Using PDEs to find a perpendicular set of curves to a vector field
$begingroup$
So I was interested in finding the set of curves which are perpendicular to this Vector field
$$vec F(x,y) = (x-y,x+y)$$
Lets say the perpendicular curves can be expressed as level curves of the function $G(x,y)$. We know that the normal of the perpendicular curves is parallel to $vec F(x,y)$. The general expression for finding perpendicular surfaces in 3 dimensions would be
$$nabla Gtimesvec F = 0$$
Of course since we're only operating in two dimensions, the "cross product" is simply a scalar and the when applied to the given $vec F$ the following equation results
$$(x-y)frac{partial G}{partial y} = (x+y)frac{partial G}{partial x}$$
This is where I stop. My only known method for solving PDEs is to assume it a product of two single variable functions, rearrange the functions of each variable to their respective side, and solve two ODEs. This isn't possible here so I am not sure what to do. Can anyone solve, recommend tips, or point me to any sources?
calculus pde vector-fields
asked Jan 4 at 7:15
BOTKBOTK
1
$endgroup$
add a comment |
$begingroup$
So I was interested in finding the set of curves which are perpendicular to this Vector field
$$vec F(x,y) = (x-y,x+y)$$
Lets say the perpendicular curves can be expressed as level curves of the function $G(x,y)$. We know that the normal of the perpendicular curves is parallel to $vec F(x,y)$. The general expression for finding perpendicular surfaces in 3 dimensions would be
$$nabla Gtimesvec F = 0$$
Of course since we're only operating in two dimensions, the "cross product" is simply a scalar and the when applied to the given $vec F$ the following equation results
$$(x-y)frac{partial G}{partial y} = (x+y)frac{partial G}{partial x}$$
This is where I stop. My only known method for solving PDEs is to assume it a product of two single variable functions, rearrange the functions of each variable to their respective side, and solve two ODEs. This isn't possible here so I am not sure what to do. Can anyone solve, recommend tips, or point me to any sources?
calculus pde vector-fields
asked Jan 4 at 7:15
BOTKBOTK
1
$endgroup$
add a comment |
$begingroup$
So I was interested in finding the set of curves which are perpendicular to this Vector field
$$vec F(x,y) = (x-y,x+y)$$
Lets say the perpendicular curves can be expressed as level curves of the function $G(x,y)$. We know that the normal of the perpendicular curves is parallel to $vec F(x,y)$. The general expression for finding perpendicular surfaces in 3 dimensions would be
$$nabla Gtimesvec F = 0$$
Of course since we're only operating in two dimensions, the "cross product" is simply a scalar and the when applied to the given $vec F$ the following equation results
$$(x-y)frac{partial G}{partial y} = (x+y)frac{partial G}{partial x}$$
This is where I stop. My only known method for solving PDEs is to assume it a product of two single variable functions, rearrange the functions of each variable to their respective side, and solve two ODEs. This isn't possible here so I am not sure what to do. Can anyone solve, recommend tips, or point me to any sources?
calculus pde vector-fields
asked Jan 4 at 7:15
BOTKBOTK
1
$endgroup$
So I was interested in finding the set of curves which are perpendicular to this Vector field
$$vec F(x,y) = (x-y,x+y)$$
Lets say the perpendicular curves can be expressed as level curves of the function $G(x,y)$. We know that the normal of the perpendicular curves is parallel to $vec F(x,y)$. The general expression for finding perpendicular surfaces in 3 dimensions would be
$$nabla Gtimesvec F = 0$$
Of course since we're only operating in two dimensions, the "cross product" is simply a scalar and the when applied to the given $vec F$ the following equation results
$$(x-y)frac{partial G}{partial y} = (x+y)frac{partial G}{partial x}$$
This is where I stop. My only known method for solving PDEs is to assume it a product of two single variable functions, rearrange the functions of each variable to their respective side, and solve two ODEs. This isn't possible here so I am not sure what to do. Can anyone solve, recommend tips, or point me to any sources?
calculus pde vector-fields
calculus pde vector-fields
asked Jan 4 at 7:15
BOTKBOTK
1
asked Jan 4 at 7:15
BOTKBOTK
1
asked Jan 4 at 7:15
BOTKBOTK
1
asked Jan 4 at 7:15
BOTKBOTK
1
asked Jan 4 at 7:15
BOTKBOTK
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
$$ (x+y)frac{partial G}{partial x}+(y-x)frac{partial G}{partial y}=0$$
The Charpit-Lagrange system of characteristic equations is :
$$frac{dx}{x+y}=frac{dy}{y-x}=frac{dG}{0}$$
A first characteristic equation comes from $frac{dx}{x+y}=frac{dy}{y-x}$ .
