Coordinate method for PDE
$begingroup$
Solving the PDE $au_x+bu_y+cu=0$ The PDE is transformed by the coordinate method via,
$begin{cases}x'=ax+by\y'=bx-ay\end{cases}$.
What I don't understand is how should I know I have to pick this as the transformation ?
How do I know that this is the transformation to be used?
Is it something to do with linear transformations in matrix?
multivariable-calculus pde
asked Mar 21 '14 at 2:50
clarksonclarkson
87111535
$endgroup$
add a comment |
$begingroup$
Solving the PDE $au_x+bu_y+cu=0$ The PDE is transformed by the coordinate method via,
$begin{cases}x'=ax+by\y'=bx-ay\end{cases}$.
What I don't understand is how should I know I have to pick this as the transformation ?
How do I know that this is the transformation to be used?
Is it something to do with linear transformations in matrix?
multivariable-calculus pde
asked Mar 21 '14 at 2:50
clarksonclarkson
87111535
$endgroup$
add a comment |
$begingroup$
Solving the PDE $au_x+bu_y+cu=0$ The PDE is transformed by the coordinate method via,
$begin{cases}x'=ax+by\y'=bx-ay\end{cases}$.
What I don't understand is how should I know I have to pick this as the transformation ?
How do I know that this is the transformation to be used?
Is it something to do with linear transformations in matrix?
multivariable-calculus pde
asked Mar 21 '14 at 2:50
clarksonclarkson
87111535
$endgroup$
Solving the PDE $au_x+bu_y+cu=0$ The PDE is transformed by the coordinate method via,
$begin{cases}x'=ax+by\y'=bx-ay\end{cases}$.
What I don't understand is how should I know I have to pick this as the transformation ?
How do I know that this is the transformation to be used?
Is it something to do with linear transformations in matrix?
multivariable-calculus pde
multivariable-calculus pde
asked Mar 21 '14 at 2:50
clarksonclarkson
87111535
asked Mar 21 '14 at 2:50
clarksonclarkson
87111535
asked Mar 21 '14 at 2:50
clarksonclarkson
87111535
asked Mar 21 '14 at 2:50
clarksonclarkson
87111535
asked Mar 21 '14 at 2:50
clarksonclarkson
87111535
87111535
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You should recognize $au_x+bu_y$ as the dot product of $nabla u$ with the vector $left<a,bright>$. Also known as the directional derivative. This makes the direction of vector $left<a,bright>$ seem special for this equation, so we take a coordinate system in which it is one of two coordinate axes. This is best done by a rotation matrix, or a multiple of it, so that the axes remain nicely orthogonal (although this isn't strictly necessary).
answered Mar 21 '14 at 3:00
user127096user127096
8,17011140
$endgroup$
$begingroup$
So one transformation is selected using the directional derivative<a,b>.The other is the orthogonal of $<a,b>$ that is $<b,-a>$.Since $<-b,a>$ is also orthogonal to $<a,b>$ I could have selected $y=-bx+ay$ as well. Couldn't I?
$endgroup$
– clarkson
Mar 21 '14 at 3:35
$begingroup$
@clarkson Yes you could. You could flip either axis, or both.
$endgroup$
– user127096
Mar 21 '14 at 3:36
add a comment |
$begingroup$
It would be useful for first order PDE of this kind to consider it as being extracted from the direction derivatives. For instance, the directional derivative of a function u(x,y) is given by
begin{equation} du=frac{partial u}{partial x}dx+frac{partial u}{partial y}dy=u_xdx+u_ydyend{equation} Rewriting your equation as
begin{equation}
au_x+bu_y=-cu
end{equation}
What your directional derivative of u tells is that begin{equation} du=-cu end{equation}
One simply gets begin{equation} frac{dx}{a}=frac{dy}{b}=frac{du}{-cu}.end{equation}
Solving the first two equations, you will just get begin{equation} bx-ay=constant,end{equation} which is one of your transformed coordinates. In fact this kind of transformation is called methods of characteristics. It roughly suggests that you assume the solution depends on both x and y and then try to go to the space where the solution is constant, just like the previous expression has shown.
Well, it doesn't exactly find the solution for you but provide you with a hint on how the solution profile looks like. In this your particular case, use the second equation with any pairs and find expression for u(x,y). Most of such problems are supplimented with initial conditions to help you find such particular u.
answered Jan 23 at 23:15
M.D.M.D.
549
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
You should recognize $au_x+bu_y$ as the dot product of $nabla u$ with the vector $left<a,bright>$. Also known as the directional derivative. This makes the direction of vector $left<a,bright>$ seem special for this equation, so we take a coordinate system in which it is one of two coordinate axes. This is best done by a rotation matrix, or a multiple of it, so that the axes remain nicely orthogonal (although this isn't strictly necessary).
answered Mar 21 '14 at 3:00
user127096user127096
8,17011140
$endgroup$
$begingroup$
So one transformation is selected using the directional derivative<a,b>.The other is the orthogonal of $<a,b>$ that is $<b,-a>$.Since $<-b,a>$ is also orthogonal to $<a,b>$ I could have selected $y=-bx+ay$ as well. Couldn't I?
