Mixing
Solving a PDE for determining a metric
$begingroup$
A friend is studying some Finsler metrics and she has come upon the following partial differential equation:
$$begin{cases}
y^ifrac{partial b(x,y)}{partial x^i}=0\
y^ifrac{partial b(x,y)}{partial y^i}=1\
end{cases},$$
where $b:mathbb{R}^{2n}tomathbb{R}, x=(x^1,dots,x^n)inmathbb{R}^n, y=(y^1,dots,y^n)inmathbb{R}^n.$
I have managed to find a family of solutions:
$$b_k(x,y)=sqrt{|y|^2-k(|x|^2|y|^2-langle x,yrangle^2)}, kinmathbb{R}.$$
but I don't see how to solve the equation in general.
differential-geometry pde
edited Feb 8 at 7:51
Daniele Tampieri
2,65221022
asked Jan 30 at 12:48
IuliaIulia
1,025415
$endgroup$
add a comment |
$begingroup$
A friend is studying some Finsler metrics and she has come upon the following partial differential equation:
$$begin{cases}
y^ifrac{partial b(x,y)}{partial x^i}=0\
y^ifrac{partial b(x,y)}{partial y^i}=1\
end{cases},$$
where $b:mathbb{R}^{2n}tomathbb{R}, x=(x^1,dots,x^n)inmathbb{R}^n, y=(y^1,dots,y^n)inmathbb{R}^n.$
I have managed to find a family of solutions:
$$b_k(x,y)=sqrt{|y|^2-k(|x|^2|y|^2-langle x,yrangle^2)}, kinmathbb{R}.$$
but I don't see how to solve the equation in general.
differential-geometry pde
edited Feb 8 at 7:51
Daniele Tampieri
2,65221022
asked Jan 30 at 12:48
IuliaIulia
1,025415
$endgroup$
$begingroup$
Are you familiar with the method of characteristics for solving 1st-order PDEs?
$endgroup$
– Amitai Yuval
Jan 30 at 14:05
$begingroup$
I do not know hot to apply it for the case $x$, $y$ $n$-dimensional. Just to be sure, $y^ifrac{partial b(x,y)}{partial y^i}$ means $sum_{i=1}^n y^ifrac{partial b(x,y)}{partial y^i}$.
$endgroup$
– Iulia
Jan 30 at 18:43
add a comment |
$begingroup$
A friend is studying some Finsler metrics and she has come upon the following partial differential equation:
$$begin{cases}
y^ifrac{partial b(x,y)}{partial x^i}=0\
y^ifrac{partial b(x,y)}{partial y^i}=1\
end{cases},$$
where $b:mathbb{R}^{2n}tomathbb{R}, x=(x^1,dots,x^n)inmathbb{R}^n, y=(y^1,dots,y^n)inmathbb{R}^n.$
I have managed to find a family of solutions:
$$b_k(x,y)=sqrt{|y|^2-k(|x|^2|y|^2-langle x,yrangle^2)}, kinmathbb{R}.$$
but I don't see how to solve the equation in general.
differential-geometry pde
edited Feb 8 at 7:51
Daniele Tampieri
2,65221022
asked Jan 30 at 12:48
IuliaIulia
1,025415
$endgroup$
A friend is studying some Finsler metrics and she has come upon the following partial differential equation:
$$begin{cases}
y^ifrac{partial b(x,y)}{partial x^i}=0\
y^ifrac{partial b(x,y)}{partial y^i}=1\
end{cases},$$
where $b:mathbb{R}^{2n}tomathbb{R}, x=(x^1,dots,x^n)inmathbb{R}^n, y=(y^1,dots,y^n)inmathbb{R}^n.$
I have managed to find a family of solutions:
$$b_k(x,y)=sqrt{|y|^2-k(|x|^2|y|^2-langle x,yrangle^2)}, kinmathbb{R}.$$
but I don't see how to solve the equation in general.
differential-geometry pde
differential-geometry pde
edited Feb 8 at 7:51
Daniele Tampieri
2,65221022
asked Jan 30 at 12:48
IuliaIulia
1,025415
edited Feb 8 at 7:51
Daniele Tampieri
2,65221022
asked Jan 30 at 12:48
IuliaIulia
1,025415
edited Feb 8 at 7:51
Daniele Tampieri
2,65221022
edited Feb 8 at 7:51
Daniele Tampieri
2,65221022
edited Feb 8 at 7:51
Daniele Tampieri
2,65221022
2,65221022
asked Jan 30 at 12:48
IuliaIulia
1,025415
asked Jan 30 at 12:48
IuliaIulia
1,025415
asked Jan 30 at 12:48
IuliaIulia
1,025415
1,025415
$begingroup$
Are you familiar with the method of characteristics for solving 1st-order PDEs?
