Mixing
Whenever Laplace's equation is solved in $Omega subset mathbbR^2$, the boundary $partialOmega$ is one-dimensional. …
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
My introductory PDE textbook has this figure in a chapter on boundary and initial data:
It then says
Note that whenever Laplace's equation is solved in $Omega subset mathbbR^2$, the boundary $partialOmega$ is one-dimensional. When solving the equation in $mathbbR^3$, the boundary is two-dimensional, so that there are two tangential derivatives and one normal derivative at every point on the boundary.
Why is this? This is not clear to me.
Please help me understand this.
real-analysis pde vector-analysis harmonic-functions boundary-value-problem
asked 2 days ago
Wyuw
966
add a comment |Â
up vote
0
down vote
favorite
My introductory PDE textbook has this figure in a chapter on boundary and initial data:
It then says
Note that whenever Laplace's equation is solved in $Omega subset mathbbR^2$, the boundary $partialOmega$ is one-dimensional. When solving the equation in $mathbbR^3$, the boundary is two-dimensional, so that there are two tangential derivatives and one normal derivative at every point on the boundary.
Why is this? This is not clear to me.
Please help me understand this.
real-analysis pde vector-analysis harmonic-functions boundary-value-problem
asked 2 days ago
Wyuw
966
2
As an example, let $Omega$ be the $xy$ plane $(x,y,0)inmathbb R^3$ with normal vector $(0,0,-1)$. Then there are three possible derivatives at $z=0$: the normal derivative $-fracpartialpartial zf(x,y,z)|_z=0$, and the two tangential derivatives $fracpartialpartial xf(x,y,0),fracpartialpartial yf(x,y,0)$.
– user254433
2 days ago
@user254433 Your example is a good one. I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My introductory PDE textbook has this figure in a chapter on boundary and initial data:
It then says
Note that whenever Laplace's equation is solved in $Omega subset mathbbR^2$, the boundary $partialOmega$ is one-dimensional. When solving the equation in $mathbbR^3$, the boundary is two-dimensional, so that there are two tangential derivatives and one normal derivative at every point on the boundary.
Why is this? This is not clear to me.
Please help me understand this.
real-analysis pde vector-analysis harmonic-functions boundary-value-problem
asked 2 days ago
Wyuw
966
My introductory PDE textbook has this figure in a chapter on boundary and initial data:
It then says
Note that whenever Laplace's equation is solved in $Omega subset mathbbR^2$, the boundary $partialOmega$ is one-dimensional. When solving the equation in $mathbbR^3$, the boundary is two-dimensional, so that there are two tangential derivatives and one normal derivative at every point on the boundary.
Why is this? This is not clear to me.
Please help me understand this.
real-analysis pde vector-analysis harmonic-functions boundary-value-problem
asked 2 days ago
Wyuw
966
asked 2 days ago
Wyuw
966
asked 2 days ago
Wyuw
966
asked 2 days ago
Wyuw
966
966
2
As an example, let $Omega$ be the $xy$ plane $(x,y,0)inmathbb R^3$ with normal vector $(0,0,-1)$. Then there are three possible derivatives at $z=0$: the normal derivative $-fracpartialpartial zf(x,y,z)|_z=0$, and the two tangential derivatives $fracpartialpartial xf(x,y,0),fracpartialpartial yf(x,y,0)$.
– user254433
2 days ago
@user254433 Your example is a good one. I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago
add a comment |Â
2
As an example, let $Omega$ be the $xy$ plane $(x,y,0)inmathbb R^3$ with normal vector $(0,0,-1)$. Then there are three possible derivatives at $z=0$: the normal derivative $-fracpartialpartial zf(x,y,z)|_z=0$, and the two tangential derivatives $fracpartialpartial xf(x,y,0),fracpartialpartial yf(x,y,0)$.
– user254433
2 days ago
@user254433 Your example is a good one. I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago
2
2
As an example, let $Omega$ be the $xy$ plane $(x,y,0)inmathbb R^3$ with normal vector $(0,0,-1)$. Then there are three possible derivatives at $z=0$: the normal derivative $-fracpartialpartial zf(x,y,z)|_z=0$, and the two tangential derivatives $fracpartialpartial xf(x,y,0),fracpartialpartial yf(x,y,0)$.
– user254433
2 days ago
As an example, let $Omega$ be the $xy$ plane $(x,y,0)inmathbb R^3$ with normal vector $(0,0,-1)$. Then there are three possible derivatives at $z=0$: the normal derivative $-fracpartialpartial zf(x,y,z)|_z=0$, and the two tangential derivatives $fracpartialpartial xf(x,y,0),fracpartialpartial yf(x,y,0)$.
