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