Uniform Cone Condition with Unit Normal Vector
I was going through this article from M. BRAMSON, K. BURDZY AND W. KENDALL
https://projecteuclid.org/download/pdfview_1/euclid.aop/1362750941
where the uniform interior cone condition is given this way:
A domain $D$ is said to satisfy a uniform interior cone condition,
based on radius $delta > 0$ and angle $alpha in (0,frac{pi}{2}]$, if, for every $x in partial D$, there is at least one unit vector $m$ such that the cone $C(m) = {z : langle z, m rangle > |z| cos(alpha)}$ satisfies
$$ (y + C(m)) cap B(x,delta) subset D, qquad forall y in D cap B(x,delta). $$
Because of the terminology and of the geometric intuition, it seems to me that this definition should be equivalent to the following:
For all $x in partial D$, there exists a radius $h$ and an angle $beta$ such that
$x + su in D$ for all $u in mathbb{S}^{n-1}, |u cdot n_x| > beta$, $s in (0,h)$, where $n_x$ the inward normal vector at $x$.
I believe the second definition is basically the first one with the choice $m = n_x$ for all $x$, and that this choice can be made at all $x in partial D$ if the uniform interior cone condition in the sense of the first definition holds.
Am I missing something here ?
differential-geometry pde boundary-value-problem
asked Nov 27 at 12:39
Gâteau-Gallois
347111
add a comment |
I was going through this article from M. BRAMSON, K. BURDZY AND W. KENDALL
https://projecteuclid.org/download/pdfview_1/euclid.aop/1362750941
where the uniform interior cone condition is given this way:
A domain $D$ is said to satisfy a uniform interior cone condition,
based on radius $delta > 0$ and angle $alpha in (0,frac{pi}{2}]$, if, for every $x in partial D$, there is at least one unit vector $m$ such that the cone $C(m) = {z : langle z, m rangle > |z| cos(alpha)}$ satisfies
$$ (y + C(m)) cap B(x,delta) subset D, qquad forall y in D cap B(x,delta). $$
Because of the terminology and of the geometric intuition, it seems to me that this definition should be equivalent to the following:
For all $x in partial D$, there exists a radius $h$ and an angle $beta$ such that
$x + su in D$ for all $u in mathbb{S}^{n-1}, |u cdot n_x| > beta$, $s in (0,h)$, where $n_x$ the inward normal vector at $x$.
I believe the second definition is basically the first one with the choice $m = n_x$ for all $x$, and that this choice can be made at all $x in partial D$ if the uniform interior cone condition in the sense of the first definition holds.
Am I missing something here ?
differential-geometry pde boundary-value-problem
asked Nov 27 at 12:39
Gâteau-Gallois
347111
In your definition, you are just checking that $(y+C(m))cap B(x,delta)subset D$ for $y=x$, not for every $yin Dcap B(x,delta)$. It is certainly (not strictly) weaker. To me, their condition seems equivalent to saying that $partial D$ is locally a graph of a $cos(alpha)$-Lipschitz function, in the direction $m$
– Federico
Nov 29 at 18:27
add a comment |
I was going through this article from M. BRAMSON, K. BURDZY AND W. KENDALL
https://projecteuclid.org/download/pdfview_1/euclid.aop/1362750941
where the uniform interior cone condition is given this way:
A domain $D$ is said to satisfy a uniform interior cone condition,
based on radius $delta > 0$ and angle $alpha in (0,frac{pi}{2}]$, if, for every $x in partial D$, there is at least one unit vector $m$ such that the cone $C(m) = {z : langle z, m rangle > |z| cos(alpha)}$ satisfies
$$ (y + C(m)) cap B(x,delta) subset D, qquad forall y in D cap B(x,delta). $$
Because of the terminology and of the geometric intuition, it seems to me that this definition should be equivalent to the following:
For all $x in partial D$, there exists a radius $h$ and an angle $beta$ such that
$x + su in D$ for all $u in mathbb{S}^{n-1}, |u cdot n_x| > beta$, $s in (0,h)$, where $n_x$ the inward normal vector at $x$.
I believe the second definition is basically the first one with the choice $m = n_x$ for all $x$, and that this choice can be made at all $x in partial D$ if the uniform interior cone condition in the sense of the first definition holds.
Am I missing something here ?
differential-geometry pde boundary-value-problem
asked Nov 27 at 12:39
Gâteau-Gallois
347111
I was going through this article from M. BRAMSON, K. BURDZY AND W. KENDALL
https://projecteuclid.org/download/pdfview_1/euclid.aop/1362750941
where the uniform interior cone condition is given this way:
A domain $D$ is said to satisfy a uniform interior cone condition,
based on radius $delta > 0$ and angle $alpha in (0,frac{pi}{2}]$, if, for every $x in partial D$, there is at least one unit vector $m$ such that the cone $C(m) = {z : langle z, m rangle > |z| cos(alpha)}$ satisfies
$$ (y + C(m)) cap B(x,delta) subset D, qquad forall y in D cap B(x,delta). $$
Because of the terminology and of the geometric intuition, it seems to me that this definition should be equivalent to the following:
For all $x in partial D$, there exists a radius $h$ and an angle $beta$ such that
$x + su in D$ for all $u in mathbb{S}^{n-1}, |u cdot n_x| > beta$, $s in (0,h)$, where $n_x$ the inward normal vector at $x$.
