) = k 11 + 2 i , 2 0 k 2 , 11 + 2 i > 0 (64)

i = 1 , , n . The third equality holds for = q , 1 , , n . The fourth equality holds for the multipeutics system f n j , i = 1 , , n j , j { 0 , 1 , , n } .

By the fundamental theorem of calculus applied twice

d * C ( t ) = 0 t ( 0 s 2 d * C t 2 ( v ) d v + d * C t ( 0 ) ) d s (65)

So

d q C ( t ) > d 1 C ( t ) > > d n C ( t ) (66)

To see the first inequality, observe that there exists a positive constant c, such that

d q C ( t ) d 1 C ( t ) = 0 t 0 s ( 2 d q C t 2 ( v ) 2 d 1 C t 2 ( v ) ) d v d s 1 2 c t 2 > 0 (67)

because a positive function on a compact interval [ 0, δ ] assumes a minimum c > 0 . The other inequalities follow in the same way. And this proves all the inequalities in (24) except C * q < C * c r . Henceforth assume

k 2 , 10 + 2 i = k 11 + 2 i , 2 = k 2 , 11 + 2 i = 0 (68)

i = 1 , , n . Finally differentiate (57) to (60) with respect to t to get

3 k 21 ( 2 d * C t 2 d * N t + d * C t 2 d * N t 2 ) (69)

3 k 23 ( 2 d * C t 2 d * D t + d * C t 2 d * D t 2 ) (70)

+ a ^ 3 d * C t 3 = 0 (71)

when t = 0. But

d * M t ( 0 ) = d * N t ( 0 ) = 0 (72)

However when = q

2 d * N t 2 ( 0 ) > 0 (73)

To see this differentiate (17) twice with respect to t to find

( k 65 + k 25 ) 2 d * M t 2 ( 0 ) = 2 k 54 d * D t ( 0 ) d * G t ( 0 ) > 0 (74)

If we now differentiate (18) twice with respect to t we find

k 26 2 d * N t 2 ( 0 ) = k 65 2 d * M t 2 ( 0 ) > 0 (75)

which is what we sought to show.

Furthermore

( k 65 + k 25 ) d * M ( t ) = k 54 d * D d * G ( t ) > 0 (76)

and

d * N ( t ) = k 65 d * M ( t ) k 21 d * C ( t ) + k 26 > 0 (77)

when t > 0 and small. Differentiate (15) twice with respect to t when = c r , q to get

k 2 , 10 2 d * D t 2 = 2 k 23 d * C t d * D t + 2 k 54 d * D t d * G t (78)

Now

a ^ 2 d * C t 2 = 2 k 23 d * C t d * D t (79)

Also

a ^ 3 d * C t 3 ( 0 ) = 3 k 21 ( d * C t k 65 k 54 2 k 26 ( k 65 + k 25 ) d * D t d * G t ) (80)

+ 3 k 23 1 a ^ ( 2 k 23 d * C t d * D t ) d * D t (81)

3 k 23 1 k 2,10 d * C t ( 2 k 23 d * C t d * D t + 2 k 54 d * D t d * G t ) (82)

where we have evaluated in t = 0 on the right hand side. So if

3 k 21 k 65 2 k 26 ( k 65 + k 25 ) 3 k 32 2 k 2 , 10 > 0 (83)

we have

3 d c r C t 3 ( t ) > 3 d q C t 3 ( t ) (84)

when t = 0 and hence also for small t > 0. The fundamental theorem of calculus applied three times gives

d * C ( t ) = 0 t ( 0 s ( 0 v 3 d * C t 3 ( u ) d u + 2 d * C t 2 ( 0 ) ) d v + d * C t ( 0 ) ) d s (85)

So

d c r C ( t ) > d q C ( t ) (86)

for small t > 0 and the theorem follows in the same way as (67). Define the matrices

B = D f * c * λ id A = D f * c * (87)

and then write the definition of the determinant

det B = σ S p ( 1 ) I ( σ ) B 1 , σ ( 1 ) B p , σ ( p ) (88)

where S p is the set of permutations of { 1, , p } and I ( σ ) = 0 if σ is even and I ( σ ) = 1 , when σ is odd. This formula shows immediately that

det B = ( 1 ) p λ p + i = 0 p 1 a i ( A 11 , , A p p ) λ i (89)

where a i is a polynomial in A 11 , , A p p , and hence is smooth in A 11 , , A p p . You can prove this by induction. Now apply the continuous dependence of roots of a polynomial on its coefficients to show that d * ( t ) is stable for f * , when t is small, see [19] . □

Remark. If the quadrapeutics chemo rate is

k 10 , 2 q = k 10 , 2 c r α ˜ α ˜ ] 0 , 1 [ (90)

and the x ray rate is

k 27 q X = k 27 c r X β ˜ β ˜ ] 0 , 1 [ (91)

where

k 10 , 2 q = t k 10 , 2 q , 0 k 10 , 2 c r = t k 10 , 2 c r , 0 (92)

while a ^ = k 98 k 28 k 27 X < 0. Here

k 10,2 q ,0 , k 10,2 c r ,0 + (93)

Then by taking k 21 big we find

3 d q C t 3 ( 0 ) < 3 d c r C t 3 ( 0 ) (94)

since

a ^ 3 d q C t 3 ( 0 ) + (95)

as k 21 + . Hence

d q C ( t ) < d c r C ( t ) (96)

for small t > 0 , when we impose (90) and (91), arguing as in the proof of theorem 1.

