Lecture notes on nonlocal equations - Revision history

imported>Luis at 05:50, 21 June 2013

2013-06-21T05:50:09Z

imported>Luis: /* Holder estimates in the parabolic case */

2012-08-30T00:12:35Z

Holder estimates in the parabolic case

imported>Luis at 20:42, 3 August 2012

2012-08-03T20:42:59Z

← Older revision		Revision as of 15:42, 3 August 2012
Line 1:		Line 1:
	Indication: A star (*) in an exercise indicates that I don't know how to solve it.		Indication: A star (*) in an exercise indicates that I don't know how to solve it.

	[[User:Luis\|Luis]] 14:00, 12 May 2012 (CDT)		By [[User:Luis\|Luis]] 14:00, 12 May 2012 (CDT)

	=Lecture 1=		=Lecture 1=

imported>Luis: Protected "Lecture notes on nonlocal equations" ([edit=sysop] (indefinite) [move=sysop] (indefinite))

2012-08-03T20:42:20Z

Protected "Lecture notes on nonlocal equations" ([edit=sysop] (indefinite) [move=sysop] (indefinite))

← Older revision	Revision as of 15:42, 3 August 2012
(No difference)

imported>Luis: moved Lecture notes of Silvestre's course to Lecture notes on nonlocal equations: generic (better) title

2012-08-03T20:41:19Z

moved Lecture notes of Silvestre's course to Lecture notes on nonlocal equations: generic (better) title

← Older revision	Revision as of 15:41, 3 August 2012
(No difference)

imported>Luis: /* Uniform ellipticity */

2012-07-06T16:41:22Z

Uniform ellipticity

← Older revision		Revision as of 11:41, 6 July 2012
Line 111:		Line 111:

	'''Big Theorems''':		'''Big Theorems''':
	* [[Krylov-Safonov]] (~~1981~~): Solutions to fully nonlinear uniformly elliptic equations are $C^{1,\alpha}$ for some $\alpha>0$.		* [[Krylov-Safonov]] (1979): Solutions to fully nonlinear uniformly elliptic equations are $C^{1,\alpha}$ for some $\alpha>0$.
	* [[Evans-Krylov]] (~~1983~~): Solutions to convex fully nonlinear uniformly elliptic equations are $C^{2,\alpha}$ for some $\alpha>0$.		* [[Evans-Krylov]] (1982): Solutions to convex fully nonlinear uniformly elliptic equations are $C^{2,\alpha}$ for some $\alpha>0$.

	At the end of this course, we should be able to understand the proof of these two theorems and their generalizations to nonlocal equations.		At the end of this course, we should be able to understand the proof of these two theorems and their generalizations to nonlocal equations.

imported>Luis: /* Holder estimates */

2012-07-03T17:51:05Z

Holder estimates

← Older revision		Revision as of 12:51, 3 July 2012
Line 323:		Line 323:
	* $\|\{v < 0\} \cap B_1\| > 1/2 \|B_1\|$.		* $\|\{v < 0\} \cap B_1\| > 1/2 \|B_1\|$.
	* $M^+_{\mathcal L} v \leq 0$ in $B_1$		* $M^+_{\mathcal L} v \leq 0$ in $B_1$
	~~Imply~~ that $v \~~geq~~ (1/2-\theta)$ in $B_{1/2}$.		imply that $v \leq (1/2-\theta)$ in $B_{1/2}$.

	If that holds for any choice of $\alpha$ and $\theta$, it also holds for smaller values. Thus, a posteriori, we can make one of them smaller so that $\alpha = \log(1-\theta)/\log(1/2)$.		If that holds for any choice of $\alpha$ and $\theta$, it also holds for smaller values. Thus, a posteriori, we can make one of them smaller so that $\alpha = \log(1-\theta)/\log(1/2)$.

imported>Weiran at 16:31, 1 July 2012

2012-07-01T16:31:12Z

← Older revision		Revision as of 11:31, 1 July 2012
Line 323:		Line 323:
	* $\|\{v < 0\} \cap B_1\| > 1/2 \|B_1\|$.		* $\|\{v < 0\} \cap B_1\| > 1/2 \|B_1\|$.
	* $M^+_{\mathcal L} v \leq 0$ in $B_1$		* $M^+_{\mathcal L} v \leq 0$ in $B_1$
	Imply that $v \~~leq~~ (1/2-\theta)$ in $B_{1/2}$.		Imply that $v \geq (1/2-\theta)$ in $B_{1/2}$.

