Test for Exactness
In this theorem apperas:"(simply conected)" i don't understand why.
Because we can proof with out use "(simply connected)":
if we assume that: exist $phi(x,y) $ s.t $phi_{x}=frac{ partialphi}{partial x}=M$ and $phi_{y}=frac{ partialphi}{partial y}=N$ then $phi_{xy}=M_y$ and $phi_{yx}=N_x $ but from continuity of $M_y$ and $N_x $ and symmetry of second derivatives (schwarz theorem) it follows $phi_{xy}=phi_{yx}$ and (1.9.5) is proven.
To proof converse is just an algorithm to find $phi$ using integration and (1.9.5)
Where we need simply connected?
differential-equations analysis
asked Nov 25 '18 at 17:11
Ica Sandu
515
add a comment |
In this theorem apperas:"(simply conected)" i don't understand why.
Because we can proof with out use "(simply connected)":
if we assume that: exist $phi(x,y) $ s.t $phi_{x}=frac{ partialphi}{partial x}=M$ and $phi_{y}=frac{ partialphi}{partial y}=N$ then $phi_{xy}=M_y$ and $phi_{yx}=N_x $ but from continuity of $M_y$ and $N_x $ and symmetry of second derivatives (schwarz theorem) it follows $phi_{xy}=phi_{yx}$ and (1.9.5) is proven.
To proof converse is just an algorithm to find $phi$ using integration and (1.9.5)
Where we need simply connected?
differential-equations analysis
asked Nov 25 '18 at 17:11
Ica Sandu
515
The standard example is the angle of the point $x,y$, the gradient makes a nice exact vector field outside the origin, but going once around the origin will increase the angle by $2pi$.
– LutzL
Nov 25 '18 at 17:22
In not sure this is correct, but I think it is: The simply-connected requirement (along with the others) is a sufficient condition. It's not the weakest sufficient condition, but it is sufficient to guarantee that you can write down a potential function $phi$. This is like if I said that running 10,000 miles is sufficient to burn 1 calorie. However running 10,000 miles is not the weakest sufficient condition. I'm in the process of collecting my thoughts for an answer but I'm not sure if I know this subject well enough to give an answer
– DWade64
Nov 25 '18 at 18:01
Whether the 2-dimensional domain is simply-connected or not simply-connected, you just need every point of your integration rectangle, to be a point of the domain. Then you can define a potential function over that integration rectangle, a subset of the full domain. I hope someone else comes along and answers the question. I would leave an answer but now I'm getting confused. In 2-dimensions, simply-connected is easy to understand. Given a point $(x_0, y_0)$ contained in the common domain, it seems like you can always find the other 3 vertices of an integration rectangle, whose area is in $D$
– DWade64
Nov 25 '18 at 18:44
add a comment |
In this theorem apperas:"(simply conected)" i don't understand why.
Because we can proof with out use "(simply connected)":
if we assume that: exist $phi(x,y) $ s.t $phi_{x}=frac{ partialphi}{partial x}=M$ and $phi_{y}=frac{ partialphi}{partial y}=N$ then $phi_{xy}=M_y$ and $phi_{yx}=N_x $ but from continuity of $M_y$ and $N_x $ and symmetry of second derivatives (schwarz theorem) it follows $phi_{xy}=phi_{yx}$ and (1.9.5) is proven.
To proof converse is just an algorithm to find $phi$ using integration and (1.9.5)
Where we need simply connected?
differential-equations analysis
asked Nov 25 '18 at 17:11
Ica Sandu
515
In this theorem apperas:"(simply conected)" i don't understand why.
Because we can proof with out use "(simply connected)":
if we assume that: exist $phi(x,y) $ s.t $phi_{x}=frac{ partialphi}{partial x}=M$ and $phi_{y}=frac{ partialphi}{partial y}=N$ then $phi_{xy}=M_y$ and $phi_{yx}=N_x $ but from continuity of $M_y$ and $N_x $ and symmetry of second derivatives (schwarz theorem) it follows $phi_{xy}=phi_{yx}$ and (1.9.5) is proven.
To proof converse is just an algorithm to find $phi$ using integration and (1.9.5)
Where we need simply connected?
differential-equations analysis
differential-equations analysis
asked Nov 25 '18 at 17:11
Ica Sandu
515
asked Nov 25 '18 at 17:11
Ica Sandu
515
asked Nov 25 '18 at 17:11
Ica Sandu
515
asked Nov 25 '18 at 17:11
Ica Sandu
515
asked Nov 25 '18 at 17:11
Ica Sandu
515
515
The standard example is the angle of the point $x,y$, the gradient makes a nice exact vector field outside the origin, but going once around the origin will increase the angle by $2pi$.
