imported>Luis |
imported>Hector |
Line 1: |
Line 1: |
| =Lecture 6=
| | {{stub}} |
|
| |
|
| ==The obstacle problem==
| | The Hele-Shaw model describes an incompressible flow lying between two nearby horizontal plates<ref name="MR0097227"/>. The following equations are given for a non-negative pressure $u$, supported in in a time dependent domain, |
| | |
| The [[obstacle problem]] consists in finding the smallest super-solution to a non local equation which is constrained to remain above a given obstacle $\varphi$. The equation reads | |
| \begin{align*} | | \begin{align*} |
| &Lu \leq 0 \ \text{ in } \Omega, \\ | | \Delta u &= 0 \text{ in } \Omega^+ = \{u>0\}\cap \Omega\\ |
| &u \geq \varphi \ \text{ in } \Omega, \\
| | \frac{\partial_t u}{|Du|} &= |Du| \text{ on } \partial \{u>0\}\cap \Omega |
| &Lu = 0 \ \text{ wherever } u>\varphi. | |
| \end{align*} | | \end{align*} |
| | The first equation expresses the incompressibility of the fluid. The second equation, also known as the free boundary condition, says that the normal speed of the inter-phase (left-hand side) is the velocity of the fluid (right-hand side). |
| | Particular solutions are given for instance by the planar profiles |
| | \[ |
| | P(x,t) = a(t)(x_n-A(t))_+ \qquad\text{where}\qquad A(t) = \int_t^0 a(s)ds \qquad\text{and}\qquad a(t)>0 |
| | \] |
|
| |
|
| In order to have a well posed problem, it should be complemented with a boundary condition. For example, we can consider the Dirichlet condition $u=\varphi$ in $\R^n \setminus \Omega$.
| | Non-local aspects of the equation can be appreciated by noticing that a given deformation of the domain $\Omega^+$ affects all the values of $|Du|$, at least in the corresponding connected component. To be more precise let us also formally show that the linearization about a planar profile leads to a fractional heat equation of order one. |
| | |
| The definition of the problem already tells us how to prove the existence of the solution: we use [[Perron's method]]. Under reasonable assumptions on the non local operator $L$, it is possible to prove the uniqueness of solutions as well.
| |
| | |
| The regularity of both the solution $u$ and the free boundary $\partial \{u=\varphi\}$ is a much more delicate subject. It is in fact not well understood for generic non local operators $L$, even linear.
| |
| | |
| Note that the equation can be written as a [[Bellman equation]],
| |
| \[ \max(Lu,\varphi-u) = 0 \ \text{ in } \Omega.\]
| |
| | |
| The equation models a problem in [[stochastic control]] known as the [[optimal stopping problem]]. We follow a [[Levy process]] $X(t)$ with generator $L$ (assume it is linear). We are allowed to stop at any time while $X(t) \in \Omega$ but we must stop if at any time $X(t) \notin \Omega$. Whenever we stop at a time $\tau$, we are given the payoff $\varphi(X(\tau))$. Our objective is to maximize the expected payoff by choosing the best stopping time
| |
| \[ u(x) := \sup_{\text{stopping time }\tau} \mathbb E \big[ \varphi(X(\tau)) \vert X(0)=x \big].\]
| |
| The stopping time $\tau$ must fulfill the probabilistic definition of ''stopping time''. That is, it has to be measurable with respect to the filtration associated with $X$. In plain words, our decision to stop or continue must be based only on the current position $X(t)$ (the past is irrelevant from the Markovian assumption and the future cannot be predicted).
| |
| | |
| Let us explain heuristically why $u$ solves the obstacle problem. For every $y \notin \Omega$, we are forced to stop the process immediately, thus naturally $u(x) = \varphi(x)$. If $x \in \Omega$, we have the choice to either stop of follow the process. If we choose to stop, we get $u(x)=\varphi(x)$. If we choose to continue, we get $Lu(x)=0$. Moreover, we make those choices because the other choice would give us a worse expectation. That is, if we choose to stop at $x$ then $Lu(x) \leq 0$, and if we choose to continue at $x$ then $u(x) \geq \varphi(x)$. These are all the conditions that define the obstacle problem.
| |
| | |
| The regularity is quite delicate even in the simplest cases. We will start by only considering the case $L = -(-\Delta)^s$ for $s \in (0,1)$. Moreover, we will take $\Omega = \R^n$ to avoid issues concerning the boundary. The problem is well posed in the full space $\R^n$ if $\varphi$ is compactly supported and $n > 1$.
