Problem of the obstacle

The problem of the obstacle is a traditional problem of mechanics. To visualize this problem, a membrane should be imagined recovering an object called then obstacle (a such film cellophane recovering a joint of beef!). Indeed, this problem consists in finding a curve solution $u$ which has a very precise position compared to the obstacle and which moreover checks a property of minimization length.

For better apprehending the problem, let us consider it in dimension 1. The membrane is then a rubber band recovering an object. This rubber band which is always with the top of the object tends to minimize its length. Moreover, for small variations, to minimize its length amounts minimizing its energy, indeed parameterizing the rubber band on the interval by

\ begin {boxes}
          X (T) = T \ \ 
          there (T) = U (T)
\ end {boxes}

for

t \ in

the length of the rubber band between $-1$ and $1$ is:

l = \ int_ {- 1} ^ {1} \ sqrt {1 + u' (T) ^2} \ mathrm dt

We would like to minimize this length, i.e. to find

\ min_ {U \ in K} \ int_ {- 1} ^ {1} \ sqrt {1 + u' (T) ^2} \ mathrm dt

. However, if

u'

is sufficiently small, we have

\ sqrt {1 + u' (T) ^2} \ sim 1 + \ tfrac {1} {2} u' (T) ^2

To minimize the length of the rubber band thus amounts finding

{\ min_ {U \ in K} {\ int_ {- 1} ^ {1} \ left (1 + {\ frac {1} {2}} u' (T) ^2 \ right) \ mathrm dt}}

, what amounts minimizing

{\ frac {1} {2}} {\ int_ {- 1} ^ {1} u' (T) ^2 \ mathrm dt} = {\ frac {1} {2}} has (U, U) = {\ frac {1} {2}} E (U)

Theoretical approach

Existence and unicity

The problem of the obstacle can be seen like an application of the Théorème of Stampacchia, by considering the following proposal:

(The demonstrations are done easily by using the inequalities of Hölder and Poincaré.)

The Théorème of Stampacchia thus applies, and thus there exists single

u \ in K = \ {U \ in H_0^1 (- 1,1) \ text {such as} U \ geq \ chi \ text {almost everywhere} \}

such as:

$a (U, considering) \ \ geq \ L (considering) \ quad \ forall \ v \ in K.$

What is equivalent to say that there exists single a $u \ in K$ such as:

{\ int_ {- 1} ^1 {u' (T) (considering) '(T) \ mathrm dt}} \ \ geq \ {\ int_ {- 1} ^1 {G (T) (considering) (T) \ mathrm dt}} \ quad \ forall \ v \ in K.

Moreover, $a$ being symmetrical, then $u$ are characterized by the property:

\ begin {boxes}
     U \ in K \ \
     \ displaystyle {\ frac {1} {2}} {\ int_ {- 1} ^1 {(u' (T))^2 \ mathrm dt}} - {\ int_ {- 1} ^1 {G (T) U (T) \ mathrm dt}} = {\ min_ {v \ in K} \ left (\ frac {1} {2} {\ int_ {- 1} ^1 {(v' (T))^2 \ mathrm dt}} - {\ int_ {- 1} ^1 {G (T) v (T) \ mathrm dt}} \ right)}
  \ end {boxes}

Properties of the solution

Let us show now some properties vérifées by the solution $u$ , by using the derivative second $u$ . However $u \ in H^1_0$ , $u'$ is then a weak derivative and poses the problem then to define the derivative second. This is why should be used a more general theory: the theory of the distributions.

Intuitive search for the solution

Let us establish a reasoning in order to find a solution explicit with the problem of the obstacle where

\ chi (T) = 1-2t^2

and

g = 0

. The obstacle is thus a symmetrical parabola.

Form solution close to the edges

Comparison of energies

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