Experimentally this is what you see, that quadrapeutics can outperform chemo radiation even if we take the chemo radiation doses smaller, see [1] [13] [14] [15] .

Consider also the system (1) to (13) with

2 C + X 0 (97)

replacing (7). The ODE s are the same (15) to (19) except (14), which is

C = k 21 C N k 23 C D + ( k 98 k 28 ) C + k 82 2 k 27 X C 2 i = 1 n k 2 , 10 + 2 i C I i (98)

We are now going to find candidates of the positive singular points of the different systems. We start with f and assume that all k i j are positive. (19) gives

I i = k 11 + 2 i , 2 k 2 , 10 + 2 i C + k 2 , 11 + 2 i (99)

when C > 0 and (18) gives

N = k 65 M k 21 C + k 26 (100)

(16) and (17) give

k 11 , 2 k 2 , 11 G ( k 65 + k 25 ) M = 0 (101)

or

G = 1 k 2 , 11 ( k 11 , 2 ( k 65 + k 25 ) M ) = α M + β (102)

Also

D = k 10 , 2 k 23 C + k 54 G + k 2 , 10 (103)

Insert these formulas in (14)

k 21 C k 65 M k 21 C + k 26 k 23 C k 10,2 k 23 C + k 54 G + k 2,10 (104)

+ ( k 98 k 28 k 27 X ( 1 + δ ( 2 C 1 ) ) ) C (105)

i = 1 n k 2,10 + 2 i C k 11 + 2 i ,2 k 2,10 + 2 i C + k 2,11 + 2 i + k 82 = 0 (106)

δ = 0 , 1 . Insert the formula for D in M = 0 to get

k 54 G k 10 , 2 k 23 C + k 54 G + k 2 , 10 ( k 65 + k 26 ) M = 0 (107)

We can isolate C in this equation to find

C = a M 2 + b M + c M (108)

where

a = ( k 65 + k 25 ) k 54 k 2 , 11 k 23 (109)

and

b = k 54 k 10 , 2 k 23 k 2 , 11 ( k 54 k 2 , 11 k 23 k 11 , 2 + k 2 , 10 k 23 ) (110)

Finally

c = k 54 k 10 , 2 k 11 , 2 k 23 k 2 , 11 ( k 65 + k 25 ) (111)

We now get by multiplying with the product of the denominators in (104), (105) and (106)

k 21 C k 65 M ( k 23 C + k 54 G + k 2 , 10 ) i = 1 n ( k 2 , 10 + 2 i C + k 2 , 11 + 2 i ) (112)

k 23 C k 10 , 2 ( k 21 C + k 26 ) i = 1 n ( k 2 , 10 + 2 i C + k 2 , 11 + 2 i ) (113)

+ ( k 82 + ( k 98 k 28 k 27 X ( 1 + δ ( 2 C 1 ) ) ) C ) ( k 21 C + k 26 ) (114)

( k 23 C + k 54 G + k 2 , 10 ) i = 1 n ( k 2 , 10 + 2 i C + k 2 , 11 + 2 i ) (115)

i = 1 n k 2 , 10 + 2 i C k 11 + 2 i , 2 ( j = 1 , j i n ( k 2 , 10 + 2 j C + k 2 , 11 + 2 j ) ) (116)

( k 23 C + k 54 G + k 2 , 10 ) ( k 21 C + k 26 ) = 0 (117)

If we now insert the formulas for G , C and multiply with M n + 3 , δ = 0 or M n + 4 , δ = 1 , we get a polynomial p of degree at most 2 ( n + 3 ) ( δ = 0 ) or 2 ( n + 4 ) ( δ = 1 ) in M, such that a positive singular point

( C , D , G , M , N , I 1 , , I n ) (118)

will have p ( M ) = 0.

Let δ = 1. Then we have

Theorem 2. For δ = 1 , there exist rate constants k i j > 0 , such that (24) holds and all equilibria are positive and stable.

Proof. Now let

a ^ = k 98 k 28 (119)

and assume that it is negative. Then

d * C t ( 0 ) = k 82 0 a ^ (120)

is the same for all treatments. But now the first coordinate of formula (48) becomes

k 21 d * C d * N k 23 d * C d * D + a ^ d * C 2 k 27 X ( d * C ) 2 + k 82 i = 1 n k 2 , 10 + 2 i d * C d * I i = 0 (121)

So

a ^ 2 d * C t 2 ( 0 ) = k 21 ( 2 d * C t 2 d * N + 2 d * C t d * N t + d * C 2 d * N t ) (122)

+ k 23 ( 2 d * C t 2 d * D + 2 d * C t d * D t + d * C 2 d * D t ) (123)

+ i = 1 n k 2 , 10 + 2 i ( 2 d * C t 2 d * I i + 2 d * C t d * I i t + d * C 2 d * I i t ) (124)

+ 4 k 27 X ( d * C t ) 2 (125)

and when k 2 , 10 + 2 i = k 2 , 11 + 2 i = k 11 + 2 i , 2 = 0 , i = 1 , , n .

