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I want to solve two coupled partial differential equations on two dimension. There are two variables v and m. The geometry is a disk. The variable v diffuses inside the disk until it reaches the boundary and then it converts to variable m. Variable m then diffuses on the boundary, on the edge of the disk. Variable m does not exist inside the disk, it only exists on the boundary. In the diagram below you see the summary of the problem: enter image description here

I use the set of equations below to define the problem: enter image description here

The first equation describes the diffusion of variable v inside the disk.

The second equation describes the conversion of variable v to variable m (the term alpha*v(x,y,t)) and the diffusion of variable m on the boundary of the disk, here it is a circle.

The last equation is the boundary condition at the boundary of the disk which accounts for the conversion of variable v to variable m. On the left ∇ is the gradient operator which indicates the flux of variable v on the boundary. It will appear as the Neumann boundary condition:

NeumannValue[-1*alpha*v[x, y, t], x^2 + y^2 == 1]

Problem:

My problem is that how I should tell Mathematica that in the system of equations below (also shown above before) the first equation applies to the disk and the second equation applies to the boundary of the disk? The way I solved it below, the value of variable m is calculated on the entire of the disk which is not desired. m has value only on the boundary while it diffuses there.

enter image description here

Here is the code in Mathematica, The symmetric initial condition of v is just for simplification, otherwise the initial distribution of v does not have to be symmetric or Gaussian and in practice it should be a random distribution. Also the Neumann boundary condition in general will depend on the value of other variables which only exist on the boundary (here for simplification it is not the case). For example protein (variable) m could detach from the boundary and converts to protein (variable) v with a rate proportional to m.:

alpha = 1.0;
geometry = Disk[];

sol = NDSolveValue[{D[v[x, y, t], t] == 
     D[v[x, y, t], x, x] + D[v[x, y, t], y, y] + 
      NeumannValue[-1*alpha*v[x, y, t], x^2 + y^2 == 1],
    D[m[x, y, t], t] == 
     D[m[x, y, t], x, x] + D[m[x, y, t], y, y] + alpha*v[x, y, t], 
    m[x, y, 0] == 0, v[x, y, 0] == Exp[-((x^2 + y^2)/0.01)]}, {v, 
    m}, {x, y} \[Element] geometry, {t, 0, 10}];

v = sol[[1]];
m = sol[[2]];

ContourPlot[v[x, y, 1], {x, y} \[Element] geometry, PlotRange -> All, 
 PlotLegends -> Automatic]

enter image description here

ContourPlot[m[x, y, 10], {x, y} \[Element] geometry, PlotRange -> All,
  PlotLegends -> Automatic]

enter image description here

Adding DirichletCondition[m[x, y, t] == 0, x^2 + y^2 < 1] to enforce the value of m inside the geometry (here the disk) gives this error:

NDSolveValue::bcnop: No places were found on the boundary where x^2+y^2<1 was True, so DirichletCondition[m==0,x^2+y^2<1] will effectively be ignored.

I hope at the end I can reproduce the results of the paper below in which several proteins diffuse inside a sphere and on its surface and convert to each other on the surface. The paper is open access:

https://journals.plos.org/ploscompbiol/article?id=10.1371/journal.pcbi.1003396

Physical interpretation

The variable v and m represent two proteins. Protein v diffuses freely inside the cytosol (inside of the cell, here represented as a disk). Protein m is a membrane-bound protein that is it attaches to the cell's membrane (here the boundary of the disk) and only can exist as a membrane-bound protein. The protein v diffuses freely inside the disk and reaches the membrane or the boundary. There it converts to protein m with a rate that is proportional to the value of protein v on the membrane. The created membrane-bound protein m then diffuses on the membrane. Protein m cannot detach from the membrane and thus it must not exist in the cytosol (inside the disk).

Edit

I added this explanation to the question: The symmetric initial condition of v is just for simplification, otherwise the initial distribution of v does not have to be symmetric or Gaussian and in practice it should be a random distribution. Also the Neumann boundary condition in general will depend on the value of other variables which only exist on the boundary (here for simplification it is not the case). For example protein (variable) m could detach from the boundary and converts to protein (variable) v with a rate proportional to m.

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  • 1
    $\begingroup$ Please add your equations as MMA code as well. $\endgroup$ – MarcoB Mar 22 at 20:52
  • $\begingroup$ I added the code. $\endgroup$ – MOON Mar 22 at 21:27
  • $\begingroup$ You can formulate the problem in polar coordinates, see mathematica.stackexchange.com/questions/190754/… $\endgroup$ – Alex Trounev Mar 24 at 21:24
  • $\begingroup$ We can solve the problem in 3D and reproduce the solution from the article. $\endgroup$ – Alex Trounev Mar 24 at 22:35
  • 2
    $\begingroup$ @user21 in nature there is no 2D at all, but that does not stop you from simulating it, does it? $\endgroup$ – Vsevolod A. Mar 27 at 10:48
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+500
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Since I have the code to solve the original problem described in the article GDI-Mediated Cell Polarization in Yeast Provides Precise Spatial and Temporal Control of Cdc42 Signaling, I will give here a modification of this code for 2D. I did not manage to find the solution described in the article, since the system rather quickly evolves to an equilibrium state with all reasonable initial data. But something similar to clusters is obtained in 3D and a 2D.

Needs["NDSolve`FEM`"]; mesh = 
 ImplicitRegion[x^2 + y^2 <= R^2, {x, y}]; mesh1 = 
 ImplicitRegion[R1^2 <= x^2 + y^2 <= R^2, {x, y}];
d2 = .03; d3 = 11 ; R = 4; R1 = 
 7/2; N42 = 3000; NB = 6500; N24 = 1000; α1 = 0.2; α2 = 
 0.12 /60; α3 = 1 ; β1 = 0.266 ; β2 = 0.28 ; \
β3 = 1; γ1 = 0.2667 ; γ2 = 0.35 ; δ1 = \
0.00297;  δ2 = 0.35;
c0 = {.3, .65, .1}; m0 = {.0, .3, .65, 0.1};
C1[0][x_, y_] := 
 c0[[1]]*(1 + 
    Sum[RandomReal[{-.01, .01}]*
      Exp[-Norm[{x, y} - RandomReal[{-R, R}, 2]]^2], {i, 1, 10}]); 
C2[0][x_, y_] := 
 c0[[2]]*(1 + 
    Sum[RandomReal[{-.01, .01}]*
      Exp[-Norm[{x, y} - RandomReal[{-R, R}, 2]]^2], {i, 1, 10}]); 
C3[0][x_, y_] := 
 c0[[3]]*(1 + 
    Sum[RandomReal[{-.01, .01}]*
      Exp[-Norm[{x, y} - RandomReal[{-R, R}, 2]]^2], {i, 1, 10}]);
M1[0][x_, y_] := 
  m0[[1]]*(1 + 
     Sum[RandomReal[{-.01, .01}]*
       Exp[-Norm[{x, y} - RandomReal[{-R, R}, 2]]^2], {i, 1, 10}]);
M2[0][x_, y_] := 
 m0[[2]]*(1 + 
    Sum[RandomReal[{-.01, .01}]*
      Exp[-Norm[{x, y} - RandomReal[{-R, R}, 2]]^2], {i, 1, 10}]); 
M3[0][x_, y_] := 
 m0[[3]]*(1 + 
    Sum[RandomReal[{-.01, .01}]*
      Exp[-Norm[{x, y} - RandomReal[{-R, R}, 2]]^2], {i, 1, 10}]); 
M4[0][x_, y_] := 
 m0[[4]]*(1 + 
    Sum[RandomReal[{-.01, .01}]*
      Exp[-Norm[{x, y} - RandomReal[{-R, R}, 2]]^2], {i, 1, 10}]);
t0 = 1/2; n = 60;
Do[{C1[t], C2[t], C3[t]} = 
   NDSolveValue[{(c1[x, y] - C1[t - t0][x, y])/t0 - 
       d3*Laplacian[c1[x, y], {x, y}] == 
      NeumannValue[-C1[t - t0][x, 
           y] (β1*M4[t - t0][x, y] + β2) + β3*
         M2[t - t0][x, y], True], (c2[x, y] - C2[t - t0][x, y])/t0 - 
       d3*Laplacian[c2[x, y], {x, y}] == 
      NeumannValue[-γ1*M1[t - t0][x, y] + γ2*
         M3[t - t0][x, y], True], (c3[x, y] - C3[t - t0][x, y])/t0 - 
       d3*Laplacian[c3[x, y], {x, y}] == 
      NeumannValue[-δ1*M3[t - t0][x, y]*
         C3[t - t0][x, y] + δ2*M4[t - t0][x, y], True]}, {c1, 
     c2, c3}, {x, y} ∈ mesh, 
    Method -> {"FiniteElement", 
      InterpolationOrder -> {c1 -> 2, c2 -> 2, c3 -> 2}, 
      "MeshOptions" -> {"MaxCellMeasure" -> 0.01, "MeshOrder" -> 2}}];
  {M1[t], M2[t], M3[t], M4[t]} = 
   NDSolveValue[{(m1[x, y] - M1[t - t0][x, y])/t0 - 
       d2*Laplacian[m1[x, y], {x, y}] == -α3 M1[t - t0][x, 
         y] + β1 C1[t - t0][x, y] M4[t - t0][x, y] + 
       M2[t - t0][x, 
         y] (α2 + α1 M4[t - t0][x, y]), (m2[x, y] - 
          M2[t - t0][x, y])/t0 - 
       d2*Laplacian[m2[x, y], {x, y}] == β2 C1[t - t0][x, 
         y] + α3 M1[t - t0][x, y] - β3 M2[t - t0][x, y] +
        M2[t - t0][x, 
         y] (-α2 - α1 M4[t - t0][x, y]), (m3[x, y] - 
          M3[t - t0][x, y])/t0 - 
       d2*Laplacian[m3[x, y], {x, y}] == γ1 C2[t - t0][x, 
         y] M1[t - t0][x, y] - γ2 M3[t - t0][x, 
         y] - δ1 C3[t - t0][x, y] M3[t - t0][x, 
         y] + δ2 M4[t - t0][x, 
         y], (m4[x, y] - M4[t - t0][x, y])/t0 - 
       d2*
        Laplacian[m4[x, y], {x, y}] == δ1 C3[t - t0][x, 
         y] M3[t - t0][x, y] - δ2 M4[t - t0][x, y]}, {m1, m2, 
     m3, m4}, {x, y} ∈ mesh1, 
    Method -> {"FiniteElement", 
      InterpolationOrder -> {m1 -> 2, m2 -> 2, m3 -> 2, m4 -> 2}, 
      "MeshOptions" -> {"MaxCellMeasure" -> 0.01, 
        "MeshOrder" -> 2}}];, {t, t0, n*t0, t0}] // Quiet