	If that holds for any choice of $\alpha$ and $\theta$, it also holds for smaller values. Thus, a posteriori, we can make one of them smaller so that $\alpha = \log(1-\theta)/\log(1/2)$.		If that holds for any choice of $\alpha$ and $\theta$, it also holds for smaller values. Thus, a posteriori, we can make one of them smaller so that $\alpha = \log(1-\theta)/\log(1/2)$.
Line 329:		Line 329:
	Let $\rho$ be a smooth radial function supported in $B_{3/4}$ such that $\rho \equiv 1$ in $B_{1/2}$.		Let $\rho$ be a smooth radial function supported in $B_{3/4}$ such that $\rho \equiv 1$ in $B_{1/2}$.

	If $v ~~\geq~~ (1/2-\theta)$ at any point in $B_{1/2}$, then $(v+\theta \rho)$ would have a local maximum at a point $x_0 \in B_{3/4}$ for which $(v+\theta \rho)(x_0) > 1/2$.		If $v > (1/2-\theta)$ at any point in $B_{1/2}$, then $(v+\theta \rho)$ would have a local maximum at a point $x_0 \in B_{3/4}$ for which $(v+\theta \rho)(x_0) > 1/2$.
	\[ \max_{B_1} (v+\theta \rho) = (v+\theta \rho)(x_0) > 1.\]		\[ \max_{B_1} (v+\theta \rho) = (v+\theta \rho)(x_0) > 1/2.\]
	In order to obtain a contradiction, we evaluate $M^+ (u+\theta \rho)(x_0)$.		In order to obtain a contradiction, we evaluate $M^+ (v+\theta \rho)(x_0)$.

	On one hand		On one hand
	\begin{align*}		\begin{align*}
	M^+ (v+\theta \rho)(x_0) &\geq M^+ v(x_0) + \theta M^- \rho(x_0) \\		M^+ (v+\theta \rho)(x_0) &\geq M^+ v(x_0) + \theta M^- \rho(x_0) \\
	&\geq \theta \ \min_{B_{3/4}} \ M^- \rho(~~x_0~~).		&\geq \theta \ \min_{B_{3/4}} \ M^- \rho(x).
	\end{align*}		\end{align*}

Line 342:		Line 342:
	\begin{align*}		\begin{align*}
	M^+ (v+\theta \rho)(x_0) &= \int_{\R^n} \frac{\Lambda (w(x_0+y)+w(x_0-y)-2w(x_0))^+ - \lambda (w(x_0+y)+w(x_0-y)-2w(x_0))^-}{\|y\|^{n+s}} dy \\		M^+ (v+\theta \rho)(x_0) &= \int_{\R^n} \frac{\Lambda (w(x_0+y)+w(x_0-y)-2w(x_0))^+ - \lambda (w(x_0+y)+w(x_0-y)-2w(x_0))^-}{\|y\|^{n+s}} dy \\
	&\leq \int_{\R^n} \frac{\Lambda (w(x_0+y)-w(x_0))^+ - \lambda (w(x_0+y)-w(x_0))^-}{\|y\|^{n+s}} dy \\		&\leq 2\int_{\R^n} \frac{\Lambda (w(x_0+y)-w(x_0))^+ - \lambda (w(x_0+y)-w(x_0))^-}{\|y\|^{n+s}} dy \\
	&\leq \int_{x_0+y \notin B_1} (\dots) + \int_{x_0+y \in B_1} (\dots)		&\leq 2\int_{x_0+y \notin B_1} (\dots) + 2\int_{x_0+y \in B_1} (\dots)
	\end{align*}		\end{align*}

	The first integral can be bounded using that $v(x) \leq (2\|x\|)^\alpha-1/2$ for all $x \notin B_1$. In fact, it is arbitrarily small if $\alpha$ is chosen close to $0$.		The first integral can be bounded using that $v(x) \leq (2\|x\|)^\alpha-1/2$ for all $x \notin B_1$. In fact, it is arbitrarily small if $\alpha$ is chosen close to $0$.
	\[ \int_{x_0+y \notin B_1} (\dots) \leq \int_{x_0+y \notin B_1} ((2\|x\|)^\alpha-1) \frac{\Lambda}{\|y\|^{n+s}} dy \ll 1.\]		\[ \int_{x_0+y \notin B_1} (\dots) \leq \int_{x_0+y \notin B_1} ((2\|x_0+y\|)^\alpha-1) \frac{\Lambda}{\|y\|^{n+s}} dy \ll 1.\]

