# Obstacle problem

### From Mwiki

(Created page with "The study of the obstacle problem originated in the context of elasticity as the equations that models the shape of an elastic membrane which is pushed by an obstacle from one si...") |
|||

Line 36: | Line 36: | ||

Following the argument above it is not hard to conclude that $\Delta u \in L^\infty$ and $u \in C^{1,\alpha}$ for all $\alpha<1$. In order to obtain the sharp $C^{1,1}$ regularity, a slightly sharper estimate is needed using the [[Harnack inequality]] centered on a free boundary point of $u$. | Following the argument above it is not hard to conclude that $\Delta u \in L^\infty$ and $u \in C^{1,\alpha}$ for all $\alpha<1$. In order to obtain the sharp $C^{1,1}$ regularity, a slightly sharper estimate is needed using the [[Harnack inequality]] centered on a free boundary point of $u$. | ||

- | The case $L = -(-\ | + | The case $L = -(-\Delta)^s$ corresponds to the [[obstacle problem for the fractional Laplacian]]. In this case the reasoning above suggests that the solution is in the class $C^{2s}$. In fact, an adaptation of the ideas of the classical case can be used to prove the $C^{2s}$ regularity of solutions. However, the optimal regularity is $C^{1,s}$. This is a somewhat surprising result because the regularity exponent is higher than the order of the equation and there is no scaling argument suggesting this class. |

- | + | ||

- | + | ||

+ | ===Regularity of the free boundary=== | ||

+ | In general one can prove that the free boundary is smooth (analytic in the case of the Laplacian) in a generic point. However, there can be some exceptional points where the free boundary forms a cusp singularity. | ||

==Parabolic version== | ==Parabolic version== | ||

+ | The parabolic version of the obstacle problem is inspired in the [[optimal stopping problem]] with a deadline $T$. It corresponds to the American option pricing problem with expiration at time $T$. Because of the model, it makes sense to consider the obstacle $\varphi$ to coincide with the initial value of $u$. The precise equation is | ||

+ | \begin{align} | ||

+ | u(x,0) &= \varphi(x) \\ | ||

+ | u(\cdot,t) &= \varphi \qquad \text{on the lateral boundary } \partial D \times [0,T],\\ | ||

+ | u &\geq \varphi \qquad \text{everywhere in the domain } D \times [0,T],\\ | ||

+ | u_t - Lu &\geq 0 \qquad \text{everywhere in the domain } D \times [0,T],\\ | ||

+ | u_t - Lu &= 0 \qquad \text{wherever } u > \varphi. | ||

+ | \end{align} | ||

+ | |||

+ | The optimal stopping problem corresponds to this equation backwards in time. |

## Revision as of 16:41, 28 January 2012

The study of the obstacle problem originated in the context of elasticity as the equations that models the shape of an elastic membrane which is pushed by an obstacle from one side affecting its shape. The resulting equation for the function whose graph represents the shape of the membrane involves two distinctive regions. In the part of the domain where the membrane does not touch the obstacle, the function will satisfy an elliptic PDE. In the part of the domain where the function touches the obstacle, the function will be a supersolution of the elliptic PDE. Everywhere, the function in constrain to stay larger or equal than the value of the obstacle.

More precisely, there is an elliptic operator $L$ and a function $\varphi$ (the obstacle) so that \begin{align} u &\geq \varphi \qquad \text{everywhere in the domain } D,\\ Lu &\leq 0 \qquad \text{everywhere in the domain } D,\\ Lu &= 0 \qquad \text{wherever } u > \varphi. \end{align}

The operator $L$ can be a classical second order elliptic operator, an integro-differential one, and even a nonlinear one.

The same equation can be derived from a stochastic control problem called optimal stopping problem. This is a model in financial mathematics used to value American options. This application made the obstacle problem very relevant in recent times in all its forms.

## Contents |

## Existence and uniqueness of solutions

The existence and uniqueness of solutions follows from appropriately understanding the equation. There are different approaches that are explained below.

### Variational inequality

When the elliptic operator $L$ is the Euler-Lagrange equation of a functional $F: H^1 \to \mathbb R$, then we can identify the solution to the obstacle problem as the minimizer of the functional $F(u)$ among all functions $u$ in the set $\{u \in H^1(D) : u \geq \varphi \text{ in } D\}$. This is a minimization problem constrained to a convex set and thus it has a minimum provided $F$ is a weakly lower semicontinuous functional. If $F$ is a convex functional, then this minimizer is unique.

### Nonvariational techniques

For any elliptic equation $L$ that satisfies a comparison principle, the solution $u$ to the obstacle problem can be identified as the minimum supersolution of the equation which is above the obstacle $\varphi$.

## Optimal stopping problem

This is an excerpt from the main article on the optimal stopping problem.

Given a Levy process $X_t$ we consider the following problem. We want to find the optimal stopping time $\tau$ to maximize the expected value of $\varphi(X_{\tau})$ for some given (payoff) function $\varphi$.

In this setting, the value of $\varphi(X_{\tau})$ depends on the initial point $x$ where the Levy process starts. We can call $u(x) = \mathbb E[\varphi(X_{\tau}) | X_0 = x]$. Under this choice, the function $u$ satisfies the obstacle problem with $L$ being the generator of the Levy process $X_t$.

## Regularity considerations

### Regularity of the solutions

For the classical obstacle problem where $L$ is just the Laplacian, the solutions are known to be $C^{1,1}$. This was originally proved by Frehse.

The intuition behind the regularity result is that where $u = \varphi$ then $\Delta u = \Delta \varphi$ (strictly speaking, we can only claim this in the interior of the contact set $\{u=\varphi\}$). On the other hand, where $u > \varphi$ we have $\Delta u = 0$. Since the Laplacian jumps from $\Delta \varphi$ to $0$ across the free boundary, the second derivatives of $u$ must have a discontinuity and $C^{1,1}$ is the maximum regularity class than can be expected.

Following the argument above it is not hard to conclude that $\Delta u \in L^\infty$ and $u \in C^{1,\alpha}$ for all $\alpha<1$. In order to obtain the sharp $C^{1,1}$ regularity, a slightly sharper estimate is needed using the Harnack inequality centered on a free boundary point of $u$.

The case $L = -(-\Delta)^s$ corresponds to the obstacle problem for the fractional Laplacian. In this case the reasoning above suggests that the solution is in the class $C^{2s}$. In fact, an adaptation of the ideas of the classical case can be used to prove the $C^{2s}$ regularity of solutions. However, the optimal regularity is $C^{1,s}$. This is a somewhat surprising result because the regularity exponent is higher than the order of the equation and there is no scaling argument suggesting this class.

### Regularity of the free boundary

In general one can prove that the free boundary is smooth (analytic in the case of the Laplacian) in a generic point. However, there can be some exceptional points where the free boundary forms a cusp singularity.

## Parabolic version

The parabolic version of the obstacle problem is inspired in the optimal stopping problem with a deadline $T$. It corresponds to the American option pricing problem with expiration at time $T$. Because of the model, it makes sense to consider the obstacle $\varphi$ to coincide with the initial value of $u$. The precise equation is \begin{align} u(x,0) &= \varphi(x) \\ u(\cdot,t) &= \varphi \qquad \text{on the lateral boundary } \partial D \times [0,T],\\ u &\geq \varphi \qquad \text{everywhere in the domain } D \times [0,T],\\ u_t - Lu &\geq 0 \qquad \text{everywhere in the domain } D \times [0,T],\\ u_t - Lu &= 0 \qquad \text{wherever } u > \varphi. \end{align}

The optimal stopping problem corresponds to this equation backwards in time.