Elements of Mathematical Oncology

Elements of Mathematical Oncology

Franco Flandoli, University of Pisa

Padova 2015, Lecture 2

Fisher-Kolmogorov-Petrovskii-Piskunov model

The complexity of the model of Lecture 1, invasive with angiogenesis, does not allow a straightforward mathematical analysis.

It is convenient to start by an easier model: FKPP.

Single quantity (density of tumor cells), only di¤usion and proliferation:

∂u

∂t =D∆u+u(u_max u), uj^t=0=u₀.

Proliferation rate u_max u reduces proliferation when u increases to the threshold u_max.

By the scaling transformation v(t, x) =u(λt, µx)we may change all constants.

By v(t, x) =1 u(t, x)we may get

∂v

∂t = ¹

2∆v+v² v .

Simulations

By simple R codes:

Simulations

We see a traveling pro…le.

Traveling waves

Assume there is a solution of the form

u(t, x) =w(x ct). Substituting into FKPP we get

cw⁰ = ¹

2w⁰⁰+w w². (1)

On the function w we impose

x!lim∞w(x) =1, lim

x!+∞w(x) =0. (2)

and also ask that w is decreasing.

Theorem If 0 c <p

2, there are no solutions of (1)-(2). If c p

2, there exists one and only one solution, denoted in the sequel by w_c(x).

Traveling waves, sketch of proof

The equation is equivalent to the system w⁰ =z

z⁰ = 2cz 2w +2w².

To have a better dynamical intuition, let us write t for x (time) and (x, y) for (w , w⁰). Hence we study the system

x⁰(t) =y(t)

y⁰(t) = 2cy(t) 2x(t) +2x(t)².

We are looking for solutions (x(t), y(t))de…ned on the whole R, such that

t!lim∞x(t) =1, lim

t!+∞x(t) =0.

Traveling waves, sketch of proof

Fixed points (x, y) satisfy

y =0 2cy 2x+2x²=0 hence 2x+2x² =0, x=0 or x =1, namely

A= (0, 0), B = (1, 0). Stability analysis around A and B is not di¢ cult.

A is a locally attractive stationary point.

B is an hyperbolic point with unstable manifold tangent to a known vector e₂^B.

Traveling waves, sketch of proof

A solution arises at time ∞ from B and converges to A as t! +^{∞. For}

c p

2:

Traveling waves, sketch of proof

But for c <p

2, the solution x(t)becomes negative:

Example of other results

Theorem

If u(0, x) =1_x<0(x), the solution converges to the traveling pro…le with c =p

2. Precisely

t!+lim∞u(t, mt +x) =w^p₂(x)

for every x 2R, where mt is the unique point such that u(t, mt) =1/2.

A probabilistic proof has been given by McKean.

One should not think that every initial condition converges to such traveling wave. One can prove, for instance, that if u0 2 [^{0, 1}]is such that for some b2 (^0,p

2]the limit lim_x!+∞e^bx(1 f (x))exists

…nite and di¤erent from zero, then lim_t!+∞u(t, ct+x) =w_c (x), with c =1/b+b/2.

A probabilistic representation of FKPP

For the equation ^∂v_∂t = ¹_2∆v+λ v² v McKean proved the following probabilistic representation

u(t, x) =1 E

"_N

∏

i=1

u0 x+X_tⁱ

#

At time t =0, from x =0 a Brownian motion X_t¹ starts.

At the random time T0 Exp(λ), independent of X¹, the process X¹ disappears and, at position X_T¹

0, two new and independent Brownian motions start, X_t¹ and X_t².

Each one lives an exponential time Exp(λ)(independent of the previous objects) then dies and generates two new Brownian motions;

and so on.

At any time t there are N_t points alive, that we call X_tⁱ, i =1, ..., N_t.

More formal scheme

Multi-indices a= (i₁, ..., i_n), i_k 2 f^{1, 2}g^{, k} =1, ..., n, n2^{N. Tree} structure.

Probability space (Ω,F^{, P}), countable family T_a, a= (i₁, ..., i_n)of independent Exp(λ)random times. T0 associated to the …rst particle.

Set τi1,...,in =T₀+T_i1 +T_i1,i2+...+T_i1,...,in: time when particle a= (i1, ..., in)duplicates.

Countable family of independent Brownian motions Ba(t), a= (i₁, ..., i_n).

More formal scheme

Countable family of processes X_a(t), a= (i₁, ..., i_n), taking values in R[δ, where δ is an auxiliary point outsideR.

