Fourier transform of product of Bessel functions
up vote
1
down vote
favorite
I need help finding the Fourier transform of the function
$$
rho(vec{r}) = alpha delta_{vec{r},0} left(lambdalambda' J_1
(beta |vec{r}|)Y_1(beta |vec{r}|) - pi^2 J_0
(beta |vec{r}|)Y_0(beta |vec{r}|) right),
$$
where $alpha,beta in mathcal{R}$ and $lambda,lambda'=pm1$. The $J_i(x)$ are the $i$-th Bessel functions of the first, the $Y_i(x)$ are the $i$-th Bessel functions of the second kind. $delta_{vec{r},0}$ denotes the Kronecker delta.
The vector $vec{r} = (x,y)^T$ is discrete because I'm working on a discrete set of points. I'm a physicist, so please don't hesitate to ask for more specific information.
I tried something like
$$
rho(vec{k}) = sum_vec{r} e^{-ivec{k}cdotvec{r}}rho(vec{r}),
$$
which should in fact not be too difficult because the only vector that contributes to the sum is $vec{r} = vec{0}$ due to the Kronecker delta in $rho(vec{r})$. The problem is that $rho(vec{r})$ diverges at this point. I need some mathematical advice here. Is there anybody who can show me how to compute the Fourier transform of $rho(vec{r})$?
physics fourier-transform divergent-series bessel-functions
add a comment |
up vote
1
down vote
favorite
I need help finding the Fourier transform of the function
$$
rho(vec{r}) = alpha delta_{vec{r},0} left(lambdalambda' J_1
(beta |vec{r}|)Y_1(beta |vec{r}|) - pi^2 J_0
(beta |vec{r}|)Y_0(beta |vec{r}|) right),
$$
where $alpha,beta in mathcal{R}$ and $lambda,lambda'=pm1$. The $J_i(x)$ are the $i$-th Bessel functions of the first, the $Y_i(x)$ are the $i$-th Bessel functions of the second kind. $delta_{vec{r},0}$ denotes the Kronecker delta.
The vector $vec{r} = (x,y)^T$ is discrete because I'm working on a discrete set of points. I'm a physicist, so please don't hesitate to ask for more specific information.
I tried something like
$$
rho(vec{k}) = sum_vec{r} e^{-ivec{k}cdotvec{r}}rho(vec{r}),
$$
which should in fact not be too difficult because the only vector that contributes to the sum is $vec{r} = vec{0}$ due to the Kronecker delta in $rho(vec{r})$. The problem is that $rho(vec{r})$ diverges at this point. I need some mathematical advice here. Is there anybody who can show me how to compute the Fourier transform of $rho(vec{r})$?
physics fourier-transform divergent-series bessel-functions
Please confirm that this problem has strict radial symmetry. The reason I ask is that $delta^2(x,y)$ and $delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform.
– Andy Walls
Aug 11 at 22:09
@AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :)
– MeMeansMe
Aug 13 at 7:50
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need help finding the Fourier transform of the function
$$
rho(vec{r}) = alpha delta_{vec{r},0} left(lambdalambda' J_1
(beta |vec{r}|)Y_1(beta |vec{r}|) - pi^2 J_0
(beta |vec{r}|)Y_0(beta |vec{r}|) right),
$$
where $alpha,beta in mathcal{R}$ and $lambda,lambda'=pm1$. The $J_i(x)$ are the $i$-th Bessel functions of the first, the $Y_i(x)$ are the $i$-th Bessel functions of the second kind. $delta_{vec{r},0}$ denotes the Kronecker delta.
The vector $vec{r} = (x,y)^T$ is discrete because I'm working on a discrete set of points. I'm a physicist, so please don't hesitate to ask for more specific information.
I tried something like
$$
rho(vec{k}) = sum_vec{r} e^{-ivec{k}cdotvec{r}}rho(vec{r}),
$$
which should in fact not be too difficult because the only vector that contributes to the sum is $vec{r} = vec{0}$ due to the Kronecker delta in $rho(vec{r})$. The problem is that $rho(vec{r})$ diverges at this point. I need some mathematical advice here. Is there anybody who can show me how to compute the Fourier transform of $rho(vec{r})$?
physics fourier-transform divergent-series bessel-functions
I need help finding the Fourier transform of the function
$$
rho(vec{r}) = alpha delta_{vec{r},0} left(lambdalambda' J_1
(beta |vec{r}|)Y_1(beta |vec{r}|) - pi^2 J_0
(beta |vec{r}|)Y_0(beta |vec{r}|) right),
$$
where $alpha,beta in mathcal{R}$ and $lambda,lambda'=pm1$. The $J_i(x)$ are the $i$-th Bessel functions of the first, the $Y_i(x)$ are the $i$-th Bessel functions of the second kind. $delta_{vec{r},0}$ denotes the Kronecker delta.