– user254433
2 days ago
@user254433 Your example is a good one. I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago
@user254433 Your example is a good one. I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Explanations with manifolds :
PDE are solved on open sets of $mathbbR^n$. Open sets are manifolds. The boundary of a manifold of dimension $n$ is a manifold of dimension $n-1$.
More intuitively, you can find coordinate charts going from an hyperplane of $mathbbR^n$ to the boundary. Coordinate charts being basically homeomorphisms (continuous bijections with continuous inverse).
Manifolds of dimension $m$ are space that can be seen locally as $mathbbR^m$. This means that locally (around a point) we can find an homeomorphism between our manifold and an open subset of $mathbbR^m$.
Often, we also assume that the boundary of the open set is $C^1$. Typically, this seems to be the case on your drawing, but would not have been if $Omega$ was a square (because of the corners), for example.
By assuming this, we always can find $n-1$ tangential derivatives and $1$ normal derivative.
Note : sorry if you are not familiar with manifolds, but they are the most rigourous way to talk about this.
answered 2 days ago
nicomezi
3,3871718
I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago
By the way, I have never studied manifolds, hahaha. I will accept your answer anyway, since others might benefit from it.
– Wyuw
2 days ago
Thanks but I am not sure you should accept answer without knowing if what has been said is reasonnable and helpfull.
– nicomezi
2 days ago
You have high reputation, so I trust that you know what you are doing. Would you like me to remove it? I can wait till people upvote your answer and then do it later?
– Wyuw
2 days ago
1
It is probably better that way indeed. Thank you.
– nicomezi
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Explanations with manifolds :
PDE are solved on open sets of $mathbbR^n$. Open sets are manifolds. The boundary of a manifold of dimension $n$ is a manifold of dimension $n-1$.
More intuitively, you can find coordinate charts going from an hyperplane of $mathbbR^n$ to the boundary. Coordinate charts being basically homeomorphisms (continuous bijections with continuous inverse).
Manifolds of dimension $m$ are space that can be seen locally as $mathbbR^m$. This means that locally (around a point) we can find an homeomorphism between our manifold and an open subset of $mathbbR^m$.
Often, we also assume that the boundary of the open set is $C^1$. Typically, this seems to be the case on your drawing, but would not have been if $Omega$ was a square (because of the corners), for example.
By assuming this, we always can find $n-1$ tangential derivatives and $1$ normal derivative.
Note : sorry if you are not familiar with manifolds, but they are the most rigourous way to talk about this.
answered 2 days ago
nicomezi
3,3871718
I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago
By the way, I have never studied manifolds, hahaha. I will accept your answer anyway, since others might benefit from it.
– Wyuw
2 days ago
Thanks but I am not sure you should accept answer without knowing if what has been said is reasonnable and helpfull.
– nicomezi
2 days ago
You have high reputation, so I trust that you know what you are doing. Would you like me to remove it? I can wait till people upvote your answer and then do it later?
– Wyuw
2 days ago
1
It is probably better that way indeed. Thank you.
– nicomezi
2 days ago
add a comment |Â
up vote
0
down vote
Explanations with manifolds :
PDE are solved on open sets of $mathbbR^n$. Open sets are manifolds. The boundary of a manifold of dimension $n$ is a manifold of dimension $n-1$.
More intuitively, you can find coordinate charts going from an hyperplane of $mathbbR^n$ to the boundary. Coordinate charts being basically homeomorphisms (continuous bijections with continuous inverse).
Manifolds of dimension $m$ are space that can be seen locally as $mathbbR^m$. This means that locally (around a point) we can find an homeomorphism between our manifold and an open subset of $mathbbR^m$.
Often, we also assume that the boundary of the open set is $C^1$. Typically, this seems to be the case on your drawing, but would not have been if $Omega$ was a square (because of the corners), for example.
By assuming this, we always can find $n-1$ tangential derivatives and $1$ normal derivative.
Note : sorry if you are not familiar with manifolds, but they are the most rigourous way to talk about this.
answered 2 days ago
nicomezi
3,3871718
I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago
By the way, I have never studied manifolds, hahaha. I will accept your answer anyway, since others might benefit from it.
– Wyuw
2 days ago
Thanks but I am not sure you should accept answer without knowing if what has been said is reasonnable and helpfull.
– nicomezi
2 days ago
You have high reputation, so I trust that you know what you are doing. Would you like me to remove it? I can wait till people upvote your answer and then do it later?
– Wyuw
2 days ago
1
It is probably better that way indeed. Thank you.
– nicomezi
2 days ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Explanations with manifolds :
PDE are solved on open sets of $mathbbR^n$. Open sets are manifolds. The boundary of a manifold of dimension $n$ is a manifold of dimension $n-1$.