a ^ 3 d * C t 3 ( 0 ) = 3 k 21 ( 2 d * C t 2 d * N t + d * C t 2 d * N t 2 ) (126)

+ 3 k 23 ( 2 d * C t 2 d * D t + d * C t 2 d * D t 2 ) (127)

+ 12 k 27 X 2 d * C t 2 d * C t (128)

We can now take k 21 big and argue as in theorem 1. □

3. The Chemo Radiation Model

Consider the mass action kinetic system

C + D 0 (129)

( 1 + δ ) C + X 0 , δ = 0 , 1 (130)

C 2 C (131)

C 0 (132)

D 0 (133)

Here the complexes are C ( 2 ) = 0 , C ( 3 ) = C + D , C ( 7 ) = ( 1 + δ ) C + X , C ( 8 ) = C , C ( 9 ) = 2 C , C ( 10 ) = D . With mass action kinetics the vector field is

f ( C , D ) = ( k 82 k 23 C D + ( k 98 k 28 k 27 X ( 1 + δ ( 2 C 1 ) ) ) C k 23 C D + k 10 , 2 k 2 , 10 D ) (134)

all k i j > 0.

This system with δ = 0 is similar to the reduced system from [17] .

G F C (135)

C + G I 0 (136)

C 2 C (137)

G F 0 (138)

G I 0 (139)

C 0 (140)

where we have added the last reaction. The complexes are C ( 1 ) = G F , C ( 2 ) = C , C ( 3 ) = C + G I , C ( 4 ) = 0 , C ( 5 ) = 2 C , C ( 6 ) = G I . The vector field is

g ( C , G F , G I ) = ( k 21 G F k 43 C G I + ( k 52 k 42 ) C + k 24 k 14 ( k 21 + k 41 ) G F k 43 C G I + k 64 k 46 G I ) (141)

We can assume that G F is at equilibrium

G F = k 14 k 21 + k 41 (142)

The reduced system is then

g ^ ( C , G I ) = ( k k 43 C G I + ( k 52 k 42 ) C k 43 C G I + k 64 k 46 G I ) (143)

k = k 21 k 14 k 21 + k 41 + k 24 (144)

We shall find the singular points of the chemo radiation system. D = 0 gives

D = k 10 , 2 k 23 C + k 2 , 10 (145)

when C > 0 and then C = 0 amounts to

k 82 k 23 C k 10 , 2 k 23 C + k 2 , 10 + ( k 98 k 28 k 27 X ( 1 + δ ( 2 C 1 ) ) ) C = 0 (146)

which is equivalent to

k 23 ( k 98 k 28 k 27 X ) C 2 (147)

+ ( k 2 , 10 ( k 98 k 28 k 27 X ) + k 23 k 82 k 23 k 10 , 2 ) C + k 82 k 2 , 10 = 0 (148)

when δ = 0 , C > 0 and when δ = 1 , C > 0 it is equivalent to

2 k 23 k 27 X C 3 + ( ( k 98 k 28 ) k 23 2 k 27 X k 2,10 ) C 2 (149)

+ ( k 2 , 10 ( k 98 k 28 ) + k 82 k 23 k 23 k 10 , 2 ) C + k 82 k 2 , 10 = 0 (150)

The linearization of f at a singular point c * = ( C , D ) is

A = D f c * = ( k 23 D + ( k 98 k 28 k 27 X ) k 23 C k 23 D k 2 , 10 k 23 C ) (151)

when δ = 0 and when δ = 1 it is

A ˜ = D f c * = ( k 23 D + ( k 98 k 28 4 k 27 X C ) k 23 C k 23 D k 2 , 10 k 23 C ) (152)

If δ = 0

a ^ = k 98 k 28 k 27 X < 0 (153)

and ( C , D ) is a positive singular point, then

σ = trace A < 0 det A = a ^ k 23 C + k 2 , 10 ( k 23 D a ^ ) > 0 (154)

hence this is a stable, positive singular point.

Consider the chemo radiation system where

k 10 , 2 = 23 k 98 = 13 X = 1 δ = 1 (155)

and all other k i j = 1. Then the cubic polynomial giving candidates of singular points is

2 C 3 + 10 C 2 10 C + 1 = 0 (156)

and this gives three singular points

( C , D ) = ( ( 0.112 , 20.68 ) ( 1.21 , 10.41 ) ( 3.677 , 4.918 ) (157)

where we have used that

D = k 10 , 2 k 23 C + k 2 , 10 (158)

It is a simple matter to check that the first and last of these are stable equilibria and the middle is an unstable saddle by computing the trace and determinant of the linearization of f q in the singular points. I have plotted a phase portrait for these values of the parameters in Figure 3.

4. The Quadrapeutics Equilibrium Equation

We shall find the quadrapeutics equilibrium equation too.

Theorem 3. Suppose δ = 0. There exists a polynomial of degree atmost six

s ( M ) = i = 0 6 q i M i q i (159)

such that if ( C , D , G , M , N ) is a positive singular point of f q then

s ( M ) = 0 (160)

Figure 3. A tristable chemo radiation system. There are three singular points marked with a circle.