In this FIG. shows how the concentration of components changes with time in volume (left) and on the membrane (right)

ListPlot[{Table[{t, C1[t][0, z] /. z -> .99*R}, {t, 0, n*t0, t0}], 
  Table[{t, C2[t][0, z] /. z -> .99*R}, {t, 0, n*t0, t0}], 
  Table[{t, C3[t][0, z] /. z -> .99*R}, {t, 0, n*t0, t0}]}, 
 PlotLegends -> Automatic]

ListPlot[{Table[{t, M1[t][0, z] /. z -> .99*R}, {t, 0, n*t0, t0}], 
  Table[{t, M2[t][0, z] /. z -> .99*R}, {t, 0, n*t0, t0}], 
  Table[{t, M3[t][0, z] /. z -> .99*R}, {t, 0, n*t0, t0}], 
  Table[{t, M4[t][0, z] /. z -> .99*R}, {t, 0, n*t0, t0}]}, 
 PlotLegends -> Automatic]

fig1

This figure shows a cluster on a membrane.

Table[DensityPlot[Evaluate[M1[t][x, y]], {x, -R, R}, {y, -R, R}, 
  PlotLegends -> Automatic, ColorFunction -> Hue, 
  PlotLabel -> Row[{"t = ", t*1.}], PlotPoints -> 50], {t, 10*t0, 
  n*t0, 10*t0}]

fig2

Simplify the code to solve the problem that MOON formulated. We use the initial data as in Henrik Schumacher answer and compare the result with his code with the options $\alpha =1,\theta =1$ and "MaxCellMeasure" -> 0.01 at `t=0.4' (points on the figure). Here we use Cartesian coordinates, and the membrane is replaced by a narrow ring

Needs["NDSolve`FEM`"]; mesh = 
 ImplicitRegion[x^2 + y^2 <= R^2, {x, y}]; mesh1 = 
 ImplicitRegion[R1^2 <= x^2 + y^2 <= R^2, {x, y}];
C0[x_, y_] := Exp[-20*Norm[{x + 1/2, y}]^2];
M0[x_, y_] := 0;
t0 = 1; d3 = 1; d2 = 1; R = 1; R1 = 9/10;
C1 = NDSolveValue[{D[c1[t, x, y], t] - 
      d3*Laplacian[c1[t, x, y], {x, y}] == 
     NeumannValue[-c1[t, x, y], True], c1[0, x, y] == C0[x, y]}, 
   c1, {t, 0, t0}, {x, y} ∈ mesh, 
   Method -> {"FiniteElement", InterpolationOrder -> {c1 -> 2}, 
     "MeshOptions" -> {"MaxCellMeasure" -> 0.01, "MeshOrder" -> 2}}];
M1 = NDSolveValue[{D[m1[t, x, y], t] - 
      d2*Laplacian[m1[t, x, y], {x, y}] == C1[t, x, y], 
    m1[0, x, y] == M0[x, y]} , 
   m1, {t, 0, t0}, {x, y} ∈ mesh1, 
   Method -> {"FiniteElement", InterpolationOrder -> {m1 -> 2}, 
     "MeshOptions" -> {"MaxCellMeasure" -> 0.01, "MeshOrder" -> 2}}];

fig4 Slightly modify the code of Michael E2 to remove osillation from the border. Compare the result with the solution of equations using the Henrik Schumacher model with $\alpha =1,\theta =1$ and "MaxCellMeasure" -> 0.01 at `t=0.4' (points on the figure) and Michael E2 model

ClearAll[b, m, v, x, y, t];
alpha = 1.0; R1 = .9;
geometry = Disk[];

sol = NDSolveValue[{D[v[x, y, t], t] == 
    D[v[x, y, t], x, x] + D[v[x, y, t], y, y] + 
     NeumannValue[-1*alpha*v[x, y, t], x^2 + y^2 == 1], 
   D[m[x, y, t], t] == 
    UnitStep[
      x^2 + y^2 - R1^2] (D[m[x, y, t], x, x] + D[m[x, y, t], y, y] + 
       alpha*v[x, y, t]), m[x, y, 0] == 0, 
   v[x, y, 0] == Exp[-20*((x + .5)^2 + y^2)]}, {v, 
   m}, {x, y} ∈ geometry, {t, 0, 10}]

vsol = sol[[1]];
msol = sol[[2]];

fig5 The concentration distribution on the membrane in our model fig6

The concentration distribution on the disk in Michael E2 model fig7

Modifier code MK, add options in NDSolve. Compare the result with the solution of equations using the Henrik Schumacher model with $\alpha =1,\theta =1$ and "MaxCellMeasure" -> 0.01 at `t=0.4' (points on the figure) and MK model. Note the good agreement of the data on the membrane (in both models, the Laplace operator on the circle is used)

alpha = 1.0;
geometry = Disk[];

{x0, y0} = {-.5, .0};

sol = NDSolve[{D[v[x, y, t], t] == 
     D[v[x, y, t], x, x] + D[v[x, y, t], y, y] + 
      NeumannValue[-1*alpha*v[x, y, t], x^2 + y^2 == 1], 
    v[x, y, 0] == Exp[-20*((x - x0)^2 + (y - y0)^2)]}, 
   v, {x, y} ∈ geometry, {t, 0, 10}, 
   Method -> {"FiniteElement", InterpolationOrder -> {v -> 2}, 
     "MeshOptions" -> {"MaxCellMeasure" -> 0.01, "MeshOrder" -> 2}}];

vsol = v /. sol[[1, 1]];

vBoundary[phi_, t_] := vsol[.99 Cos[phi], .99 Sin[phi], t]

sol = NDSolve[{D[m[phi, t], t] == 
     D[m[phi, t], {phi, 2}] + alpha*vBoundary[phi, t], 
    PeriodicBoundaryCondition[m[phi, t], phi == 2 π, 
     Function[x, x - 2 π]], m[phi, 0] == 0}, 
   m, {phi, 0, 2 π}, {t, 0, 10}];

msol = m /. sol[[1, 1]];

fig8

Finally, back to our source code. Compare the result with the solution of equations using the Henrik Schumacher model with $\alpha =1,\theta =1$ and "MaxCellMeasure" -> 0.01 at `t=0.4' (points on the figure) and our model. We note a good coincidence of data on the membrane (in both models, an explicit Euler in time is used):

Needs["NDSolve`FEM`"]; mesh = 
 ImplicitRegion[x^2 + y^2 <= R^2, {x, y}]; mesh1 = 
 ImplicitRegion[R1^2 <= x^2 + y^2 <= R^2, {x, y}];
d2 = 1; d3 = 1 ; R = 1; R1 = 9/10; 
C1[0][x_, y_] := Exp[-20*Norm[{x + 1/2, y}]^2];
M1[0][x_, y_] := 0;

t0 = 1/50; n = 20;
Do[C1[t] = 
   NDSolveValue[(c1[x, y] - C1[t - t0][x, y])/t0 - 
      d3*Laplacian[c1[x, y], {x, y}] == NeumannValue[-c1[x, y], True],
     c1, {x, y} ∈ mesh, 
    Method -> {"FiniteElement", InterpolationOrder -> {c1 -> 2}, 
      "MeshOptions" -> {"MaxCellMeasure" -> 0.01, "MeshOrder" -> 2}}];
  M1[t] = 
   NDSolveValue[(m1[x, y] - M1[t - t0][x, y])/t0 - 
      d2*Laplacian[m1[x, y], {x, y}] == C1[t][x, y] , 
    m1, {x, y} ∈ mesh1, 
    Method -> {"FiniteElement", InterpolationOrder -> {m1 -> 2}, 
      "MeshOptions" -> {"MaxCellMeasure" -> 0.01, 
        "MeshOrder" -> 2}}];, {t, t0, n*t0, t0}] // Quiet

fig9

As I promised, let's move on to the 3D model. We consider a system of 7 nonlinear equations for seven functions depending on four variables [t,x,y,z]. Three functions are defined in the whole region and four functions are defined on the border (membrane). We use an approximate model in which the membrane is replaced by a spherical layer. We have shown that in the case of 2D this approximation agrees well with other models. Initial system of equations and boundary conditions I took from the article as

fig11

We use the following notation {C1, C2, C3} = {cD, cB, cG}; {M1, M2, M3, M4} = {mT, mD, mB, mBG}. Functions {c1,c2,c3,m1,m2,m3,m4} are used at each time step. Here is the working code, but there are warnings that the solution in 3D is not unique. This example shows the formation of a cluster on a membrane. The initial data for each function are given as a constant + 10 Gaussian distribution with random parameters. The number of random parameters has little effect on the dynamics, but affects the number of clusters on the membrane.

