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# Variational Inequality

We present, a classical example of variational inequality.

Let us denote $$\mathcal{C} = \{ u\in H^1_0(\Omega), u \le g \}$$

The problem is:

$u = arg \min_{u\in \mathcal{C}} J(u) = \frac{1}{2} \int_\Omega \nabla u . \nabla u - \int_\Omega f u$

where $$f$$ and $$g$$ are given function.

The solution is a projection on the convex $$\mathcal{C}$$ of $$f^\star$$ for the scalar product $$((v,w)) = \int_\Omega \nabla v . \nabla w$$ of $$H^1_0(\Omega)$$ where $$f^\star$$ is solution of:

$(f^\star, v ) = \int_{\Omega}{f v}, \forall v \in H^1_0(\Omega)$

The projection on a convex satisfy clearly $$\forall v \in \mathcal{C}, \quad (( u -v , u - \tilde{f} )) \leq 0$$, and after expanding, we get the classical inequality:

$\forall v \in \mathcal{C}, \quad \int_\Omega \nabla(u -v) \nabla u \leq \int_\Omega (u-v) f$

We can also rewrite the problem as a saddle point problem:

Find $$\lambda, u$$ such that:

$\max_{\lambda\in L^2(\Omega), \lambda\geq 0} \min_{u\in H^1_0(\Omega)} \mathcal{L}(u,\lambda) = \frac{1}{2} \int_\Omega \nabla u . \nabla u - \int_\Omega f u + \int_{\Omega} \lambda (u-g)^+$

where $$((u-g)^+ = max(0,u-g)$$.

This saddle point problem is equivalent to find $$u, \lambda$$ such that:

$\left\{ \begin{array}{cc} \displaystyle \int_\Omega \nabla u . \nabla v + \lambda v^+ \,d\omega= \int_\Omega f u , &\forall v \in H^1_0(\Omega) \cr \displaystyle \int_\Omega \mu (u-g)^+ = 0 , & \forall \mu \in L^2(\Omega) , \mu \geq 0, \lambda \geq 0, \end{array}\right.$

An algorithm to solve the previous problem is:

1. k=0, and choose $$\lambda_0$$ belong $$H^{-1}(\Omega)$$

2. Loop on $$k = 0, .....$$

• set $$\mathcal{I}_{k} = \{ x \in \Omega / \lambda_{k} + c * ( u_{k+1} - g) \leq 0 \}$$

• $$V_{g,k+1} = \{ v\in H^1_0(\Omega) / v = g$$ on $${I}_{k} \}$$,

• $$V_{0,k+1} = \{ v\in H^1_0(\Omega) / v = 0$$ on $${I}_{k} \}$$,

• Find $$u_{k+1} \in V_{g,k+1}$$ and $$\lambda_{k+1} \in H^{-1}(\Omega)$$ such that

$\left\{\begin{array}{cc} \displaystyle \int_\Omega \nabla u_{k+1}. \nabla v_{k+1} \,d\omega = \int_\Omega f v_{k+1} , &\forall v_{k+1} \in V_{0,k+1} \cr \displaystyle <\lambda_{k+1},v> = \int_\Omega \nabla u_{k+1}. \nabla v - f v \,d\omega & \end{array}\right.$

where $$<,>$$ is the duality bracket between $$H^{1}_0(\Omega)$$ and $$H^{-1}(\Omega)$$, and $$c$$ is a penalty constant (large enough).

Now how to do that in FreeFEM? The full example is:

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Variational inequality

  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 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 load "medit" // Parameters real eps = 1e-5; real c = 1000; //penalty parameter of the algoritm real tgv = 1e30; //a huge value for exact penalization func f = 1; //right hand side function func fd = 0; //Dirichlet boundary condition function // Mesh mesh Th = square(20, 20); // Fespace fespace Vh(Th, P1); int n = Vh.ndof; //number of degree of freedom Vh uh, uhp; //u^n+1 and u^n Vh Ik; //to define the set where the containt is reached. Vh g = 0.05; //discret function g Vh lambda = 0; // Problem varf a (uh, vh) = int2d(Th)( dx(uh)*dx(vh) + dy(uh)*dy(vh) ) - int2d(Th)( f*vh ) + on(1, 2, 3, 4, uh=fd) ; //the mass Matrix construction varf vM (uh, vh) = int2d(Th)(uh*vh); //two versions of the matrix of the problem matrix A = a(Vh, Vh, tgv=tgv, solver=CG); //one changing matrix AA = a(Vh, Vh, solver=CG); //one for computing residual matrix M = vM(Vh, Vh); //to do a fast computing of L^2 norm : sqrt(u'*(w=M*u)) real[int] Aiin(n); real[int] Aii = A.diag; //get the diagonal of the matrix real[int] rhs = a(0, Vh, tgv=tgv); // Initialization Ik = 0; uhp = -tgv; // Loop for(int iter = 0; iter < 100; ++iter){ // Update real[int] b = rhs; //get a copy of the Right hand side real[int] Ak(n); //the complementary of Ik ( !Ik = (Ik-1)) Ak = 1.; Ak -= Ik[]; //adding new locking condition on b and on the diagonal if (Ik ==1 ) b = Ik[] .* g[]; b *= tgv; b -= Ak .* rhs; Aiin = Ik[] * tgv; Aiin += Ak .* Aii; //set Aii= tgv i in Ik A.diag = Aiin; //set the matrix diagonal set(A, solver=CG); //important to change preconditioning for solving // Solve uh[] = A^-1* b; //solve the problem with more locking condition // Residual lambda[] = AA * uh[]; //compute the residual (fast with matrix) lambda[] += rhs; //remark rhs = -\int f v Ik = (lambda + c*( g- uh)) < 0.; //the new locking value // Plot plot(Ik, wait=true, cmm=" lock set ", value=true, fill=true); plot(uh, wait=true, cmm="uh"); // Error //trick to compute L^2 norm of the variation (fast method) real[int] diff(n), Mdiff(n); diff = uh[] - uhp[]; Mdiff = M*diff; real err = sqrt(Mdiff'*diff); cout << "|| u_{k=1} - u_{k} ||_2 = " << err << endl; // Stop test if(err < eps) break; // Update uhp[] = uh[]; } // Plot medit("uh", Th, uh); 

Note

As you can see on this example, some vector, or matrix operator are not implemented so a way is to skip the expression and we use operator +=, -=` to merge the result.

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