Solving the Gierer-Meinhardt model using Julia

Thu, 31 Aug 2023 00:00:00 +0000

Solving the Gierer-Meinhardt model using Julia

The Gierer-Meinhardt model is defined as follows:

$$ \frac{\partial u}{\partial t} = D_u \Delta u + \rho\frac{u^2}{v} - \mu_u u + \rho_u $$

$$ \frac{\partial v}{\partial t} = D_v \Delta v + \rho u^2 - \mu_v v + \rho_v, $$

with

$u$ being a short-range autocatalytic substance, or in other words, an activator,
$v$ being its long-range antagonist, or in other words, an inhibitor, and,
$\Delta = \sum\limits_{i = 1}^{n} \frac{\partial^2}{\partial x_i^2}$ being the n-dimensional Laplace operator.

In the Gierer-Meinhardt equations, the autocatalytic substance activates both itself and the inhibitor substance (with rate $\rho u^2)$, whereas the inhibitor function inhibits the growth of the autocatalytic substance (with rate $\frac{1}{v})$. Both substances have a natural decay rate of the form $\mu_u u$ and $\mu_v v$ respectively. Finally, both substances have an activator-independent production rate ($\rho_u$ and $\rho_v$).

For the right choice of parameters, pattern formation can be observed in the solution of the Gierer-Meinhardt model.

More info on the Gierer-Meinhardt model can be found in this scholarpedia article.

Solving the system

We’ll solve the following IBVP:

$$ \frac{\partial u}{\partial t} = D_u \Delta u + \rho\frac{u^2}{v} - \mu_u u + \rho_u $$

$$ \frac{\partial v}{\partial t} = D_v \Delta v + \rho u^2 - \mu_v v + \rho_v $$

$$ u(x,y,0) = \exp{\left(-(x-a)^2-(y-a)^2\right)}, \quad \forall (x,y) \in (0,L)^2$$ $$ v(x,y,0) = {\rm rand}(), \quad \forall (x,y) \in (0,L)^2$$ $$ \frac{\partial u}{\partial n}(x,0,t) = \frac{\partial v}{\partial n}(x,0,t) = 0, \quad \forall x \in [0,L]$$ $$ \frac{\partial u}{\partial n}(x,L,t) = \frac{\partial v}{\partial n}(x,L,t) = 0, \quad \forall x \in [0,L]$$ $$ \frac{\partial u}{\partial n}(0,y,t) = \frac{\partial v}{\partial n}(0,y,t) = 0, \quad \forall y \in [0,L]$$ $$ \frac{\partial u}{\partial n}(L,y,t) = \frac{\partial v}{\partial n}(L,y,t) = 0, \quad \forall y \in [0,L] .$$

We start by initializing the parameters

Du = 1;
Dv = 100;

ρ_u = 0.5;
ρ_v = 0;
ρ = 1;
μ_u = 1;
μ_v = 6.1;

a = 5;

L = 100;

Then, we discretize the 2D space grid

sizez = L; # size of the 2D grid
dx = 100.0 / sizez; # space step

and then we discretize time

T = 150.0; # total time
dt = 0.0003; # time step
n = floor(Int, (T / dt)); # number of iterations

nvis = 500; # saves every nvis time steps

Next we use a random seed in order for the results to be the same across multiple runs and set the initial conditions

using Random
Random.seed!(1)
U = [exp(-(x-a)^2-(y-a)^2) for x in 1:1:sizez, y in 1:1:sizez];
V = rand(sizez, sizez);

Consequently, we initialize two matrices for the inside of the grid, and five matrices that will help us compute the Laplacian

Uc = zeros(Float64, sizez - 2, sizez - 2);
Vc = copy(Uc);
Ztop = copy(Uc);
Zleft = copy(Uc);
Zbottom = copy(Uc);
Zright = copy(Uc);
Zcenter = copy(Uc);

Finally, we initialize a 3D matrix, in order to save $u$ every $nvis$ time steps

UR = zeros(sizez, sizez, floor(Int, n/nvis));

We define a function in order to discretize the Laplacian, using the broadcasting abilities of Julia and the @views macro.

@views function DLaplacian(Z) # Centered differences discretization of the Laplacian
 Ztop .= Z[1:end-2, 2:end-1]
 Zleft .= Z[2:end-1, 1:end-2]
 Zbottom .= Z[3:end, 2:end-1]
 Zright .= Z[2:end-1, 3:end]
 Zcenter .= Z[2:end-1, 2:end-1]
 return (Ztop .+ Zleft .+ Zbottom .+ Zright .- 4 .* Zcenter) ./ (dx^2)
end

DLaplacian (generic function with 1 method)

We then proceed to iterate the solution beginning from the second timestep and ending in timestep $n$.