Proof. Isolate D, G, M, N from (15), (16), (17), (18) and insert it in f q 1 ( C , D , G , M , N ) = 0 to find

p ( C , G , M ) = k 21 C k 65 M ( k 23 C + k 54 G + k 2 , 10 ) (161)

k 23 C k 10 , 2 ( k 21 C + k 26 ) (162)

+ ( k 82 + ( k 98 k 28 k 27 X ( 1 + δ ( 2 C 1 ) ) ) C ) (163)

( k 21 C + k 26 ) ( k 23 C + k 54 G + k 2 , 10 ) = 0 (164)

where we have multiplied with

( k 21 C + k 26 ) ( k 23 C + k 54 G + k 2,10 ) (165)

We shall now insert

C = a M 2 + b M + c M G = α M + β (166)

where

α = k 65 + k 25 k 2 , 11 (167)

β = k 11 , 2 k 2 , 11 (168)

in the polynomial p. Compute

M 3 C 3 = i = 0 6 a ˜ i M i (169)

where

a ˜ 6 = a 3 (170)

a ˜ 5 = 3 a 2 b (171)

a ˜ 4 = 3 ( a b 2 + a 2 c ) (172)

a ˜ 3 = 6 a b c + b 3 (173)

a ˜ 2 = 3 ( a c 2 + b 2 c ) (174)

a ˜ 1 = 3 b c 2 (175)

a ˜ 0 = c 3 (176)

Now define

p 1 ( C , G , M ) = M 3 ( k 21 C k 65 M ( k 23 C + k 54 G + k 2 , 10 ) ) (177)

and this is

s 1 ( M ) = p 1 ( a M 2 + b M + c M , α M + β , M ) (178)

Also put

p 2 ( C , G , M ) = M 3 ( k 23 C k 10 , 2 ( k 21 C + k 26 ) ) (179)

and this is

s 2 ( M ) = p 2 ( a M 2 + b M + c M , α M + β , M ) (180)

Finally set

p 3 ( C , G , M ) = M 3 ( k 82 + a ^ C ) ( k 21 C + k 26 ) ( k 23 C + k 54 G + k 2 , 10 ) (181)

which is

s 3 ( M ) = p 3 ( a M 2 + b M + c M , α M + β , M ) (182)

So

s ( M ) = s 1 ( M ) + s 2 ( M ) + s 3 ( M ) (183)

Now compute

#Math_332# (184)

k 21 ( a M 2 + b M + c ) k 65 M 3 ( k 54 ( α M + β ) ) + k 2 , 10 ) (185)

= k 21 k 65 k 23 ( a 2 M 6 + b 2 M 4 + c 2 M 2 + 2 a b M 5 + 2 a c M 4 + 2 b c M 3 ) (186)

k 21 k 65 ( a M 2 + b M + c ) M 3 k 54 α M k 21 k 65 ( a M 2 + b M + c ) M 3 ( k 54 β + k 2 , 10 ) (187)

Write

s 1 ( M ) = i = 0 6 a i M i (188)

Then by the previous computation

a 6 = k 21 a k 65 ( k 23 a + k 54 α ) (189)

a 5 = k 21 b k 65 ( k 23 a + k 54 α ) k 21 a k 65 ( k 54 β + k 23 b + k 2 , 10 ) (190)

a 4 = k 21 c k 65 ( k 23 a + k 54 α ) k 21 b k 65 ( k 54 β + k 23 b + k 2 , 10 ) k 21 k 65 k 23 a c (191)

a 3 = k 21 c k 65 ( k 54 β + k 23 b + k 2 , 10 ) k 21 k 65 k 23 b c (192)

a 2 = k 21 c 2 k 65 k 23 (193)

a 1 = 0 (194)

a 0 = 0 (195)

Now compute

s 2 ( M ) = M 3 ( k 23 C k 10 , 2 ( k 21 C + k 26 ) ) (196)

= M 3 k 23 k 10 , 2 ( k 21 a 2 M 4 + b 2 M 2 + c 2 + 2 a b M 3 + 2 a c M 2 + 2 b c M M 2 + k 26 a M 2 + b M + c M ) (197)

= k 23 k 10 , 2 k 21 ( a 2 M 5 + b 2 M 3 + c 2 M + 2 a b M 4 + 2 a c M 3 + 2 b c M 2 ) (198)

k 23 k 10 , 2 k 26 ( a M 4 + b M 3 + c M 2 ) (199)

Write

s 2 ( M ) = i = 0 6 b i M i (200)

and by the computation above

b 6 = 0 (201)

b 5 = k 23 a 2 k 10 , 2 k 21 (202)

b 4 = k 23 b k 10 , 2 k 21 a k 23 a k 10 , 2 ( k 21 b + k 26 ) (203)

b 3 = 2 k 23 k 10 , 2 k 21 a c k 23 k 10 , 2 b ( k 21 b + k 26 ) (204)

b 2 = k 23 c k 10 , 2 ( k 21 b + k 26 ) k 23 b k 10 , 2 k 21 c (205)

b 1 = k 23 c 2 k 10 , 2 k 21 (206)

b 0 = 0 (207)

Now consider

p 3 ( C , G , M ) = k 21 k 23 a ^ C 3 + k 21 k 54 a ^ C 2 G (208)