The vector $vec{r} = (x,y)^T$ is discrete because I'm working on a discrete set of points. I'm a physicist, so please don't hesitate to ask for more specific information.
I tried something like
$$
rho(vec{k}) = sum_vec{r} e^{-ivec{k}cdotvec{r}}rho(vec{r}),
$$
which should in fact not be too difficult because the only vector that contributes to the sum is $vec{r} = vec{0}$ due to the Kronecker delta in $rho(vec{r})$. The problem is that $rho(vec{r})$ diverges at this point. I need some mathematical advice here. Is there anybody who can show me how to compute the Fourier transform of $rho(vec{r})$?
physics fourier-transform divergent-series bessel-functions
physics fourier-transform divergent-series bessel-functions
edited Nov 27 at 16:54
Robert Howard
1,9181822
1,9181822
asked Aug 10 at 10:04
MeMeansMe
1354
1354
Please confirm that this problem has strict radial symmetry. The reason I ask is that $delta^2(x,y)$ and $delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform.
– Andy Walls
Aug 11 at 22:09
@AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :)
– MeMeansMe
Aug 13 at 7:50
add a comment |
Please confirm that this problem has strict radial symmetry. The reason I ask is that $delta^2(x,y)$ and $delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform.
– Andy Walls
Aug 11 at 22:09
@AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :)
– MeMeansMe
Aug 13 at 7:50
Please confirm that this problem has strict radial symmetry. The reason I ask is that $delta^2(x,y)$ and $delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform.
– Andy Walls
Aug 11 at 22:09
Please confirm that this problem has strict radial symmetry. The reason I ask is that $delta^2(x,y)$ and $delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform.
– Andy Walls
Aug 11 at 22:09
@AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :)
– MeMeansMe
Aug 13 at 7:50
@AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :)
– MeMeansMe
Aug 13 at 7:50
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
I'm almost certain the Fourier Transform doesn't exist.
I'll rewrite your radially symmetric (density?) function, $rho(vec{r})$, to be continuous, as opposed to discrete, in 2D polar coordinates:
$$d(r) = {}^2delta(r) alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$
Where ${}^2delta(r)$ is the 2D Dirac Delta function at the origin of the polar coordinate system. (For the impulse at the origin, $theta$ doesn't matter, so it doesn't appear as an argument.) In this analysis, ${}^2delta(r)$ represents the following limiting sequence of functions (that are asymmetrical about 0):
$${}^2delta(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{piepsilon^2} quad 0 < r < epsilon$$
so that
$$int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta = 1$$
To separate the $r$ and $theta$ portions of ${}^2delta(r)$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:
$$delta_a(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{epsilon} quad 0 < r < epsilon$$
and then note
$$begin{align}\
int_0^{infty} delta_a(r) space mathrm{d}r &= 1\
&= int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta\
&= int_0^{infty} {}^2delta(r) space 2pi r space mathrm{d}r\
end{align}$$
so we have
$${}^2delta(r) = dfrac{delta_a(r)}{2pi r}$$
$d(r)$ can now be written without any dependence on $theta$ as
$$d(r) = dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$
The 2D Fourier Transform of a radially symmetric function can be expressed in terms of the Hankel Transform of order 0:
$$F(rho) = mathcal{F}left{f(r)right} = 2pi mathcal{H_0}left{f(r)right} = 2pi int_0^{infty} f(r) J_0(rho r) space r space mathrm{d}r$$
When attempting to compute the Fourier Transform
$$begin{align}mathcal{F}left{d(r)right} = D(rho) &= 2pi int_0^{infty} dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) space r space mathrm{d}r\
\
&= alpha int_0^{infty} delta_a(r) left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) mathrm{d}r\
\
&= alphalambda lambda' int_0^{infty} delta_a(r) J_1(beta r) Y_1(beta r) J_0(rho r) mathrm{d}r - alpha pi^2 int_0^{infty} delta_a(r)J_0(beta r) Y_0(beta r) J_0(rho r) mathrm{d}r\
\
&= alphalambda lambda' lim_{r rightarrow 0+} J_1(beta r) Y_1(beta r) J_0(rho r) - alpha pi^2 lim_{r rightarrow 0+} J_0(beta r) Y_0(beta r) J_0(rho r) \
\
&= alphalambda lambda'left(-dfrac1{pi}right) - alpha pi^2 (-infty) \
end{align}$$
it diverges.