Process X_a takes the values δ except on a random time interval Ia. For t 2 [^τi1,...,in, τi1,...,in,in+1)set

Xi1,...,in,in+1(t) =Xi1,...,in(τi1,...,in) +Bi1,...,in,in+1(t τi1,...,in). Denote by Λt the set of multi-indices a= (i₁, ..., i_n)such that Xa(t)6=δ. The precise meaning of the formula above is now

u(t, x) =1 E

"

a

∏

u0(x+Xa(t))

# .

A probabilistic representation of FKPP

Sketch of proof of the formula: disintegrate the expected value with respect to T₀: v(t, x):=

E

"

a

∏

u₀(x+X_a(t))

#

=P(T₀>t)E

"

a

∏

u₀(x+X_a(t))j^T0 >t

#

+

Z _t

0

λe ^λsE

"

a

∏

u0(x+Xa(t))j^T0 =s

# ds

=e ^λtE[u0(x+B0(t))]

+

Z t

0 λe ^λsE 2

4

∏

a2Λ¹t

u₀(x+X_a(t))

∏

a2Λ²t

u₀(x+X_a(t))j^T0 =s 3 5 ds

whereΛ^kt is the set of elements(i₁, ..., i_n)2^Λt such that i₁ =k, k =1, 2.

A probabilistic representation of FKPP

e ^λtE[u₀(x+B₀(t))]

+

Z t

0 λe ^λsE 2

4

∏

a2Λ¹t

u₀(x+X_a(t))

∏

a2Λ²t

u₀(x+X_a(t))j^T0=s 3 5 ds

=e ^λt e^12∆tu₀ (x)

+

Z _t

0

λe ^λsE 2 4E

"

a2

∏

u0 x⁰+Xa(t s)

#2

x⁰=x+B0(s)

3 5 ds

= e(¹2∆ λ)^t_u

0 (x) +

Z t 0

e(¹2∆ λ)^s_λv²(t s, ) (x)ds and this is the PDE in the mild sense.

Is FKPP the macroscopic equation of the tree structure?

Consider an family of N independent copies of the processes X_aⁱ(t) described above, i =1, ..., N

The intuition is: we start from N independent cells; each one proliferates, as described above.

The question is:

S_t^N := ¹ N

∑

∑

a2Λⁱt

δ_Xaⁱ_(t) !^{? u} where u is the solution of FKPP?

The answer is negative. The previous scheme, in the sense of macroscopic limit, converges to an exponential proliferation.

Probabilistic formulae ad macroscopic limits are di¤erent procedures.

Exponential proliferation

Theorem

S_t^N weakly converges to a measure-valued solution of the equation

∂u

∂t = ¹

2∆u+λu. (3)

Proof. By the LLN:

D S_t^N, φE

= ¹ N

∑

∑

a2Λⁱt

φ X_aⁱ(t) !^E

"

a

∑

φ(X_a(t))

#

=:h^{ut, φ}i^.

Now we have to …nd the equation satis…ed by this time-dependent probability measure u_t. Notice it is a di¤erent expected value from E

"

a

∏

u0(x+Xa(t))

# .

Exponential proliferation

Introduce ρφ(t, x) (so that ρφ(t, 0) = h^{ut, φ}i⁾

ρ_φ(t, x):=E

"

a

∑

φ(x+X_a(t))

#

=P(T₀ >t)E

"

a

∑

φ(x+X_a(t))j^T0 >t

#

+

Z t 0

λe ^λsE

"

a

∑

φ(x+Xa(t))j^T0 =s

# ds

=e ^λtE[φ(x+B0(t))]

+

Z _t

0 λe ^λsE 2

4

∑

a2Λ¹t

φ(x+X_a(t)) +

∑

a2Λ²t

φ(x+X_a(t))j^T0 =s 3 5 ds

Exponential proliferation

=e ^λt e^12∆tφ (x)

+

Z _t

0 λe ^λsE 2 42E

"

a2

∑

φ x⁰+X_a(t s)

#

x⁰=x+B0(s)

3 5 ds

= e(¹2∆ λ)^t_φ (x) +2 Z _t

0

e(¹2∆ λ)^s_λρ

φ(t s, ) (x)ds.

which implies

∂

∂tρ_φ(t, x) = ¹

2∆ρφ(t, x) λρ_φ(t, x) +2λρφ(t, x)

= ¹

2∆ρφ(t, x) +λρφ(t, x) ρφ(0, x) =φ(x).

