Cancer grows if g = 0 and

α 1 G F 1 0 G I 1 0 + ⋯ + α p G F p 0 G I 1 0 > − β (1)

and is eliminated if the reverse inequality holds. Here α i ∈ ℝ + , β ∈ ℝ − and G F i 0 , G I 1 0 are initial conditions in C = 0 , see section three for definitions and also [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 p = 1 implied that cancer grows and is eliminated if the reverse inequality holds. In the present paper we prove, that cancer grows if g = 0 , p ≥ q and

∑ i = 1 p     α i G F i 0 + ∑ j = 1 q     β j G I j 0 > 0 (2)

and is eliminated if the reverse inequality holds. In [1] we also considered a mass action kinetic system with vector field f like the one in Section 4 with p = q = 1 and proved, that there is a relationship between such a model and the model T of Section 3. Namely if you linearize f at a singular point and then discretize the flow then you get a mapping T of Section 3. See section 4 for details.

Consider now the cancer model from [1]

T ( y ) = ( 1 + γ α β δ ( 1 + μ F ) 0 σ 0 ( 1 + μ I ) ) ( C G F G I ) + g (3)

Here g = ( g C , g F , g I ) T , y = ( C , G F , G I ) T ∈ ℝ 3 , where T denotes a transpose. If you fit my model to measurements, you will get some information about the particular cancer. γ ∈ ℝ is the cancer agressiveness parameter. If this parameter is high cancer initially proliferates rapidly. α ∈ ℝ + is the carcinogen severity. β ∈ ℝ − is the fitness of the immune system, its response to cancer. μ F , μ I ∈ ℝ − are decay rates. g is a vector of birth rates. δ ∈ ℝ gives the growth factor response to cancer and σ ∈ ℝ 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

G I ≤ P ∈ ℝ + (4)

then we can keep the system at the toxicology limit by requiring

P = σ C + ( 1 + μ I ) P + g I (5)

which is equivalent to

g I = − σ C − μ I P (6)

If σ , μ I < 0 , then we can give chemo therapy at this rate. Then we get the induced system

S : ℝ 2 → ℝ 2 (7)

( C , G F ) ↦ ( 1 + γ α δ 1 + μ F ) ( C G F ) + ( g C + P β g F ) (8)

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

G I = η P ,       η ∈ ] 0 , 1 [ (9)

Then looking at the third coordinate of T we see that we shall require

P = σ C + ( 1 + μ I ) η P + g I (10)

which implies that

g I = P − ( 1 + μ I ) η P − σ C       = ( 1 − η − η μ I ) P − σ C (11)

We can also fit the ODE model of section 4 with p = q = 1 , by defining the Euler map

H ( c ) = c + ϵ ( k 21 G F − k 43 C ⋅ G I + a C + k 24 − ( k 21 + k 41 ) G F + k 14 − k 43 C ⋅ G I + k 64 − k 46 G I ) (12)

c = ( C , G F , G I ) ∈ ℝ 3 . Iterating this map will give an approximation to the flow. Then k 64 is the rate at which you give chemo therapy. If we have the constraint

G I ≤ 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

P = P + ϵ ( − k 43 C ⋅ P + k 64 − k 46 P ) (14)

Solving for k 64 we get

k 64 = k 43 C ⋅ P + k 46 P (15)

Since the k i j are positive we can give the chemo therapy at this rate. To get this system to the toxicology limit we shall require

P = H 3 ( C , G F , η P ) = η P + ϵ ( − k 43 C η P + k 64 − k 46 η P ) (16)

which means, that

k 64 = P ( 1 − η ) 1 ϵ + k 43 C η P + k 46 η P (17)

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

C ( t ) = exp ( 0.5 ( 1 − exp ( − 0.5 t ) ) ) (18)

G F ( t ) = exp ( 0.3 ( 1 − exp ( − 0.3 t ) ) ) (19)

G I ( t ) = exp ( 0.4 ( 1 − exp ( − 0.4 t ) ) ) (20)

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

E 1 = ∑ i = 1 n ( C i + 1 − ( ( 1 + γ ) C i + α G F i + β G I i + g C ) ) 2 (21)

E 2 = ∑ i = 1 n ( G F i + 1 − ( δ C i + ( 1 + μ F ) G F i + g F ) ) 2 (22)

E 3 = ∑ i = 1 n ( G I i + 1 − ( σ C i + ( 1 + μ I ) G I i + g I ) ) 2 (23)

where C i , G F i , G I i , i = 1 , ⋯ , n + 1 are measurements of C , G F , G I at equidistant time points t i = ϵ i , i = 1 , ⋯ , n + 1 , ϵ > 0 . We set C i = C ( t i ) , G F i = G F ( t i ) , G I i = G I ( t i ) Then solve the equations

