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Pure Convection : The Rotating Hill

Summary: Here we will present two methods for upwinding for the simplest convection problem. We will learn about Characteristics-Galerkin and Discontinuous-Galerkin Finite Element Methods.

Let \(\Omega\) be the unit disk centered at \((0,0)\); consider the rotation vector field

\[\mathbf{u} = [u1,u2], \qquad u_1 = y,\quad u_2 = -x\]

Pure convection by \(\mathbf{u}\) is

\[\begin{split}\begin{array}{rcl} \partial_t c + \mathbf{u}.\nabla c &= 0 &\hbox{ in } \Omega\times(0,T)\\ c (t=0) &= c ^0 &\hbox{ in } \Omega. \end{array}\end{split}\]

The exact solution \(c(x_t,t)\) at time \(t\) en point \(x_t\) is given by:

\[c(x_t,t)=c^0(x,0)\]

where \(x_t\) is the particle path in the flow starting at point \(x\) at time \(0\). So \(x_t\) are solutions of

\[\dot{x_t} = u(x_t), \quad\ x_{t=0} =x , \quad\mbox{where}\quad \dot{x_t} = \frac{\text{d} ( x_t }{\text{d} t}\]

The ODE are reversible and we want the solution at point \(x\) at time \(t\) ( not at point \(x_t\)) the initial point is \(x_{-t}\), and we have

\[c(x,t)=c^0(x_{-t},0)\]

The game consists in solving the equation until \(T=2\pi\), that is for a full revolution and to compare the final solution with the initial one; they should be equal.

Solution by a Characteristics-Galerkin Method

In FreeFEM there is an operator called convect([u1,u2], dt, c) which compute \(c\circ X\) with \(X\) is the convect field defined by \(X(x)= x_{dt}\) and where \(x_\tau\) is particule path in the steady state velocity field \(\mathbf{u}=[u1,u2]\) starting at point \(x\) at time \(\tau=0\), so \(x_\tau\) is solution of the following ODE:

\[\dot{x}_\tau = u(x_\tau), {x}_{\tau=0}=x.\]

When \(\mathbf{u}\) is piecewise constant; this is possible because \(x_\tau\) is then a polygonal curve which can be computed exactly and the solution exists always when \(\mathbf{u}\) is divergence free; convect returns \(c(x_{df})=C\circ X\).

 1// Parameters
 2real dt = 0.17;
 3
 4// Mesh
 5border C(t=0., 2.*pi) {x=cos(t); y=sin(t);};
 6mesh Th = buildmesh(C(100));
 7
 8// Fespace
 9fespace Uh(Th, P1);
10Uh cold, c = exp(-10*((x-0.3)^2 +(y-0.3)^2));
11Uh u1 = y, u2 = -x;
12
13// Time loop
14real t = 0;
15for (int m = 0; m < 2.*pi/dt; m++){
16    t += dt;
17    cold = c;
18    c = convect([u1, u2], -dt, cold);
19    plot(c, cmm=" t="+t +", min="+c[].min+", max="+c[].max);
20}

Note

3D plots can be done by adding the qualifyer dim=3 to the plot instruction.

The method is very powerful but has two limitations:

  • it is not conservative

  • it may diverge in rare cases when \(|\mathbf{u}|\) is too small due to quadrature error.

Solution by Discontinuous-Galerkin FEM

Discontinuous Galerkin methods take advantage of the discontinuities of \(c\) at the edges to build upwinding. There are may formulations possible. We shall implement here the so-called dual-\(P_1^{DC}\) formulation (see [ERN2006]):

\[\int_\Omega(\frac{c^{n+1}-c^n}{\delta t} +u\cdot\nabla c)w +\int_E(\alpha|n\cdot u|-\frac 12 n\cdot u)[c]w =\int_{E_\Gamma^-}|n\cdot u| cw~~~\forall w\]

where \(E\) is the set of inner edges and \(E_\Gamma^-\) is the set of boundary edges where \(u\cdot n<0\) (in our case there is no such edges). Finally \([c]\) is the jump of \(c\) across an edge with the convention that \(c^+\) refers to the value on the right of the oriented edge.