+ k 21 a ^ C 2 k 2,10 + k 26 k 23 a ^ C 2 (209)

+ k 26 k 54 G a ^ C + k 26 k 2,10 a ^ C (210)

+ k 82 k 21 k 23 C 2 + k 82 k 21 k 54 C G (211)

+ k 82 k 21 k 2,10 C + k 82 k 26 k 23 C (212)

+ k 82 k 26 k 54 G + k 82 k 26 k 2,10 (213)

Name these 12 summands d 1 , , d 12 . Now

M 3 d 1 = k 21 k 23 a ^ i = 0 6 a ˜ i M i (214)

and

M 3 d 2 = k 21 k 54 a ^ ( α ( a 2 M 6 + b 2 M 4 + c 2 M 2 + 2 a b M 5 + 2 a c M 4 + 2 b c M 3 ) (215)

+ β ( a 2 M 5 + b 2 M 3 + c 2 M + 2 a b M 4 + 2 a c M 3 + 2 b c M 2 ) ) (216)

Also

#Math_366# (217)

Similarly

M 3 d 4 = k 26 a ^ k 23 ( a 2 M 5 + b 2 M 3 + c 2 M + 2 a b M 4 + 2 a c M 3 + 2 b c M 2 ) (218)

Then

#Math_368# (219)

And

M 3 d 6 = k 26 k 2 , 10 a ^ ( a M 4 + b M 3 + c M 2 ) (220)

We also find

M 3 d 7 = k 82 k 21 k 23 ( a 2 M 4 + b 2 M 2 + c 2 + 2 a b M 3 + 2 a c M 2 + 2 b c M ) M (221)

#Math_371# (222)

Combine

M 3 ( d 9 + d 10 ) = ( k 82 k 21 k 2 , 10 + k 82 k 26 k 23 ) ( a M 4 + b M 3 + c M 2 ) (223)

The last two ds give

M 3 d 11 = k 82 k 26 k 54 ( α M 4 + β M 3 ) (224)

M 3 d 12 = k 82 k 26 k 2 , 10 M 3 (225)

Write

s 3 ( M ) = i = 0 6 c i M i (226)

Then we find from the above computations

c 6 = a ^ a 2 k 21 ( k 23 a + k 54 α ) (227)

and

c 5 = k 21 k 23 a ^ 3 a 2 b + k 21 k 54 a ^ ( α 2 a b + β a 2 ) + k 21 a ^ k 2 , 10 a 2 (228)

+ k 26 k 23 a ^ a 2 + k 82 k 21 k 23 a 2 + k 82 k 21 k 54 a α + k 26 k 54 a ^ α a (229)

Now collect terms

c 4 = k 21 k 23 a ^ 3 ( a b 2 + a 2 c ) + k 21 k 54 a ^ α ( b 2 + 2 a c ) (230)

+ k 21 k 54 a ^ β 2 a b + k 21 a ^ k 2,10 2 a b + k 26 k 23 a ^ 2 a b (231)

+ k 26 k 54 a ^ ( α b + β a ) + k 26 k 2,10 a ^ a (232)

+ k 82 k 21 k 23 2 a b + k 82 k 21 k 54 ( α b + β a ) (233)

+ ( k 82 k 21 k 2,10 + k 82 k 26 k 23 ) a + k 82 k 26 k 54 α (234)

We can also find

c 3 = k 21 k 23 a ^ ( 6 a b c + b 3 ) + k 21 k 54 a ^ ( 2 b c α + β ( b 2 + 2 a c ) ) (235)

+ k 21 a ^ k 2,10 ( b 2 + 2 a c ) + k 26 k 23 a ^ ( b 2 + 2 a c ) (236)

+ k 26 k 54 a ^ ( c α + β b ) + k 26 k 2,10 a ^ b (237)

+ k 82 k 21 k 23 ( b 2 + 2 a c ) + k 82 k 21 k 54 ( α c + β b ) (238)

+ ( k 82 k 21 k 2,10 + k 82 k 26 k 23 ) b + k 82 k 26 k 54 β + k 82 k 26 k 2,10 (239)

Now

c 2 = k 21 k 23 a ^ 3 ( a c 2 + b 2 c ) + k 21 k 54 a ^ ( α c 2 + β 2 b c ) (240)

+ ( k 21 a ^ k 2,10 + k 26 k 23 a ^ ) 2 b c + k 26 k 54 a ^ β c (241)

+ k 26 k 2,10 a ^ c + k 82 k 21 k 23 2 b c (242)

+ k 82 k 21 k 54 β c + ( k 82 k 21 k 2,10 + k 82 k 26 k 23 ) c (243)

Also

c 1 = k 21 k 23 a ^ 3 b c 2 + k 21 k 54 a ^ c 2 β + k 21 a ^ k 2 , 10 c 2 (244)

+ k 82 k 21 k 23 c 2 + k 23 c 2 a ^ k 26 (245)

Finally

c 0 = k 21 k 23 a ^ c 3 (246)

The quadrapeutics polynomial is

s ( M ) = i = 0 6 ( a i + b i + c i ) M i (247)

If all rate constants k i j = 1 , X = 1 then

s ( M ) = 2 M 5 8 M 4 + 10 M 3 23 4 M 2 + 3 2 M 1 8 = 0 (248)

This polynomium has three real roots

M = 0.148638 , M = 1 2 , M = 2.32234 (249)

and two imaginary roots

0.514511 ± 0.312089 i (250)

There are thus two candidates of singular points ( M = 1 2 does not give a singular point).