	The second integral has a non negative integrand just because $(v+\theta \rho)$ takes its maximum in $B_1$ at $x_0$. But we can say more using the set $G = \{v < 0\} \cap B_1$.		The second integral has a non negative integrand just because $(v+\theta \rho)$ takes its maximum in $B_1$ at $x_0$. But we can say more using the set $G = \{v < 0\} \cap B_1$.

imported>Luis: /* Holder estimates */

2012-06-25T18:10:44Z

Holder estimates

← Older revision		Revision as of 13:10, 25 June 2012
Line 329:		Line 329:
	Let $\rho$ be a smooth radial function supported in $B_{3/4}$ such that $\rho \equiv 1$ in $B_{1/2}$.		Let $\rho$ be a smooth radial function supported in $B_{3/4}$ such that $\rho \equiv 1$ in $B_{1/2}$.

	If $v \geq (1/2-\theta)$ at any point in $B_{1/2}$, then $(v+\theta \rho)$ would have a local maximum at a point $x_0 \in B_{3/4}$ for which $(u+\theta \rho)(x_0) > 1/2$.		If $v \geq (1/2-\theta)$ at any point in $B_{1/2}$, then $(v+\theta \rho)$ would have a local maximum at a point $x_0 \in B_{3/4}$ for which $(v+\theta \rho)(x_0) > 1/2$.
	\[ \max_{B_1} (v+\theta \rho) = (v+\theta \rho)(x_0) > 1.\]		\[ \max_{B_1} (v+\theta \rho) = (v+\theta \rho)(x_0) > 1.\]
	In order to obtain a contradiction, we evaluate $M^+ (u+\theta \rho)(x_0)$.		In order to obtain a contradiction, we evaluate $M^+ (u+\theta \rho)(x_0)$.

imported>Luis: /* Holder estimates */

2012-06-25T18:09:51Z

Holder estimates

← Older revision		Revision as of 13:09, 25 June 2012
Line 329:		Line 329:
	Let $\rho$ be a smooth radial function supported in $B_{3/4}$ such that $\rho \equiv 1$ in $B_{1/2}$.		Let $\rho$ be a smooth radial function supported in $B_{3/4}$ such that $\rho \equiv 1$ in $B_{1/2}$.

	If $v \geq (1/2-\theta)$ at any point in $B_{1/2}$, then $(u+\theta \rho)$ would have a local maximum at a point $x_0 \in B_{3/4}$ for which $(u+\theta \rho)(x_0) > 1/2$.		If $v \geq (1/2-\theta)$ at any point in $B_{1/2}$, then $(v+\theta \rho)$ would have a local maximum at a point $x_0 \in B_{3/4}$ for which $(u+\theta \rho)(x_0) > 1/2$.
	\[ \max_{B_1} (u+\theta \rho) = (u+\theta \rho)(x_0) > 1.\]		\[ \max_{B_1} (v+\theta \rho) = (v+\theta \rho)(x_0) > 1.\]
	In order to obtain a contradiction, we evaluate $M^+ (u+\theta \rho)(x_0)$.		In order to obtain a contradiction, we evaluate $M^+ (u+\theta \rho)(x_0)$.

	On one hand		On one hand
	\begin{align*}		\begin{align*}
	M^+ (u+\theta \rho)(x_0) &\geq M^+ u(x_0) + \theta M^- \rho(x_0) \\		M^+ (v+\theta \rho)(x_0) &\geq M^+ v(x_0) + \theta M^- \rho(x_0) \\
	&\geq \theta \ \min_{B_{3/4}} \ M^- \rho(x_0).		&\geq \theta \ \min_{B_{3/4}} \ M^- \rho(x_0).
	\end{align*}		\end{align*}

