Elliptic operator
How to show that the operator
begin{align*}
L[u]equiv frac {partial^2 u}{partial x^2}+frac {partial^2 u}{partial y^2}-frac {partial^2 u}{partial z^2}
end{align*}
is not elliptic?
Define $L[u]equiv Afrac {partial^2 u}{partial x^2}+Bfrac {partial^2 u}{partial y^2}+Cfrac {partial^2 u}{partial z^2}+Dfrac {partial^2 u}{partial x partial y}+Efrac {partial^2 u}{partial xpartial z}+Ffrac {partial^2 u}{partial y partial z}+Gfrac {partial u}{partial x}+Hfrac {partial u}{partial y}+Jfrac {partial u}{partial z}+Ku$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $J$ may depend upon $x$, $y$, and $z$ to be elliptic if there is a linear transformation to new coordinates $(xi, eta, zeta)$ such that in these coordinates,
begin{align*}
L[u]equiv alpha[frac {partial^2 u}{partial xi^2}+frac {partial^2 u}{partial eta^2}+frac {partial^2 u}{partial zeta^2}]+betafrac {partial u}{partial xi}+gammafrac {partial u}{partial eta}+deltafrac {partial u}{partial zeta}+Ku
end{align*}
pde elliptic-operators
asked Nov 27 '18 at 8:53
Jiexiong687691
705
add a comment |
How to show that the operator
begin{align*}
L[u]equiv frac {partial^2 u}{partial x^2}+frac {partial^2 u}{partial y^2}-frac {partial^2 u}{partial z^2}
end{align*}
is not elliptic?
Define $L[u]equiv Afrac {partial^2 u}{partial x^2}+Bfrac {partial^2 u}{partial y^2}+Cfrac {partial^2 u}{partial z^2}+Dfrac {partial^2 u}{partial x partial y}+Efrac {partial^2 u}{partial xpartial z}+Ffrac {partial^2 u}{partial y partial z}+Gfrac {partial u}{partial x}+Hfrac {partial u}{partial y}+Jfrac {partial u}{partial z}+Ku$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $J$ may depend upon $x$, $y$, and $z$ to be elliptic if there is a linear transformation to new coordinates $(xi, eta, zeta)$ such that in these coordinates,
begin{align*}
L[u]equiv alpha[frac {partial^2 u}{partial xi^2}+frac {partial^2 u}{partial eta^2}+frac {partial^2 u}{partial zeta^2}]+betafrac {partial u}{partial xi}+gammafrac {partial u}{partial eta}+deltafrac {partial u}{partial zeta}+Ku
end{align*}
pde elliptic-operators
asked Nov 27 '18 at 8:53
Jiexiong687691
705
2
What are your thoughts on the problem? To start, maybe you could write out what it means for an operator to be elliptic.
– Carmeister
Nov 27 '18 at 9:15
If I assume it's an elliptic operator, I don't know what to do next.@Carmeister
– Jiexiong687691
Nov 27 '18 at 9:35
add a comment |
How to show that the operator
begin{align*}
L[u]equiv frac {partial^2 u}{partial x^2}+frac {partial^2 u}{partial y^2}-frac {partial^2 u}{partial z^2}
end{align*}
is not elliptic?
Define $L[u]equiv Afrac {partial^2 u}{partial x^2}+Bfrac {partial^2 u}{partial y^2}+Cfrac {partial^2 u}{partial z^2}+Dfrac {partial^2 u}{partial x partial y}+Efrac {partial^2 u}{partial xpartial z}+Ffrac {partial^2 u}{partial y partial z}+Gfrac {partial u}{partial x}+Hfrac {partial u}{partial y}+Jfrac {partial u}{partial z}+Ku$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $J$ may depend upon $x$, $y$, and $z$ to be elliptic if there is a linear transformation to new coordinates $(xi, eta, zeta)$ such that in these coordinates,
begin{align*}
L[u]equiv alpha[frac {partial^2 u}{partial xi^2}+frac {partial^2 u}{partial eta^2}+frac {partial^2 u}{partial zeta^2}]+betafrac {partial u}{partial xi}+gammafrac {partial u}{partial eta}+deltafrac {partial u}{partial zeta}+Ku
end{align*}
pde elliptic-operators
asked Nov 27 '18 at 8:53
Jiexiong687691
705
How to show that the operator
begin{align*}
L[u]equiv frac {partial^2 u}{partial x^2}+frac {partial^2 u}{partial y^2}-frac {partial^2 u}{partial z^2}
end{align*}
is not elliptic?