This is an homogeneous ODE easy to solve : change of variable $y(x)=xu(x)$. The solution is:
$$ln(x^2+y^2)+2arctan(frac{y}{x})=c_1$$
A second characteristic equation comes from $frac{dG}{0}neq 0$ which implies $G=c_2$.
From the relationship $Phi(c_1,c_2)=0$ where $Phi$ is an arbitrary function, or equivalently $c_2=F(c_1)$ where $F$ is an arbitrary function :
$$G(x,y)=Fleft(ln(x^2+y^2)+2arctan(frac{y}{x}) right)$$
$F$ is an arbitrary function which could be determined insofar some boundary conditions where specified.
Note : Don't confuse this notation of function $F$ with the notation $vec F(x,y)$ used in the wording of the question.
answered Jan 4 at 8:34
JJacquelinJJacquelin
44.3k21854
$endgroup$
$begingroup$
Thank you so much. What kind of boundary condition would you need to determine $F$?
$endgroup$
– BOTK
Jan 4 at 16:36
$begingroup$
Boundary conditions are necessary in a general context of solving PDE. I suppose that you only want the "level curves" where $G(x,y)=$constant. Then doesn't mater $F$ since the function of a constant is a constant. $$ln(x^2+y^2)+2arctan(y/x)=text{constant}$$
$endgroup$
– JJacquelin
Jan 4 at 18:54
$begingroup$
I understand that. I'm asking what kind of information you would need in order to determine a specific $F$.
$endgroup$
– BOTK
Jan 5 at 8:25
$begingroup$
They are several kind of boundary conditions. For example : a specified relationship between $x$ and $y$ along a specified line. But depending on the context, the condition might also involve specified relationship with partial derivative, or something else. mathworld.wolfram.com/BoundaryConditions.html .
$endgroup$
– JJacquelin
Jan 5 at 8:39
$begingroup$
en.wikipedia.org/wiki/Boundary_value_problem
$endgroup$
– JJacquelin
Jan 5 at 8:44
add a comment |
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1 Answer
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$begingroup$
$$ (x+y)frac{partial G}{partial x}+(y-x)frac{partial G}{partial y}=0$$
The Charpit-Lagrange system of characteristic equations is :
$$frac{dx}{x+y}=frac{dy}{y-x}=frac{dG}{0}$$
A first characteristic equation comes from $frac{dx}{x+y}=frac{dy}{y-x}$ .
This is an homogeneous ODE easy to solve : change of variable $y(x)=xu(x)$. The solution is:
$$ln(x^2+y^2)+2arctan(frac{y}{x})=c_1$$
A second characteristic equation comes from $frac{dG}{0}neq 0$ which implies $G=c_2$.
From the relationship $Phi(c_1,c_2)=0$ where $Phi$ is an arbitrary function, or equivalently $c_2=F(c_1)$ where $F$ is an arbitrary function :
$$G(x,y)=Fleft(ln(x^2+y^2)+2arctan(frac{y}{x}) right)$$
$F$ is an arbitrary function which could be determined insofar some boundary conditions where specified.
Note : Don't confuse this notation of function $F$ with the notation $vec F(x,y)$ used in the wording of the question.
answered Jan 4 at 8:34
JJacquelinJJacquelin
44.3k21854
$endgroup$
$begingroup$
Thank you so much. What kind of boundary condition would you need to determine $F$?
$endgroup$
– BOTK
Jan 4 at 16:36
$begingroup$
Boundary conditions are necessary in a general context of solving PDE. I suppose that you only want the "level curves" where $G(x,y)=$constant. Then doesn't mater $F$ since the function of a constant is a constant. $$ln(x^2+y^2)+2arctan(y/x)=text{constant}$$
$endgroup$
– JJacquelin
Jan 4 at 18:54
$begingroup$
I understand that. I'm asking what kind of information you would need in order to determine a specific $F$.
$endgroup$
– BOTK
Jan 5 at 8:25
$begingroup$
They are several kind of boundary conditions. For example : a specified relationship between $x$ and $y$ along a specified line. But depending on the context, the condition might also involve specified relationship with partial derivative, or something else. mathworld.wolfram.com/BoundaryConditions.html .
$endgroup$
– JJacquelin
Jan 5 at 8:39
$begingroup$
en.wikipedia.org/wiki/Boundary_value_problem
$endgroup$
– JJacquelin
Jan 5 at 8:44
add a comment |
$begingroup$
$$ (x+y)frac{partial G}{partial x}+(y-x)frac{partial G}{partial y}=0$$
The Charpit-Lagrange system of characteristic equations is :
$$frac{dx}{x+y}=frac{dy}{y-x}=frac{dG}{0}$$
A first characteristic equation comes from $frac{dx}{x+y}=frac{dy}{y-x}$ .