( C , D , G , M , N ) = ( ( 0.661153,0.423634,0.702724,0.148638,0.089479 ) ( 1.85998, 1.27437, 3.64468,2.32234,0.812013 ) (251)

and they are both singular points (numerical evidence). We have used, that

G = α M + β (252)

C = a M 2 + b M + c M (253)

D = k 10 , 2 k 23 C + k 54 G + k 2 , 10 (254)

N = k 65 M k 21 C + k 26 (255)

5. The Extended Quadrapeutics Model

This is the system (1) to (11) and the reactions

D ˜ + C 0 (256)

N + D D ˜ (257)

D ˜ 0 (258)

D D ˜ (259)

taking into account, that plasmonic nanobubbles destroy the liposomes D and thus the chemo therapeutic drug D ˜ is injected into the cytoplasm. But some of the liposomes decay, producing chemo therapeutic drug. The complexes here are C ( 12 ) = N + D , C ( 13 ) = D ˜ , C ( 14 ) = D ˜ + C . The ODEs are

C = k 21 C N k 23 C D + ( k 98 k 28 k 27 X ) C + k 82 k 2 , 14 D ˜ C (260)

D ˜ = k 2 , 14 D ˜ C + k 13 , 12 N D k 2 , 13 D ˜ + k 13 , 10 D (261)

D = k 23 C D k 54 D G + k 10 , 2 k 2 , 10 D k 13 , 12 N D k 13 , 10 D (262)

G = k 54 D G + k 11 , 2 k 2 , 11 G (263)

M = k 54 D G ( k 65 + k 25 ) M (264)

N = k 21 C N + k 65 M k 26 N k 13 , 12 N D (265)

The vector field is denoted g q . Denote by g c r the vector field in (260), (261) and (262) with the rate constants in (263), (264) and (265) all equal to zero and the rest positive.

Theorem 4. There exist positive values of the rate constants such that

C * q < C * c r (266)

holds and the singular points are both positive and stable.

Proof. Define for = q

k 1 = ( k 82 , k 10 , 2 , k 11 , 2 ) (267)

#Math_425# (268)

And for = c r

k 1 c r = ( k 82 , k 10 , 2 ) (269)

k 2 c r = ( k 23 , k 98 , k 28 , k ˜ 27 , k 2 , 14 , k 2 , 13 , k 2 , 10 , k 13 , 10 ) (270)

Define

K c r = ( k 1 c r , k 2 c r ) K q = ( k 1 q , k 2 q ) (271)

and

K ^ * = ( k 1 * , k 2 * ) k 1 * = 0 (272)

Use the implicit function theorem to find a mapping

c q : V q 18 6 (273)

and

c c r : V c r 10 3 (274)

such that

g * ( c * ( K * ) , K * ) = 0 c * ( K ^ * ) = 0 (275)

and define

d * ( t ) = c * ( t k 1 * , 0 , k 2 * ) = ( c r q (276)

V q , V c r are open subsets of 18 , 10 respectively. For = c r let d * G = d * M Math_439# By the implicit function theorem

d * D ˜ ( 0 ) = 0 (277)

Differentiate (262) and (261) with respect to t to get

d * D t = k 10 , 2 0 k 2 , 10 + k 13 , 10 (278)

and

d * D ˜ t = k 10 , 2 0 k 13 , 10 k 2 , 13 ( k 2 , 10 + k 13 , 10 ) (279)

Now differentiate (261) twice with respect to t to find

k 2 , 13 2 d * D ˜ t 2 = k 13 , 10 2 d * D t 2 2 k 2 , 14 d * D ˜ t d * C t (280)

and differentiate (262) with respect to t to get

2 d * D t 2 = 2 k 2 , 10 + k 13 , 10 ( k 23 d * C t d * D t + k 54 d * D t d * G t ) (281)

We also have

d * C t = k 82 0 a ^ d * G t = k 11 , 2 0 k 2 , 11 (282)

the last equality when = q . We now get

2 d * C t 2 = 1 a ^ ( 2 k 23 d * C t d * D t + 2 k 2 , 14 d * D ˜ t d * C t + k 21 d * C t d * N t ) (283)

so

2 d c r C t 2 ( 0 ) = 2 d q C t 2 ( 0 ) (284)

and

2 d * N t 2 = k 65 2 k 54 k 26 ( k 65 + k 25 ) d * D t d * G t (285)

where we have used that

d * M t ( 0 ) = 0 (286)

hence

d * N t ( 0 ) = 0 (287)

Finally

a ^ 3 d * C t 3 = 3 k 23 ( 2 d * C t 2 d * D t + d * C t 2 d * D t 2 ) (288)

+ 3 k 2,14 ( 2 d * D ˜ t 2 d * C t + d * D ˜ t 2 d * C t 2 ) (289)

