$nabla cdot (b nabla c) = 0$ where $b$ and $c$ are unknown
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Description:
I want to solve the equation
$$ nabla cdot left( b nabla c right) = 0 $$
for $b(x,y)$ and $c(x,y)$ in the domain $x,y geq 0$. This partial differential equation is supplemented by
$$ b nabla c = vec{g} $$
where $vec{g}(x,y)$ is a known vector that has non-zero curl. Furthermore, the values of $b$ and $c$ are known on the boundary $(x = 0, y = 0)$. As a consequence, the normal derivatives of $c(x,y)$ on the boundary are also known:
$$ nabla times vec{g} = R(x,y) ne 0 \
c = 0 hspace{4mm} text{at} hspace{4mm} x=0 \
c = 0 hspace{4mm} text{at} hspace{4mm} y=0 \ $$
What solution methods to this problem exist? Can you point me to any resources? If I try to solve this numerically, which software packages can you recommend? Thanks.
It is possible to derive the first-order pde's
$$ vec{g} times nabla c = 0 \ nabla ln(b) times vec{g} = R ; . $$
Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$. So the interior of the domain is not connected to the boundary. I was hoping there are other approaches.
pde vectors vector-analysis boundary-value-problem
asked Oct 30 at 8:57
Crenguta
406
add a comment |
up vote
1
down vote
favorite
Description:
I want to solve the equation
$$ nabla cdot left( b nabla c right) = 0 $$
for $b(x,y)$ and $c(x,y)$ in the domain $x,y geq 0$. This partial differential equation is supplemented by
$$ b nabla c = vec{g} $$
where $vec{g}(x,y)$ is a known vector that has non-zero curl. Furthermore, the values of $b$ and $c$ are known on the boundary $(x = 0, y = 0)$. As a consequence, the normal derivatives of $c(x,y)$ on the boundary are also known:
$$ nabla times vec{g} = R(x,y) ne 0 \
c = 0 hspace{4mm} text{at} hspace{4mm} x=0 \
c = 0 hspace{4mm} text{at} hspace{4mm} y=0 \ $$
What solution methods to this problem exist? Can you point me to any resources? If I try to solve this numerically, which software packages can you recommend? Thanks.
It is possible to derive the first-order pde's
$$ vec{g} times nabla c = 0 \ nabla ln(b) times vec{g} = R ; . $$
Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$. So the interior of the domain is not connected to the boundary. I was hoping there are other approaches.
pde vectors vector-analysis boundary-value-problem
asked Oct 30 at 8:57
Crenguta
406
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Description:
I want to solve the equation
$$ nabla cdot left( b nabla c right) = 0 $$
for $b(x,y)$ and $c(x,y)$ in the domain $x,y geq 0$. This partial differential equation is supplemented by
$$ b nabla c = vec{g} $$
where $vec{g}(x,y)$ is a known vector that has non-zero curl. Furthermore, the values of $b$ and $c$ are known on the boundary $(x = 0, y = 0)$. As a consequence, the normal derivatives of $c(x,y)$ on the boundary are also known:
$$ nabla times vec{g} = R(x,y) ne 0 \
c = 0 hspace{4mm} text{at} hspace{4mm} x=0 \
c = 0 hspace{4mm} text{at} hspace{4mm} y=0 \ $$
What solution methods to this problem exist? Can you point me to any resources? If I try to solve this numerically, which software packages can you recommend? Thanks.
It is possible to derive the first-order pde's
$$ vec{g} times nabla c = 0 \ nabla ln(b) times vec{g} = R ; . $$
Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$. So the interior of the domain is not connected to the boundary. I was hoping there are other approaches.
pde vectors vector-analysis boundary-value-problem
asked Oct 30 at 8:57
Crenguta
406
Description:
I want to solve the equation
$$ nabla cdot left( b nabla c right) = 0 $$
for $b(x,y)$ and $c(x,y)$ in the domain $x,y geq 0$. This partial differential equation is supplemented by
$$ b nabla c = vec{g} $$
where $vec{g}(x,y)$ is a known vector that has non-zero curl. Furthermore, the values of $b$ and $c$ are known on the boundary $(x = 0, y = 0)$. As a consequence, the normal derivatives of $c(x,y)$ on the boundary are also known:
$$ nabla times vec{g} = R(x,y) ne 0 \
c = 0 hspace{4mm} text{at} hspace{4mm} x=0 \
c = 0 hspace{4mm} text{at} hspace{4mm} y=0 \ $$
What solution methods to this problem exist? Can you point me to any resources? If I try to solve this numerically, which software packages can you recommend? Thanks.
It is possible to derive the first-order pde's
$$ vec{g} times nabla c = 0 \ nabla ln(b) times vec{g} = R ; . $$
Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$. So the interior of the domain is not connected to the boundary. I was hoping there are other approaches.
pde vectors vector-analysis boundary-value-problem
pde vectors vector-analysis boundary-value-problem
asked Oct 30 at 8:57
Crenguta
406
asked Oct 30 at 8:57
Crenguta
406
edited Nov 22 at 20:49
asked Oct 30 at 8:57
Crenguta
406
asked Oct 30 at 8:57
Crenguta
406
asked Oct 30 at 8:57
Crenguta
406
406
add a comment |
add a comment |
1 Answer
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You can start by noticing that we need
$$
begin{split}
0 &= nablatimesnabla c = nablatimesfrac gb
= nablafrac1b times g + frac1b nablatimes g
= -frac{nabla b}{b^2} times g + frac Rb .
end{split}
$$
Multiplying by $b$ you get
$$
nabla(log b) times g = R.
$$
This is a linear first order partial differential equation, whose solution is of the form $b=exp(tilde f+f_0)$, where $tilde f$ is a particular solution to the inhomogeneous PDE $nabla f times g = R$ and $f_0$ is a solution to the homogeneous one. The PDE is solvable as long as $g$ is not orthogonal to $partialOmega$ (where $Omega$ is your domain, the positive quadrant) and boundary value for $b$ is compatible with the characteristic lines.
