The symbol of a pseudifferential operator: how to reconstruct the symbol from an operator?
$begingroup$
To simplify let us assume that $U subset mathbb{R}^n$ is an open set. Let $P$ be a (scalar) psuedodifferential operator of order $d$ and $p$ be its symbol. Therefore it may be expressed as follows:
$Pf(x)=int e^{ix xi}p(x,xi)hat{f}(xi)dxi$ where $f in C^{infty}_c(U)$ is a smooth function with compact support. The fact that $P$ is of order $d$ imposes some growth condition on the function $p$.
Why the symbol may be reconstructed as $p(x,xi)=e^{-ix xi}P(e^{ix xi})$.
This should be simple but I have a problem in understanding this formula since most likely it should interpreted in the distributional sense. I would be grateful for any help.
ordinary-differential-equations fourier-transform distribution-theory pseudo-differential-operators
asked Dec 18 '18 at 23:32
truebarantruebaran
2,2102723
$endgroup$
add a comment |
$begingroup$
To simplify let us assume that $U subset mathbb{R}^n$ is an open set. Let $P$ be a (scalar) psuedodifferential operator of order $d$ and $p$ be its symbol. Therefore it may be expressed as follows:
$Pf(x)=int e^{ix xi}p(x,xi)hat{f}(xi)dxi$ where $f in C^{infty}_c(U)$ is a smooth function with compact support. The fact that $P$ is of order $d$ imposes some growth condition on the function $p$.
Why the symbol may be reconstructed as $p(x,xi)=e^{-ix xi}P(e^{ix xi})$.
This should be simple but I have a problem in understanding this formula since most likely it should interpreted in the distributional sense. I would be grateful for any help.
ordinary-differential-equations fourier-transform distribution-theory pseudo-differential-operators
asked Dec 18 '18 at 23:32
truebarantruebaran
2,2102723
$endgroup$
add a comment |
$begingroup$
To simplify let us assume that $U subset mathbb{R}^n$ is an open set. Let $P$ be a (scalar) psuedodifferential operator of order $d$ and $p$ be its symbol. Therefore it may be expressed as follows:
$Pf(x)=int e^{ix xi}p(x,xi)hat{f}(xi)dxi$ where $f in C^{infty}_c(U)$ is a smooth function with compact support. The fact that $P$ is of order $d$ imposes some growth condition on the function $p$.
Why the symbol may be reconstructed as $p(x,xi)=e^{-ix xi}P(e^{ix xi})$.
This should be simple but I have a problem in understanding this formula since most likely it should interpreted in the distributional sense. I would be grateful for any help.
ordinary-differential-equations fourier-transform distribution-theory pseudo-differential-operators
asked Dec 18 '18 at 23:32
truebarantruebaran
2,2102723
$endgroup$
To simplify let us assume that $U subset mathbb{R}^n$ is an open set. Let $P$ be a (scalar) psuedodifferential operator of order $d$ and $p$ be its symbol. Therefore it may be expressed as follows:
$Pf(x)=int e^{ix xi}p(x,xi)hat{f}(xi)dxi$ where $f in C^{infty}_c(U)$ is a smooth function with compact support. The fact that $P$ is of order $d$ imposes some growth condition on the function $p$.
Why the symbol may be reconstructed as $p(x,xi)=e^{-ix xi}P(e^{ix xi})$.
This should be simple but I have a problem in understanding this formula since most likely it should interpreted in the distributional sense. I would be grateful for any help.