∂ E 1 ∂ γ = 0 (24)

∂ E 1 ∂ α = 0 (25)

∂ E 1 ∂ β = 0 (26)

∂ E 1 ∂ g C = 0 (27)

in unknowns γ , α , β , g C and

∂ E 2 ∂ δ = 0 (28)

∂ E 2 ∂ μ F = 0 (29)

∂ E 2 ∂ g F = 0 (30)

in unknowns δ , μ F , g F and

∂ E 3 ∂ σ = 0 (31)

∂ E 3 ∂ μ I = 0 (32)

∂ E 3 ∂ g I = 0 (33)

in unknowns σ , μ I , g I . For instance

∂ E 1 ∂ γ = 0 (34)

gives

( 1 + γ ) ∑ i = 1 n     C i 2 + α ∑ i = 1 n     C i ⋅ G F i + ∑ i = 1 n     C i ⋅ G I i + g C ∑ i = 1 n     C i = ∑ i = 1 n     C i + 1 C i (35)

The result is

γ = − 0.1344 (36)

α = 0.1656 (37)

β = − 0.4023 (38)

g C = 0.598 (39)

δ = 0.017 (40)

σ = 0.06485 (41)

μ F = − 0.2664 (42)

μ I = − 0.3815 (43)

g F = 0.3312 (44)

g I = 0.4622 (45)

I have also fitted S to two Gompertz functions

C ( t ) = exp ( 0.5 ( 1 − exp ( − 0.5 t ) ) ) (46)

G F ( t ) = exp ( 0.3 ( 1 − exp ( − 0.3 t ) ) ) (47)

See Figure 2 and Figure 3. Define error functions

and measurements. Solve

in unknowns and

in unknowns. The result is

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 publications by the author impacting on the present paper, see [9] - [15] .

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

implies

, and

implies

Define

Then

when lies in the interior of. Also

This shows, that g is a bijective map with inverse. Let

denote the characteristic polynomial of the two by two matrix A in (7). Note, that if and and then and. Here

Define

So if then we have the polynomial

If this polynomial is a Routh Hurwitz polynomial, i. e. the roots lie in, then the roots of

lie in the interior of the unit circle. Now compute

and

Also

If

and the fixed point of S is stable, then by the Routh Hurwitz criterion

But this implies, by adding these two inequalities, that

However, then

A contradiction to (80). So if is stable, then. Assume now that. If is stable, then

by the Routh Hurwitz criterion. We shall find the fixed points of S, with. From the second coordinate find

Then the first coordinate gives

But the denominator is positive if is stable, by the above. Hence the treatment benefits the patient, because we assume that, and

and we are lowering by.

Now suppose

The assumption implies that -1 is a root of

Since, we have

We claim that is stable, when. But then and are the distinct eigenvalues of A. So there is a change of basis matrix D such that

Clearly both

and

have unique fixed points and and we clearly have

We need the following definition.

Definition. A fixed point of is stable if given an open neighbourhood of there exists an open neighbourhood of such that for all

for all. A fixed point is unstable if it is not stable.

Now observe, that

Notice that

in the max norm

because

But now stability follows from the estimate

and this implies that is stable, because S and are conjugate:

If, then we get the estimate when

as. So is unstable and since and S are conjugate, is unstable.

Consider the mapping

where

and

The matrix here is denoted A. T maps to itself. g is a vector of birth rates and. The. Finally . Also put

C is cancer are growth factors and are growth inhibitors,.

Proposition 1 The characteristic polynomial of A is

Proof. With,

Decompose after the last column to obtain, assuming the formula for holds

Suppose henceforth, that

Then the characteristic polynomial of A is

So the eigenvalues are and since

is a factor of, then

are eigenvalues of A, where

For the moment assume. Define the matrix of eigenvectors of A by

We shall find formulas for the complements

of D and the determinant of D,.

Proposition 2 For we have

For

Proof. Suppose. We are deleting row r. So in column there is only one nonzero element. Decomposing after this column and then after row one we get a matrix with zeroes under the diagonal. The signs here are

is the sign on the complement and is the sign on the complement to. It is in row and column and we delete two rows. is the sign on. Hence

which is what we wanted to prove.