This is an homogeneous ODE easy to solve : change of variable $y(x)=xu(x)$. The solution is:
$$ln(x^2+y^2)+2arctan(frac{y}{x})=c_1$$
A second characteristic equation comes from $frac{dG}{0}neq 0$ which implies $G=c_2$.
From the relationship $Phi(c_1,c_2)=0$ where $Phi$ is an arbitrary function, or equivalently $c_2=F(c_1)$ where $F$ is an arbitrary function :
$$G(x,y)=Fleft(ln(x^2+y^2)+2arctan(frac{y}{x}) right)$$
$F$ is an arbitrary function which could be determined insofar some boundary conditions where specified.
Note : Don't confuse this notation of function $F$ with the notation $vec F(x,y)$ used in the wording of the question.
answered Jan 4 at 8:34
JJacquelinJJacquelin
44.3k21854
$endgroup$
$begingroup$
Thank you so much. What kind of boundary condition would you need to determine $F$?
$endgroup$
– BOTK
Jan 4 at 16:36
$begingroup$
Boundary conditions are necessary in a general context of solving PDE. I suppose that you only want the "level curves" where $G(x,y)=$constant. Then doesn't mater $F$ since the function of a constant is a constant. $$ln(x^2+y^2)+2arctan(y/x)=text{constant}$$
$endgroup$
– JJacquelin
Jan 4 at 18:54
$begingroup$
I understand that. I'm asking what kind of information you would need in order to determine a specific $F$.
$endgroup$
– BOTK
Jan 5 at 8:25
$begingroup$
They are several kind of boundary conditions. For example : a specified relationship between $x$ and $y$ along a specified line. But depending on the context, the condition might also involve specified relationship with partial derivative, or something else. mathworld.wolfram.com/BoundaryConditions.html .
$endgroup$
– JJacquelin
Jan 5 at 8:39
$begingroup$
en.wikipedia.org/wiki/Boundary_value_problem
$endgroup$
– JJacquelin
Jan 5 at 8:44
add a comment |
$begingroup$
$$ (x+y)frac{partial G}{partial x}+(y-x)frac{partial G}{partial y}=0$$
The Charpit-Lagrange system of characteristic equations is :
$$frac{dx}{x+y}=frac{dy}{y-x}=frac{dG}{0}$$
A first characteristic equation comes from $frac{dx}{x+y}=frac{dy}{y-x}$ .
This is an homogeneous ODE easy to solve : change of variable $y(x)=xu(x)$. The solution is:
$$ln(x^2+y^2)+2arctan(frac{y}{x})=c_1$$
A second characteristic equation comes from $frac{dG}{0}neq 0$ which implies $G=c_2$.
From the relationship $Phi(c_1,c_2)=0$ where $Phi$ is an arbitrary function, or equivalently $c_2=F(c_1)$ where $F$ is an arbitrary function :
$$G(x,y)=Fleft(ln(x^2+y^2)+2arctan(frac{y}{x}) right)$$
$F$ is an arbitrary function which could be determined insofar some boundary conditions where specified.
Note : Don't confuse this notation of function $F$ with the notation $vec F(x,y)$ used in the wording of the question.
answered Jan 4 at 8:34
JJacquelinJJacquelin
44.3k21854
$endgroup$
$$ (x+y)frac{partial G}{partial x}+(y-x)frac{partial G}{partial y}=0$$
The Charpit-Lagrange system of characteristic equations is :
$$frac{dx}{x+y}=frac{dy}{y-x}=frac{dG}{0}$$
A first characteristic equation comes from $frac{dx}{x+y}=frac{dy}{y-x}$ .
This is an homogeneous ODE easy to solve : change of variable $y(x)=xu(x)$. The solution is:
$$ln(x^2+y^2)+2arctan(frac{y}{x})=c_1$$
A second characteristic equation comes from $frac{dG}{0}neq 0$ which implies $G=c_2$.
From the relationship $Phi(c_1,c_2)=0$ where $Phi$ is an arbitrary function, or equivalently $c_2=F(c_1)$ where $F$ is an arbitrary function :
$$G(x,y)=Fleft(ln(x^2+y^2)+2arctan(frac{y}{x}) right)$$
$F$ is an arbitrary function which could be determined insofar some boundary conditions where specified.
Note : Don't confuse this notation of function $F$ with the notation $vec F(x,y)$ used in the wording of the question.
answered Jan 4 at 8:34
JJacquelinJJacquelin
44.3k21854
edited Jan 4 at 8:43
answered Jan 4 at 8:34
JJacquelinJJacquelin
44.3k21854
answered Jan 4 at 8:34
JJacquelinJJacquelin
44.3k21854
answered Jan 4 at 8:34
JJacquelinJJacquelin
44.3k21854
44.3k21854
$begingroup$
Thank you so much. What kind of boundary condition would you need to determine $F$?