Needs["NDSolve`FEM`"]; mesh = ImplicitRegion[x^2 + y^2 + z^2 <= R^2, {x, y, z}]; mesh1 = ImplicitRegion[(9*(R/10))^2 <= x^2 + y^2 + z^2 <= R^2, {x, y, z}]; 
d2 = 0.03; d3 = 11; R = 4; N42 = 3000; NB = 6500; N24 = 1000; α1 = 0.2; α2 = 0.12/60; α3 = 1; β1 = 0.266; β2 = 0.28; β3 = 1; γ1 = 0.2667; γ2 = 0.35; 
  δ1 = 0.00297; δ2 = 0.35; 
c0 = {3, 6.5, 1}; m0 = {3, 3, 6.5, 1}; a = 1/30; 
C1[0][x_, y_, z_] := c0[[1]] + Sum[RandomReal[{-a, a}]*Exp[-Norm[{x, y, z} - RandomReal[{-R, R}, 3]]^2], {i, 1, 10}]; 
  C2[0][x_, y_, z_] := c0[[2]] + Sum[RandomReal[{-a, a}]*Exp[-Norm[{x, y, z} - RandomReal[{-R, R}, 3]]^2], {i, 1, 10}]; 
  C3[0][x_, y_, z_] := c0[[3]] + Sum[RandomReal[{-a, a}]*Exp[-Norm[{x, y, z} - RandomReal[{-R, R}, 3]]^2], {i, 1, 10}]; 
M1[0][x_, y_, z_] := m0[[1]] + Sum[RandomReal[{-a, a}]*Exp[-Norm[{x, y, z} - RandomReal[{-R, R}, 3]]^2], {i, 1, 10}]; 
  M2[0][x_, y_, z_] := m0[[2]] + Sum[RandomReal[{-a, a}]*Exp[-Norm[{x, y, z} - RandomReal[{-R, R}, 3]]^2], {i, 1, 10}]; 
  M3[0][x_, y_, z_] := m0[[3]] + Sum[RandomReal[{-a, a}]*Exp[-Norm[{x, y, z} - RandomReal[{-R, R}, 3]]^2], {i, 1, 10}]; 
  M4[0][x_, y_, z_] := m0[[4]] + Sum[RandomReal[{-a, a}]*Exp[-Norm[{x, y, z} - RandomReal[{-R, R}, 3]]^2], {i, 1, 10}]; 
t0 = 1/10; n = 40; 
Quiet[Do[{C1[t], C2[t], C3[t]} = NDSolveValue[{(c1[x, y, z] - C1[t - t0][x, y, z])/t0 - d3*Laplacian[c1[x, y, z], {x, y, z}] == 
        NeumannValue[(-C1[t - t0][x, y, z])*(β1*M4[t - t0][x, y, z] + β2) + β3*M2[t - t0][x, y, z], True], 
       (c2[x, y, z] - C2[t - t0][x, y, z])/t0 - d3*Laplacian[c2[x, y, z], {x, y, z}] == NeumannValue[(-γ1)*M1[t - t0][x, y, z] + γ2*M3[t - t0][x, y, z], True], 
       (c3[x, y, z] - C3[t - t0][x, y, z])/t0 - d3*Laplacian[c3[x, y, z], {x, y, z}] == NeumannValue[(-δ1)*M3[t - t0][x, y, z]*C3[t - t0][x, y, z] + 
          δ2*M4[t - t0][x, y, z], True]}, {c1, c2, c3}, Element[{x, y, z}, mesh], 
      Method -> {"FiniteElement", InterpolationOrder -> {c1 -> 2, c2 -> 2, c3 -> 2}}]; {M1[t], M2[t], M3[t], M4[t]} = 
     NDSolveValue[{(m1[x, y, z] - M1[t - t0][x, y, z])/t0 - d2*Laplacian[m1[x, y, z], {x, y, z}] == (-α3)*M1[t - t0][x, y, z] + 
         β1*C1[t - t0][x, y, z]*M4[t - t0][x, y, z] + M2[t - t0][x, y, z]*(α2 + α1*M4[t - t0][x, y, z]), 
       (m2[x, y, z] - M2[t - t0][x, y, z])/t0 - d2*Laplacian[m2[x, y, z], {x, y, z}] == β2*C1[t - t0][x, y, z] + α3*M1[t - t0][x, y, z] - 
         β3*M2[t - t0][x, y, z] + M2[t - t0][x, y, z]*(-α2 - α1*M4[t - t0][x, y, z]), 
       (m3[x, y, z] - M3[t - t0][x, y, z])/t0 - d2*Laplacian[m3[x, y, z], {x, y, z}] == γ1*C2[t - t0][x, y, z]*M1[t - t0][x, y, z] - γ2*M3[t - t0][x, y, z] - 
         δ1*C3[t - t0][x, y, z]*M3[t - t0][x, y, z] + δ2*M4[t - t0][x, y, z], (m4[x, y, z] - M4[t - t0][x, y, z])/t0 - d2*Laplacian[m4[x, y, z], {x, y, z}] == 
        δ1*C3[t - t0][x, y, z]*M3[t - t0][x, y, z] - δ2*M4[t - t0][x, y, z]}, {m1, m2, m3, m4}, Element[{x, y, z}, mesh1], 
      Method -> {"FiniteElement", InterpolationOrder -> {m1 -> 2, m2 -> 2, m3 -> 2, m4 -> 2}}]; , {t, t0, n*t0, t0}]]  

The distribution of $m_T,m_D$ on the membrane

Table[DensityPlot3D[
  Evaluate[M1[t][x, y, z]], {x, -R, R}, {y, -R, R}, {z, -R, R}, 
  PlotLegends -> Automatic, ColorFunction -> Hue, 
  PlotLabel -> Row[{"t = ", t*1.}]], {t, 2*t0, n*t0, 6*t0}]

Table[DensityPlot3D[
  Evaluate[M2[t][x, y, z]], {x, -R, R}, {y, -R, R}, {z, -R, R}, 
  PlotLegends -> Automatic, ColorFunction -> Hue, 
  PlotLabel -> Row[{"t = ", t*1.}]], {t, 2*t0, n*t0, 6*t0}]

fig12

The distribution of $m_T,m_D$ on the membrane with multiple clusters fig13

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  • $\begingroup$ Do I understand correctly that you have the comsol file for this simulation? Could you upload that some place for people to inspect? $\endgroup$ – user21 Mar 27 at 6:50
  • $\begingroup$ @user21 No, it's just a Wolfram Math code that I wrote for this problem. But I see that even MOON does not show interest in the real problem, but prefers a toy model. $\endgroup$ – Alex Trounev Mar 27 at 10:48
  • 1
    $\begingroup$ @AlexTrounev Thanks for your answer. It would be nice if you could upload the code you wrote for the paper. I simplified the real problem so that others can provide answers without unnecessary complication and also I could understand their answers. $\endgroup$ – MOON Mar 28 at 14:00
  • $\begingroup$ @MOON I will prepare an answer with an explanation of the 3D model. Although it differs from 2D only in an extra variable z. You can test the 2D model and ask questions. $\endgroup$ – Alex Trounev Mar 28 at 14:48
  • $\begingroup$ @Alex Wow, you put a lot of effort into this. It is indeed reassuring to see that all these methods generate comparable results. $\endgroup$ – Henrik Schumacher Mar 29 at 9:23
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+250
$\begingroup$

Denote the disk by $\varOmega$ and its boundary by $\varGamma = \partial \varOmega$. I'd prefer to denote the function residing on the boundary by $u \colon \varGamma \to \mathbb{R}$; the function on the whole disk is called $v \colon \varOmega \to \mathbb{R}$.