I believe the second definition is basically the first one with the choice $m = n_x$ for all $x$, and that this choice can be made at all $x in partial D$ if the uniform interior cone condition in the sense of the first definition holds.
Am I missing something here ?
differential-geometry pde boundary-value-problem
differential-geometry pde boundary-value-problem
asked Nov 27 at 12:39
Gâteau-Gallois
347111
asked Nov 27 at 12:39
Gâteau-Gallois
347111
asked Nov 27 at 12:39
Gâteau-Gallois
347111
asked Nov 27 at 12:39
Gâteau-Gallois
347111
asked Nov 27 at 12:39
Gâteau-Gallois
347111
347111
In your definition, you are just checking that $(y+C(m))cap B(x,delta)subset D$ for $y=x$, not for every $yin Dcap B(x,delta)$. It is certainly (not strictly) weaker. To me, their condition seems equivalent to saying that $partial D$ is locally a graph of a $cos(alpha)$-Lipschitz function, in the direction $m$
– Federico
Nov 29 at 18:27
add a comment |
In your definition, you are just checking that $(y+C(m))cap B(x,delta)subset D$ for $y=x$, not for every $yin Dcap B(x,delta)$. It is certainly (not strictly) weaker. To me, their condition seems equivalent to saying that $partial D$ is locally a graph of a $cos(alpha)$-Lipschitz function, in the direction $m$
– Federico
Nov 29 at 18:27
In your definition, you are just checking that $(y+C(m))cap B(x,delta)subset D$ for $y=x$, not for every $yin Dcap B(x,delta)$. It is certainly (not strictly) weaker. To me, their condition seems equivalent to saying that $partial D$ is locally a graph of a $cos(alpha)$-Lipschitz function, in the direction $m$
– Federico
Nov 29 at 18:27
In your definition, you are just checking that $(y+C(m))cap B(x,delta)subset D$ for $y=x$, not for every $yin Dcap B(x,delta)$. It is certainly (not strictly) weaker. To me, their condition seems equivalent to saying that $partial D$ is locally a graph of a $cos(alpha)$-Lipschitz function, in the direction $m$
– Federico
Nov 29 at 18:27
add a comment |
1 Answer
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There are at least two reasons why the two definitions are different:
- In your second definition, you use the unit normal vector, which is not defined if the domain boundary is not sufficiently smooth. In fact, bounded Lipschitz domains already satisfy a uniform cone condition.
- Why we call it a uniform cone condition is that just one set of parameters $alpha$ are $delta$ should work everywhere. Take the planar domain ${(x,y):x>1,,0<y<1/x}$. This would fulfill your second definition but would not satisfy a uniform cone condition.
answered Dec 4 at 0:48
timur
11.8k1943
1. Is it possible to have a Lipschitz domain without a unit normal vector ? Or do we just lose uniqueness ? 2. Your answer suggests that my condition is actually weaker (i.e. satisfied by more domains). Is it actually true ?
– Gâteau-Gallois
Dec 4 at 9:49
1. I don't know what you mean by a unit vector for non smooth domains, but consider convex domains, and then non-convex domains. One of these might eliminate what you mean by a normal vector. 2. As stated, your condition is neither weaker nor stronger. Your condition is weaker, if you allow arbitrary vector m instead of requiring it to be the normal vector.
– timur
Dec 4 at 16:47
add a comment |
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There are at least two reasons why the two definitions are different:
- In your second definition, you use the unit normal vector, which is not defined if the domain boundary is not sufficiently smooth. In fact, bounded Lipschitz domains already satisfy a uniform cone condition.
- Why we call it a uniform cone condition is that just one set of parameters $alpha$ are $delta$ should work everywhere. Take the planar domain ${(x,y):x>1,,0<y<1/x}$. This would fulfill your second definition but would not satisfy a uniform cone condition.
answered Dec 4 at 0:48
timur
11.8k1943
1. Is it possible to have a Lipschitz domain without a unit normal vector ? Or do we just lose uniqueness ? 2. Your answer suggests that my condition is actually weaker (i.e. satisfied by more domains). Is it actually true ?
– Gâteau-Gallois
Dec 4 at 9:49
1. I don't know what you mean by a unit vector for non smooth domains, but consider convex domains, and then non-convex domains. One of these might eliminate what you mean by a normal vector. 2. As stated, your condition is neither weaker nor stronger. Your condition is weaker, if you allow arbitrary vector m instead of requiring it to be the normal vector.
– timur
Dec 4 at 16:47
add a comment |
There are at least two reasons why the two definitions are different:
- In your second definition, you use the unit normal vector, which is not defined if the domain boundary is not sufficiently smooth. In fact, bounded Lipschitz domains already satisfy a uniform cone condition.
