_{1}

^{*}

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 GF
_{i} ,
*i*= 1,
,p, (oncogene, tumor suppressor gene or carcinogen) and growth inhibitor
GF
_{i}
,
*i*
= 1,
,
p, (cells of the immune system or chemo or immune therapy). For q =1 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∈N, p ≥ q . In the present paper, we propose to apply personalized treatment using the simple model presented in the introduction.

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 [

In [

∑ 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 [

Consider now the cancer model from [

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

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 [

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

and measurements

in unknowns

in unknowns

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 [

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

implies

implies

Define

Then

when

This shows, that g is a bijective map with inverse

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

Define

So if

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

lie in the interior

and

Also

If

and

But this implies, by adding these two inequalities, that

However, then

A contradiction to (80). So if

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

Then the first coordinate gives

But the denominator is positive if

and we are lowering

Now suppose

The assumption

Since

We claim that

Clearly both

and

have unique fixed points

We need the following definition.

Definition. A fixed point

for all

Now observe, that

Notice that

in the max norm

because

But now stability follows from the estimate

and this implies that

If

as

Consider the mapping

where

and

The matrix here is denoted A. T maps

C is cancer

Proposition 1 The characteristic polynomial

Proof. With

Decompose

Suppose henceforth, that

Then the characteristic polynomial of A is

So the eigenvalues are

is a factor of

are eigenvalues of A, where

For the moment assume

We shall find formulas for the complements

of D and the determinant of D,

Proposition 2 For

For

Proof. Suppose

which is what we wanted to prove.

Now suppose that

Now consider the case

Decompose after rows

which gives

The proposition follows, because

Proposition 3

Proof. We have

Initially let

when

when

If

(

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

The proposition follows.

The aim of our computations is to show that there exists an affine vector field X on

Let

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

where we denote the last vector

Now we get

It follows that

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

We shall write

and then we have

When

while for

Notice that

Continuing from (184)

So this gives the first term in

Note that

Now we have

Hence the

and the

So

The theorem follows.

Now suppose that

The first column is denoted

Now as in [

and similarly

So

because we assume that

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

Thus

Remember the formula

Define when

where

where the last vector is denoted

Now we also get

So

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

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

The

where

The theorem follows.

Now assume that

Proposition 7 For q odd and

and for

For q even and

and for

Also

when q is odd and

when q is even.

Proof. (277) q odd. We are deleting the row r with

The first sign here is the sign when decomposing after row

We have the sign

on

on column

For

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

Define for

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 [

Here the complexes are

the forward reaction rate is denoted

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

From

where

for

We can then multiply with

and define the constants

to obtain the polynomial of degree

if we assume that

and the discrete dynamical system of section three, see also [

Also define the Euler map

for

and

then you obtain a discrete model T of section three.

Example Let

In this paper, we considered a discrete mathematical model and an ODE model of cancer growth in the variables

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