Our aim is to solve the system of parabolic equations $$ \left\{ \begin{aligned} \partial_t u - c_2 \varDelta_{\varGamma} u &= \alpha \, v && \text{on $\varGamma$,} \\ \partial_t v - c_1 \varDelta_{\varOmega} v &= 0 && \text{in $\varOmega$,} \\ N v - \alpha v &= 0 && \text{on $\varGamma$.} \end{aligned} \right.$$

Spatial discretization

We integrate against the test functions $\varphi \colon \varGamma \to \mathbb{R}$ and $\psi \colon \varOmega \to \mathbb{R}$ with $\psi|_{\partial \varOmega} = 0$ and $N \psi = 0$.

(I assume that $\alpha$, $c_1$ and $c_2$ are constant.)

This leads to the following weak formulation of the PDE: $$ \begin{aligned}\frac{\mathrm{d}}{\mathrm{d}t}\int_{\varGamma} u(t,x) \, \varphi(x) \, \mathrm{vol}_{\partial \varOmega}(x) + c_2 \, \int_{\varGamma} \langle \mathrm{d} u(t,x) , \mathrm{d} \varphi(x) \rangle \, \mathrm{vol}_{\varGamma} (x) &= \alpha \int_{\varGamma} v(t,x) \, \varphi(x)\, \mathrm{vol}_{\varGamma} (x) \\ \frac{\mathrm{d}}{\mathrm{d}t}\int_{\varOmega} v(t,x) \, \psi(x) \, \mathrm{vol}_{\varOmega}(x) + c_1 \, \int_{\varOmega} \langle \mathrm{d} v(t,x) , \mathrm{d} \psi(x) \rangle \, \mathrm{vol}_{\varOmega} (x) &= 0 \\ \int_{\varGamma} \big(\tfrac{\partial v}{\partial \nu}(t,x) + \alpha v(t,x)\big) \, \varphi(x) \, \mathrm{vol}_{\varGamma} (x) &= 0 \end{aligned} $$

We discretization this in space by finite elements leading to the following entities ($\mathrm{b}$ stands for boundary):

  • time-dependent vectors $\mathbf{u}(t)$ and $\mathbf{v}(t)$,
  • stiffness matrices $\mathbf{A}$ and $\mathbf{A}_{\mathrm{b}}$; Mathematica's FEM tools can produce $\mathbf{A}$ but not $\mathbf{A}_{\mathrm{b}}$ in general; I will provide code for it below.
  • mass matrices $\mathbf{M}$ and $\mathbf{M}_{\mathrm{b}}$; same here: $\mathbf{M}$ can be produced easily; $\mathbf{M}_{\mathrm{b}}$ requires special treatment).
  • the matrix $\mathbf{N}$ encoding the Neumann operator times boundary mass matrix; this consists of those rows of $\mathbf{A}$ that belong to boundary degrees of freedom.
  • the matrix $\mathbf{D}$ encoding the Dirichlet boundary operator; this consists of those rows of the identity matrix that belong to boundary degrees of freedom, multiplied by $\mathbf{M}_{\mathrm{b}}$.

Then this reads as the following system of ODEs:

$$ \begin{aligned} \tfrac{\mathrm{d}}{\mathrm{d}t} \mathbf{M}_{\mathrm{b}} \, \mathbf{u}(t) + c_2 \, \mathbf{A}_{\mathrm{b}} \, \mathbf{u}(t) &= \alpha \, \mathbf{D} \, \mathbf{v}(t) \quad \text{for boundary vertices} \\ \tfrac{\mathrm{d}}{\mathrm{d}t} \mathbf{M} \, \mathbf{v}(t) + c_1 \, \mathbf{A} \, \mathbf{v}(t) &= 0 \quad \text{for interior(!) vertices} \\ (\mathbf{N} + \alpha \, \mathbf{D})\, \mathbf{v}(t) &= 0 \quad \text{for boundary vertices} \end{aligned} $$

Time discretization

I am going to supply code for the $\theta$-method with $\theta \in {[1/2,1]}$. For $\theta = 1/2$, this is the Crank-Nicolson scheme, while for $\theta = 1$, this boils down to the implicit Euler scheme.

We pick a time step $\tau > 0$ and set $\mathbf{u}_i = \mathbf{u}(i \, \tau)$ and $\mathbf{v}_i = \mathbf{v}(i \, \tau)$. One may think of $\mathbf{u}(t)$ and $\mathbf{v}(t)$ being the piecewise-linear interpolations of the $\mathbf{u}_i$ and the $\mathbf{v}_i$, repectively. (Purists from numerical analysis won't like this because of nuances between several Petrov-Galerkin schemes, but I am not going to argue with zealots here.)

$$ \begin{aligned} \tfrac{1}{\tau} (\mathbf{M}_{\mathrm{b}} \, \mathbf{u}_{i+1} - \mathbf{M}_{\mathrm{b}} \, \mathbf{u}_{i}) + c_2 \, (1-\theta) \, \mathbf{A}_{\mathrm{b}} \, \mathbf{u}_{i} + c_2 \, \theta \, \mathbf{A}_{\mathrm{b}} \, \mathbf{u}_{i+1} &= \alpha \, (1-\theta)\, \mathbf{D} \, \mathbf{v}_{i} + \alpha \, \theta \, \mathbf{D} \, \mathbf{v}_{i+1} &&\text{for boundary vertices} \\ \tfrac{1}{\tau}(\mathbf{M} \, \mathbf{v}_{i+1} - \mathbf{M} \, \mathbf{v}_{i}) + c_1 \, (1-\theta) \, \mathbf{A} \, \mathbf{v}_i + c_1 \, \theta \, \mathbf{A} \, \mathbf{v}_{i+1} &= 0 && \text{for interior(!) vertices} \\ (\mathbf{N} + \alpha \, \mathbf{D}) \, \mathbf{v}_{i+1} &= 0 &&\text{for boundary vertices} \end{aligned} $$ This provides us with a linear system to determine $\mathbf{u}_{i+1}$ and $\mathbf{v}_{i+1}$ from $\mathbf{u}_{i}$ and $\mathbf{v}_{i}$.

Side remark

Actually, I am not 100% sure whether the last line shouldn't better read as $$ (1-\theta) \, (\mathbf{N} + \alpha \, \mathbf{D}) \, \mathbf{v}_{i} + \theta \, (\mathbf{N} + \alpha \, \mathbf{D}) \, \mathbf{v}_{i+1} = 0. $$ However, I guess this may lead to spurious oscillations for $\theta \approx 1/2$. So I better leave it as it is.

Let's multiply by $\tau$ and let's put all expressions containing the "new" time steps $\mathbf{u}_{i+1}$ and $\mathbf{v}_{i+1}$ to the left of the equality sign and all the other terms to the right:

$$ \begin{aligned} (\mathbf{M}_{\mathrm{b}} + c_2 \, \tau \, \theta \, \mathbf{A}_{\mathrm{b}} )\, \mathbf{u}_{i+1} - \tau \, \alpha \, \theta \, \mathbf{D} \, \mathbf{v}_{i+1} &= ( \mathbf{M}_{\mathrm{b}} - c_2 \, \tau \, (1-\theta) \, \mathbf{A}_{\mathrm{b}} ) \, \mathbf{u}_{i} + \tau \, \alpha \, (1-\theta)\, \mathbf{D} \, \mathbf{v}_{i} &&\text{for boundary vertices} \\ (\mathbf{M} + c_1 \, \tau \, \theta \, \mathbf{A}) \, \mathbf{v}_{i+1} &= (\mathbf{M}- c_1 \, \tau \, (1-\theta) \, \mathbf{A}) \, \mathbf{v}_i && \text{for interior(!) vertices} \\ (\mathbf{N} + \alpha \, \mathbf{D}) \, \mathbf{v}_{i+1} &= 0 && \text{for boundary vertices} \end{aligned} $$

We may write this as a single linear system $$\mathbf{L}_+ \begin{pmatrix}\mathbf{u}_{i+1}\\\mathbf{v}_{i+1}\end{pmatrix} = \mathbf{L}_- \, \begin{pmatrix}\mathbf{u}_{i}\\\mathbf{v}_{i}\end{pmatrix} $$ with the block matrices $$ \mathbf{L}_+ = \begin{pmatrix} ( \mathbf{M}_{\mathrm{b}} + c_2 \, \tau \, \theta \, \mathbf{A}_{\mathrm{b}} ) & - \tau \, \alpha \, \theta \, \mathbf{D} \\ 0 & \mathbf{B}_+ \end{pmatrix} $$ and $$ \mathbf{L}_- = \begin{pmatrix} ( \mathbf{M}_{\mathrm{b}} - c_2 \, \tau \, (1-\theta) \, \mathbf{A}_{\mathrm{b}} ) & \tau \, \alpha \, (1-\theta)\, \mathbf{D} \\ 0 & \mathbf{B}_- \end{pmatrix} $$ where $\mathbf{B}_+$ and $\mathbf{B}_-$ encode the second and third equations: This is done by overwriting those rows of the second equations that belong to boundary degrees of freedom by the Robin-boundary conditions from the third equations; see also the implementation below.

Implementation - 2D case nD

First, we need to execute the first code block from the section "Code Dump" in this post the following code block. It provides us with tools to assemble mass and stiffness matrices for general MeshRegions.