$endgroup$
– clarkson
Mar 21 '14 at 3:35
$begingroup$
@clarkson Yes you could. You could flip either axis, or both.
$endgroup$
– user127096
Mar 21 '14 at 3:36
add a comment |
$begingroup$
You should recognize $au_x+bu_y$ as the dot product of $nabla u$ with the vector $left<a,bright>$. Also known as the directional derivative. This makes the direction of vector $left<a,bright>$ seem special for this equation, so we take a coordinate system in which it is one of two coordinate axes. This is best done by a rotation matrix, or a multiple of it, so that the axes remain nicely orthogonal (although this isn't strictly necessary).
answered Mar 21 '14 at 3:00
user127096user127096
8,17011140
$endgroup$
$begingroup$
So one transformation is selected using the directional derivative<a,b>.The other is the orthogonal of $<a,b>$ that is $<b,-a>$.Since $<-b,a>$ is also orthogonal to $<a,b>$ I could have selected $y=-bx+ay$ as well. Couldn't I?
$endgroup$
– clarkson
Mar 21 '14 at 3:35
$begingroup$
@clarkson Yes you could. You could flip either axis, or both.
$endgroup$
– user127096
Mar 21 '14 at 3:36
add a comment |
$begingroup$
You should recognize $au_x+bu_y$ as the dot product of $nabla u$ with the vector $left<a,bright>$. Also known as the directional derivative. This makes the direction of vector $left<a,bright>$ seem special for this equation, so we take a coordinate system in which it is one of two coordinate axes. This is best done by a rotation matrix, or a multiple of it, so that the axes remain nicely orthogonal (although this isn't strictly necessary).
answered Mar 21 '14 at 3:00
user127096user127096
8,17011140
$endgroup$
You should recognize $au_x+bu_y$ as the dot product of $nabla u$ with the vector $left<a,bright>$. Also known as the directional derivative. This makes the direction of vector $left<a,bright>$ seem special for this equation, so we take a coordinate system in which it is one of two coordinate axes. This is best done by a rotation matrix, or a multiple of it, so that the axes remain nicely orthogonal (although this isn't strictly necessary).
answered Mar 21 '14 at 3:00
user127096user127096
8,17011140
answered Mar 21 '14 at 3:00
user127096user127096
8,17011140
answered Mar 21 '14 at 3:00
user127096user127096
8,17011140
answered Mar 21 '14 at 3:00
user127096user127096
8,17011140
8,17011140
$begingroup$
So one transformation is selected using the directional derivative<a,b>.The other is the orthogonal of $<a,b>$ that is $<b,-a>$.Since $<-b,a>$ is also orthogonal to $<a,b>$ I could have selected $y=-bx+ay$ as well. Couldn't I?
$endgroup$
– clarkson
Mar 21 '14 at 3:35
$begingroup$
@clarkson Yes you could. You could flip either axis, or both.
$endgroup$
– user127096
Mar 21 '14 at 3:36
add a comment |
$begingroup$
So one transformation is selected using the directional derivative<a,b>.The other is the orthogonal of $<a,b>$ that is $<b,-a>$.Since $<-b,a>$ is also orthogonal to $<a,b>$ I could have selected $y=-bx+ay$ as well. Couldn't I?
$endgroup$
– clarkson
Mar 21 '14 at 3:35
$begingroup$
@clarkson Yes you could. You could flip either axis, or both.
$endgroup$
– user127096
Mar 21 '14 at 3:36
$begingroup$
So one transformation is selected using the directional derivative<a,b>.The other is the orthogonal of $<a,b>$ that is $<b,-a>$.Since $<-b,a>$ is also orthogonal to $<a,b>$ I could have selected $y=-bx+ay$ as well. Couldn't I?
$endgroup$
– clarkson
Mar 21 '14 at 3:35
$begingroup$
So one transformation is selected using the directional derivative<a,b>.The other is the orthogonal of $<a,b>$ that is $<b,-a>$.Since $<-b,a>$ is also orthogonal to $<a,b>$ I could have selected $y=-bx+ay$ as well. Couldn't I?
$endgroup$
– clarkson
Mar 21 '14 at 3:35
$begingroup$
@clarkson Yes you could. You could flip either axis, or both.
$endgroup$
– user127096
Mar 21 '14 at 3:36
$begingroup$
@clarkson Yes you could. You could flip either axis, or both.