| |
| | |
| === Lipschitz regularity and semiconvexity ===
| |
| | |
| The next two propositions hold for generic non local operators $L$. In fact not even the linearity of $L$ is used, only the convexity of $L$ is necessary for the second proposition (the one on semiconvexity).
| |
| | |
| The first proposition says in particular that if $\varphi$ is Lipschitz, then also $u$ is Lipschitz with the same seminorm.
| |
|
| |
|
| {{Proposition}}
| | Let $u = P + \varepsilon v$. Then $u$ and $P$ harmonic in their positivity sets imply $v$ harmonic in the intersection, notice that as $\varepsilon\searrow0$, $v$ becomes harmonic in $\{x_n>A(t)\}$. On the other hand, the free boundary relation over $\{x_n=A(t)\}$ gives |
| Assume $\varphi$ has a modulus of continuity $\omega$, then also $u$ has the modulus of continuity $\omega$.
| | \[ |
| {{end}}
| | \frac{a^2+\varepsilon \partial_t v}{|ae_n+\varepsilon Dv|} = |ae_n+\varepsilon Dv| \qquad\Rightarrow\qquad \partial_t v = 2a\partial_n v+\varepsilon |Dv|^2 |
| | | \] |
| {{proof}}
| | By taking the reparametrization $w(x,t) = v(x+Ae_n,t)$ and letting $\varepsilon\searrow0$ we get that $w$ satisfies |
| We know that for all $x$ and $y$ in $\R^n$,
| |
| \[ \varphi(x+y) + \omega(|y|) \geq \varphi(x).\]
| |
| Since $u \geq \varphi$, we also have
| |
| \[ u(x+y) + \omega(|y|) \geq \varphi(x).\]
| |
| Fixing $y$, we can take the left hand side of the inequality as a function of $x$, and we realize it is a supersolution of the equation which is above $\varphi$. Therefore, from the definition of the obstacle problem, it is also above $u$ (recall that $u$ was the minimum of such supersolutions). Then
| |
| \[ u(x+y) + \omega(|y|) \geq u(x).\]
| |
| The fact that this holds for any values of $x,y \in \R^n$ means that $u$ has the modulus of continuity $\omega$.
| |
| {{endproof}}
| |
| | |
| The following proposition implies that if $\varphi$ is smooth, then $u$ is semiconvex in the sense that $D^2 u \geq -C \, \mathrm{I}$. By this inequality we mean that the function $u(x) + C \frac{|x|^2}2$ is convex.
| |
| | |
| {{proposition}}
| |
| Assume $D^2 \varphi \geq -C \, \mathrm I$. Then also $D^2 u \geq -C \, \mathrm I$.
| |
| {{end}} | |
| | |
| {{proof}}
| |
| For any $x,y \in \R^n$ we have
| |
| \[ \varphi(x+y) + \varphi(x-y) - 2 \varphi(x) \geq -C |y|^2.\]
| |
| This can be rewritten as
| |
| \[ \frac{\varphi(x+y)+\varphi(x-y)+C|y|^2} 2 \geq \varphi(x).\]
| |
| Since $u \geq \varphi$,
| |
| \[ \frac{u(x+y)+u(x-y)+C|y|^2} 2 \geq \varphi(x).\] | |
| This is the point in which the convexity of the equation plays a role. Notice that the obstacle problem can be written as a Bellman equation. It is, itself, a convex problem, meaning that the average of two solutions is a super-solution. Therefore, the left hand side of the inequality is a super-solution above $\varphi$, and therefore must be larger than $u$.
| |
| \[ \frac{u(x+y)+u(x-y)+C|y|^2} 2 \geq u(x).\]
| |
| This is precisely the fact that $D^2 u \geq -C \, \mathrm I$.
| |
| {{endproof}}
| |
| | |
| === $C^{2s}$ regularity ===
| |
| | |
| For the classical Laplacian, the optimal regularity is $C^{1,1}$, which was originally proved by Frehse. The proof goes like this
| |
| * On one hand, we have the semiconvexity: $D^2 u \geq -C \mathrm I$.
| |
| * On the other hand, we have from the equation: $\Delta u \leq 0$.
