21 tumor sizes Flashcards
Here, we shall focus on two types of
non-dangerous tumours that are found in human bodies:
Tumours consisting entirely of living, but abnormal, cells, see
- Tumours with a “necrotic core”. In the core, cells are dead because the concentration of
nutrients is too low to sustain them
tumors dont produce oxygen does not transport oxygen, but rely on that of the healthy tissues which do transport it and then oxygen permeates the tissues by diffusion
(distance means less oxygen cells die)
instead of modelling the tumour cells, we model the..
model variables
assumptions
instead of modelling the tumour cells, we model the
nutrient concentration c(r, t) inside the tumor as a function of the distance from the core and time
r_2 larger radius includes living cells
r_1 radius of necrotic core
nutrient concentration in healthy tissue: c_2 (outside the tumor) and is CONSTANT
nutrients diffuse inside the tumour (avascular tumour: no production/transport inside the tumour)
tumour living cells (red) assumed consume at constant rate k > 0
minimal nutrient concentration to sustain cells: c₁ < c₂
if concentration goes below this the cells die
necrotic cells (grey) do not consume nutrients
Tumor models:
Hence, the diffusion(-like) equation obeyed by the concentration c(r) is…
∂c(r, t)/∂t =
{D∇²c(r, t) in zones with no living cells
{D∇²c(r, t) − k in zones with living cells .
(no consumption)
Tumours with no necrotic core: Modelling
Under what conditions can the tumour be formed by all living cells?
BCs?
∂C(r,t)/∂t =
diffusion- consumption
here as no necrotic core r∈[0,r₂]
(diffusion modelled with laplacian where D is the diffusion coefficient inside the tumor; consumption assumed constant)
∂C(r,t)/∂t =
** D∇²c(r,t) **- K
PDE 2nd order we need BCs:
∂c(0,t)/∂r = 0 (no nutrients exchange at the core, no flux)
c(r₂,t)=c₂ concentration in healthy tissue outside is the same
Tumor model:
∂C(r,t)/∂t =
** D∇²c(r,t) **- K
PDE 2nd order we need BCs:
∂c(0,t)/∂r = 0
c(r₂,t)=c₂
meaning
BCs
∂c(0,t)/∂r = 0 (no nutrients exchange at the core, rate of change is 0 and nutrients are constant/steady
c(r₂,t)=c₂
concentration in healthy tissue at boundary of tumor is the same as the nutrient concentration in the healthy tissue
oxygen concentration diffuses inside the tumor
Tumor model:
∂C(r,t)/∂t =
** D∇²c(r,t) **- K
PDE 2nd order we need BCs:
∂c(0,t)/∂r = 0
c(r₂,t)=c₂
meaning
Model terms
oxygen concentration diffuses inside the tumor
similar to diffusion eq
∂C(r,t)/∂t =
diffusion- consumption
is a reaction diffusion eq
diffusion from the concentration higher outside
consumption process as the cells are living
Tumours with no necrotic core: Modelling
spherical?
Tumors assumed to be spherical so spacial dependence of c(r,t) varies only with distance r from the centre with laplacian
∇²c(r,t) = (1/r²)(∂/∂r)(r² ∂c(r,t)/∂r))
Tumours with no necrotic core: stationary solution assumptions
We assume that the tumour is stationary
(i.e., stable in time, not expanding):
Under what conditions can we have a stationary tumour formed by living cells only?
stationary (time-independent) concentration c(r) = lim_t→∞ c(r, t)
Even if tumor growing in the body, the time scale at which this growth happens is much larger than time scale of diffusion of oxygen. so its slow enough to ignore at times.
condition
assumption that the tumor is stationary gives
∂c(r,t)/∂t=0
so it becomes: that c(r,t)=c(r) as it does not dep on time
concentration only depends on r
0= D∇²c(r) -k
spherical coords
∇²c(r) = (1/r²)(d/dr)(r² dc(r,t)/dr))
D(1/r²)(d/dr)(r² dc(r)/dr)) - k = 0
time not involved!
with bcs
c’(0)=0 c(r_2)=c_2
gives a solution…
Tumours with no necrotic core: stationary solution
stationary (time-independent) D(1/r²)(d/dr)(r² dc(r)/dr)) - k = 0
time not involved!
with bcs
c’(0)=0 c(r_2)=c_2
gives a solution…
valid for?
c(r) = c₂ - k/6D
quadratic function
minimal cocnentration at centre r=0
c(0)=cₘᵢₙ = c₂ - r₂²[k/6D]
graph against r
parabola increases
bounded above by c₂ at r₂
intercepts y-axis at r=0cₘᵢₙ
only valid for r in[0,r₂]
as expected from diffusion: you expect to decrease to the core inversely proportional to r².
bounded below by c₁? Yes bc we are considering healthy tissue in the tumor, all concentration inside must be above c₁.
Ie we require cₘᵢₙ≥c₁
c₂ - r₂²[k/6D]≥c₁
[6D/k] (c₂ -c₁) ≥ r₂² =R²𝒸ᵣᵢₜ
sqrt of this gives us the critical values
R²𝒸ᵣᵢₜ
tumor model:
We want healthy tissue, using our solution under stationary assumptions solution is
c(r) = c₂ - k/6D
bounded below by c₁? Yes bc we are considering healthy tissue in the tumor, all concentration inside must be above c₁.