Now suppose that. For we get a matrix with zeroes under the diagonal, so

Now consider the case. Write. After decomposing after rows and row one column two in we are left with

Decompose after rows, to get

which gives

The proposition follows, because

Proposition 3

Proof. We have

Initially let. Now

when and

when. Now decompose after the last column to get

If

(in the statement of the proposition) we get

Now we shall use induction over q to prove the formula in the statement of the proposition. Decompose after the last row

In B we have decomposed after and then after the rows and in the remaining matrix decomposed after row p and column. The signs here are

The proposition follows.

The aim of our computations is to show that there exists an affine vector field X on such that the time one map is

Let denote an integral curve of X through. Then we shall find a formula for

First notice that

if

and

which we assume. We have used that

So the eigenvectors in D are linearly independent, hence

Now define when

where. The flow of Y is

where we denote the last vector. This is readily shown by differentiating with respect to t

Now we get

It follows that is the flow of Y. Now require

that is

We shall require

because then the time one map of Y is

Now define

Then the flows of X and Y are related by

But then the time one map of X is

which is what we wanted.

Theorem 4 Assume, that

and

We have the formula

Proof. We use the formula

We have

and then

for and for

We shall write

and then we have

When then

while for we have

Notice that

Continuing from (184)

So this gives the first term in

Note that

Now we have

Hence the contribution is from (184)

and the contribution is

So

The theorem follows.

Now suppose that. Then

. We shall require that. Now define the matrix

The first column is denoted, the second is denoted. Here, both in. Notice that

Now as in [1] we get

and similarly

So

because we assume that. Exactly as before we get

Proposition 5 For

and for

Also

Finally

since

Proof. The proposition follows immediately from proposition 2 and 3.

The flow of

is, for

We want to have that this equals for, the matrix

Thus

Remember the formula

Define when.

where. The flow of Y is

where the last vector is denoted. To see this compute

Now we also get

So is the flow of Y. We need to have

because then the time one map of Y is

Then define

The flows are related by

But then the time one map of X is

which is what we intended to find.

Theorem 6 When then

Proof. We have the following computation

And we want to have

that is

But we have arranged that

so we get

Denote the two by two matrix in the last line

We can also compute the first term in

omitting the factor

hence the first term in

Now

The contribution is omitting the factor

The contribution is, omitting the factor

where

The theorem follows.

Now assume that. Define

Proposition 7 For q odd and

and for and q odd

For q even and

and for and q even

Also

when q is odd and

when q is even.

Proof. (277) q odd. We are deleting the row r with. Decompose after that column with in it and row one column two. The sign on is

The first sign here is the sign when decomposing after row, except the row with. The second sign is the sign on the complement to. The third sign is the sign on. The last sign is the sign on in column r and row one. (277) follows. Now let. Write. Decomposing after rows to give

We have the sign

on. And we have the sign

on column and row one. Hence the formula. is obvious. And the formulas for q even follow similarly. (281) q odd. First let and. Then

For we get

We have, decomposing after the last column

Here

For q even we get

Now decompose after the last column

Now we get decomposing after the last column

where

In the determinant B, we have decomposed after row 2 to. In the remaining determinant decompose after row one and column. The proposition follows.

Define for and

From Proposition 7, we get

Proposition 8 For q odd and

and for q odd and

For q even and

and for q even and

Also for q odd and

and for q odd and

For q even and

and for q even and

Finally for q odd

and

for q even.

In [1] we also considered a three dimensional ODE model of cancer growth in the variables cancer, growth factors and growth inhibitors, respectively. Analogous to what we did in section three define a mass action kinetic system

Here the complexes are, , , , , This defines the rate constants. For a reaction

the forward reaction rate is denoted and the reverse reaction rate is denoted. The differential equations are

We shall find a polynomial giving candidates of singular points of this vector field.

From, we find

where. From, we find

for. Inserted into we get

We can then multiply with

and define the constants

to obtain the polynomial of degree

if we assume that. There is a relation between the ODE model of this chapter, with vector field

and the discrete dynamical system of section three, see also [1] . Linearize the vector field at a singular point and set

Also define the Euler map

for. This is an approximation to the flow of h. If we let

and

then you obtain a discrete model T of section three.

Example Let and define the rate constants and,. Then there are two positive singular points.

In this paper, we considered a discrete mathematical model and an ODE model of cancer growth in the variables cancer, growth factors and growth inhibitors, respectively. We have shown that this model is a threshold model. If and

then cancer grows, and if the reverse inequality holds, cancer is eliminated. We also proposed personalized treatment using the simple model of cancer growth in the introduction and the ODE model of section four.

Larsen, J.C. (2018) Models of Cancer Growth Revisited. Applied Mathematics, 9, 418-447. https://doi.org/10.4236/am.2018.94031