Uniqueness of Neumann conditions for Laplace equations
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I know that generally Neumann problem only has unique solutions up to constants. But what about this case (which was brought to me by my friend): If $Omega$ is a bounded region and its boundary is $C^2$, $gin C(partial Omega)$ is such that $partial_nu u=g$ on $partialOmega$, $Delta u=0$ on $Omega$. Then does it admit a unique solution?
pde
asked Dec 8 '13 at 4:51
XuxuXuxu
561413
$endgroup$
add a comment |
$begingroup$
I know that generally Neumann problem only has unique solutions up to constants. But what about this case (which was brought to me by my friend): If $Omega$ is a bounded region and its boundary is $C^2$, $gin C(partial Omega)$ is such that $partial_nu u=g$ on $partialOmega$, $Delta u=0$ on $Omega$. Then does it admit a unique solution?
pde
asked Dec 8 '13 at 4:51
XuxuXuxu
561413
$endgroup$
1
$begingroup$
Why isn't $u+c$ not also a solution if $c$ is constant? Am I missing something? Or are you asking about existence instead of uniqueness?
$endgroup$
– Stephen Montgomery-Smith
Dec 8 '13 at 5:13
$begingroup$
@StephenMontgomery-Smith I already know this. I am merely wondering whether the condition on boundary of the domain has effects on the uniqueness or not. If not, what conditions else might guarantee uniqueness?
$endgroup$
– Xuxu
Dec 8 '13 at 7:03
1
$begingroup$
The point I was making is that is it very easy to see that $u+c$ is also a solution. You should be able to immediately see that your conditions don't imply uniqueness.
$endgroup$
– Stephen Montgomery-Smith
Dec 8 '13 at 15:43
add a comment |
$begingroup$
I know that generally Neumann problem only has unique solutions up to constants. But what about this case (which was brought to me by my friend): If $Omega$ is a bounded region and its boundary is $C^2$, $gin C(partial Omega)$ is such that $partial_nu u=g$ on $partialOmega$, $Delta u=0$ on $Omega$. Then does it admit a unique solution?
pde
asked Dec 8 '13 at 4:51
XuxuXuxu
561413
$endgroup$
I know that generally Neumann problem only has unique solutions up to constants. But what about this case (which was brought to me by my friend): If $Omega$ is a bounded region and its boundary is $C^2$, $gin C(partial Omega)$ is such that $partial_nu u=g$ on $partialOmega$, $Delta u=0$ on $Omega$. Then does it admit a unique solution?
pde
pde
asked Dec 8 '13 at 4:51
XuxuXuxu
561413
asked Dec 8 '13 at 4:51
XuxuXuxu
561413
asked Dec 8 '13 at 4:51
XuxuXuxu
561413
asked Dec 8 '13 at 4:51
XuxuXuxu
561413
asked Dec 8 '13 at 4:51
XuxuXuxu
561413
561413
1
$begingroup$
Why isn't $u+c$ not also a solution if $c$ is constant? Am I missing something? Or are you asking about existence instead of uniqueness?
$endgroup$
– Stephen Montgomery-Smith
Dec 8 '13 at 5:13
$begingroup$
@StephenMontgomery-Smith I already know this. I am merely wondering whether the condition on boundary of the domain has effects on the uniqueness or not. If not, what conditions else might guarantee uniqueness?
$endgroup$
– Xuxu
Dec 8 '13 at 7:03
1
$begingroup$
The point I was making is that is it very easy to see that $u+c$ is also a solution. You should be able to immediately see that your conditions don't imply uniqueness.
$endgroup$
– Stephen Montgomery-Smith
Dec 8 '13 at 15:43
add a comment |
1
$begingroup$
Why isn't $u+c$ not also a solution if $c$ is constant? Am I missing something? Or are you asking about existence instead of uniqueness?
$endgroup$
– Stephen Montgomery-Smith
Dec 8 '13 at 5:13
$begingroup$
@StephenMontgomery-Smith I already know this. I am merely wondering whether the condition on boundary of the domain has effects on the uniqueness or not. If not, what conditions else might guarantee uniqueness?
$endgroup$
– Xuxu
Dec 8 '13 at 7:03
1
$begingroup$
The point I was making is that is it very easy to see that $u+c$ is also a solution. You should be able to immediately see that your conditions don't imply uniqueness.
$endgroup$
– Stephen Montgomery-Smith
Dec 8 '13 at 15:43
1
1
$begingroup$
Why isn't $u+c$ not also a solution if $c$ is constant? Am I missing something? Or are you asking about existence instead of uniqueness?
$endgroup$
– Stephen Montgomery-Smith
Dec 8 '13 at 5:13
$begingroup$
Why isn't $u+c$ not also a solution if $c$ is constant? Am I missing something? Or are you asking about existence instead of uniqueness?
$endgroup$
– Stephen Montgomery-Smith
Dec 8 '13 at 5:13
$begingroup$
@StephenMontgomery-Smith I already know this. I am merely wondering whether the condition on boundary of the domain has effects on the uniqueness or not. If not, what conditions else might guarantee uniqueness?
$endgroup$
– Xuxu
Dec 8 '13 at 7:03
$begingroup$
@StephenMontgomery-Smith I already know this. I am merely wondering whether the condition on boundary of the domain has effects on the uniqueness or not. If not, what conditions else might guarantee uniqueness?
$endgroup$
– Xuxu
Dec 8 '13 at 7:03
1
1
$begingroup$
The point I was making is that is it very easy to see that $u+c$ is also a solution. You should be able to immediately see that your conditions don't imply uniqueness.