I completely reworked this section in order to provide more convenient user interface by caching frequently used results in PropertyValues of MeshRegions.

SetAttributes[AssemblyFunction, HoldAll];

Assembly::expected = "Values list has `2` elements. Expected are `1` elements. Returning prototype.";

Assemble[pat_?MatrixQ, dims_, background_: 0.] := 
  Module[{pa, c, ci, rp, pos}, 
   pa = SparseArray`SparseArraySort@SparseArray[pat -> _, dims];
   rp = pa["RowPointers"];
   ci = pa["ColumnIndices"];
   c = Length[ci];
   pos = cLookupAssemblyPositions[Range[c], rp, Flatten[ci], pat];
   Module[{a}, 
    a = <|"Dimensions" -> dims, "Positions" -> pos, "RowPointers" -> rp, "ColumnIndices" -> ci, "Background" -> background, "Length" -> c|>;
    AssemblyFunction @@ {a}]];

AssemblyFunction /: a_AssemblyFunction[vals0_] := 
  Module[{len, expected, dims, u, vals, dat}, 
   dat = a[[1]];
   If[VectorQ[vals0], vals = vals0, vals = Flatten[vals0]];
   len = Length[vals];
   expected = Length[dat[["Positions"]]];
   dims = dat[["Dimensions"]];
   If[len === expected, 
    If[Length[dims] == 1, u = ConstantArray[0., dims[[1]]];
     u[[dat[["ColumnIndices"]]]] = AssembleDenseVector[dat[["Positions"]], vals, {dat[["Length"]]}];
     u, 
     SparseArray @@ {Automatic, dims, dat[["Background"]], {1, {dat[["RowPointers"]], dat[["ColumnIndices"]]}, AssembleDenseVector[dat[["Positions"]], vals, {dat[["Length"]]}]}}], 
    Message[Assembly::expected, expected, len];
    Abort[]]];

cLookupAssemblyPositions = Compile[{{vals, _Integer, 1}, {rp, _Integer, 1}, {ci, _Integer, 1}, {pat, _Integer, 1}}, 
   Block[{k, c, i, j}, 
    i = Compile`GetElement[pat, 1];
    j = Compile`GetElement[pat, 2];
    k = Compile`GetElement[rp, i] + 1;
    c = Compile`GetElement[rp, i + 1];
    While[k < c + 1 && Compile`GetElement[ci, k] != j, ++k];
    Compile`GetElement[vals, k]], 
   RuntimeAttributes -> {Listable}, 
   Parallelization -> True, 
   CompilationTarget -> "C", 
   RuntimeOptions -> "Speed"
   ];

AssembleDenseVector = 
  Compile[{{ilist, _Integer, 1}, {values, _Real, 1}, {dims, _Integer, 1}}, 
   Block[{A}, 
    A = Table[0., {Compile`GetElement[dims, 1]}];
    Do[
     A[[Compile`GetElement[ilist, i]]] += Compile`GetElement[values, i],
     {i, 1, Length[values]}
     ];
    A],
   CompilationTarget -> "C",
   RuntimeOptions -> "Speed"
   ];


getRegionLaplacianCombinatorics = Compile[{{ff, _Integer, 1}},
   Flatten[
    Table[
     Table[{Compile`GetElement[ff, i], Compile`GetElement[ff, j]}, {i,
        1, Length[ff]}], {j, 1, Length[ff]}],
    1],
   CompilationTarget -> "C",
   RuntimeAttributes -> {Listable},
   Parallelization -> True,
   RuntimeOptions -> "Speed"
   ];

SetAttributes[RegionLaplacianCombinatorics, HoldFirst]
RegionLaplacianCombinatorics[R_] /; Region`Mesh`Utilities`SimplexMeshQ[R] := Module[{result},
   result = PropertyValue[R, "RegionLaplacianCombinatorics"];
   If[result === $Failed,
    result = Assemble[
      Flatten[
       getRegionLaplacianCombinatorics[
        MeshCells[R, RegionDimension[R], "Multicells" -> True][[1, 1]]],
       1
       ], {1, 1} MeshCellCount[R, 0]
      ];
    R = SetProperty[R, "RegionLaplacianCombinatorics" -> result];
    ];
   result
   ];

SetAttributes[RegionElementData, HoldFirst]
RegionElementData[R_] /; Region`Mesh`Utilities`SimplexMeshQ[R] := 
  Module[{result},
   result = PropertyValue[R, "RegionElementData"];
   If[result === $Failed,
    result = Partition[ MeshCoordinates[R][[Flatten[ MeshCells[R, RegionDimension[R], "Multicells" -> True][[1, 1]]]]], RegionDimension[R] + 1
      ];
    R = SetProperty[R, "RegionElementData" -> result];
    ];
   result
   ];

SetAttributes[RegionBoundaryFaces, HoldFirst]
RegionBoundaryFaces[R_] /; Region`Mesh`Utilities`SimplexMeshQ[R] := 
  Module[{result},
   result = PropertyValue[R, "RegionBoundaryFaces"];
   If[result === $Failed,
    result = With[{n = RegionDimension[R]},
      MeshCells[R, n - 1, "Multicells" -> True][[1, 1,Random`Private`PositionsOf[Length /@ R["ConnectivityMatrix"[n - 1, n]]["AdjacencyLists"],1]]]
      ];
    R = SetProperty[R, "RegionBoundaryFaces" -> result];
    ];
   result
   ];

SetAttributes[RegionBoundaryVertices, HoldFirst]
RegionBoundaryVertices[R_] /; Region`Mesh`Utilities`SimplexMeshQ[R] :=
   Module[{result},
   result = PropertyValue[R, "RegionBoundaryVertices"];
   If[result === $Failed,
    result = DeleteDuplicates[Sort[Flatten[RegionBoundaryFaces[R]]]];
    R = SetProperty[R, "RegionBoundaryVertices" -> result];
    ];
   result
   ];

getRegionMassMatrix[n_, m_] := getRegionMassMatrix[n, m] = 
   Block[{xx, x, PP, P, UU, U, VV, V, f, Df, u, Du, g, integrand, quadraturepoints, quadratureweight, λ, simplex, center}, 
    λ = 1 - 1/Sqrt[2 + n];
    xx = Table[Indexed[x, i], {i, 1, n}];
    PP = Table[Compile`GetElement[P, i, j], {i, 1, n + 1}, {j, 1, m}];
    UU = Table[Indexed[U, i], {i, 1, n + 1}];
    f = x \[Function] Evaluate[PP[[1]] + Sum[Indexed[x, i] (PP[[i + 1]] - PP[[1]]), {i, 1, n}]];
    Df = x \[Function] Evaluate[D[f[xx], {xx}]];
    (*the Riemannian pullback metric with respect to f*)
    g = x \[Function] Evaluate[Df[xx]\[Transpose].Df[xx]];
    (*affine function u and its derivatives*)
    u = x \[Function] Evaluate[ UU[[1]] + Sum[Indexed[x, i] (UU[[i + 1]] - UU[[1]]), {i, 1, n}]];
    Du = x \[Function] Evaluate[D[u[xx], {xx}]];
    integrand = x \[Function] Evaluate[1/2 D[u[xx] u[xx] Sqrt[Abs[Det[g[xx]]]], {UU, 2}]];
    simplex = Join[ConstantArray[0, {1, n}], IdentityMatrix[n]];
    center = Mean[simplex];
    quadraturepoints = Table[λ center + (1 - λ) y, {y, simplex}];
    quadratureweight = 1/(n + 1)!;
    With[{code = N[quadratureweight Total[integrand /@ quadraturepoints]]}, 
     Compile[{{P, _Real, 2}}, code, CompilationTarget -> "C", 
      RuntimeAttributes -> {Listable}, Parallelization -> True, 
      RuntimeOptions -> "Speed"]
     ]
    ];

SetAttributes[RegionMassMatrix, HoldFirst]
RegionMassMatrix[R_] /; Region`Mesh`Utilities`SimplexMeshQ[R] := 
  Module[{result},
   result = PropertyValue[R, "RegionMassMatrix"];
   If[result === $Failed,
    result = RegionLaplacianCombinatorics[R][
      Flatten[ getRegionMassMatrix[RegionDimension[R], RegionEmbeddingDimension[R]][RegionElementData[R]]]
      ];
    R = SetProperty[R, "RegionMassMatrix" -> result];
    ];
   result
   ];


getRegionLaplacian[n_, m_] := getRegionLaplacian[n, m] = 
   Block[{xx, x, PP, P, UU, U, VV, V, f, Df, u, Du, g, integrand, quadraturepoints, quadratureweight, λ, simplex, center}, 
    λ = 1 - 1/Sqrt[2 + n];
    xx = Table[Indexed[x, i], {i, 1, n}];
    PP = Table[Compile`GetElement[P, i, j], {i, 1, n + 1}, {j, 1, m}];
    UU = Table[Indexed[U, i], {i, 1, n + 1}];
    f = x \[Function] Evaluate[PP[[1]] + Sum[Indexed[x, i] (PP[[i + 1]] - PP[[1]]), {i, 1, n}]];
    Df = x \[Function] Evaluate[D[f[xx], {xx}]];
    (*the Riemannian pullback metric with respect to f*)
    g = x \[Function] Evaluate[Df[xx]\[Transpose].Df[xx]];
    (*affine function u and its derivatives*)
    u = x \[Function] Evaluate[UU[[1]] + Sum[Indexed[x, i] (UU[[i + 1]] - UU[[1]]), {i, 1, n}]];
    Du = x \[Function] Evaluate[D[u[xx], {xx}]];
    integrand = x \[Function] Evaluate[ 1/2 D[Du[xx].Inverse[g[xx]].Du[xx] Sqrt[Abs[Det[g[xx]]]], {UU, 2}]];
    simplex = Join[ConstantArray[0, {1, n}], IdentityMatrix[n]];
    center = Mean[simplex];
    quadraturepoints = Table[λ center + (1 - λ) y, {y, simplex}];
    quadratureweight = 1/(n + 1)!;
    With[{code = N[quadratureweight Total[integrand /@ quadraturepoints]]}, 
     Compile[{{P, _Real, 2}}, 
      code, 
      CompilationTarget -> "C", 
      RuntimeAttributes -> {Listable}, 
      Parallelization -> True, 
      RuntimeOptions -> "Speed"
      ]
     ]
    ];