 1// Parameters
 2real al=0.5;
 3real dt = 0.05;
 4
 5// Mesh
 6border C(t=0., 2.*pi) {x=cos(t); y=sin(t);};
 7mesh Th = buildmesh(C(100));
 8
 9// Fespace
10fespace Vh(Th,P1dc);
11Vh w, ccold, v1 = y, v2 = -x, cc = exp(-10*((x-0.3)^2 +(y-0.3)^2));
12
13// Macro
14macro n() (N.x*v1 + N.y*v2) // Macro without parameter
15
16// Problem
17problem Adual(cc, w)
18    = int2d(Th)(
19          (cc/dt+(v1*dx(cc)+v2*dy(cc)))*w
20    )
21    + intalledges(Th)(
22          (1-nTonEdge)*w*(al*abs(n)-n/2)*jump(cc)
23    )
24    - int2d(Th)(
25          ccold*w/dt
26    )
27    ;
28
29// Time iterations
30for (real t = 0.; t < 2.*pi; t += dt){
31    ccold = cc;
32    Adual;
33    plot(cc, fill=1, cmm="t="+t+", min="+cc[].min+", max="+ cc[].max);
34}
35
36// Plot
37real [int] viso = [-0.2, -0.1, 0., 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1., 1.1];
38plot(cc, wait=1, fill=1, ps="ConvectCG.eps", viso=viso);
39plot(cc, wait=1, fill=1, ps="ConvectDG.eps", viso=viso);

Note

New keywords: intalledges to integrate on all edges of all triangles

\[\mathtt{intalledges}(\mathtt{Th}) \equiv \sum_{T\in\mathtt{Th}}\int_{\partial T }\]

(so all internal edges are see two times), nTonEdge which is one if the triangle has a boundary edge and two otherwise, jump to implement \([c]\).

Results of both methods are shown on Fig. 17 nad Fig. 18 with identical levels for the level line; this is done with the plot-modifier viso.

Notice also the macro where the parameter \(\mathbf{u}\) is not used (but the syntax needs one) and which ends with a //; it simply replaces the name n by (N.x*v1+N.y*v2). As easily guessed N.x,N.y is the normal to the edge.

convectCG

Fig. 17 The rotating hill after one revolution with Characteristics-Galerkin

convectDG

Fig. 18 The rotating hill after one revolution with Discontinuous \(P_1\) Galerkin

Rotating hill

Now if you think that DG is too slow try this:

 1// Mesh
 2border C(t=0., 2.*pi) {x=cos(t); y=sin(t);};
 3mesh Th = buildmesh(C(100));
 4
 5fespace Vh(Th,P1);//P1,P2,P0,P1dc,P2dc, uncond stable
 6
 7Vh vh,vo,u1 = y, u2 = -x, v = exp(-10*((x-0.3)^2 +(y-0.3)^2));
 8real dt = 0.03,t=0, tmax=2*pi, al=0.5, alp=200;
 9
10problem  A(v,vh) = int2d(Th)(v*vh/dt-v*(u1*dx(vh)+u2*dy(vh)))
11  + intalledges(Th)(vh*(mean(v)*(N.x*u1+N.y*u2)
12                   +alp*jump(v)*abs(N.x*u1+N.y*u2)))
13  + int1d(Th,1)(((N.x*u1+N.y*u2)>0)*(N.x*u1+N.y*u2)*v*vh)
14  - int2d(Th)(vo*vh/dt);
15
16varf  Adual(v,vh) = int2d(Th)((v/dt+(u1*dx(v)+u2*dy(v)))*vh)
17  + intalledges(Th)((1-nTonEdge)*vh*(al*abs(N.x*u1+N.y*u2)
18                             -(N.x*u1+N.y*u2)/2)*jump(v));
19
20varf rhs(vo,vh)= int2d(Th)(vo*vh/dt);
21
22real[int] viso=[-0.1,0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1,1.1];
23
24matrix AA=Adual(Vh,Vh,solver=GMRES);
25matrix BB=rhs(Vh,Vh);
26
27for ( t=0; t< tmax ; t+=dt)
28{
29   vo[]=v[];
30   vh[]=BB*vo[];
31   v[]=AA^-1*vh[];
32   plot(v,fill=0,viso=viso,cmm=" t="+t + ", min=" + v[].min + ", max=" +  v[].max);
33};
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