Correlation between $u$ and $partial_nu u$ on $partial Omega$ for a give PDEDecompose solution of pde in...
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Correlation between $u$ and $partial_nu u$ on $partial Omega$ for a give PDE
Decompose solution of pde in harmonic and non-harmonic partOne problem about harmonic functionsWhat is precisely the definition of Elliptic Partial Differential Equation?Why $-int_{partial Omega }|u(x)|^2ds-int_Omega |nabla u|^2=0implies nabla u=0$?Prove that $|v|_{H^1(Omega )}leq C(|f|_{L^2(Omega )}+|v|_{H^{1/2}(partial Omega )}+|partial _nu v|_{H^{1/2}(partial Omega )}))$Prove $|v|_{H^1(Omega )}leq C(|f|_{L^2(Omega )}+|v|_{H^{1/2}(partial Omega )}+|partial _nu v|_{H^{-1/2}(partial Omega )})$Change of Variables for PDE posed in the bounded domain $Omega subset mathbb{R}^n$A bound for the PDE $frac{partial}{partial t}u=Delta u+ au$Critical points of solutions to PDEs on convex domainsA continuous extension operator $L^2(partial Omega) to H^1(Omega)$
$begingroup$
Given a bounded area $Omega subset mathbb{R}^n$ which is at least of class $C^3$. If now there is a function $u$ satisfying the PDE
begin{align*}
Delta u &= f~~~text{in }Omega \
u &= 0 ~~~text{in } partialOmega,
end{align*}
for some smooth function $f$. Is it then possible to show that $$partial_nu u overset{?}{=} 0$$
on $partial Omega$?
real-analysis pde
asked 20 hours ago
BaraBara
5510
$endgroup$
add a comment |
$begingroup$
Given a bounded area $Omega subset mathbb{R}^n$ which is at least of class $C^3$. If now there is a function $u$ satisfying the PDE
begin{align*}
Delta u &= f~~~text{in }Omega \
u &= 0 ~~~text{in } partialOmega,
end{align*}
for some smooth function $f$. Is it then possible to show that $$partial_nu u overset{?}{=} 0$$
on $partial Omega$?
real-analysis pde
asked 20 hours ago
BaraBara
5510
$endgroup$
add a comment |
$begingroup$
Given a bounded area $Omega subset mathbb{R}^n$ which is at least of class $C^3$. If now there is a function $u$ satisfying the PDE
begin{align*}
Delta u &= f~~~text{in }Omega \
u &= 0 ~~~text{in } partialOmega,
end{align*}
for some smooth function $f$. Is it then possible to show that $$partial_nu u overset{?}{=} 0$$
on $partial Omega$?
real-analysis pde
asked 20 hours ago
BaraBara
5510
$endgroup$
Given a bounded area $Omega subset mathbb{R}^n$ which is at least of class $C^3$. If now there is a function $u$ satisfying the PDE
begin{align*}
Delta u &= f~~~text{in }Omega \
u &= 0 ~~~text{in } partialOmega,
end{align*}
for some smooth function $f$. Is it then possible to show that $$partial_nu u overset{?}{=} 0$$
on $partial Omega$?
real-analysis pde
real-analysis pde
asked 20 hours ago
BaraBara
5510
asked 20 hours ago
BaraBara
5510
edited 20 hours ago
Bara
asked 20 hours ago
BaraBara
5510
asked 20 hours ago
BaraBara
5510
asked 20 hours ago
BaraBara
5510
5510
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is not possible to show that result because it is in general not true. Think for instance in the PDE
$$
left{
begin{array}{rl}
-Delta u = 4,& textrm{in } Omega \
u = 0,& textrm{on } partial Omega
end{array}right.
$$
where $Omega={(x,y)in mathbb{R}^2: x^2+y^2 < 1}$.
The exact solution is $u=1-x^2-y^2$, which does not have zero normal derivative on $partial Omega$.
answered 20 hours ago
PierreCarrePierreCarre
1,243211
$endgroup$
$begingroup$
I guess you are missing "$<$" inside the definition of $Omega$.