SetAttributes[RegionLaplacian, HoldFirst]
RegionLaplacian[R_] /; Region`Mesh`Utilities`SimplexMeshQ[R] := 
  Module[{result},
   result = PropertyValue[R, "RegionLaplacian"];
   If[result === $Failed,
    result = RegionLaplacianCombinatorics[R][
      Flatten[getRegionLaplacian[RegionDimension[R], RegionEmbeddingDimension[R]][RegionElementData[R]]]
      ];
    R = SetProperty[R, "RegionLaplacian" -> result];
    ];
   result
   ];

SetAttributes[RegionDirichletOperator, HoldFirst]
RegionDirichletOperator[R_] /; Region`Mesh`Utilities`SimplexMeshQ[R] :=
   Module[{result},
   result = PropertyValue[R, "RegionDirichletOperator"];
   If[result === $Failed,
    result = IdentityMatrix[
       MeshCellCount[R, 0],
       SparseArray,
       WorkingPrecision -> MachinePrecision
       ][[RegionBoundaryVertices[R]]];
    R = SetProperty[R, "RegionDirichletOperator" -> result];
    ];
   result
   ];

SetAttributes[RegionNeumannOperator, HoldFirst]
RegionNeumannOperator[R_] /; Region`Mesh`Utilities`SimplexMeshQ[R] := 
  Module[{result},
   result = PropertyValue[R, "RegionNeumannOperator"];
   If[result === $Failed,
    result = RegionLaplacian[R][[RegionBoundaryVertices[R]]];
    R = SetProperty[R, "RegionNeumannOperator" -> result];
    ];
   result
   ];

getRegionReactionMatrix[n_, m_] := getRegionReactionMatrix[n, m] = 
   Block[{xx, x, PP, P, UU, U, VV, V, f, Df, u, v, w, g, integrand, quadraturepoints, quadratureweights, λ, ω, simplex, center},
    xx = Table[Indexed[x, i], {i, 1, n}];
    PP = Table[Compile`GetElement[P, i, j], {i, 1, n + 1}, {j, 1, m}];
    UU = Table[Compile`GetElement[U, i], {i, 1, n + 1}];
    VV = Table[Compile`GetElement[V, i], {i, 1, n + 1}];
    f = x \[Function] Evaluate[PP[[1]] + Sum[Indexed[x, i] (PP[[i + 1]] - PP[[1]]), {i, 1, n}]];
    Df = x \[Function] Evaluate[D[f[xx], {xx}]];
    (*the Riemannian pullback metric with respect to f*)

    g = x \[Function] Evaluate[Df[xx]\[Transpose].Df[xx]];
    (*affine function u and its derivatives*)
    u = x \[Function] Evaluate[UU[[1]] + Sum[Indexed[x, i] (UU[[i + 1]] - UU[[1]]), {i, 1, n}]];
    v = x \[Function] Evaluate[VV[[1]] + Sum[Indexed[x, i] (VV[[i + 1]] - VV[[1]]), {i, 1, n}]];
    integrand = 
     x \[Function] Evaluate[1/2! D[u[xx]^2 v[xx] Sqrt[Abs[Det[g[xx]]]], {UU, 2}]];

    (*Gauss quadrature of order 3*)
    λ = (1 + n)/(3 + n);
    ω = -(1 + n)^2/4 /(2 + n);
    simplex = Join[ConstantArray[0, {1, n}], IdentityMatrix[n]];
    center = Mean[simplex];
    quadraturepoints = Join[{center}, ConstantArray[center, n + 1] λ + (1 - λ) simplex];
    quadratureweights = Join[{ω/n!}, ConstantArray[(1 - ω)/(n + 1)!, n + 1]];
    With[{code = N[Dot[quadratureweights, integrand /@ quadraturepoints]]},
     Compile[{{P, _Real, 2}, {V, _Real, 1}},
      code, 
      CompilationTarget -> "C",
      RuntimeAttributes -> {Listable},
      Parallelization -> True,
      RuntimeOptions -> "Speed"
      ]
     ]];

SetAttributes[RegionReactionMatrix, HoldFirst]
RegionReactionMatrix[R_, u_?VectorQ] /; 
   Region`Mesh`Utilities`SimplexMeshQ[R] := Module[{result},
   result = RegionLaplacianCombinatorics[R][
     Flatten[
      getRegionReactionMatrix[RegionDimension[R], RegionEmbeddingDimension[R]][
       RegionElementData[R],
       Partition[
        u[[Flatten[ MeshCells[R, RegionDimension[R], "Multicells" -> True][[1, 1]]]]],
        RegionDimension[R] + 1
        ]
       ]
      ]
     ];
   result
   ];

getRegionReactionVector[n_, m_] := getRegionReactionVector[n, m] = 
   Block[{xx, x, PP, P, UU, U, VV, V, WW, W, f, Df, u, v, w, g, integrand, quadraturepoints, quadratureweights, λ, ω, simplex, center},
    xx = Table[Indexed[x, i], {i, 1, n}];
    PP = Table[Compile`GetElement[P, i, j], {i, 1, n + 1}, {j, 1, m}];
    UU = Table[Compile`GetElement[U, i], {i, 1, n + 1}];
    VV = Table[Compile`GetElement[V, i], {i, 1, n + 1}];
    WW = Table[Compile`GetElement[W, i], {i, 1, n + 1}];
    f = x \[Function] Evaluate[PP[[1]] + Sum[Indexed[x, i] (PP[[i + 1]] - PP[[1]]), {i, 1, n}]];
    Df = x \[Function] Evaluate[D[f[xx], {xx}]];
    (*the Riemannian pullback metric with respect to f*)

    g = x \[Function] Evaluate[Df[xx]\[Transpose].Df[xx]];
    (*affine function u and its derivatives*)
    u = x \[Function] Evaluate[UU[[1]] + Sum[Indexed[x, i] (UU[[i + 1]] - UU[[1]]), {i, 1, n}]];
    v = x \[Function] Evaluate[VV[[1]] + Sum[Indexed[x, i] (VV[[i + 1]] - VV[[1]]), {i, 1, n}]];
    w = x \[Function] Evaluate[WW[[1]] + Sum[Indexed[x, i] (WW[[i + 1]] - WW[[1]]), {i, 1, n}]];
    integrand = x \[Function] Evaluate[D[u[xx] v[xx] w[xx] Sqrt[Abs[Det[g[xx]]]], {UU, 1}]];

    (*Gauss quadrature of order 3*)
    λ = (1 + n)/(3 + n);
    ω = -(1 + n)^2/4 /(2 + n);
    simplex = Join[ConstantArray[0, {1, n}], IdentityMatrix[n]];
    center = Mean[simplex];
    quadraturepoints = Join[{center}, ConstantArray[center, n + 1] λ + (1 - λ) simplex];
    quadratureweights = Join[{ω/n!}, ConstantArray[(1 - ω)/(n + 1)!, n + 1]];
    With[{code = N[Dot[quadratureweights, integrand /@ quadraturepoints]]},
     Compile[{{P, _Real, 2}, {V, _Real, 1}, {W, _Real, 1}},
      code, CompilationTarget -> "C",
      RuntimeAttributes -> {Listable},
      Parallelization -> True,
      RuntimeOptions -> "Speed"
      ]
     ]];

SetAttributes[RegionReactionVector, HoldFirst]
RegionReactionVector[R_, u_?VectorQ, v_?VectorQ] /; 
   Region`Mesh`Utilities`SimplexMeshQ[R] := Module[{result},
   result = With[{
      n = RegionDimension[R],
      flist = Flatten[MeshCells[R, RegionDimension[R], "Multicells" -> True][[1, 1]]]
      },
     AssembleDenseVector[
      flist,
      Flatten[
       getRegionReactionVector[RegionDimension[R], RegionEmbeddingDimension[R]][
        RegionElementData[R],
        Partition[u[[flist]], n + 1],
        Partition[v[[flist]], n + 1]
        ]
       ],
      {MeshCellCount[R, 0]}
      ]
     ];
   result
   ];

Application

dim = 2;
Ω = DiscretizeRegion[Ball[ConstantArray[0., dim]], MaxCellMeasure -> {1 -> 0.05}];
Ωb = RegionBoundary[Ω];

This generates the Laplacian, mass, Neumann, and Dirichlet matrices:

A = RegionLaplacian[Ω];
M = RegionMassMatrix[Ω];

Ab = RegionLaplacian[Ωb];
Mb = RegionMassMatrix[Ωb];

Dir = RegionMassMatrix[Ωb].RegionDirichletOperator[Ω];
Neu = RegionNeumannOperator[Ω];

Setting some constants...

c1 = 1.;
c2 = 1.;

h = Max[PropertyValue[{Ω, 1}, MeshCellMeasure]];
τ = 0.5 h^2;

θ = 0.5;
α = 0.1;

I made a rather conservative choice for τ; it should lead to stable evolution and maximal convergence rates for all values of θ between 0.5 and 1.. However, it might also be chosen significantly larger, in particular for θ close to 0.5.