Define $L[u]equiv Afrac {partial^2 u}{partial x^2}+Bfrac {partial^2 u}{partial y^2}+Cfrac {partial^2 u}{partial z^2}+Dfrac {partial^2 u}{partial x partial y}+Efrac {partial^2 u}{partial xpartial z}+Ffrac {partial^2 u}{partial y partial z}+Gfrac {partial u}{partial x}+Hfrac {partial u}{partial y}+Jfrac {partial u}{partial z}+Ku$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $J$ may depend upon $x$, $y$, and $z$ to be elliptic if there is a linear transformation to new coordinates $(xi, eta, zeta)$ such that in these coordinates,
begin{align*}
L[u]equiv alpha[frac {partial^2 u}{partial xi^2}+frac {partial^2 u}{partial eta^2}+frac {partial^2 u}{partial zeta^2}]+betafrac {partial u}{partial xi}+gammafrac {partial u}{partial eta}+deltafrac {partial u}{partial zeta}+Ku
end{align*}
pde elliptic-operators
pde elliptic-operators
asked Nov 27 '18 at 8:53
Jiexiong687691
705
asked Nov 27 '18 at 8:53
Jiexiong687691
705
edited Nov 27 '18 at 9:32
asked Nov 27 '18 at 8:53
Jiexiong687691
705
asked Nov 27 '18 at 8:53
Jiexiong687691
705
asked Nov 27 '18 at 8:53
Jiexiong687691
705
705
2
What are your thoughts on the problem? To start, maybe you could write out what it means for an operator to be elliptic.
– Carmeister
Nov 27 '18 at 9:15
If I assume it's an elliptic operator, I don't know what to do next.@Carmeister
– Jiexiong687691
Nov 27 '18 at 9:35
add a comment |
2
What are your thoughts on the problem? To start, maybe you could write out what it means for an operator to be elliptic.
– Carmeister
Nov 27 '18 at 9:15
If I assume it's an elliptic operator, I don't know what to do next.@Carmeister
– Jiexiong687691
Nov 27 '18 at 9:35
2
2
What are your thoughts on the problem? To start, maybe you could write out what it means for an operator to be elliptic.
– Carmeister
Nov 27 '18 at 9:15
What are your thoughts on the problem? To start, maybe you could write out what it means for an operator to be elliptic.
– Carmeister
Nov 27 '18 at 9:15
If I assume it's an elliptic operator, I don't know what to do next.@Carmeister
– Jiexiong687691
Nov 27 '18 at 9:35
If I assume it's an elliptic operator, I don't know what to do next.@Carmeister
– Jiexiong687691
Nov 27 '18 at 9:35
add a comment |
1 Answer
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A linear differential operator with constant coefficients is elliptic iff it's characteristic polynomial is non-zero for non-zero inputs.
The characteristic polynomial of your operator is $p(x,y,z)=x^2+y^2-z^2$ and the equation $p(x,y,z)=0$ has plenty of non-zero solutions.
answered Nov 27 '18 at 9:26
Bernard W
1,516512
add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
A linear differential operator with constant coefficients is elliptic iff it's characteristic polynomial is non-zero for non-zero inputs.
The characteristic polynomial of your operator is $p(x,y,z)=x^2+y^2-z^2$ and the equation $p(x,y,z)=0$ has plenty of non-zero solutions.
answered Nov 27 '18 at 9:26
Bernard W
1,516512
add a comment |
A linear differential operator with constant coefficients is elliptic iff it's characteristic polynomial is non-zero for non-zero inputs.
The characteristic polynomial of your operator is $p(x,y,z)=x^2+y^2-z^2$ and the equation $p(x,y,z)=0$ has plenty of non-zero solutions.
answered Nov 27 '18 at 9:26
Bernard W
1,516512
add a comment |
A linear differential operator with constant coefficients is elliptic iff it's characteristic polynomial is non-zero for non-zero inputs.
The characteristic polynomial of your operator is $p(x,y,z)=x^2+y^2-z^2$ and the equation $p(x,y,z)=0$ has plenty of non-zero solutions.
answered Nov 27 '18 at 9:26
Bernard W
1,516512
A linear differential operator with constant coefficients is elliptic iff it's characteristic polynomial is non-zero for non-zero inputs.
The characteristic polynomial of your operator is $p(x,y,z)=x^2+y^2-z^2$ and the equation $p(x,y,z)=0$ has plenty of non-zero solutions.
answered Nov 27 '18 at 9:26
Bernard W
1,516512
answered Nov 27 '18 at 9:26
Bernard W
1,516512
answered Nov 27 '18 at 9:26
Bernard W
1,516512
answered Nov 27 '18 at 9:26
Bernard W
1,516512
1,516512
add a comment |
add a comment |
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2
What are your thoughts on the problem? To start, maybe you could write out what it means for an operator to be elliptic.
– Carmeister
Nov 27 '18 at 9:15
If I assume it's an elliptic operator, I don't know what to do next.@Carmeister
– Jiexiong687691
Nov 27 '18 at 9:35