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

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

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

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