- Why we call it a uniform cone condition is that just one set of parameters $alpha$ are $delta$ should work everywhere. Take the planar domain ${(x,y):x>1,,0<y<1/x}$. This would fulfill your second definition but would not satisfy a uniform cone condition.
answered Dec 4 at 0:48
timur
11.8k1943
1. Is it possible to have a Lipschitz domain without a unit normal vector ? Or do we just lose uniqueness ? 2. Your answer suggests that my condition is actually weaker (i.e. satisfied by more domains). Is it actually true ?
– Gâteau-Gallois
Dec 4 at 9:49
1. I don't know what you mean by a unit vector for non smooth domains, but consider convex domains, and then non-convex domains. One of these might eliminate what you mean by a normal vector. 2. As stated, your condition is neither weaker nor stronger. Your condition is weaker, if you allow arbitrary vector m instead of requiring it to be the normal vector.
– timur
Dec 4 at 16:47
add a comment |
There are at least two reasons why the two definitions are different:
- In your second definition, you use the unit normal vector, which is not defined if the domain boundary is not sufficiently smooth. In fact, bounded Lipschitz domains already satisfy a uniform cone condition.
- Why we call it a uniform cone condition is that just one set of parameters $alpha$ are $delta$ should work everywhere. Take the planar domain ${(x,y):x>1,,0<y<1/x}$. This would fulfill your second definition but would not satisfy a uniform cone condition.
answered Dec 4 at 0:48
timur
11.8k1943
There are at least two reasons why the two definitions are different:
- In your second definition, you use the unit normal vector, which is not defined if the domain boundary is not sufficiently smooth. In fact, bounded Lipschitz domains already satisfy a uniform cone condition.
- Why we call it a uniform cone condition is that just one set of parameters $alpha$ are $delta$ should work everywhere. Take the planar domain ${(x,y):x>1,,0<y<1/x}$. This would fulfill your second definition but would not satisfy a uniform cone condition.
answered Dec 4 at 0:48
timur
11.8k1943
edited Dec 4 at 0:55
answered Dec 4 at 0:48
timur
11.8k1943
answered Dec 4 at 0:48
timur
11.8k1943
answered Dec 4 at 0:48
timur
11.8k1943
11.8k1943
1. Is it possible to have a Lipschitz domain without a unit normal vector ? Or do we just lose uniqueness ? 2. Your answer suggests that my condition is actually weaker (i.e. satisfied by more domains). Is it actually true ?
– Gâteau-Gallois
Dec 4 at 9:49
1. I don't know what you mean by a unit vector for non smooth domains, but consider convex domains, and then non-convex domains. One of these might eliminate what you mean by a normal vector. 2. As stated, your condition is neither weaker nor stronger. Your condition is weaker, if you allow arbitrary vector m instead of requiring it to be the normal vector.
– timur
Dec 4 at 16:47
add a comment |
1. Is it possible to have a Lipschitz domain without a unit normal vector ? Or do we just lose uniqueness ? 2. Your answer suggests that my condition is actually weaker (i.e. satisfied by more domains). Is it actually true ?
– Gâteau-Gallois
Dec 4 at 9:49
1. I don't know what you mean by a unit vector for non smooth domains, but consider convex domains, and then non-convex domains. One of these might eliminate what you mean by a normal vector. 2. As stated, your condition is neither weaker nor stronger. Your condition is weaker, if you allow arbitrary vector m instead of requiring it to be the normal vector.
– timur
Dec 4 at 16:47
1. Is it possible to have a Lipschitz domain without a unit normal vector ? Or do we just lose uniqueness ? 2. Your answer suggests that my condition is actually weaker (i.e. satisfied by more domains). Is it actually true ?
– Gâteau-Gallois
Dec 4 at 9:49
1. Is it possible to have a Lipschitz domain without a unit normal vector ? Or do we just lose uniqueness ? 2. Your answer suggests that my condition is actually weaker (i.e. satisfied by more domains). Is it actually true ?
– Gâteau-Gallois
Dec 4 at 9:49
1. I don't know what you mean by a unit vector for non smooth domains, but consider convex domains, and then non-convex domains. One of these might eliminate what you mean by a normal vector. 2. As stated, your condition is neither weaker nor stronger. Your condition is weaker, if you allow arbitrary vector m instead of requiring it to be the normal vector.
– timur
Dec 4 at 16:47
1. I don't know what you mean by a unit vector for non smooth domains, but consider convex domains, and then non-convex domains. One of these might eliminate what you mean by a normal vector. 2. As stated, your condition is neither weaker nor stronger. Your condition is weaker, if you allow arbitrary vector m instead of requiring it to be the normal vector.
– timur
Dec 4 at 16:47
add a comment |
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In your definition, you are just checking that $(y+C(m))cap B(x,delta)subset D$ for $y=x$, not for every $yin Dcap B(x,delta)$. It is certainly (not strictly) weaker. To me, their condition seems equivalent to saying that $partial D$ is locally a graph of a $cos(alpha)$-Lipschitz function, in the direction $m$
– Federico
Nov 29 at 18:27