Ie we require cₘᵢₙ≥c₁
c₂ - r₂²[k/6D]≥c₁
[6D/k] (c₂ -c₁) ≥ r₂² =R²𝒸ᵣᵢₜ
sqrt of this gives us the critical values for which the healthy tumor is still healthy (viable)
Maximal viable size
of a tumor relates to R𝒸ᵣᵢₜ
Which gives us the maximal size for a tumour composed of living cells only, given the nutrient concentration parameters:
c1, c2, D and the uptake k
What happens if the tumor r_2 is larger than R𝒸ᵣᵢₜ
we will have a tumor with a necrotic core.
Then diffusion of nutrients is not sufficient to guarantee enough nutrients at the core
Tumours with necrotic core
model: when is it true?
model
tumor r_2 is larger than R𝒸ᵣᵢₜ
then we have
∂c(r, t)/∂t =
{D∇²c(r, t) − k if r ∈ [r1, r2] (living cells)
{D∇²c(r, t) if r ∈ [0, r1) (necrotic core)
notice when modelling for the outer healthy tissue part we have the original eq, necrotic core is just diffusion no consumption
Can we say anything about the size r1 of the necrotic core for a
‘large’ tumour (r_2 > R𝒸ᵣᵢₜ)?
.
Tumours with necrotic core
model:
BCs
stationary solution
stationary so assumed….
c(r₂) = c₂ (concentration in healthy tissue outiside)
c(r₁) = c₁ (minimum concentration at the necrotic boundary so need this at the start of the red region assumed to be = as starts dying as soon as youre less than r₁)
c′(r₁) = 0 (no nutrients exchange with the necrotic core)
stationary sol: no time dep so set time derivative to 0
c(r, t) ≡ c(r)
0 = ∂c/∂t = D∇²c(r, t) − k, r ∈ [r₁, r₂]
(as working in the red healthy region only)
Tumours with necrotic core
model:
stationary solution if
If r₁ is known
If r₁ is unknown
If r₁ were known, two BCs determine the solution
If r₁ is unknown, three BCs impose conditions on r₁
Tumours with necrotic core
model:
stationary solution assumed,
solving to find it
c(r₂) = c₂
c(r₁) = c₁
c′(r₁) = 0
c(r, t) ≡ c(r)
0 = ∂c/∂t = D∇²c(r, t) − k, r ∈ [r₁, r₂]
(as working in the red healthy region only solution is for this)
You can solve the system (exercise) and find
c(r) = c₁ +[k/6D][r² +(2r₁³/r)− 3r₁²]
with the extra condition(from the 3rd BC)
r₂² ( 1 + 2r₁/r₂)(1 - r₁/r₂)²
= [6D/k] (c₂-c₁)
= R𝒸ᵣᵢₜ ²
RHS seen before
RHS shows a ratio of size of necrotic against total size and total size itself
or in terms of x=r₁/r₂ necrotic part to total ratio
(1+2x)(1-x)² = R𝒸ᵣᵢₜ ² / r₂²
relating r₁/r₂ wth R𝒸ᵣᵢₜ ² / r₂²
relating r₁/r₂ wth R𝒸ᵣᵢₜ ² / r₂²
why?
Helps us to see how the ratio
behaves as the tumor r_2 becomes larger and larger.
plotting the function for the solution
plot x=r₁/r₂ against (r₂/R𝒸ᵣᵢₜ which is larger than 1))
When tumor is at R𝒸ᵣᵢₜ we have no necrotic core
so if r_2=R𝒸ᵣᵢₜ
ratio of x = 0 when r₂/R𝒸ᵣᵢ=1
so plot(1,0) increases and approaches 1 as it tends to infinity (bounded above)
as r₂ becomes larger the ratio becomes larger and x= r₁/r₂ tends to 1
ie necrotic core grows also to equal the whole tumor, r_2
as tumor gets larger we get a thinner layer of red healthy part as the necrotic core grows also due to the concentration being below c_1 as tumor getting too big
also r₂/R=0.5
Exercise: might come up in the exam
Looking at the concentration of nutrients diffusing in a tumor
with a constant consumption of nutrients
in R³:
∂C(r,t)/∂t =
** D∇²c(r,t) **- K
with no necrotic core
stationary tumor
in R³:D∇²c(r,t) =
= D(1/r²) ∂/∂r(r²∂c(r,t)/∂r) -K
pde as time dependence
no necrotic core:
r∈[0, r₂]
c(r₂,t)=c₂
∂c(0,t)/∂r =0 (c’(r)=0 as assumed stationary not dep on time)
stationary tumor: ∂c/∂t =0
this now only dep on r not on time so no longer PDE:
0 = ∂c/∂t =
(D/r²) d/dr (r² dc(r)/dr) -k
giving (as r>0, no longer at origin)
d/dr(r² dc(r)/dr) = kr²/D
integrate once wrt r: (r>0)
r² dc(r)/dr = kr³/3D + A
dc(r)/dr= kr/3D + A/r²
integrate again:
c(r) = kr²/6D -A/r +B
solve for A and B using Bcs:
c(r₂,t)=c₂
c’(r)=0
A=0
B=c₂ - [kr₂²/6D]
c(r) = kr²/6D + c₂ - [kr₂²/6D]
upside down parabola,
maximum outside of tumor c₂ and less and less concentration further inside