$endgroup$
– user127096
Mar 21 '14 at 3:36
add a comment |
$begingroup$
It would be useful for first order PDE of this kind to consider it as being extracted from the direction derivatives. For instance, the directional derivative of a function u(x,y) is given by
begin{equation} du=frac{partial u}{partial x}dx+frac{partial u}{partial y}dy=u_xdx+u_ydyend{equation} Rewriting your equation as
begin{equation}
au_x+bu_y=-cu
end{equation}
What your directional derivative of u tells is that begin{equation} du=-cu end{equation}
One simply gets begin{equation} frac{dx}{a}=frac{dy}{b}=frac{du}{-cu}.end{equation}
Solving the first two equations, you will just get begin{equation} bx-ay=constant,end{equation} which is one of your transformed coordinates. In fact this kind of transformation is called methods of characteristics. It roughly suggests that you assume the solution depends on both x and y and then try to go to the space where the solution is constant, just like the previous expression has shown.
Well, it doesn't exactly find the solution for you but provide you with a hint on how the solution profile looks like. In this your particular case, use the second equation with any pairs and find expression for u(x,y). Most of such problems are supplimented with initial conditions to help you find such particular u.
answered Jan 23 at 23:15
M.D.M.D.
549
$endgroup$
add a comment |
$begingroup$
It would be useful for first order PDE of this kind to consider it as being extracted from the direction derivatives. For instance, the directional derivative of a function u(x,y) is given by
begin{equation} du=frac{partial u}{partial x}dx+frac{partial u}{partial y}dy=u_xdx+u_ydyend{equation} Rewriting your equation as
begin{equation}
au_x+bu_y=-cu
end{equation}
What your directional derivative of u tells is that begin{equation} du=-cu end{equation}
One simply gets begin{equation} frac{dx}{a}=frac{dy}{b}=frac{du}{-cu}.end{equation}
Solving the first two equations, you will just get begin{equation} bx-ay=constant,end{equation} which is one of your transformed coordinates. In fact this kind of transformation is called methods of characteristics. It roughly suggests that you assume the solution depends on both x and y and then try to go to the space where the solution is constant, just like the previous expression has shown.
Well, it doesn't exactly find the solution for you but provide you with a hint on how the solution profile looks like. In this your particular case, use the second equation with any pairs and find expression for u(x,y). Most of such problems are supplimented with initial conditions to help you find such particular u.
answered Jan 23 at 23:15
M.D.M.D.
549
$endgroup$
add a comment |
$begingroup$
It would be useful for first order PDE of this kind to consider it as being extracted from the direction derivatives. For instance, the directional derivative of a function u(x,y) is given by
begin{equation} du=frac{partial u}{partial x}dx+frac{partial u}{partial y}dy=u_xdx+u_ydyend{equation} Rewriting your equation as
begin{equation}
au_x+bu_y=-cu
end{equation}
What your directional derivative of u tells is that begin{equation} du=-cu end{equation}
One simply gets begin{equation} frac{dx}{a}=frac{dy}{b}=frac{du}{-cu}.end{equation}
Solving the first two equations, you will just get begin{equation} bx-ay=constant,end{equation} which is one of your transformed coordinates. In fact this kind of transformation is called methods of characteristics. It roughly suggests that you assume the solution depends on both x and y and then try to go to the space where the solution is constant, just like the previous expression has shown.
Well, it doesn't exactly find the solution for you but provide you with a hint on how the solution profile looks like. In this your particular case, use the second equation with any pairs and find expression for u(x,y). Most of such problems are supplimented with initial conditions to help you find such particular u.
answered Jan 23 at 23:15
M.D.M.D.
549
$endgroup$
It would be useful for first order PDE of this kind to consider it as being extracted from the direction derivatives. For instance, the directional derivative of a function u(x,y) is given by
begin{equation} du=frac{partial u}{partial x}dx+frac{partial u}{partial y}dy=u_xdx+u_ydyend{equation} Rewriting your equation as
begin{equation}
au_x+bu_y=-cu
end{equation}
What your directional derivative of u tells is that begin{equation} du=-cu end{equation}
One simply gets begin{equation} frac{dx}{a}=frac{dy}{b}=frac{du}{-cu}.end{equation}
Solving the first two equations, you will just get begin{equation} bx-ay=constant,end{equation} which is one of your transformed coordinates. In fact this kind of transformation is called methods of characteristics. It roughly suggests that you assume the solution depends on both x and y and then try to go to the space where the solution is constant, just like the previous expression has shown.
Well, it doesn't exactly find the solution for you but provide you with a hint on how the solution profile looks like. In this your particular case, use the second equation with any pairs and find expression for u(x,y). Most of such problems are supplimented with initial conditions to help you find such particular u.
answered Jan 23 at 23:15
M.D.M.D.
549
edited Feb 12 at 8:47
answered Jan 23 at 23:15
M.D.M.D.
549
answered Jan 23 at 23:15
M.D.M.D.
549
answered Jan 23 at 23:15
M.D.M.D.
549
549
add a comment |
add a comment |
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