$nabla cdot (b nabla c) = 0$ where $b$ and $c$ are unknown
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
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
1
active
oldest
votes
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
4,178512
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2977349%2fnabla-cdot-b-nabla-c-0-where-b-and-c-are-unknown%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
4,178512
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
4,178512
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
4,178512
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
4,178512
answered Nov 5 at 14:17
Federico
4,178512
answered Nov 5 at 14:17
Federico
4,178512
answered Nov 5 at 14:17
Federico
4,178512
4,178512
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2977349%2fnabla-cdot-b-nabla-c-0-where-b-and-c-are-unknown%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
