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3.6 Variational inequalities

Slide : [ obstacle problem - complementarity form - variational form || VIDEO login]

Consider the simplest obstacle problem, which arises when an elastic string held fixed at both ends is pulled over a smooth object and an equilibrium is sought without knowing a priori where are the regions of contact between the string and this object. Define the function $ S(x)\in\mathcal{C}^1(\Omega)$ measuring the elevation of the string in the interval $ \Omega=[x_L;x_R]$ and the function $ O(x)\in\mathcal{C}^1(\Omega)$ modeling the shape of the object.

Mathematically, the obstacle problem amounts to finding a solution $ S(x)$ satisfying the conditions:

  1. the string always remains above the obstacle $ S(x)\ge O(x)$ ,
  2. the string satisfies the equilibrium equation. Neglecting the inertia, this says that the string has either zero curvature (straight line between the points of contact) or a negative curvature $ S^{\prime\prime}\le 0$ (line of contact - in other words, the obstacle can only push the string up, not pull it down).
This section considers two methods for solving problems that involve inequalities.
* Linear complementary formulation.
Reassemble all the constraints in a form

$\displaystyle \mathcal{A}\mathcal{B}=0, \hspace{1cm} \mathcal{A}\ge 0, \hspace{1cm} \mathcal{B}\ge 0$ (1)

After discretization, assuming $ \mathbf{A}$ invertible and positive definite, the linear problem

$\displaystyle (\mathbf{x}-\mathbf{c})(\mathbf{A}\mathbf{x}-\mathbf{b})=0, \hspace{1cm} \mathbf{A}\mathbf{x}\ge\mathbf{b}, \hspace{1cm} \mathbf{x}\ge\mathbf{c}$ (2)

can be solved with the so-called projected SOR method, by replacing (3.5#eq.10) with

$\displaystyle \mathbf{x_{k+1}} = \max[\mathbf{c}, \omega \mathbf{x_{k}^{GS}} +(1-\omega) \mathbf{x_{k}} ]$ (3)

and making sure that the iteration starts from an initial guess $ \mathbf{x_0}\ge\mathbf{c}$ .

For the obstacle problem, the string follows either a straight line above the obstacle $ S^{\prime\prime}=0$ or fits exactly the object $ S-O=0$ ; this leads to a complementary problem

$\displaystyle S^{\prime\prime}(S-O)=0 \hspace{1cm} S-O\ge 0 \hspace{1cm} -S^{\prime\prime}\ge 0$ (4)

which be discretized using finite differences and solved with the projected SOR method (exercise 3.05).

* Variational formulation.
This approach is particularly well suited for a discretization with finite elements is and best illustrated directly with the example. Choose a test function $ \forall w\in\mathcal{V}\in\mathcal{C}^1(\Omega)$ that satisfies the same constraints as the solution $ (w-c)\ge 0$ . Having already $ (S-c)\ge 0$ and $ -S^{\prime\prime}\ge 0$ , write two inequalities
$\displaystyle \int_{x_L}^{x_R} -S^{\prime\prime}(w-c)\ge 0$      
$\displaystyle \int_{x_L}^{x_R} -S^{\prime\prime}(S-c)\ge 0$      

and subtract them

$\displaystyle \int_{x_L}^{x_R} -S^{\prime\prime}(w-S)\ge 0$ (5)

The constraint $ c$ now appears only implicitly through the fact that $ w$ and $ S$ are members of the same sub-space $ \mathcal{V}$ ; all what remains to be done is to prove that the inequality (3.6#eq.5) in fact is a strict equality if the solution satisfies the constraint. After integration by parts and discretization with linear FEMs, the problem can therefore be solved with projected SOR iterations (3.6#eq.3, exercises 3.05 and 3.07).

To prove that (3.6#eq.5) satisfies a strict equality, integrate by part and a discretize $ S(x)=\sum_j s_j e_j(x)$ and $ w(x)=\sum_j w_j e_j(x)$ with linear finite elements

$\displaystyle \int dx \left(\sum_i s_i e_i^\prime\right) \left(\sum_j (w_j-s_j)...
...m_j s_i \underbrace{\int dx \; e_i^\prime e_j^\prime}_\mathbf{M}(w_j-s_j) \ge 0$    

If the matrix is symmetric $ \mathbf{M^T}=\mathbf{M}$ as is the case with diffusion, this simply yields

$\displaystyle \mathbf{s^T M (w-s)} = \mathbf{(w-s)^T M^T s} = \mathbf{(w-s)^T M s} \ge 0$ (6)

If the problem is positive definite $ \sum_j M_{ij} s_j\ge 0$ as is the case with diffusion and the solution satisfies the constraint $ s_i-c_i\ge 0$ , the product satisfies the inequality

$\displaystyle \mathbf{(s-c)^T M s} \ge 0$ (7)

Since (3.6#eq.6) must hold for the particular choice where the test function coincides with the constraint $ \mathbf{w}=\mathbf{c}$ , the last two inequalities can only be satisfied simultaneously if they both satisfy the strict equality.$ \square$

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