$endgroup$
– Amitai Yuval
Jan 30 at 14:05
$begingroup$
I do not know hot to apply it for the case $x$, $y$ $n$-dimensional. Just to be sure, $y^ifrac{partial b(x,y)}{partial y^i}$ means $sum_{i=1}^n y^ifrac{partial b(x,y)}{partial y^i}$.
$endgroup$
– Iulia
Jan 30 at 18:43
add a comment |
$begingroup$
Are you familiar with the method of characteristics for solving 1st-order PDEs?
$endgroup$
– Amitai Yuval
Jan 30 at 14:05
$begingroup$
I do not know hot to apply it for the case $x$, $y$ $n$-dimensional. Just to be sure, $y^ifrac{partial b(x,y)}{partial y^i}$ means $sum_{i=1}^n y^ifrac{partial b(x,y)}{partial y^i}$.
$endgroup$
– Iulia
Jan 30 at 18:43
$begingroup$
Are you familiar with the method of characteristics for solving 1st-order PDEs?
$endgroup$
– Amitai Yuval
Jan 30 at 14:05
$begingroup$
Are you familiar with the method of characteristics for solving 1st-order PDEs?
$endgroup$
– Amitai Yuval
Jan 30 at 14:05
$begingroup$
I do not know hot to apply it for the case $x$, $y$ $n$-dimensional. Just to be sure, $y^ifrac{partial b(x,y)}{partial y^i}$ means $sum_{i=1}^n y^ifrac{partial b(x,y)}{partial y^i}$.
$endgroup$
– Iulia
Jan 30 at 18:43
$begingroup$
I do not know hot to apply it for the case $x$, $y$ $n$-dimensional. Just to be sure, $y^ifrac{partial b(x,y)}{partial y^i}$ means $sum_{i=1}^n y^ifrac{partial b(x,y)}{partial y^i}$.
$endgroup$
– Iulia
Jan 30 at 18:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Change the $y$ variables to spherical coordinates $(r,theta)=(r,theta_1,ldots,theta_{n-1})$. Then
$$1=y^ifrac{partial b(x,y)}{partial y^i}=rfrac{partial b(x,r,theta)}{partial r}$$
Hence $b(x,r,theta)=log r+F(x,theta)$. By the first PDE, $F$ must satisfy:
$$0=y^ifrac{partial b(x,y)}{partial x^i}=y^i(theta)frac{partial F(x,theta)}{partial x^i}$$ Thus it is of the form $F(x,theta)=F(x^iv_i(theta),theta)$ where $v$ can be functions of $theta$ which are orthogonal to $y(theta)$: $y^i(theta)v_i(theta)=0$ for all $theta$. For example when $n=2$ and $(y^1,y^2)=(rcostheta,rsintheta)$:
$$b(x,y)=log r+F(x^2costheta-x^1sintheta,theta)$$
As you can see, the above solution does not cover the solutions you have found. I believe the solutions you suggested do not satisfy the second PDE but rather:
$$y^ifrac{partial b(x,y)}{partial y^i}=b(x,y)$$
answered Feb 8 at 9:41
MOMOMOMO
717312
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Change the $y$ variables to spherical coordinates $(r,theta)=(r,theta_1,ldots,theta_{n-1})$. Then
$$1=y^ifrac{partial b(x,y)}{partial y^i}=rfrac{partial b(x,r,theta)}{partial r}$$
Hence $b(x,r,theta)=log r+F(x,theta)$. By the first PDE, $F$ must satisfy:
$$0=y^ifrac{partial b(x,y)}{partial x^i}=y^i(theta)frac{partial F(x,theta)}{partial x^i}$$ Thus it is of the form $F(x,theta)=F(x^iv_i(theta),theta)$ where $v$ can be functions of $theta$ which are orthogonal to $y(theta)$: $y^i(theta)v_i(theta)=0$ for all $theta$. For example when $n=2$ and $(y^1,y^2)=(rcostheta,rsintheta)$:
$$b(x,y)=log r+F(x^2costheta-x^1sintheta,theta)$$
As you can see, the above solution does not cover the solutions you have found. I believe the solutions you suggested do not satisfy the second PDE but rather:
$$y^ifrac{partial b(x,y)}{partial y^i}=b(x,y)$$
answered Feb 8 at 9:41
MOMOMOMO
717312
$endgroup$
add a comment |