Note that if the problematic term had converged, the Fourier Transform would simply be an infinite disk of constant height above the $rhophi$ plane. That should be no surprise for the transform of a Dirac Delta function.
The problematic term would converge, if there were a factor of $1/ln(r)$:
$$lim_{r rightarrow 0+} dfrac{J_0(beta r) Y_0(beta r)}{ln(r)} J_0(rho r) = dfrac{2}{pi}$$
I don't know the physical situation you are analyzing, and whether inclusion of such a term could ever be justified.
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',
autoActivateHeartbeat: false,
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%2f2878201%2ffourier-transform-of-product-of-bessel-functions%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
I'm almost certain the Fourier Transform doesn't exist.
I'll rewrite your radially symmetric (density?) function, $rho(vec{r})$, to be continuous, as opposed to discrete, in 2D polar coordinates:
$$d(r) = {}^2delta(r) alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$
Where ${}^2delta(r)$ is the 2D Dirac Delta function at the origin of the polar coordinate system. (For the impulse at the origin, $theta$ doesn't matter, so it doesn't appear as an argument.) In this analysis, ${}^2delta(r)$ represents the following limiting sequence of functions (that are asymmetrical about 0):
$${}^2delta(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{piepsilon^2} quad 0 < r < epsilon$$
so that
$$int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta = 1$$
To separate the $r$ and $theta$ portions of ${}^2delta(r)$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:
$$delta_a(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{epsilon} quad 0 < r < epsilon$$
and then note
$$begin{align}\
int_0^{infty} delta_a(r) space mathrm{d}r &= 1\
&= int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta\
&= int_0^{infty} {}^2delta(r) space 2pi r space mathrm{d}r\
end{align}$$
so we have
$${}^2delta(r) = dfrac{delta_a(r)}{2pi r}$$
$d(r)$ can now be written without any dependence on $theta$ as
$$d(r) = dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$
The 2D Fourier Transform of a radially symmetric function can be expressed in terms of the Hankel Transform of order 0:
$$F(rho) = mathcal{F}left{f(r)right} = 2pi mathcal{H_0}left{f(r)right} = 2pi int_0^{infty} f(r) J_0(rho r) space r space mathrm{d}r$$
When attempting to compute the Fourier Transform
$$begin{align}mathcal{F}left{d(r)right} = D(rho) &= 2pi int_0^{infty} dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) space r space mathrm{d}r\
\
&= alpha int_0^{infty} delta_a(r) left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) mathrm{d}r\
\
&= alphalambda lambda' int_0^{infty} delta_a(r) J_1(beta r) Y_1(beta r) J_0(rho r) mathrm{d}r - alpha pi^2 int_0^{infty} delta_a(r)J_0(beta r) Y_0(beta r) J_0(rho r) mathrm{d}r\
\
&= alphalambda lambda' lim_{r rightarrow 0+} J_1(beta r) Y_1(beta r) J_0(rho r) - alpha pi^2 lim_{r rightarrow 0+} J_0(beta r) Y_0(beta r) J_0(rho r) \
\
&= alphalambda lambda'left(-dfrac1{pi}right) - alpha pi^2 (-infty) \
end{align}$$
it diverges.
Note that if the problematic term had converged, the Fourier Transform would simply be an infinite disk of constant height above the $rhophi$ plane. That should be no surprise for the transform of a Dirac Delta function.
The problematic term would converge, if there were a factor of $1/ln(r)$:
$$lim_{r rightarrow 0+} dfrac{J_0(beta r) Y_0(beta r)}{ln(r)} J_0(rho r) = dfrac{2}{pi}$$
I don't know the physical situation you are analyzing, and whether inclusion of such a term could ever be justified.
add a comment |
up vote
1
down vote
I'm almost certain the Fourier Transform doesn't exist.