| |
| These two things combined give an estimate $|D^2 u| \leq C$.
| |
| | |
| A similar argument works for the fractional Laplacian to prove $u \in C^{2s}$. However, this regularity is not optimal as soon as $s<1$. Instead $u \in C^{1+s}$ would be the optimal one. The argument goes like this.
| |
| * On one hand, $u$ is bounded and semiconvex: $D^2 u \geq -C \mathrm I$ and $u \in L^\infty$. Therefore $(-\Delta)^s u \leq C$.
| |
| * On the other hand, we have from the equation: $(-\Delta)^s u \geq 0$.
| |
| These two things combined say that $|(-\Delta)^s u| \leq C$. This is almost like $u \in C^{2s}$.
| |
| | |
| The precise estimate $u \in C^{2s}$ follows with a little extra work, but we will not need it.
| |
| | |
| '''Exercise 13.''' Let $u$ be a function such that for some constant $C$, $|u| \leq C$, $D^2 u \geq -C \, \mathrm I$ and $(-\Delta)^s u \geq 0$ in $\R^n$. Prove that there is a constant $\tilde C$ (depending only on $C$) so that for all $x \in \R^n$,
| |
| \[ \int_{\R^n} \frac{|u(x+y) - u(x)|}{|y|^{n+2s}} dy \leq \tilde C.\]
| |
| Recall from Exercise 11 that the conclusion above implies that $u \in C^{2s}$.
| |
| | |
| === $C^{2s+\alpha}$ regularity ===
| |
| The next step is to prove that $u$ is a little bit more regular than $C^{2s}$. We will now show that $w := (-\Delta)^s u$ is $C^\alpha$ for some $\alpha>0$ small (the optimal $\alpha$ will appear in the next section).
| |
| | |
| Note that the function $w$ solves the following Dirichlet problem for $(-\Delta)^{1-s}$,
| |
| \begin{align*} | | \begin{align*} |
| (-\Delta)^{1-s} w &= -\Delta \varphi \ \text{ inside } \{u=\varphi\},\\
| | \Delta w &= 0 \text{ in } \{x_n>0\}\\ |
| w &= 0 \ \text{ in } \{u > \varphi\}. | | \partial_t w &= a\partial_n w \text{ on } \{x_n=0\} |
| \end{align*} | | \end{align*} |
| | Or in terms of the half-laplacian in $\mathbb R^{n-1} = \{x_n=0\}$, |
| | \[ |
| | \partial_t w = a\Delta_{\mathbb R^{n-1}}^{1/2} w |
| | \] |
|
| |
|
| Here is an important distinction between the cases $s=1$ and $s<1$. If $s=1$, the function $w = \Delta u$ does not satisfy any useful equation, whereas here we can obtain some extra regularity from the equation for $w$.
| | == References == |
| | {{reflist|refs= |
|
| |
|
| We can observe here a heuristic reason for the optimal regularity. If we expect that $w$ will be continuous and the contact set $\{u=\varphi\}$ will have a smooth boundary, then the regularity of $w$ on the boundary $\partial \{u=\varphi\}$ is determined by the equation and should be $w \in C^{1-s}$. This corresponds precisely to $u \in C^{1+s}$. The proof will be long though, because we do not know a priori either that $\{u=\varphi\}$ has a smooth boundary or that $w$ is continuous across $\partial \{u=\varphi\}$.
| | <ref name="MR0097227">{{Citation | last1=Saffman | first1= P. G. | last2=Taylor | first2= Geoffrey | title=The penetration of a fluid into a porous medium or Hele-Shaw cell containing a more viscous liquid | journal=Proc. Roy. Soc. London. Ser. A | issn=0962-8444 | year=1958 | volume=245 | pages=312--329. (2 plates)}}</ref> |
|
| |
|
| It is convenient to rewrite the problem in terms of $u-\varphi$ instead of $u$. We will abuse notation and still call it $u$, but now it solves an obstacle problem with obstacle $\varphi=0$ but with a non zero right hand side $\varphi$.
| | }} |
| \begin{align*}
| |
| u &\geq 0 \ \text{ in } \R^n, \\
| |
| (-\Delta)^s u &\geq \phi \ \text{ in } \R^n, \\
| |
| (-\Delta)^s u &= \phi \ \text{ wherever } u>0.
| |
| \end{align*}
| |
| We also call $w = (-\Delta)^s u$.
| |