by But then the time one map of X is which is what we wanted.
Proof. We use the formula We have So this gives the first term in Now we have ( ) Hence the 2 r ≥ contribution is from (184) and the 1 r = contribution is The theorem follows. Now suppose that 0 ∆ < . Then a ib λ ± = ± (210) 0 b ≠ . We shall require that 0 a > . Now define the matrix The first column is denoted Here ( ) 1,0, ,0 , 0,1,0, ,0 e e = =   , both in Now as in [1] we get ( ) ( ) because we assume that det 0 U ≠ . Exactly as before we get and for 2, , 1 r p p q = + + +  ( ) Proof. The proposition follows immediately from proposition 2 and 3.
The flow of We want to have that this equals for 1 t = , the matrix Remember the formula Define when 1 0 µ The flows are related by But then the time one map of X is which is what we intended to find.   We can also compute the first term in ( ) omitting the factor 1 detU The 12 U contribution is omitting the factor ( ) the forward reaction rate is denoted sr k and the reverse reaction rate is denoted rs k . The differential equations are