	On the other hand, the other estimate is more delicate. Let $w = (u+\theta \rho)$.		On the other hand, the other estimate is more delicate. Let $w = (v+\theta \rho)$.
	\begin{align*}		\begin{align*}
	M^+ (u+\theta \rho)(x_0) &= \int_{\R^n} \frac{\Lambda (w(x_0+y)+w(x_0-y)-2w(x_0))^+ - \lambda (w(x_0+y)+w(x_0-y)-2w(x_0))^-}{\|y\|^{n+s}} dy \\		M^+ (v+\theta \rho)(x_0) &= \int_{\R^n} \frac{\Lambda (w(x_0+y)+w(x_0-y)-2w(x_0))^+ - \lambda (w(x_0+y)+w(x_0-y)-2w(x_0))^-}{\|y\|^{n+s}} dy \\
	&\leq \int_{\R^n} \frac{\Lambda (w(x_0+y)-w(x_0))^+ - \lambda (w(x_0+y)-w(x_0))^-}{\|y\|^{n+s}} dy \\		&\leq \int_{\R^n} \frac{\Lambda (w(x_0+y)-w(x_0))^+ - \lambda (w(x_0+y)-w(x_0))^-}{\|y\|^{n+s}} dy \\
	&\leq \int_{x_0+y \notin B_1} (\dots) + \int_{x_0+y \in B_1} (\dots)		&\leq \int_{x_0+y \notin B_1} (\dots) + \int_{x_0+y \in B_1} (\dots)
Line 349:		Line 349:
	\[ \int_{x_0+y \notin B_1} (\dots) \leq \int_{x_0+y \notin B_1} ((2\|x\|)^\alpha-1) \frac{\Lambda}{\|y\|^{n+s}} dy \ll 1.\]		\[ \int_{x_0+y \notin B_1} (\dots) \leq \int_{x_0+y \notin B_1} ((2\|x\|)^\alpha-1) \frac{\Lambda}{\|y\|^{n+s}} dy \ll 1.\]

	The second integral has a non negative integrand just because $(u+\theta \rho)$ takes its maximum in $B_1$ at $x_0$. But we can say more using the set $G = \{v < 0\} \cap B_1$.		The second integral has a non negative integrand just because $(v+\theta \rho)$ takes its maximum in $B_1$ at $x_0$. But we can say more using the set $G = \{v < 0\} \cap B_1$.
	\begin{align*}		\begin{align*}
	\int_{x_0+y \in B_1} (\dots) &\leq \int_{x_0+y \in G} (\dots) + \int_{x_0+y \in B_1 \setminus G} (\dots) \\		\int_{x_0+y \in B_1} (\dots) &\leq \int_{x_0+y \in G} (\dots) + \int_{x_0+y \in B_1 \setminus G} (\dots) \\

← Older revision		Revision as of 19:12, 29 August 2012
Line 482:		Line 482:
	m'(t) &= c_0 \| \{x \in B_1: v(x,t) \leq 0\}\| - C_1 m(t).		m'(t) &= c_0 \| \{x \in B_1: v(x,t) \leq 0\}\| - C_1 m(t).
	\end{align*}		\end{align*}
	for constants $c_0$ and $C$ to be chosen later.		for constants $c_0$ and $C_1$ to be chosen later.