$endgroup$
– Stephen Montgomery-Smith
Dec 8 '13 at 15:43
$begingroup$
The point I was making is that is it very easy to see that $u+c$ is also a solution. You should be able to immediately see that your conditions don't imply uniqueness.
$endgroup$
– Stephen Montgomery-Smith
Dec 8 '13 at 15:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
whether the condition on boundary of the domain has effects on the uniqueness or not.
We need a smoothness condition on the boundary just so that $partial_nu u$ makes sense. This is the probably reason why you see that condition.
Otherwise, your question is unclear. You asked
Then does it admit a unique solution?
to which Stephen Montgomery-Smith gave the negative answer, to which you replied that you already know that. What exactly do you want?
If you want to know how one proves the uniqueness up to constants: let $u$ be the difference of two solutions, let $v=u^2$ and notice that $Delta v = |nabla u|^2$. Since the normal derivative of $u$ vanishes on the boundary, so does the normal derivative of $v$. By the divergence theorem, $int_{Omega} Delta v =0$. Hence $nabla uequiv 0$.
answered Dec 9 '13 at 13:26
Post No BullsPost No Bulls
6,64121783
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add a comment |
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$begingroup$
whether the condition on boundary of the domain has effects on the uniqueness or not.
We need a smoothness condition on the boundary just so that $partial_nu u$ makes sense. This is the probably reason why you see that condition.
Otherwise, your question is unclear. You asked
Then does it admit a unique solution?
to which Stephen Montgomery-Smith gave the negative answer, to which you replied that you already know that. What exactly do you want?
If you want to know how one proves the uniqueness up to constants: let $u$ be the difference of two solutions, let $v=u^2$ and notice that $Delta v = |nabla u|^2$. Since the normal derivative of $u$ vanishes on the boundary, so does the normal derivative of $v$. By the divergence theorem, $int_{Omega} Delta v =0$. Hence $nabla uequiv 0$.
answered Dec 9 '13 at 13:26
Post No BullsPost No Bulls
6,64121783
$endgroup$
add a comment |
$begingroup$
whether the condition on boundary of the domain has effects on the uniqueness or not.
We need a smoothness condition on the boundary just so that $partial_nu u$ makes sense. This is the probably reason why you see that condition.
Otherwise, your question is unclear. You asked
Then does it admit a unique solution?
to which Stephen Montgomery-Smith gave the negative answer, to which you replied that you already know that. What exactly do you want?
If you want to know how one proves the uniqueness up to constants: let $u$ be the difference of two solutions, let $v=u^2$ and notice that $Delta v = |nabla u|^2$. Since the normal derivative of $u$ vanishes on the boundary, so does the normal derivative of $v$. By the divergence theorem, $int_{Omega} Delta v =0$. Hence $nabla uequiv 0$.
answered Dec 9 '13 at 13:26
Post No BullsPost No Bulls
6,64121783
$endgroup$
add a comment |
$begingroup$
whether the condition on boundary of the domain has effects on the uniqueness or not.
We need a smoothness condition on the boundary just so that $partial_nu u$ makes sense. This is the probably reason why you see that condition.
Otherwise, your question is unclear. You asked
Then does it admit a unique solution?
to which Stephen Montgomery-Smith gave the negative answer, to which you replied that you already know that. What exactly do you want?
If you want to know how one proves the uniqueness up to constants: let $u$ be the difference of two solutions, let $v=u^2$ and notice that $Delta v = |nabla u|^2$. Since the normal derivative of $u$ vanishes on the boundary, so does the normal derivative of $v$. By the divergence theorem, $int_{Omega} Delta v =0$. Hence $nabla uequiv 0$.
answered Dec 9 '13 at 13:26
Post No BullsPost No Bulls
6,64121783
$endgroup$
whether the condition on boundary of the domain has effects on the uniqueness or not.
We need a smoothness condition on the boundary just so that $partial_nu u$ makes sense. This is the probably reason why you see that condition.
Otherwise, your question is unclear. You asked
Then does it admit a unique solution?
to which Stephen Montgomery-Smith gave the negative answer, to which you replied that you already know that. What exactly do you want?
If you want to know how one proves the uniqueness up to constants: let $u$ be the difference of two solutions, let $v=u^2$ and notice that $Delta v = |nabla u|^2$. Since the normal derivative of $u$ vanishes on the boundary, so does the normal derivative of $v$. By the divergence theorem, $int_{Omega} Delta v =0$. Hence $nabla uequiv 0$.
answered Dec 9 '13 at 13:26
Post No BullsPost No Bulls
6,64121783
answered Dec 9 '13 at 13:26
Post No BullsPost No Bulls
6,64121783
answered Dec 9 '13 at 13:26
Post No BullsPost No Bulls
6,64121783
answered Dec 9 '13 at 13:26
Post No BullsPost No Bulls
6,64121783
6,64121783
add a comment |
add a comment |
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$begingroup$
Why isn't $u+c$ not also a solution if $c$ is constant? Am I missing something? Or are you asking about existence instead of uniqueness?
$endgroup$
– Stephen Montgomery-Smith
Dec 8 '13 at 5:13
$begingroup$
@StephenMontgomery-Smith I already know this. I am merely wondering whether the condition on boundary of the domain has effects on the uniqueness or not. If not, what conditions else might guarantee uniqueness?
$endgroup$
– Xuxu
Dec 8 '13 at 7:03
1
$begingroup$
The point I was making is that is it very easy to see that $u+c$ is also a solution. You should be able to immediately see that your conditions don't imply uniqueness.
$endgroup$
– Stephen Montgomery-Smith
Dec 8 '13 at 15:43