– LutzL
Nov 25 '18 at 17:22
In not sure this is correct, but I think it is: The simply-connected requirement (along with the others) is a sufficient condition. It's not the weakest sufficient condition, but it is sufficient to guarantee that you can write down a potential function $phi$. This is like if I said that running 10,000 miles is sufficient to burn 1 calorie. However running 10,000 miles is not the weakest sufficient condition. I'm in the process of collecting my thoughts for an answer but I'm not sure if I know this subject well enough to give an answer
– DWade64
Nov 25 '18 at 18:01
Whether the 2-dimensional domain is simply-connected or not simply-connected, you just need every point of your integration rectangle, to be a point of the domain. Then you can define a potential function over that integration rectangle, a subset of the full domain. I hope someone else comes along and answers the question. I would leave an answer but now I'm getting confused. In 2-dimensions, simply-connected is easy to understand. Given a point $(x_0, y_0)$ contained in the common domain, it seems like you can always find the other 3 vertices of an integration rectangle, whose area is in $D$
– DWade64
Nov 25 '18 at 18:44
add a comment |
The standard example is the angle of the point $x,y$, the gradient makes a nice exact vector field outside the origin, but going once around the origin will increase the angle by $2pi$.
– LutzL
Nov 25 '18 at 17:22
In not sure this is correct, but I think it is: The simply-connected requirement (along with the others) is a sufficient condition. It's not the weakest sufficient condition, but it is sufficient to guarantee that you can write down a potential function $phi$. This is like if I said that running 10,000 miles is sufficient to burn 1 calorie. However running 10,000 miles is not the weakest sufficient condition. I'm in the process of collecting my thoughts for an answer but I'm not sure if I know this subject well enough to give an answer
– DWade64
Nov 25 '18 at 18:01
Whether the 2-dimensional domain is simply-connected or not simply-connected, you just need every point of your integration rectangle, to be a point of the domain. Then you can define a potential function over that integration rectangle, a subset of the full domain. I hope someone else comes along and answers the question. I would leave an answer but now I'm getting confused. In 2-dimensions, simply-connected is easy to understand. Given a point $(x_0, y_0)$ contained in the common domain, it seems like you can always find the other 3 vertices of an integration rectangle, whose area is in $D$
– DWade64
Nov 25 '18 at 18:44
The standard example is the angle of the point $x,y$, the gradient makes a nice exact vector field outside the origin, but going once around the origin will increase the angle by $2pi$.
– LutzL
Nov 25 '18 at 17:22
The standard example is the angle of the point $x,y$, the gradient makes a nice exact vector field outside the origin, but going once around the origin will increase the angle by $2pi$.
– LutzL
Nov 25 '18 at 17:22
In not sure this is correct, but I think it is: The simply-connected requirement (along with the others) is a sufficient condition. It's not the weakest sufficient condition, but it is sufficient to guarantee that you can write down a potential function $phi$. This is like if I said that running 10,000 miles is sufficient to burn 1 calorie. However running 10,000 miles is not the weakest sufficient condition. I'm in the process of collecting my thoughts for an answer but I'm not sure if I know this subject well enough to give an answer
– DWade64
Nov 25 '18 at 18:01
In not sure this is correct, but I think it is: The simply-connected requirement (along with the others) is a sufficient condition. It's not the weakest sufficient condition, but it is sufficient to guarantee that you can write down a potential function $phi$. This is like if I said that running 10,000 miles is sufficient to burn 1 calorie. However running 10,000 miles is not the weakest sufficient condition. I'm in the process of collecting my thoughts for an answer but I'm not sure if I know this subject well enough to give an answer
– DWade64
Nov 25 '18 at 18:01
Whether the 2-dimensional domain is simply-connected or not simply-connected, you just need every point of your integration rectangle, to be a point of the domain. Then you can define a potential function over that integration rectangle, a subset of the full domain. I hope someone else comes along and answers the question. I would leave an answer but now I'm getting confused. In 2-dimensions, simply-connected is easy to understand. Given a point $(x_0, y_0)$ contained in the common domain, it seems like you can always find the other 3 vertices of an integration rectangle, whose area is in $D$
– DWade64
Nov 25 '18 at 18:44
Whether the 2-dimensional domain is simply-connected or not simply-connected, you just need every point of your integration rectangle, to be a point of the domain. Then you can define a potential function over that integration rectangle, a subset of the full domain. I hope someone else comes along and answers the question. I would leave an answer but now I'm getting confused. In 2-dimensions, simply-connected is easy to understand. Given a point $(x_0, y_0)$ contained in the common domain, it seems like you can always find the other 3 vertices of an integration rectangle, whose area is in $D$
– DWade64
Nov 25 '18 at 18:44
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The standard example is the angle of the point $x,y$, the gradient makes a nice exact vector field outside the origin, but going once around the origin will increase the angle by $2pi$.
– LutzL
Nov 25 '18 at 17:22
In not sure this is correct, but I think it is: The simply-connected requirement (along with the others) is a sufficient condition. It's not the weakest sufficient condition, but it is sufficient to guarantee that you can write down a potential function $phi$. This is like if I said that running 10,000 miles is sufficient to burn 1 calorie. However running 10,000 miles is not the weakest sufficient condition. I'm in the process of collecting my thoughts for an answer but I'm not sure if I know this subject well enough to give an answer
– DWade64
Nov 25 '18 at 18:01
Whether the 2-dimensional domain is simply-connected or not simply-connected, you just need every point of your integration rectangle, to be a point of the domain. Then you can define a potential function over that integration rectangle, a subset of the full domain. I hope someone else comes along and answers the question. I would leave an answer but now I'm getting confused. In 2-dimensions, simply-connected is easy to understand. Given a point $(x_0, y_0)$ contained in the common domain, it seems like you can always find the other 3 vertices of an integration rectangle, whose area is in $D$
– DWade64
Nov 25 '18 at 18:44