Skip to main content

differential equations - Numerically solving two coupled nonlinear PDEs


I'm interested in numerically studying vortex solutions to PDEs. By this I mean solutions for PDEs for a field $\theta$ which satisfy $\oint\nabla\theta\cdot d\vec{\ell}=2\pi m$ where $m=...,-2,-1,0,1,2,...$ and $d\vec{\ell}=rd\phi$ is a vector along a curve made of a circle of fixed radius $r$ where $\phi\in(0,2\pi)$. A solution to this is $\theta=m\phi$. As a first step in this direction, I asked about the vortex solution to the Laplace equation for $\theta$ here: Numerically solving the Laplace equation in a 2d cylinder.


The next step I'm hoping to learn is how to code two coupled, nonlinear PDEs for two fields where one of them has two opposite vortex solutions. Hopefully, after understanding, this I could generalize the code and my understanding to other similar problems.


Here I've chosen, as a toy model, the stationary equations for two fields: $\theta$, whose vortex solutions are of interest, and an additional field $\rho$ that's coupled to $\theta$. The two nonlinear PDEs are $$0=\nabla\rho\cdot\nabla\theta+\frac{\rho}{2}\nabla^{2}\theta$$ and $$0=\nabla^{2}\rho(1-(\nabla\theta)^{2})-2\rho(\nabla^{2}\theta)^{2}.$$


These are derived from the Hamiltonian $\mathcal{H}=\int dr^{2}[\frac{1}{2}\rho^{2}((\nabla\theta)^{2}-1)]$.



To avoid singularities, I want to solve them on a domain that looks like this enter image description here


where the outer radius is $R$, the inner radii are the same and equal to $R_0$, the origin is dead center between the two inner circles and the distance from the origin to the center of either of the inner circles is $d$.


The boundary conditions I'm interested in are: $$\theta(x,y)=\arctan(\frac{y}{x})$$ for $(x-d)^2+y^2=R_{0}^2$ and $$\theta(x,y)=-\arctan(\frac{y}{x})$$ for $(x+d)^2+y^2=R_{0}^2$ and $$\rho(x,y)=\rho_{0}$$ for $x^2+y^2=R^2$ where $\rho_0$ is a constant (which can be set to 1). If additional conditions are required on $\rho$ on the two inner circles then these should be two different constants.



Answer



To solve a system of nonlinear equations with FEM we need to build a converging iterative process. I use the method of the false transient.The solution of the problem converges quickly, but at the first step instability arises.


R = 4; r = 1/2; d = 1; A = 
ImplicitRegion[
x^2 + y^2 <= R^2 && (x - d)^2 + y^2 >= r^2 && (x + d)^2 + y^2 >=
r^2, {x, y}];
DiscretizeRegion[A]


rho[0][x_, y_] := 1;
theta[0][x_, y_] := 0;
t0 = 1/5; k = 5;
Do[{theta[i], rho[i]} =
NDSolveValue[{Laplacian[u[x, y], {x, y}] +
2*Grad[rho[i - 1][x, y], {x, y}].Grad[
theta[i - 1][x, y], {x, y}]/rho[i - 1][x, y] == (u[x, y] -
theta[i - 1][x, y])/t0,
Laplacian[v[x, y], {x, y}] -

2*rho[i - 1][x,
y]*(2*Grad[rho[i - 1][x, y], {x, y}].Grad[
theta[i - 1][x, y], {x, y}]/rho[i - 1][x, y])^2/(1 -
Norm[Grad[theta[i - 1][x, y], {x, y}]]^2) == (v[x, y] -
rho[i - 1][x, y])/t0,
DirichletCondition[v[x, y] == 1, x^2 + y^2 == R^2],
DirichletCondition[
u[x, y] == ArcTan[x - d, y], (x - d)^2 + y^2 == r^2],
DirichletCondition[
u[x, y] == -ArcTan[x + d, y], (x + d)^2 + y^2 == r^2]}, {u,

v}, {x, y} \[Element] A,
Method -> {"FiniteElement",
"InterpolationOrder" -> {u -> 2, v -> 2},
"MeshOptions" -> {"MaxCellMeasure" -> 0.001}}], {i, 1,
k}]; // Quiet




{ContourPlot[theta[k][x, y], {x, y} \[Element] A, Contours -> 20,

ColorFunction -> Hue, PlotRange -> All, PlotLegends -> Automatic],
ContourPlot[rho[k][x, y], {x, y} \[Element] A, Contours -> 20,
ColorFunction -> Hue, PlotRange -> All, PlotPoints -> 50,
PlotLegends -> Automatic]}

Table[DensityPlot[theta[i][x, y], {x, y} \[Element] A,
ColorFunction -> Hue, PlotRange -> All], {i, 1, k}]
Table[DensityPlot[rho[i][x, y], {x, y} \[Element] A,
ColorFunction -> Hue, PlotRange -> All, PlotPoints -> 50], {i, 1,
k}]


fig1


{Plot[theta[k][-d, y], {y, r, R}, PlotRange -> All, 
AxesLabel -> {"y", "\[Theta]"}],
Plot[rho[k][-d, y], {y, r, R}, PlotRange -> All,
AxesLabel -> {"y", "\[Rho]"}],
Plot[theta[k][d, y], {y, r, R}, PlotRange -> All,
AxesLabel -> {"y", "\[Theta]"}],
Plot[rho[k][d, y], {y, r, R}, PlotRange -> All,
AxesLabel -> {"y", "\[Rho]"}]}


fig2


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

Is there a way to do conditional matrix loop using 'continue'

I have the following: n = 3; m = 5; ww = RandomReal[{0, 0.1}, {n, n}]; uu = RandomReal[{0, 1}, {m, n}]; pp = RandomReal[{0, 1}, {n, n}]; ss = RandomInteger[{0, 5}, {m, n}]; Grid[{{"ww", "uu", "pp", "ss"}, {ww // TableForm, uu // TableForm, pp // TableForm, ss // TableForm}}, Spacings -> {5, 2}, Dividers -> All] where I would like to look at every element of matrix ss and produce a matrix tt , with zeroes at the locations in ss which have zeroes, and in all other positions do the following: tt = (-1/Subscript[ww, m]) Log[(1 - uu)/(Subscript[pp, m - 1])], where Subscript[ww, m] is the value at index of ww matrix and where Subscript[pp, m - 1] is the value at index-1 of pp matrix. So for example if the first value ever read from matrix ss happens to be 2, then value taken from matrix ww would be from the row 2, but from pp would be from row 1. Also how to tell difference between a 0 as a valid value from within the matrix elemen...