You can find more pieces of information here, here, and here.
Once you have found $b$, $g/b$ is irrotational, therefore you find $c$ such that $nabla c=g/b$.
answered Nov 5 at 14:17
Federico
4,203512
Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
– Crenguta
Nov 6 at 7:47
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
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up vote
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down vote
You can start by noticing that we need
$$
begin{split}
0 &= nablatimesnabla c = nablatimesfrac gb
= nablafrac1b times g + frac1b nablatimes g
= -frac{nabla b}{b^2} times g + frac Rb .
end{split}
$$
Multiplying by $b$ you get
$$
nabla(log b) times g = R.
$$
This is a linear first order partial differential equation, whose solution is of the form $b=exp(tilde f+f_0)$, where $tilde f$ is a particular solution to the inhomogeneous PDE $nabla f times g = R$ and $f_0$ is a solution to the homogeneous one. The PDE is solvable as long as $g$ is not orthogonal to $partialOmega$ (where $Omega$ is your domain, the positive quadrant) and boundary value for $b$ is compatible with the characteristic lines.
You can find more pieces of information here, here, and here.
Once you have found $b$, $g/b$ is irrotational, therefore you find $c$ such that $nabla c=g/b$.
answered Nov 5 at 14:17
Federico
4,203512
Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
– Crenguta
Nov 6 at 7:47
add a comment |
up vote
1
down vote
You can start by noticing that we need
$$
begin{split}
0 &= nablatimesnabla c = nablatimesfrac gb
= nablafrac1b times g + frac1b nablatimes g
= -frac{nabla b}{b^2} times g + frac Rb .
end{split}
$$
Multiplying by $b$ you get
$$
nabla(log b) times g = R.
$$
This is a linear first order partial differential equation, whose solution is of the form $b=exp(tilde f+f_0)$, where $tilde f$ is a particular solution to the inhomogeneous PDE $nabla f times g = R$ and $f_0$ is a solution to the homogeneous one. The PDE is solvable as long as $g$ is not orthogonal to $partialOmega$ (where $Omega$ is your domain, the positive quadrant) and boundary value for $b$ is compatible with the characteristic lines.
You can find more pieces of information here, here, and here.
Once you have found $b$, $g/b$ is irrotational, therefore you find $c$ such that $nabla c=g/b$.
answered Nov 5 at 14:17
Federico
4,203512
Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
– Crenguta
Nov 6 at 7:47
add a comment |
up vote
1
down vote
up vote
1
down vote
You can start by noticing that we need
$$
begin{split}
0 &= nablatimesnabla c = nablatimesfrac gb
= nablafrac1b times g + frac1b nablatimes g
= -frac{nabla b}{b^2} times g + frac Rb .
end{split}
$$
Multiplying by $b$ you get
$$
nabla(log b) times g = R.
$$
This is a linear first order partial differential equation, whose solution is of the form $b=exp(tilde f+f_0)$, where $tilde f$ is a particular solution to the inhomogeneous PDE $nabla f times g = R$ and $f_0$ is a solution to the homogeneous one. The PDE is solvable as long as $g$ is not orthogonal to $partialOmega$ (where $Omega$ is your domain, the positive quadrant) and boundary value for $b$ is compatible with the characteristic lines.
You can find more pieces of information here, here, and here.
Once you have found $b$, $g/b$ is irrotational, therefore you find $c$ such that $nabla c=g/b$.
answered Nov 5 at 14:17
Federico
4,203512
You can start by noticing that we need
$$
begin{split}
0 &= nablatimesnabla c = nablatimesfrac gb
= nablafrac1b times g + frac1b nablatimes g
= -frac{nabla b}{b^2} times g + frac Rb .
end{split}
$$
Multiplying by $b$ you get
$$
nabla(log b) times g = R.
$$
This is a linear first order partial differential equation, whose solution is of the form $b=exp(tilde f+f_0)$, where $tilde f$ is a particular solution to the inhomogeneous PDE $nabla f times g = R$ and $f_0$ is a solution to the homogeneous one. The PDE is solvable as long as $g$ is not orthogonal to $partialOmega$ (where $Omega$ is your domain, the positive quadrant) and boundary value for $b$ is compatible with the characteristic lines.
You can find more pieces of information here, here, and here.
Once you have found $b$, $g/b$ is irrotational, therefore you find $c$ such that $nabla c=g/b$.
answered Nov 5 at 14:17
Federico
4,203512
answered Nov 5 at 14:17
Federico
4,203512
answered Nov 5 at 14:17
Federico
4,203512
answered Nov 5 at 14:17
Federico
4,203512
4,203512
Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
– Crenguta
Nov 6 at 7:47
add a comment |
Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
– Crenguta
Nov 6 at 7:47
Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
– Crenguta
Nov 6 at 7:47
Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
– Crenguta
Nov 6 at 7:47
add a comment |
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