More intuitively, you can find coordinate charts going from an hyperplane of $mathbbR^n$ to the boundary. Coordinate charts being basically homeomorphisms (continuous bijections with continuous inverse).
Manifolds of dimension $m$ are space that can be seen locally as $mathbbR^m$. This means that locally (around a point) we can find an homeomorphism between our manifold and an open subset of $mathbbR^m$.
Often, we also assume that the boundary of the open set is $C^1$. Typically, this seems to be the case on your drawing, but would not have been if $Omega$ was a square (because of the corners), for example.
By assuming this, we always can find $n-1$ tangential derivatives and $1$ normal derivative.
Note : sorry if you are not familiar with manifolds, but they are the most rigourous way to talk about this.
answered 2 days ago
nicomezi
3,3871718
Explanations with manifolds :
PDE are solved on open sets of $mathbbR^n$. Open sets are manifolds. The boundary of a manifold of dimension $n$ is a manifold of dimension $n-1$.
More intuitively, you can find coordinate charts going from an hyperplane of $mathbbR^n$ to the boundary. Coordinate charts being basically homeomorphisms (continuous bijections with continuous inverse).
Manifolds of dimension $m$ are space that can be seen locally as $mathbbR^m$. This means that locally (around a point) we can find an homeomorphism between our manifold and an open subset of $mathbbR^m$.
Often, we also assume that the boundary of the open set is $C^1$. Typically, this seems to be the case on your drawing, but would not have been if $Omega$ was a square (because of the corners), for example.
By assuming this, we always can find $n-1$ tangential derivatives and $1$ normal derivative.
Note : sorry if you are not familiar with manifolds, but they are the most rigourous way to talk about this.
answered 2 days ago
nicomezi
3,3871718
answered 2 days ago
nicomezi
3,3871718
answered 2 days ago
nicomezi
3,3871718
answered 2 days ago
nicomezi
3,3871718
3,3871718
I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago
By the way, I have never studied manifolds, hahaha. I will accept your answer anyway, since others might benefit from it.
– Wyuw
2 days ago
Thanks but I am not sure you should accept answer without knowing if what has been said is reasonnable and helpfull.
– nicomezi
2 days ago
You have high reputation, so I trust that you know what you are doing. Would you like me to remove it? I can wait till people upvote your answer and then do it later?
– Wyuw
2 days ago
1
It is probably better that way indeed. Thank you.
– nicomezi
2 days ago
add a comment |Â
I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago
By the way, I have never studied manifolds, hahaha. I will accept your answer anyway, since others might benefit from it.
– Wyuw
2 days ago
Thanks but I am not sure you should accept answer without knowing if what has been said is reasonnable and helpfull.
– nicomezi
2 days ago
You have high reputation, so I trust that you know what you are doing. Would you like me to remove it? I can wait till people upvote your answer and then do it later?
– Wyuw
2 days ago
1
It is probably better that way indeed. Thank you.
– nicomezi
2 days ago
I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago
I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago
By the way, I have never studied manifolds, hahaha. I will accept your answer anyway, since others might benefit from it.
– Wyuw
2 days ago
By the way, I have never studied manifolds, hahaha. I will accept your answer anyway, since others might benefit from it.
– Wyuw
2 days ago
Thanks but I am not sure you should accept answer without knowing if what has been said is reasonnable and helpfull.
– nicomezi
2 days ago
Thanks but I am not sure you should accept answer without knowing if what has been said is reasonnable and helpfull.
– nicomezi
2 days ago
You have high reputation, so I trust that you know what you are doing. Would you like me to remove it? I can wait till people upvote your answer and then do it later?
– Wyuw
2 days ago
You have high reputation, so I trust that you know what you are doing. Would you like me to remove it? I can wait till people upvote your answer and then do it later?
– Wyuw
2 days ago
1
1
It is probably better that way indeed. Thank you.
– nicomezi
2 days ago
It is probably better that way indeed. Thank you.
– nicomezi
2 days ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871761%2fwhenever-laplaces-equation-is-solved-in-omega-subset-mathbbr2-the-boun%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
As an example, let $Omega$ be the $xy$ plane $(x,y,0)inmathbb R^3$ with normal vector $(0,0,-1)$. Then there are three possible derivatives at $z=0$: the normal derivative $-fracpartialpartial zf(x,y,z)|_z=0$, and the two tangential derivatives $fracpartialpartial xf(x,y,0),fracpartialpartial yf(x,y,0)$.
– user254433
2 days ago
@user254433 Your example is a good one. I just read it again and it makes complete sense. I'm not sure why I didn't understand it in the first place. Must be sleepy? Of course if $Omega subset mathbbR^2$, then the boundary $partialOmega$ is one-dimensional! And analogously for $mathbbR^3$... This is so obvious... Sorry everyone!
– Wyuw
2 days ago