Obstacle problem - Revision history

imported>Xavi: /* Regularity of the free boundary */

2015-05-15T19:23:44Z

Regularity of the free boundary

← Older revision		Revision as of 14:23, 15 May 2015
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	===Regularity of the free boundary===		===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 <ref name="C"/>.
			In general, the contact set $\{u=\varphi\}$ could be a very bad set (like a Cantor set), even if $\varphi$ is $C^\infty$. In the classical case of the Laplacian $L=\Delta$, one rules out this possibility by assuming that the obstacle $\varphi$ satisfies $\Delta\varphi\leq c<0$. This is equivalent to studying the equation $\Delta u=1$ (and obstacle 0). Under this assumption, one can prove that the free boundary is smooth near any regular point. There can be other points, called singular points, at which the contact set have zero density. At these points the free boundary could form a cusp singularity. However, these points are known to be locally contained in a union of $C^1$ manifolds <ref name="C"/>.

			For the fractional Laplacian, the free boundary is known to be $C^{1,\alpha}$ near any regular point (those at which the function is $C^{1,s}$ an not better). Other free boundary points are classified as singular (those at which the contact set has zero density), and for $s=\frac12$ they are known to be contained in a lower dimensional $C^1$ surface. However, there may be other points on the free boundary that do not fall under those two categories.

	==Parabolic version==		==Parabolic version==

imported>Luis: /* Parabolic version */

2012-04-25T18:07:24Z

Parabolic version

← Older revision		Revision as of 13:07, 25 April 2012
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	\begin{align}		\begin{align}
	u(x,0) &= \varphi(x) \\		u(x,0) &= \varphi(x) \\
	u(\cdot,t) &= \varphi \qquad \text{on the lateral boundary } ~~\partial~~ D \times [0,T],\\		u(\cdot,t) &= \varphi \qquad \text{on the lateral boundary } D^c \times [0,T],\\
	u &\geq \varphi \qquad \text{everywhere in the domain } 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 &\geq 0 \qquad \text{everywhere in the domain } D \times [0,T],\\

imported>Luis at 17:07, 28 January 2012

2012-01-28T17:07:40Z

← Older revision		Revision as of 12:07, 28 January 2012
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	The operator $L$ can be a classical second order elliptic operator, an integro-differential one, and even a nonlinear one.		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.		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<ref name="CT"/>. This application made the obstacle problem very relevant in recent times in all its forms.

	==Existence and uniqueness of solutions==		==Existence and uniqueness of solutions==
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	===Variational inequality===		===Variational inequality===
			For equations in divergence form, the obstacle problems can be characterized by a variational inequality <ref name="KS"/>.

	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.		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.

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	==Regularity considerations==		==Regularity considerations==
	===Regularity of the solutions===		===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.		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 <ref name="F"/>.

	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.		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 $u$.		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. One can obtain it for example using the [[Harnack inequality]] centered on a free boundary point of $u$ <ref name="C"/>.

	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.		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===		===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.		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 <ref name="C"/>.

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

			== Bibliography ==

			{{reflist\|refs=
			<ref name="F">{{Citation \| last1=Frehse \| first1=J. \| title=On the regularity of the solution of a second order variational inequality \| year=1972 \| journal=Boll. Un. Mat. Ital. (4) \| volume=6 \| pages=312–315}}</ref>
			<ref name="CT">{{Citation \| last1=Cont \| first1=Rama \| last2=Tankov \| first2=Peter \| title=Financial modelling with jump processes \| publisher=Chapman & Hall/CRC, Boca Raton, FL \| series=Chapman & Hall/CRC Financial Mathematics Series \| isbn=978-1-58488-413-2 \| year=2004}}</ref>
			<ref name="C">{{Citation \| last1=Caffarelli \| first1=Luis \| title=The obstacle problem revisited \| year=1998 \| journal=The Journal of Fourier Analysis and Applications \| volume=4 \| issue=4 \| pages=383–402}}</ref>
			<ref name="KS">{{Citation \| last1=Kinderlehrer \| first1=David \| last2=Stampacchia \| first2=Guido \| title=An introduction to variational inequalities and their applications \| publisher=Society for Industrial Mathematics \| year=2000 \| volume=31}}</ref>
			}}

imported>Luis at 16:41, 28 January 2012

2012-01-28T16:41:58Z

← Older revision		Revision as of 11:41, 28 January 2012
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	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 $L = -(-\~~lap~~)^s$ 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.		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 ~~freeboundary~~===		===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}

	~~==Parabolic version==~~		The optimal stopping problem corresponds to this equation backwards in time.

imported>Luis: 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..."

2012-01-28T16:34:55Z

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..."

New page

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.

==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 $L = -(-\lap)^s$ 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 freeboundary===

==Parabolic version==