$endgroup$
– GaC
19 hours ago
$begingroup$
Thanks, I'll correct it.
$endgroup$
– PierreCarre
19 hours ago
add a comment |
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1 Answer
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1 Answer
1
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oldest
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oldest
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votes
$begingroup$
It is not possible to show that result because it is in general not true. Think for instance in the PDE
$$
left{
begin{array}{rl}
-Delta u = 4,& textrm{in } Omega \
u = 0,& textrm{on } partial Omega
end{array}right.
$$
where $Omega={(x,y)in mathbb{R}^2: x^2+y^2 < 1}$.
The exact solution is $u=1-x^2-y^2$, which does not have zero normal derivative on $partial Omega$.
answered 20 hours ago
PierreCarrePierreCarre
1,243211
$endgroup$
$begingroup$
I guess you are missing "$<$" inside the definition of $Omega$.
$endgroup$
– GaC
19 hours ago
$begingroup$
Thanks, I'll correct it.
$endgroup$
– PierreCarre
19 hours ago
add a comment |
$begingroup$
It is not possible to show that result because it is in general not true. Think for instance in the PDE
$$
left{
begin{array}{rl}
-Delta u = 4,& textrm{in } Omega \
u = 0,& textrm{on } partial Omega
end{array}right.
$$
where $Omega={(x,y)in mathbb{R}^2: x^2+y^2 < 1}$.
The exact solution is $u=1-x^2-y^2$, which does not have zero normal derivative on $partial Omega$.
answered 20 hours ago
PierreCarrePierreCarre
1,243211
$endgroup$
$begingroup$
I guess you are missing "$<$" inside the definition of $Omega$.
$endgroup$
– GaC
19 hours ago
$begingroup$
Thanks, I'll correct it.
$endgroup$
– PierreCarre
19 hours ago
add a comment |
$begingroup$
It is not possible to show that result because it is in general not true. Think for instance in the PDE
$$
left{
begin{array}{rl}
-Delta u = 4,& textrm{in } Omega \
u = 0,& textrm{on } partial Omega
end{array}right.
$$
where $Omega={(x,y)in mathbb{R}^2: x^2+y^2 < 1}$.
The exact solution is $u=1-x^2-y^2$, which does not have zero normal derivative on $partial Omega$.
answered 20 hours ago
PierreCarrePierreCarre
1,243211
$endgroup$
It is not possible to show that result because it is in general not true. Think for instance in the PDE
$$
left{
begin{array}{rl}
-Delta u = 4,& textrm{in } Omega \
u = 0,& textrm{on } partial Omega
end{array}right.
$$
where $Omega={(x,y)in mathbb{R}^2: x^2+y^2 < 1}$.
The exact solution is $u=1-x^2-y^2$, which does not have zero normal derivative on $partial Omega$.
answered 20 hours ago
PierreCarrePierreCarre
1,243211
edited 19 hours ago
answered 20 hours ago
PierreCarrePierreCarre
1,243211
answered 20 hours ago
PierreCarrePierreCarre
1,243211
answered 20 hours ago
PierreCarrePierreCarre
1,243211
1,243211
$begingroup$
I guess you are missing "$<$" inside the definition of $Omega$.
$endgroup$
– GaC
19 hours ago
$begingroup$
Thanks, I'll correct it.
$endgroup$
– PierreCarre
19 hours ago
add a comment |
$begingroup$
I guess you are missing "$<$" inside the definition of $Omega$.
$endgroup$
– GaC
19 hours ago
$begingroup$
Thanks, I'll correct it.
$endgroup$
– PierreCarre
19 hours ago
$begingroup$
I guess you are missing "$<$" inside the definition of $Omega$.
$endgroup$
– GaC
19 hours ago
$begingroup$
I guess you are missing "$<$" inside the definition of $Omega$.
$endgroup$
– GaC
19 hours ago
$begingroup$
Thanks, I'll correct it.
$endgroup$
– PierreCarre
19 hours ago
$begingroup$
Thanks, I'll correct it.
$endgroup$
– PierreCarre
19 hours ago
add a comment |
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