+ 3 k 21 d * C t 2 d * N t 2 (290)

and taking k 21 big, this implies the theorem, arguing as in the proof of theorem 1. □

From D = 0 isolate

C = 1 k 23 D ( k 54 D G + k 10 , 2 ( k 2 , 10 + k 13 , 10 ) D k 13 , 12 N D ) (291)

and insert this in N = 0 to get using

G = α M + β α < 0 β > 0 (292)

and

D = α k 2 , 11 M k 54 ( α M + β ) (293)

(this equation follows from (263)), that

k 13,12 k 21 k 23 N 2 + N ( k 54 k 21 k 23 ( α M + β ) + k 21 k 10,2 k 54 ( α M + β ) k 23 k 2,11 α M (294)

+ k 21 k 2 , 10 + k 13 , 10 k 23 k 26 + k 13 , 12 k 2 , 11 α M k 54 ( α M + β ) ) + k 65 M = 0 (295)

M > 0 , M β α . Let a ˜ be the coefficient to N 2 and let b ˜ ( M ) be the coefficient to N and c ˜ ( M ) = k 65 M . Then we get when

Δ = b ˜ ( M ) 2 4 a ˜ c ˜ ( M ) > 0 (296)

N ± ( M ) = b ˜ ( M ) ± Δ 2 a ˜ (297)

a ˜ > 0. Thus D = 0 gives

C ± = 1 k 23 D ( M ) ( k 54 D G + k 10 , 2 ( k 2 , 10 + k 13 , 10 ) D k 13 , 12 N ± D ) (298)

and

D ˜ ± = k 13 , 12 N ± D + k 13 , 10 D k 2 , 14 C ± + k 2 , 13 (299)

D = k 2 , 11 α M k 54 ( α M + β ) (300)

So let

p * ( M ) = p * ± ( M ) = k 21 C ± N ± k 23 C ± D + a ^ C ± + k 82 k 2 , 14 D ˜ ± C ± (301)

If ( C , D ˜ , D , G , M , N ) is a positive singular point of g q , then

p * ± ( M ) = 0 (302)

Define h by

1 + x 1 = 0 1 s 1 + x s d s (303)

= x 0 1 1 2 1 + x s d s = x h ( x ) (304)

where h is smooth and x ] 1, + [ .

Proposition 5. The function

N ( M ) 2 c ˜ ( M ) b ˜ ( M ) h ( 4 a ˜ c ˜ b ˜ 2 ) (305)

satisfies (294) and (295), when

4 a ˜ c ˜ b ˜ 2 ] 1 , + [ , b ˜ ( M ) < 0 (306)

Proof. If b ˜ ( M ) < 0 then

N ( M ) = 2 c ˜ ( M ) b ˜ ( M ) h ( 4 a ˜ c ˜ ( M ) b ˜ ( M ) 2 ) (307)

is smooth and

N + ( M ) = b ˜ ( M ) + Δ ( M ) 2 a ˜ (308)

when a ˜ > 0. If b ˜ ( M ) > 0 then

N + ( M ) = 2 c ˜ ( M ) b ˜ ( M ) h ( 4 a ˜ c ˜ ( M ) b ˜ ( M ) 2 ) (309)

is smooth and

N ( M ) = b ˜ ( M ) Δ ( M ) 2 a ˜ (310)

when a ˜ > 0. These are both negative, so b ˜ ( M ) > 0 is not interesting. □

Proposition 6. If k 13 , 12 = k 2 , 13 = k 2 , 14 = k 13 , 10 = 0 and (306) holds, then

s ( M ) = q ( M ) p * ( M ) (311)

where

q ( M ) = M 3 ( k 21 C + k 26 ) ( k 23 C + k 54 G + k 2 , 10 ) (312)

and

C = a M 2 + b M + c M (313)

G = α M + β (314)

Proof. We have

k 23 C + k 54 G + k 2,10 (315)

= k 23 a M 2 + b M + c M + k 54 ( α M + β ) + k 2 , 10 (316)

and here the terms with a , α cancel out to give

k 23 b + k 54 β + k 2,10 + k 23 c M (317)

= k 54 k 10 , 2 k 2 , 11 + k 23 c M (318)

So

k 10,2 k 23 C + k 54 G + k 2,10 (319)

= k 10 , 2 k 54 k 10 , 2 k 2 , 11 + k 23 c M (320)

which becomes

( k 65 + k 25 ) M k 54 k 11,2 k 2,11 k 54 k 65 + k 25 k 2,11 M (321)

= k 2 , 11 α M k 54 ( α M + β ) (322)

We also can compute

b ˜ ( M ) = k 54 k 21 k 23 ( α M + β ) + k 21 k 10 , 2 k 54 k 23 k 2 , 11 α M + β α M (323)

+ k 21 k 2,10 k 23 k 26 (324)

= k 54 k 21 k 23 ( k 65 + k 25 k 2 , 11 M + k 11 , 2 k 2 , 11 ) (325)

+ k 21 k 2,10 k 23 + k 21 k 10,2 k 54 k 23 k 2,11 k 21 k 10,2 k 54 M k 23 k 2,11 k 11,2 k 65 + k 25 k 26 (326)

= k 21 a M 2 + b M + c M k 26 = k 21 C k 26 (327)