ordinary-differential-equations fourier-transform distribution-theory pseudo-differential-operators
ordinary-differential-equations fourier-transform distribution-theory pseudo-differential-operators
asked Dec 18 '18 at 23:32
truebarantruebaran
2,2102723
asked Dec 18 '18 at 23:32
truebarantruebaran
2,2102723
asked Dec 18 '18 at 23:32
truebarantruebaran
2,2102723
asked Dec 18 '18 at 23:32
truebarantruebaran
2,2102723
asked Dec 18 '18 at 23:32
truebarantruebaran
2,2102723
2,2102723
add a comment |
add a comment |
1 Answer
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$begingroup$
First, note that
begin{align}
widehat{e^{ixeta}}(xi) = int dx e^{-ix(xi-eta)} = delta(xi-eta)
end{align}
then it follows
begin{align}
P(e^{ixeta})(x, eta) =& int dxi e^{ix xi} p(x, xi) widehat{e^{iyeta}}(xi)\
=& int dxi e^{ixxi}p(x, xi) delta(xi-eta)= e^{ixeta}p(x, eta).
end{align}
Finally, we see that
begin{align}
e^{-ixeta}P(e^{ixeta}) = p(x, eta).
end{align}
answered Dec 19 '18 at 0:14
Jacky ChongJacky Chong
18.6k21128
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, note that
begin{align}
widehat{e^{ixeta}}(xi) = int dx e^{-ix(xi-eta)} = delta(xi-eta)
end{align}
then it follows
begin{align}
P(e^{ixeta})(x, eta) =& int dxi e^{ix xi} p(x, xi) widehat{e^{iyeta}}(xi)\
=& int dxi e^{ixxi}p(x, xi) delta(xi-eta)= e^{ixeta}p(x, eta).
end{align}
Finally, we see that
begin{align}
e^{-ixeta}P(e^{ixeta}) = p(x, eta).
end{align}
answered Dec 19 '18 at 0:14
Jacky ChongJacky Chong
18.6k21128
$endgroup$
add a comment |
$begingroup$
First, note that
begin{align}
widehat{e^{ixeta}}(xi) = int dx e^{-ix(xi-eta)} = delta(xi-eta)
end{align}
then it follows
begin{align}
P(e^{ixeta})(x, eta) =& int dxi e^{ix xi} p(x, xi) widehat{e^{iyeta}}(xi)\
=& int dxi e^{ixxi}p(x, xi) delta(xi-eta)= e^{ixeta}p(x, eta).
end{align}
Finally, we see that
begin{align}
e^{-ixeta}P(e^{ixeta}) = p(x, eta).
end{align}
answered Dec 19 '18 at 0:14
Jacky ChongJacky Chong
18.6k21128
$endgroup$
add a comment |
$begingroup$
First, note that
begin{align}
widehat{e^{ixeta}}(xi) = int dx e^{-ix(xi-eta)} = delta(xi-eta)
end{align}
then it follows
begin{align}
P(e^{ixeta})(x, eta) =& int dxi e^{ix xi} p(x, xi) widehat{e^{iyeta}}(xi)\
=& int dxi e^{ixxi}p(x, xi) delta(xi-eta)= e^{ixeta}p(x, eta).
end{align}
Finally, we see that
begin{align}
e^{-ixeta}P(e^{ixeta}) = p(x, eta).
end{align}
answered Dec 19 '18 at 0:14
Jacky ChongJacky Chong
18.6k21128
$endgroup$
First, note that
begin{align}
widehat{e^{ixeta}}(xi) = int dx e^{-ix(xi-eta)} = delta(xi-eta)
end{align}
then it follows
begin{align}
P(e^{ixeta})(x, eta) =& int dxi e^{ix xi} p(x, xi) widehat{e^{iyeta}}(xi)\
=& int dxi e^{ixxi}p(x, xi) delta(xi-eta)= e^{ixeta}p(x, eta).
end{align}
Finally, we see that
begin{align}
e^{-ixeta}P(e^{ixeta}) = p(x, eta).
end{align}
answered Dec 19 '18 at 0:14
Jacky ChongJacky Chong
18.6k21128
answered Dec 19 '18 at 0:14
Jacky ChongJacky Chong
18.6k21128
answered Dec 19 '18 at 0:14
Jacky ChongJacky Chong
18.6k21128
answered Dec 19 '18 at 0:14
Jacky ChongJacky Chong
18.6k21128
18.6k21128
add a comment |
add a comment |
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