Writing the two helper matrices Lplus and Lminus and factorizing Lplus by creating a LinearSolveFunction object S.

bvertices = RegionBoundaryVertices[Ω];
Lplus = Module[{Bplus}, 
   Bplus = M + (τ θ c1) A;
   Bplus[[bvertices]] = (Neu + α Dir);
   ArrayFlatten[{{Mb + (τ θ c2) Ab, (-α τ θ) Dir}, {0., Bplus}}]
   ];

Lminus = Module[{Bminus}, 
   Bminus = M + (-τ (1 - θ) c1) A;
   Bminus[[bvertices]] *= 0.;
   ArrayFlatten[{{(Mb + (-τ (1 - θ) c2) Ab), (α τ (1 - θ)) Dir}, {0., Bminus}}]
   ];
S = LinearSolve[Lplus];

Next, we set initial conditions, solve the evolution problem with NestList and separate the solution parts.

u0 = ConstantArray[0., Length[bvertices]];
v0 = Map[X \[Function] Exp[-20 ((X[[1]] + 1/2)^2 + (X[[2]])^2)], MeshCoordinates[Ω]];
x0 = Join[u0, v0];

x = NestList[S[Lminus.#] &, x0, 5000]; // AbsoluteTiming // First
u = x[[;; , ;; Length[bvertices]]];
v = x[[;; , Length[bvertices] + 1 ;;]];

2.12089

Up to this point, things should work well for both dim = 2 and dim = 3 (apart from generating the initial condition as one might want to use a 3D Gaussian for dim = 3).

Visualization

I haven't checked this against an analytical solution, yet (who can provide one?), but the results look quite plausible. Here is an animation showing the evolution of the functions $u$ and $v$; notice that $u$ has to be scaled up quite a bit to make it visible; so this might appear a bit unnatural at the first glance.

pts = MeshCoordinates[Ω];
bfaces = RegionBoundaryFaces[Ω];
faces = MeshCells[Ω, 2, "Multicells" -> True][[1, 1]];
maxu = Max[u];
plot[i_] := Module[{p, q}, 
   p = q = Join[pts, Partition[v[[i]], 1], 2];
   q[[bvertices, 3]] = u[[i]]/(2 maxu);
   Show[Graphics3D[{Thick, ColorData[97][1], 
      GraphicsComplex[q, Line[bfaces]], EdgeForm[], 
      FaceForm[ColorData[97][2]], Specularity[White, 30], 
      GraphicsComplex[p, Polygon[faces]]}], Axes -> True, 
    AxesLabel -> {"x", "y", "values"}, Lighting -> "Neutral", 
    PlotRange -> {0, 1}]];
Manipulate[plot[i], {i, 1, Length[v], 1}]

enter image description here

Likewise, I have not checked the 3D case for correctness, yet.

Towards the nonlinear problem

With more than two reactants, this is going to become quite messy, so I merely sketch how one should proceed from here.

The resulting chemical reaction systems typically contain parabolic equations with bilinear terms of the following form $$\left\{ \begin{aligned} \partial_t u_i - c^{(2)}_{i} \, \Delta_{\partial \varOmega} u_i &= \sum_j \alpha_{i,j}\, v_j + \sum_{j,k} C^{\varGamma,\varGamma}_{i,j,k} \, u_j \, u_k + \sum_{j,k} C^{\varGamma, \varOmega}_{i,j,k} \, u_j \, v_k && \text{on $\partial \varOmega$,} \\ \partial_t v_i - c^{(1)}_{i} \, \Delta_{\varOmega} v_i &= \sum_{j,k} C^{\varOmega,\varOmega}_{i,j,k} \, v_j \, v_k && \text{in $\varOmega$,} \\ N \, v_i + \sum_j \alpha_{j,i} \, v_i &= 0 && \text{on $\partial \varOmega$.} \end{aligned} \right. $$ That means that in the weak formulation of this system, terms of the form $$ \int_{\varGamma} u_j \, u_k \, \varphi \, \mathrm{vol}_{\varGamma}, \quad \int_{\varGamma} u_j \, v_k \, \varphi \, \mathrm{vol}_{\varGamma} \quad \text{and} \quad \int_{\varOmega} v_j \, v_k \, \psi \, \mathrm{vol}_{\varOmega} $$ will show up. Hence, one has to discretize expressions of the form $$ T(u,v,w) = \int_{M} u \, v \, w \, \mathrm{vol}_{M}, $$ where $M \subset \mathbb{R}^d$ is a submanifold and $u$, $v$, $w \colon M \to \mathbb{R}$ are functions. Thus, one needs vector representations $$ \mathbf{R}(\mathbf{v},\mathbf{w}), \quad \mathbf{R}(\mathbf{u},\mathbf{w}), \quad \text{and} \quad \mathbf{R}(\mathbf{u},\mathbf{v}) $$ of the linear forms $$ T(\cdot,v,w), \quad T(u,\cdot,w), \quad \text{and} \quad T(u,v,\cdot). $$ These are provided by the routines RegionReactionVector in the section "Implementation". The usage scheme is as simple as

RegionReactionVector[Ω, v, w]

and

RegionReactionVector[Ωb, vb, wb]

for vectors v, w and vb, wb representing functions on Ω and Ωb, respectively.

In order to compute the evolution of the system, it is also desirable to use (at least semi-)implicit methods. And for that, matrix representations $$ \mathbf{R}(\mathbf{u}), \quad \mathbf{R}(\mathbf{v}), \quad \text{and} \quad \mathbf{R}(\mathbf{w}) $$ of the bilinear forms $$ T(u,\cdot,\cdot), \quad T(\cdot,v,\cdot), \quad \text{and} \quad T(\cdot,\cdot,w) $$ are required. These are provided by the routines RegionReactionMatrix in the section "Implementation". The usage scheme is as simple as

RegionReactionMatrix[Ω, w]

and

RegionReactionMatrix[Ωb, wb]

I'd like to point out that RegionReactionMatrix has to be reassembled in each each time iteration and that I therefore also included the speeding techniques from this post of mine.

With the nonlinear terms, there is now a plethora of possibilities for the time discretization. One wouldn't try to make the time stepping fully implicit as this would require a non-linear solve in each time iteration. So one has to fiddle around a bit with semi-implicit methods. Maybe it already suffices to treat the reaction terms explicitly: This would correspond to setting $\theta = 0$ for those terms while keeping $\theta \geq \frac{1}{2}$ for all the other (linear) terms. But there are also other ways and I do not feel competent enough to tell in advance, which method will work best. Unfortunately, I also do not have the time to try it for myself.

Depending on the time discretization, also Lplus and Lminus might have to be rebuilt in each time iteration. This can be done in essentially the same fashion as I did it above by utilizing ArrayFlatten to piece the various mass-, diffusion-, and reaction matrices together.

If Lplus changes over time, a one-time factorization with LinearSolve won't be efficient anymore, and it will probably be better to employ a interative solver based on Krylov space techniques (see this thread for example).

$\endgroup$
  • $\begingroup$ The code works. It is necessary to check on the real problem in which the three functions in the volume and the four at the boundary are nonlinearly connected. $\endgroup$ – Alex Trounev Mar 27 at 11:44
  • $\begingroup$ I checked your 2D model in comparison with 4 other models (see my answer). Now check out the 3D model. $\endgroup$ – Alex Trounev Mar 28 at 16:20
  • $\begingroup$ To solve a non-linear problem it is necessary to organize an iteration process, as in this problem mathematica.stackexchange.com/questions/193700/… $\endgroup$ – Alex Trounev Mar 30 at 11:21
  • $\begingroup$ Yes, @Alex, I know. I am going to sketch that but I can do only one thing at a time... =) $\endgroup$ – Henrik Schumacher Mar 30 at 11:25
  • 1
    $\begingroup$ I think the interface is quite clear: NDSolve[{-Laplacian[u[x, y], {x, y}] == 1, DirichletCondition[u[x, y] == 0, x == 1 && y == 0]}, u, {x, y} \[Element] Disk[]]; You'd just specify a boundary only region. And in fact I suspect that some of the code already exists: NeumannValue[u[x,y]....] does operate on the boundary and contributes to the stiffness matrix. Don't hold your breath for this, but it is not completely out of the questions. $\endgroup$ – user21 Apr 3 at 5:32
8
$\begingroup$

Since your initial value for m is zero, make the derivative zero inside the disk.