$endgroup$
– BOTK
Jan 4 at 16:36
$begingroup$
Boundary conditions are necessary in a general context of solving PDE. I suppose that you only want the "level curves" where $G(x,y)=$constant. Then doesn't mater $F$ since the function of a constant is a constant. $$ln(x^2+y^2)+2arctan(y/x)=text{constant}$$
$endgroup$
– JJacquelin
Jan 4 at 18:54
$begingroup$
I understand that. I'm asking what kind of information you would need in order to determine a specific $F$.
$endgroup$
– BOTK
Jan 5 at 8:25
$begingroup$
They are several kind of boundary conditions. For example : a specified relationship between $x$ and $y$ along a specified line. But depending on the context, the condition might also involve specified relationship with partial derivative, or something else. mathworld.wolfram.com/BoundaryConditions.html .
$endgroup$
– JJacquelin
Jan 5 at 8:39
$begingroup$
en.wikipedia.org/wiki/Boundary_value_problem
$endgroup$
– JJacquelin
Jan 5 at 8:44
add a comment |
$begingroup$
Thank you so much. What kind of boundary condition would you need to determine $F$?
$endgroup$
– BOTK
Jan 4 at 16:36
$begingroup$
Boundary conditions are necessary in a general context of solving PDE. I suppose that you only want the "level curves" where $G(x,y)=$constant. Then doesn't mater $F$ since the function of a constant is a constant. $$ln(x^2+y^2)+2arctan(y/x)=text{constant}$$
$endgroup$
– JJacquelin
Jan 4 at 18:54
$begingroup$
I understand that. I'm asking what kind of information you would need in order to determine a specific $F$.
$endgroup$
– BOTK
Jan 5 at 8:25
$begingroup$
They are several kind of boundary conditions. For example : a specified relationship between $x$ and $y$ along a specified line. But depending on the context, the condition might also involve specified relationship with partial derivative, or something else. mathworld.wolfram.com/BoundaryConditions.html .
$endgroup$
– JJacquelin
Jan 5 at 8:39
$begingroup$
en.wikipedia.org/wiki/Boundary_value_problem
$endgroup$
– JJacquelin
Jan 5 at 8:44
$begingroup$
Thank you so much. What kind of boundary condition would you need to determine $F$?
$endgroup$
– BOTK
Jan 4 at 16:36
$begingroup$
Thank you so much. What kind of boundary condition would you need to determine $F$?
$endgroup$
– BOTK
Jan 4 at 16:36
$begingroup$
Boundary conditions are necessary in a general context of solving PDE. I suppose that you only want the "level curves" where $G(x,y)=$constant. Then doesn't mater $F$ since the function of a constant is a constant. $$ln(x^2+y^2)+2arctan(y/x)=text{constant}$$
$endgroup$
– JJacquelin
Jan 4 at 18:54
$begingroup$
Boundary conditions are necessary in a general context of solving PDE. I suppose that you only want the "level curves" where $G(x,y)=$constant. Then doesn't mater $F$ since the function of a constant is a constant. $$ln(x^2+y^2)+2arctan(y/x)=text{constant}$$
$endgroup$
– JJacquelin
Jan 4 at 18:54
$begingroup$
I understand that. I'm asking what kind of information you would need in order to determine a specific $F$.
$endgroup$
– BOTK
Jan 5 at 8:25
$begingroup$
I understand that. I'm asking what kind of information you would need in order to determine a specific $F$.
$endgroup$
– BOTK
Jan 5 at 8:25
$begingroup$
They are several kind of boundary conditions. For example : a specified relationship between $x$ and $y$ along a specified line. But depending on the context, the condition might also involve specified relationship with partial derivative, or something else. mathworld.wolfram.com/BoundaryConditions.html .
$endgroup$
– JJacquelin
Jan 5 at 8:39
$begingroup$
They are several kind of boundary conditions. For example : a specified relationship between $x$ and $y$ along a specified line. But depending on the context, the condition might also involve specified relationship with partial derivative, or something else. mathworld.wolfram.com/BoundaryConditions.html .
$endgroup$
– JJacquelin
Jan 5 at 8:39
$begingroup$
en.wikipedia.org/wiki/Boundary_value_problem
$endgroup$
– JJacquelin
Jan 5 at 8:44
$begingroup$
en.wikipedia.org/wiki/Boundary_value_problem
$endgroup$
– JJacquelin
Jan 5 at 8:44
add a comment |
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