@views begin
 j = 1
 for i in 2:n
 Uc .= U[2:end-1, 2:end-1]
 Vc .= V[2:end-1, 2:end-1]

 U[2:end-1, 2:end-1] .= Uc .+ dt .* (Du .* DLaplacian(U) .+ ρ_u .- μ_u .* Uc .+ ρ.* (Uc.^2)./Vc )
 V[2:end-1, 2:end-1] .= Vc .+ dt .* (Dv .* DLaplacian(V) .+ ρ_v .- μ_v .* Vc .+ ρ.* Uc.^2 )

 for Z in (U, V) # Neumann boundary conditions
 Z[1, :] .= Z[2, :]
 Z[end, :] .= Z[end-1, :]
 Z[:, 1] .= Z[:, 2]
 Z[:, end] .= Z[:, end-1]
 end

 if i%nvis == 0
 UR[:,:,j] .= U
 j += 1
 end
 end
end

We notice interesting spot-like patterns in the solution for $t = 1000 \cdot dt \cdot nvis$

using CairoMakie

joint_limits = (0, 17)

time_step = 1000
time = time_step * dt * nvis

fig = Figure(backgroundcolor = "#9B89B3",
 resolution = (600, 600))
ax = Axis(fig[1,1],
 title = "Solution of the Gierer-Meinhardt model \n for t=$(time)",
 xlabel = "x",
 ylabel = "y")
hmr = CairoMakie.heatmap!(ax, UR[:,:,time_step])
Colorbar(fig[:,end+1], colorrange = joint_limits, label = L"U")
fig

Of course, if we want to speed things up, we’ll want to wrap the above code in a function

using Random

@views function GiererMeinhardt()
 Du = 1;
 Dv = 100;

 ρ_u = 0.5;
 ρ_v = 0;
 ρ = 1;
 μ_u = 1;
 μ_v = 6.1;

 a = 5;

 L = 100;

 nvis = 1000

 sizez = 100 # size of the 2D grid
 dx = 100.0 / sizez # space step

 T = 150.0 # total time
 dt = 0.0003 # time step
 n = floor(Int, (T / dt)) # number of iterations

 Uc = zeros(Float64, sizez - 2, sizez - 2)
 Vc = copy(Uc)
 Ztop = copy(Uc)
 Zleft = copy(Uc)
 Zbottom = copy(Uc)
 Zright = copy(Uc)
 Zcenter = copy(Uc)

 Random.seed!(1)
 U = [exp(-(x-a)^2-(y-a)^2) for x in 1:1:sizez, y in 1:1:sizez]
 V = fill(1.0, sizez, sizez)

 UR = zeros(sizez, sizez, floor(Int, n/nvis))

 function DLaplacian(Z) # Centered differences discretization of the Laplacian
 Ztop .= Z[1:end-2, 2:end-1]
 Zleft .= Z[2:end-1, 1:end-2]
 Zbottom .= Z[3:end, 2:end-1]
 Zright .= Z[2:end-1, 3:end]
 Zcenter .= Z[2:end-1, 2:end-1]
 return (Ztop .+ Zleft .+ Zbottom .+ Zright .- 4 .* Zcenter) ./ (dx^2)
 end

 j = 1
 for i in 2:n
 Uc .= U[2:end-1, 2:end-1]
 Vc .= V[2:end-1, 2:end-1]

 U[2:end-1, 2:end-1] .= Uc .+ dt .* (Du .* DLaplacian(U) .+ ρ_u .- μ_u .* Uc .+ ρ.* (Uc.^2)./Vc )
 V[2:end-1, 2:end-1] .= Vc .+ dt .* (Dv .* DLaplacian(V) .+ ρ_v .- μ_v .* Vc .+ ρ.* Uc.^2 )

 for Z in (U, V)
 Z[1, :] .= Z[2, :]
 Z[end, :] .= Z[end-1, :]
 Z[:, 1] .= Z[:, 2]
 Z[:, end] .= Z[:, end-1]
 end

 if i%nvis == 0
 UR[:,:,j] .= U
 j += 1
 end
 end
 return UR
end

resG = GiererMeinhardt();

Finally, we make a nice little video illustrating how the pattern formed

using CairoMakie

joint_limits = (0, 17)

fig = Figure(backgroundcolor = "#9B89B3",
 resolution = (600, 600))
ax = Axis(fig[1,1],
 title = "Solution of the Gierer-Meinhardt model",
 xlabel = "x",
 ylabel = "y")
hmr = CairoMakie.heatmap!(ax, resG[:,:,1])
Colorbar(fig[:,end+1], colorrange = joint_limits, label = L"U")
fig

nframes = 10
framerate = 30
iterator = 1:2:500

output_gif = record(fig, "GiererMeinhardt.mp4", iterator;
 framerate = framerate) do t
 CairoMakie.heatmap!(ax, resG[:,:,t], colorrange = joint_limits)
end

Of course, different initial conditions and parameter values, will change the behavior of the model. So, download the notebook and do some experimenting!

Posts | Vasilis Tsilidis' page

Solving the Gierer-Meinhardt model using Julia

Solving the Gierer-Meinhardt model using Julia

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