ClearAll[b, m, v, x, y, t];
alpha = 1.0;
geometry = Disk[];

sol = NDSolveValue[{D[v[x, y, t], t] == 
    D[v[x, y, t], x, x] + D[v[x, y, t], y, y] + 
     NeumannValue[-1*alpha*v[x, y, t], x^2 + y^2 == 1], 
   D[m[x, y, t], t] == 
    UnitStep[
      x^2 + y^2 - 1] (D[m[x, y, t], x, x] + D[m[x, y, t], y, y] + 
       alpha*v[x, y, t]),
   m[x, y, 0] == 0, v[x, y, 0] == Exp[-((x^2 + y^2)/0.01)]}, {v, 
   m}, {x, y} ∈ geometry, {t, 0, 10}]

vsol = sol[[1]];
msol = sol[[2]];

ContourPlot[msol[x, y, 10], {x, y} ∈ geometry, 
 PlotRange -> All, PlotLegends -> Automatic]

enter image description here

With a better geometry, you should be able to control m better (here m is constrained to be zero for a disk of radius 1/2):

bm1 = ToBoundaryMesh[Disk[{0, 0}, 1], 
   MaxCellMeasure -> {"Length" -> 0.05}];
bm2 = ToBoundaryMesh[Disk[{0, 0}, 1/2], 
   MaxCellMeasure -> {"Length" -> 0.02}];
bele1 = bm1["BoundaryElements"];
bele2 = bm2["BoundaryElements"];
bmesh = ToBoundaryMesh[
  "Coordinates" -> Join[bm1["Coordinates"], bm2["Coordinates"]], 
  "BoundaryElements" -> 
   Join[bele1, 
    MapThread[#1[#2] &, {Head /@ bele2, 
      Length[bm1["Coordinates"]] + ElementIncidents[bele2]}]]]

ClearAll[b, m, v, x, y, t];
alpha = 1.0;
geometry = ToElementMesh[bmesh, MaxCellMeasure -> 0.01];

sol = NDSolveValue[{D[v[x, y, t], t] == 
    D[v[x, y, t], x, x] + D[v[x, y, t], y, y] + 
     NeumannValue[-1*alpha*v[x, y, t], x^2 + y^2 == 1], 
   D[m[x, y, t], t] == 
    UnitStep[
      x^2 + y^2 - 1/4] (D[m[x, y, t], x, x] + D[m[x, y, t], y, y] + 
       alpha*v[x, y, t]),
   m[x, y, 0] == 0, v[x, y, 0] == Exp[-((x^2 + y^2)/0.01)]}, {v, 
   m}, {x, y} ∈ geometry, {t, 0, 10}]

vsol = sol[[1]];
msol = sol[[2]];

ContourPlot[msol[x, y, 0.1], {x, y} ∈ geometry, 
 PlotRange -> All, PlotLegends -> Automatic]

enter image description here

$\endgroup$
  • $\begingroup$ Thank you for your answer. The solution msol obtained by your answer is zero on the boundary for all the time. I think it should be non-zero because the variable v converts to variable m on the boundary and there is no decay term for m. Even if somehow m at the end should be zero at least initially it should be non-zero because the first time that v reaches the boundary it converts to m. $\endgroup$ – MOON Mar 24 at 22:01
  • $\begingroup$ @MOON I adapted what you indicated: "Adding` DirichletCondition[m[x, y, t] == 0, x^2 + y^2 < 1]` to enforce the value of m inside the geometry (here the disk) gives this error:" I don't know what you meant, but you included t and I thought you meant for all time. $\endgroup$ – Michael E2 Mar 24 at 22:03
  • $\begingroup$ I tried to use DirichletCondition[m[x, y, t] == 0, x^2 + y^2 < 1] to keep m inside the disk zero while it should be non-zero at the boundary. The second part of your answer is closer to what I need. However, there is still a small problem. While m is zero in the disk with radius 1/2 and non-zero from r = 1/2 to r = 1, in the region r=1/2 to r=1 both m and v coexist. v is converted to m with a rate proportional to the value of v on the boundary (r = 1). If m exist for r<1 then m coverts from v with a rate that depends on the value of v for r<1. $\endgroup$ – MOON Mar 24 at 22:39
  • 1
    $\begingroup$ @MOON I should admit that I answered without understanding what you could mean by v is converted to m: (1) In your TeX equations, there's no PDE governing the evolution of m except on the boundary. (2) The PDE in your code has m depending on both m and v, so both would have to coexist (even though I understood you to say you don't want them to). (3) It sounds to me that you might want a single solution that behaves piecewise, differently inside the small disk than outside it. The points (1) - (3) seem to be inconsistent, so I'm hoping you can clarify. $\endgroup$ – Michael E2 Mar 24 at 22:51
  • $\begingroup$ In Latex equations (now numbered) equation 1 governs the evolution of variable v inside the disk. Equation 2 governs the evolution of the m on the boundary and conversion of v to m on the boundary with a rate proportional to the value of v on the boundary. The affix of on the boundary in equation 2 tries to tell that m exist only on the boundary and it also depends on the value of v on the boundary. I will add the physical meaning of the system to the question. $\endgroup$ – MOON Mar 25 at 9:39
8
$\begingroup$

What about:

  1. Noting that equations 1 and 3 form a complete set, and solve them first, treating the remaining equation 2 for m afterwards.

  2. Noting that your imposed initial conditions for v do not satisfy boundary conditions, i.e., they violate eq(3). If you insist to use Gaussian distribution, in this particular example the factor in the exponential can be easily calculated by hand.

  3. Writing eq(2) solely in terms of boundary parametrisation, in this case, a polar angle phi. The tricky part here for curved surfaces in more dimensions would be to express the Laplacian, however, there are recipes how to do it n-dimesnions. Anyway, for a circle this is straightforwardly done by hand.

  4. Note, that not surprisingly, our solution does not depend on 'phi' as the whole problem is rotational-symmetric.

  5. Due to numerical reasons, I have defined vBoundary on a circle with a radius slightly smaller than 1. Alternatively, one could use as a boundary an approximation of a unit circle used in the InterpolatingFunction, which would be necessary for more complex geometries anyhow.

I hope that helps with your investigations.

alpha = 1.0;
geometry = Disk[];

{x0, y0} = {.0, .0};

sol = NDSolve[{D[v[x, y, t], t] == 
D[v[x, y, t], x, x] + D[v[x, y, t], y, y] + 
NeumannValue[-1*alpha*v[x, y, t], x^2 + y^2 == 1], 
v[x, y, 0] == Exp[-(((x - x0)^2 + (y - y0)^2)/(2/alpha))]}, 
v, {x, y} \[Element] geometry, {t, 0, 10}]

sol[[1, 1]]

ContourPlot[v[x, y, 1] /. sol[[1, 1]], {x, y} \[Element] geometry, 
PlotRange -> All, PlotLegends -> Automatic]

vsol = v /. sol[[1, 1]];

vBoundary[phi_, t_] := vsol[.99 Cos[phi], .99 Sin[phi], t]

sol = NDSolve[
{D[m[phi, t], t] == D[m[phi, t], {phi, 2}] + alpha*vBoundary[phi, t],
PeriodicBoundaryCondition[m[phi, t], phi == 2 \[Pi], 
Function[x, x - 2 \[Pi]]],
m[phi, 0] == 0
},
m, {phi, 0, 2 \[Pi]}, {t, 0, 10}]

msol = m /. sol[[1, 1]]

huePlot[t_] := 
PolarPlot[1, {phi, 0, 2 Pi}, PlotStyle -> Thick, 
ColorFunction -> 
Function[{x, y, phi, r}, Hue[msol[phi, t]/msol[0, t]]], 
ColorFunctionScaling -> False]

huePlot[1]
$\endgroup$
  • $\begingroup$ Looks like a viable approach. In 1) you mean: 'treating the remaining equation 2 for m afterwards', not equation 3, right? $\endgroup$ – user21 Mar 26 at 5:00
  • $\begingroup$ @MK. Thank you. Your approach is interesting and I will keep it in mind. Unfortunately, the real problem involves boundary conditions which also depends on the value of variables which only live on the boundary. For example protein (variable) m could detach from the boundary and converts to protein (variable) v with a rate proportional to m. Then solving first only for v would not be enough because there is another flux term for v which now depends on m. $\endgroup$ – MOON Mar 26 at 10:53
  • $\begingroup$ In such case other answers posted here may be more applicable $\endgroup$ – MK. Mar 26 at 20:41
  • $\begingroup$ @user21 yes indeed, I've corrected that. $\endgroup$ – MK. Mar 26 at 20:41
  • $\begingroup$ @MK. I compared the solution of the problem according to your model and the Henrik Schumacher model (see my answer). $\endgroup$ – Alex Trounev Mar 28 at 16:26

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