Recalling that ρφ(t, 0) = h^{ut, φ}i, it is easy to complete the proof.

Existence, uniqueness, invariant regions

We want to prove existence and uniqueness by analytical tools.

More important, we want to prove the invariance of "regions": if u0(x)2 [^{0, 1}] for all x, then u(t, x)2 [^{0, 1}] for all (t, x). Invariant regions are related to global existence.

But, more important, they are related to the biological meaning of u and of the model.

Consider the equation

∂u

∂t =_∆u+u(1 u).

Classical solutions: UC_b² R^d in space (uniformly continuous with

…rst and second derivatives).

Existence, uniqueness, invariant regions

Theorem

FKPP equation with UC_b² R^d -initial condition u₀2 [^{0, 1}], has one and only one classical solution u 2 [^{0, 1}](uniqueness holds in the larger class of mild solutions).

Proof. Step 1. We prove the result for the auxiliary equation

∂u

∂t = _∆u+u(1 u) +h(u) (4) where h( ) is a smooth function, strictly negative for u=1, strictly positive for u=0. This step is divided in three sub-steps.

The introduction of the auxiliary term h(u)is not needed if the invariance of the region [0, 1]is proved in a suitable way.

Existence, uniqueness, invariant regions

Step 1a. For a general equation of the form

∂u

∂t =∆u+f (u) (5)

with smooth f , one can prove local existence and uniqueness of mild solutions (UC_b⁰ R^d -functions satisfying the mild form of the equation).

If we then prove an a priori bound on solutions in uniform norm, this will lead to global existence (of mild solutions).

Step 1b. One can prove that mild solutions are also classical ones, when u₀ is regular.

Step 1c. For the particular equation (4) an a priori bound in uniform norm is proved (this is the main conceptual step). Precisely, we prove that, if u is a classical solution with u₀(x)2 [^{0, 1}], then u(t, x)2 [^{0, 1}].

Existence, uniqueness, invariant regions

Step 2. (This step is needed only if we had to introduce h(u)above.) Consider now the problem

∂u^δ

∂t =_∆u^δ+b(t, x) r^uδ+u^δ 1 u^δ δh u^δ

where h(u) is like above. We have u^δ 2 [^{0, 1}]by Step 1. One can prove that u^δ converges uniformly on compact sets (…rst locally in time, then globally), as δ!0. Then one can prove that the limit u is a solution of the original equation. Therefore there exists a solution u 2 [^{0, 1}]. Step 3. Uniqueness holds locally by Step 1a. This completes the proof.

Invariant regions by linearization

Let u be a local classical solution. Setting λ(t, x) =1 u(t, x), we have

∂u

∂t =_∆u+λu.

Lemma

If u0 0, then u 0 on [0, T].

Proof by Feynman-Kac formula: the result follows from u(t, x) =Eh

e^R0tλ(^{t s ,x+p2Bs})^ds_u

0 x+p

2B_t i

where B_t is an auxiliary Brownian motion in R^d. Proof of this formula:

given T0 >0, apply Itô formula on[0, T0] to du T0 t, x+p

2Bt and take expectation.

Corollary

If u₀ 1, then u 1 on [0, T].

Invariant regions by a systematic approach

More details on this approach can be found in the book of J. Smoller.

Let u be a classical solution of

∂u

∂t =∆u+u(1 u) +h(u)

with u₀ 2 [^{0, 1}]. By contradiction, assume that there is some(t⁰, x⁰)with u(t⁰, x⁰)2 [/ ^{0, 1}].

Assume that we can prove that there is t0 2 [^{0, T}] with the following properties:

1 t₀ >0

2 u(t, x)2 [^{0, 1}] for every x 2^Rd and every t 2 [^{0, t0}]

3 u(t₀, x₀)2 f^{0, 1}g^{for some x0} 2^Rd.

Invariant regions by a systematic approach

Let us analyze the case u(t₀, x₀) =1. We deduce:

1 r^u(t₀, x₀) =0 and ∆u(t₀, x₀) 0 (because u(t₀, x) 1 for all x and u(t0, x0) =1, and u(t0, )is C²)

2 ∂u

∂t (t₀, x₀) <0 (from the equation, because ∆u(t₀, x₀) 0, u(t0, x0) (1 u(t0, x0)) =0, h(u(t0, x0)) <0)

3 ∂u

∂t (t0, x0) 0 (because u(t, x0) 1 for t 2 [^{0, t0}]and u(t0, x0) =1).

Thus there is a contradiction. The case u(t₀, x₀) =0 is similar.