Notice that

N ( M ) = c ˜ ( M ) b ˜ ( M ) = c ˜ ( M ) k 21 C + k 26 (328)

and

D = k 10 , 2 k 23 C + k 54 G + k 2 , 10 (329)

Now we get

p * ( M ) q ( M ) = ( k 21 C N k 23 C D + ( k 82 + a ^ C ) ) q ( M ) (330)

= k 65 M C ( k 23 C + k 54 G + k 2 , 10 ) M 3 (331)

#Math_516# (332)

If we assume there exists M i > 0 such that

s ( M i ) = 0 (333)

s M ( M i ) 0 (334)

and q ( M i ) > 0 (this is so if G ( M i ) , C ( M i ) > 0 ). Then

p * ( M i , k 0 ) = 0 (335)

p * M ( M i , k 0 ) 0 (336)

where

k 0 = K q | k 13 , 12 = k 13 , 10 = k 2 , 13 = k 2 , 14 = k 13 , 10 = 0 (337)

Then there exists a C 1 function

M * i : U (338)

such that

p * ( M * i ( k ) , k ) = 0 k U (339)

M * i ( k 0 ) = M i (340)

by the implicit function theorem. Here U is an open neighbourhood of k 0 in 18 . So the roots of s are nearly zeroes of p * , when

k 2,14 , k 13,12 , k 2,13 , k 13,10 (341)

are small and (333) and (334) holds. The example in the end of section 4 applies here.

6. Summary

Consider the extended quadrapeutics system and the additional reactions

L a + C 0 (342)

L a 0 (343)

Here the complexes are C ( 15 ) = L a + C , C ( 16 ) = L a , modelling adoptive T cell therapy, see [4] . L a are activated and expanded tumor infiltrating lymphocytes. They are injected back into the patient. Then the differential equations are

d C d t = k 21 C N k 23 C D + a ^ C + k 82 k 2 , 14 D ˜ C k 2 , 15 L a C (344)

D ˜ = k 2 , 14 D ˜ C + k 13 , 12 N D k 2 , 13 D ˜ + k 13 , 10 D (345)

D = k 23 C D k 54 D G + k 10 , 2 k 2 , 10 D k 13 , 12 N D k 13 , 10 D (346)

G = k 54 D G + k 11 , 2 k 2 , 11 G (347)

M = k 54 D G ( k 65 + k 25 ) M (348)

N = k 21 C N + k 65 M k 26 N k 13 , 12 N D (349)

d L a d t = k 2 , 15 L a C + k 16 , 2 k 2 , 16 L a (350)

Denote this vector field g q and let g c r denote the vector field in (344), (345), (346) where the rate constants in (347), (348), (349), (350) are all zero and the rest positive. Use the implicit function theorem to find maps

c c r : V c r 10 3 (351)

and

c q : V q 21 7 (352)

such that

g * ( c * ( K * ) , K * ) = 0 c * ( K ^ * ) = 0 (353)

Here

K c r = ( k 1 c r , k 2 c r ) (354)

is the same as in section 5. Also V c r , V q are open subsets of 10 and 21 respectively. And

k 1 q = ( k 82 , k 10 , 2 , k 11 , 2 , k 16 , 2 ) (355)

Also

k 2 q = ( k 21 , k 23 , k 98 , k 28 , k ˜ 27 , k 2 , 14 , k 13 , 12 , k 2 , 13 , k 54 , k 2 , 10 , k 2 , 11 , k 65 , k 25 , (356)

k 26 , k 2,15 , k 2,16 , k 13,10 ) (357)

Finally set

K * = ( k 1 * , k 2 * ) K ^ * = ( k 1 * , k 2 * ) k 1 * = 0 (358)

Define

d * ( t ) = c * ( t k 1 * , 0 , k 2 * ) (359)

Let

a ^ = k 98 k 28 k 27 X (360)

and assume that it is negative. Then

a ^ 2 d * C t 2 = 2 k 23 d * C t d * D t + 2 k 2 , 15 d * L a t d * C t + 2 k 2 , 14 d * D ˜ t d * C t (361)

But

d * C t = k 82 0 a ^ (362)

d * D t = k 10 , 2 0 k 2 , 10 (363)

d * L a t = k 16 , 2 0 k 2 , 16 (364)

d * D ˜ t = k 10 , 2 0 k 13 , 10 k 2 , 13 ( k 2 , 10 + k 13 , 10 ) (365)

suggesting, that in this model multipeutics can be more potent than quadrapeutics.

In the present paper, we considered quadrapeutics and multipeutics cancer therapies. We proved the multipeutics theorem, stating that the more treatments we apply the smaller the cancer burden. We also found a polynomial of degree at most 6, giving candidates of singular points for the quadrapeutics system.

The cells of the immune system have a plasma membrane repair system and it turns out that this system is much more efficient in cancer cells than in normal cells. The two proteins S100A11 and Annexin A2 are involved in this plasma membrane repair system. They are commonly upregulated in cancer cells, see [20] . So we can ask: can cancer cells repair nanobubble injury.

Cite this paper
Larsen, J. (2017) A Study on Multipeutics. Applied Mathematics, 8, 746-773. doi: 10.4236/am.2017.85059.
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