$begingroup$
Change the $y$ variables to spherical coordinates $(r,theta)=(r,theta_1,ldots,theta_{n-1})$. Then
$$1=y^ifrac{partial b(x,y)}{partial y^i}=rfrac{partial b(x,r,theta)}{partial r}$$
Hence $b(x,r,theta)=log r+F(x,theta)$. By the first PDE, $F$ must satisfy:
$$0=y^ifrac{partial b(x,y)}{partial x^i}=y^i(theta)frac{partial F(x,theta)}{partial x^i}$$ Thus it is of the form $F(x,theta)=F(x^iv_i(theta),theta)$ where $v$ can be functions of $theta$ which are orthogonal to $y(theta)$: $y^i(theta)v_i(theta)=0$ for all $theta$. For example when $n=2$ and $(y^1,y^2)=(rcostheta,rsintheta)$:
$$b(x,y)=log r+F(x^2costheta-x^1sintheta,theta)$$
As you can see, the above solution does not cover the solutions you have found. I believe the solutions you suggested do not satisfy the second PDE but rather:
$$y^ifrac{partial b(x,y)}{partial y^i}=b(x,y)$$
answered Feb 8 at 9:41
MOMOMOMO
717312
$endgroup$
add a comment |
$begingroup$
Change the $y$ variables to spherical coordinates $(r,theta)=(r,theta_1,ldots,theta_{n-1})$. Then
$$1=y^ifrac{partial b(x,y)}{partial y^i}=rfrac{partial b(x,r,theta)}{partial r}$$
Hence $b(x,r,theta)=log r+F(x,theta)$. By the first PDE, $F$ must satisfy:
$$0=y^ifrac{partial b(x,y)}{partial x^i}=y^i(theta)frac{partial F(x,theta)}{partial x^i}$$ Thus it is of the form $F(x,theta)=F(x^iv_i(theta),theta)$ where $v$ can be functions of $theta$ which are orthogonal to $y(theta)$: $y^i(theta)v_i(theta)=0$ for all $theta$. For example when $n=2$ and $(y^1,y^2)=(rcostheta,rsintheta)$:
$$b(x,y)=log r+F(x^2costheta-x^1sintheta,theta)$$
As you can see, the above solution does not cover the solutions you have found. I believe the solutions you suggested do not satisfy the second PDE but rather:
$$y^ifrac{partial b(x,y)}{partial y^i}=b(x,y)$$
answered Feb 8 at 9:41
MOMOMOMO
717312
$endgroup$
Change the $y$ variables to spherical coordinates $(r,theta)=(r,theta_1,ldots,theta_{n-1})$. Then
$$1=y^ifrac{partial b(x,y)}{partial y^i}=rfrac{partial b(x,r,theta)}{partial r}$$
Hence $b(x,r,theta)=log r+F(x,theta)$. By the first PDE, $F$ must satisfy:
$$0=y^ifrac{partial b(x,y)}{partial x^i}=y^i(theta)frac{partial F(x,theta)}{partial x^i}$$ Thus it is of the form $F(x,theta)=F(x^iv_i(theta),theta)$ where $v$ can be functions of $theta$ which are orthogonal to $y(theta)$: $y^i(theta)v_i(theta)=0$ for all $theta$. For example when $n=2$ and $(y^1,y^2)=(rcostheta,rsintheta)$:
$$b(x,y)=log r+F(x^2costheta-x^1sintheta,theta)$$
As you can see, the above solution does not cover the solutions you have found. I believe the solutions you suggested do not satisfy the second PDE but rather:
$$y^ifrac{partial b(x,y)}{partial y^i}=b(x,y)$$
answered Feb 8 at 9:41
MOMOMOMO
717312
answered Feb 8 at 9:41
MOMOMOMO
717312
answered Feb 8 at 9:41
MOMOMOMO
717312
answered Feb 8 at 9:41
MOMOMOMO
717312
717312
add a comment |
add a comment |
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$begingroup$
Are you familiar with the method of characteristics for solving 1st-order PDEs?
$endgroup$
– Amitai Yuval
Jan 30 at 14:05
$begingroup$
I do not know hot to apply it for the case $x$, $y$ $n$-dimensional. Just to be sure, $y^ifrac{partial b(x,y)}{partial y^i}$ means $sum_{i=1}^n y^ifrac{partial b(x,y)}{partial y^i}$.
$endgroup$
– Iulia
Jan 30 at 18:43