I'll rewrite your radially symmetric (density?) function, $rho(vec{r})$, to be continuous, as opposed to discrete, in 2D polar coordinates:
$$d(r) = {}^2delta(r) alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$
Where ${}^2delta(r)$ is the 2D Dirac Delta function at the origin of the polar coordinate system. (For the impulse at the origin, $theta$ doesn't matter, so it doesn't appear as an argument.) In this analysis, ${}^2delta(r)$ represents the following limiting sequence of functions (that are asymmetrical about 0):
$${}^2delta(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{piepsilon^2} quad 0 < r < epsilon$$
so that
$$int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta = 1$$
To separate the $r$ and $theta$ portions of ${}^2delta(r)$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:
$$delta_a(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{epsilon} quad 0 < r < epsilon$$
and then note
$$begin{align}\
int_0^{infty} delta_a(r) space mathrm{d}r &= 1\
&= int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta\
&= int_0^{infty} {}^2delta(r) space 2pi r space mathrm{d}r\
end{align}$$
so we have
$${}^2delta(r) = dfrac{delta_a(r)}{2pi r}$$
$d(r)$ can now be written without any dependence on $theta$ as
$$d(r) = dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$
The 2D Fourier Transform of a radially symmetric function can be expressed in terms of the Hankel Transform of order 0:
$$F(rho) = mathcal{F}left{f(r)right} = 2pi mathcal{H_0}left{f(r)right} = 2pi int_0^{infty} f(r) J_0(rho r) space r space mathrm{d}r$$
When attempting to compute the Fourier Transform
$$begin{align}mathcal{F}left{d(r)right} = D(rho) &= 2pi int_0^{infty} dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) space r space mathrm{d}r\
\
&= alpha int_0^{infty} delta_a(r) left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) mathrm{d}r\
\
&= alphalambda lambda' int_0^{infty} delta_a(r) J_1(beta r) Y_1(beta r) J_0(rho r) mathrm{d}r - alpha pi^2 int_0^{infty} delta_a(r)J_0(beta r) Y_0(beta r) J_0(rho r) mathrm{d}r\
\
&= alphalambda lambda' lim_{r rightarrow 0+} J_1(beta r) Y_1(beta r) J_0(rho r) - alpha pi^2 lim_{r rightarrow 0+} J_0(beta r) Y_0(beta r) J_0(rho r) \
\
&= alphalambda lambda'left(-dfrac1{pi}right) - alpha pi^2 (-infty) \
end{align}$$
it diverges.
Note that if the problematic term had converged, the Fourier Transform would simply be an infinite disk of constant height above the $rhophi$ plane. That should be no surprise for the transform of a Dirac Delta function.
The problematic term would converge, if there were a factor of $1/ln(r)$:
$$lim_{r rightarrow 0+} dfrac{J_0(beta r) Y_0(beta r)}{ln(r)} J_0(rho r) = dfrac{2}{pi}$$
I don't know the physical situation you are analyzing, and whether inclusion of such a term could ever be justified.
add a comment |
up vote
1
down vote
up vote
1
down vote
I'm almost certain the Fourier Transform doesn't exist.
I'll rewrite your radially symmetric (density?) function, $rho(vec{r})$, to be continuous, as opposed to discrete, in 2D polar coordinates:
$$d(r) = {}^2delta(r) alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$
Where ${}^2delta(r)$ is the 2D Dirac Delta function at the origin of the polar coordinate system. (For the impulse at the origin, $theta$ doesn't matter, so it doesn't appear as an argument.) In this analysis, ${}^2delta(r)$ represents the following limiting sequence of functions (that are asymmetrical about 0):
$${}^2delta(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{piepsilon^2} quad 0 < r < epsilon$$
so that
$$int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta = 1$$
To separate the $r$ and $theta$ portions of ${}^2delta(r)$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:
$$delta_a(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{epsilon} quad 0 < r < epsilon$$
and then note
$$begin{align}\
int_0^{infty} delta_a(r) space mathrm{d}r &= 1\
&= int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta\
&= int_0^{infty} {}^2delta(r) space 2pi r space mathrm{d}r\
end{align}$$
so we have
$${}^2delta(r) = dfrac{delta_a(r)}{2pi r}$$
$d(r)$ can now be written without any dependence on $theta$ as
$$d(r) = dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$
The 2D Fourier Transform of a radially symmetric function can be expressed in terms of the Hankel Transform of order 0:
$$F(rho) = mathcal{F}left{f(r)right} = 2pi mathcal{H_0}left{f(r)right} = 2pi int_0^{infty} f(r) J_0(rho r) space r space mathrm{d}r$$
When attempting to compute the Fourier Transform
$$begin{align}mathcal{F}left{d(r)right} = D(rho) &= 2pi int_0^{infty} dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) space r space mathrm{d}r\
\
&= alpha int_0^{infty} delta_a(r) left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) mathrm{d}r\
\
&= alphalambda lambda' int_0^{infty} delta_a(r) J_1(beta r) Y_1(beta r) J_0(rho r) mathrm{d}r - alpha pi^2 int_0^{infty} delta_a(r)J_0(beta r) Y_0(beta r) J_0(rho r) mathrm{d}r\
\
&= alphalambda lambda' lim_{r rightarrow 0+} J_1(beta r) Y_1(beta r) J_0(rho r) - alpha pi^2 lim_{r rightarrow 0+} J_0(beta r) Y_0(beta r) J_0(rho r) \
\
&= alphalambda lambda'left(-dfrac1{pi}right) - alpha pi^2 (-infty) \
end{align}$$
it diverges.