	We show that the inequality holds by proving that it can never be invalidated for the first time. Indeed, assume there was a point $(x_0,t_0)$ where equality holds. This point must be in the support of $\rho$ (strict inequality holds in the rest since $v \leq 1/2$), thus $x_0 \in B_{3/4}$.		We show that the inequality holds by proving that it can never be invalidated for the first time. Indeed, assume there was a point $(x_0,t_0)$ where equality holds. This point must be in the support of $\rho$ (strict inequality holds in the rest since $v \leq 1/2$), thus $x_0 \in B_{3/4}$.
Line 510:		Line 510:
	\[ v_t(x_0,t_0) - M^+_{\mathcal L} v(x_0,t_0) \geq -m'(t_0) \rho(x_0) + \delta - C(\alpha) + C m(t_0) M^+_{\mathcal L} \rho(x_0,t_0) + c_0 \|G(t_0)\| \]		\[ v_t(x_0,t_0) - M^+_{\mathcal L} v(x_0,t_0) \geq -m'(t_0) \rho(x_0) + \delta - C(\alpha) + C m(t_0) M^+_{\mathcal L} \rho(x_0,t_0) + c_0 \|G(t_0)\| \]
	Recall that $m'(t) = c_0 \|G(t)\| - C_1 m(t)$ by definition (this is when $c_0$ is chosen). Since $\rho \leq 1$, we have		Recall that $m'(t) = c_0 \|G(t)\| - C_1 m(t)$ by definition (this is when $c_0$ is chosen). Since $\rho \leq 1$, we have
	\[ v_t(x_0,t_0) - M^+_{\mathcal L} v(x_0,t_0) \geq \delta - C(\alpha) + C m(t_0) M^+_{\mathcal L} \rho(x_0,t_0) + ~~C_0~~ m(t_0) \rho(x_0). \]		\[ v_t(x_0,t_0) - M^+_{\mathcal L} v(x_0,t_0) \geq \delta - C(\alpha) + C m(t_0) M^+_{\mathcal L} \rho(x_0,t_0) + C_1 m(t_0) \rho(x_0). \]
	We choose $\alpha$ small so that $C(\alpha) < \delta$, so we have		We choose $\alpha$ small so that $C(\alpha) < \delta$, so we have
	\[ v_t(x_0,t_0) - M^+_{\mathcal L} v(x_0,t_0) \geq + C m(t_0) M^+_{\mathcal L} \rho(x_0,t_0) + ~~C_0~~ m(t_0) \rho(x_0). \]		\[ v_t(x_0,t_0) - M^+_{\mathcal L} v(x_0,t_0) \geq + C m(t_0) M^+_{\mathcal L} \rho(x_0,t_0) + C_1 m(t_0) \rho(x_0). \]
	Now we have to choose $~~C_0~~$ appropriately to make this right hand side positive and contradict the equation for $v$.		Now we have to choose $C_1$ appropriately to make this right hand side positive and contradict the equation for $v$.

	This is clearly possible if we know a lower bound for $\rho(x_0)$. However, we must also consider that $x_0$ may be a point where $\rho$ is very small. It turns out that $M^+_{\mathcal L} \rho > 0$ where $\rho$ is small since trivially $M^+_{\mathcal L} \rho(x) > 0$ if $\rho(x)=0$ (from the formula for $M^+_{\mathcal L}$) and $M^+_{\mathcal L} \rho$ is a continuous function. Thus, where $\rho$ is small, the right hand side is automatically positive. We choose $~~C_0~~$ large so that this right hand side is also positive where $\rho$ is large. This gives us a contradiction with the equation and proves that then $v$ must stay below the function $b$.		This is clearly possible if we know a lower bound for $\rho(x_0)$. However, we must also consider that $x_0$ may be a point where $\rho$ is very small. It turns out that $M^+_{\mathcal L} \rho > 0$ where $\rho$ is small since trivially $M^+_{\mathcal L} \rho(x) > 0$ if $\rho(x)=0$ (from the formula for $M^+_{\mathcal L}$) and $M^+_{\mathcal L} \rho$ is a continuous function. Thus, where $\rho$ is small, the right hand side is automatically positive. We choose $C_1$ large so that this right hand side is also positive where $\rho$ is large. This gives us a contradiction with the equation and proves that then $v$ must stay below the function $b$.

	To finish the proof, all we need is to show that $b \leq 1/2-\theta$ in $Q_{1/2}$. We analyze the ODE that defines $m(t)$ and we realize that $m(t)$ is going to be bounded below in for $t \in [-1/2^s,0]$ in terms of the measure of the set $\{u \leq 0\} \cap (B_1 \times [-1,-1/2^s])$. In fact, an explicit formula can be given for $b$. Let $\theta$ be half of this lower bound for $m(t)$ and let us choose $\delta < \theta/4$. We will then have $b = 1/2 - \epsilon + \delta(t+1) - m(t) b(x,t) \leq 1/2-\theta$ in $Q_{1/2}$, which finishes the proof.		To finish the proof, all we need is to show that $b \leq 1/2-\theta$ in $Q_{1/2}$. We analyze the ODE that defines $m(t)$ and we realize that $m(t)$ is going to be bounded below in for $t \in [-1/2^s,0]$ in terms of the measure of the set $\{u \leq 0\} \cap (B_1 \times [-1,-1/2^s])$. In fact, an explicit formula can be given for $b$. Let $\theta$ be half of this lower bound for $m(t)$ and let us choose $\delta < \theta/4$. We will then have $b = 1/2 - \epsilon + \delta(t+1) - m(t) b(x,t) \leq 1/2-\theta$ in $Q_{1/2}$, which finishes the proof.