Note that if the problematic term had converged, the Fourier Transform would simply be an infinite disk of constant height above the $rhophi$ plane. That should be no surprise for the transform of a Dirac Delta function.
The problematic term would converge, if there were a factor of $1/ln(r)$:
$$lim_{r rightarrow 0+} dfrac{J_0(beta r) Y_0(beta r)}{ln(r)} J_0(rho r) = dfrac{2}{pi}$$
I don't know the physical situation you are analyzing, and whether inclusion of such a term could ever be justified.
I'm almost certain the Fourier Transform doesn't exist.
I'll rewrite your radially symmetric (density?) function, $rho(vec{r})$, to be continuous, as opposed to discrete, in 2D polar coordinates:
$$d(r) = {}^2delta(r) alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$
Where ${}^2delta(r)$ is the 2D Dirac Delta function at the origin of the polar coordinate system. (For the impulse at the origin, $theta$ doesn't matter, so it doesn't appear as an argument.) In this analysis, ${}^2delta(r)$ represents the following limiting sequence of functions (that are asymmetrical about 0):
$${}^2delta(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{piepsilon^2} quad 0 < r < epsilon$$
so that
$$int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta = 1$$
To separate the $r$ and $theta$ portions of ${}^2delta(r)$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:
$$delta_a(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{epsilon} quad 0 < r < epsilon$$
and then note
$$begin{align}\
int_0^{infty} delta_a(r) space mathrm{d}r &= 1\
&= int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta\
&= int_0^{infty} {}^2delta(r) space 2pi r space mathrm{d}r\
end{align}$$
so we have
$${}^2delta(r) = dfrac{delta_a(r)}{2pi r}$$
$d(r)$ can now be written without any dependence on $theta$ as
$$d(r) = dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$
The 2D Fourier Transform of a radially symmetric function can be expressed in terms of the Hankel Transform of order 0:
$$F(rho) = mathcal{F}left{f(r)right} = 2pi mathcal{H_0}left{f(r)right} = 2pi int_0^{infty} f(r) J_0(rho r) space r space mathrm{d}r$$
When attempting to compute the Fourier Transform
$$begin{align}mathcal{F}left{d(r)right} = D(rho) &= 2pi int_0^{infty} dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) space r space mathrm{d}r\
\
&= alpha int_0^{infty} delta_a(r) left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) mathrm{d}r\
\
&= alphalambda lambda' int_0^{infty} delta_a(r) J_1(beta r) Y_1(beta r) J_0(rho r) mathrm{d}r - alpha pi^2 int_0^{infty} delta_a(r)J_0(beta r) Y_0(beta r) J_0(rho r) mathrm{d}r\
\
&= alphalambda lambda' lim_{r rightarrow 0+} J_1(beta r) Y_1(beta r) J_0(rho r) - alpha pi^2 lim_{r rightarrow 0+} J_0(beta r) Y_0(beta r) J_0(rho r) \
\
&= alphalambda lambda'left(-dfrac1{pi}right) - alpha pi^2 (-infty) \
end{align}$$
it diverges.
Note that if the problematic term had converged, the Fourier Transform would simply be an infinite disk of constant height above the $rhophi$ plane. That should be no surprise for the transform of a Dirac Delta function.
The problematic term would converge, if there were a factor of $1/ln(r)$:
$$lim_{r rightarrow 0+} dfrac{J_0(beta r) Y_0(beta r)}{ln(r)} J_0(rho r) = dfrac{2}{pi}$$
I don't know the physical situation you are analyzing, and whether inclusion of such a term could ever be justified.
edited Aug 20 at 14:19
answered Aug 18 at 12:32
Andy Walls
1,479127
1,479127
add a comment |
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%2f2878201%2ffourier-transform-of-product-of-bessel-functions%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
Please confirm that this problem has strict radial symmetry. The reason I ask is that $delta^2(x,y)$ and $delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform.
– Andy Walls
Aug 11 at 22:09
@AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :)
– MeMeansMe
Aug 13 at 7:50