Mixing
Dirac Delta and its evaluation in a complicated integral
$begingroup$
I would like to better understand how to use and manipulate the Dirac Delta function.
It seems to me that whenever the delta function appears in an integral, it reduces the dimension of the domain of integration by $1$, i.e.
$$int limits _{Omega in mathcal{R}^2}f(x_1,x_2) delta(x_1-y) mathrm{d}Omega rightarrow int limits _{Omega {x_1rightarrow y} in mathcal{R}}f(y,x_2) mathrm{d}x_2$$
For very easy functions and domains I was able to manually check this.
I would like to exploit this property in more complicated situations, for example:
$$int limits _{Omega in mathcal{R}^4}f(x_1,x_2,x_3,x_4) delta(x_1-x_2+1) delta(x_2) mathrm{d}Omega rightarrow int limits _{Omega {x_2rightarrow 0,x_1rightarrow-1} in mathcal{R}^2}f(-1,0,x_3,x_4) mathrm{d}Omega$$
but it appears that this is not always possible (I have seen how certain examples work, but others do not$^{(1)}$). Can anyone explain to me why this is wrong? I was hoping to use this "trick" to numerically integrate functions that have a Dirac Delta inside.
$^{(1)}$ I have tried the following case using Mathematica v$11.0$:
1)The function is: $$f(x,y,z,zeta)=frac{z^4}{2 (y^2+z^2)^{5/2}}cdot delta (x+y-8) $$
2)The domain of integration is:$$[0leq xleq25]cup[-5leq yleq 5]cup[-0.5leq zleq0.5]cup[0leqzetaleq1]cup[0leq x+yleq25]cup[0leq z+zetaleq1] $$
I obtain different results when doing the domain reduction for $xrightarrow8-y$ or $yrightarrow8-x$
integration dirac-delta
asked Jan 15 at 18:25
RiccardoRiccardo
314
$endgroup$
add a comment |
$begingroup$
I would like to better understand how to use and manipulate the Dirac Delta function.
It seems to me that whenever the delta function appears in an integral, it reduces the dimension of the domain of integration by $1$, i.e.
$$int limits _{Omega in mathcal{R}^2}f(x_1,x_2) delta(x_1-y) mathrm{d}Omega rightarrow int limits _{Omega {x_1rightarrow y} in mathcal{R}}f(y,x_2) mathrm{d}x_2$$
For very easy functions and domains I was able to manually check this.
I would like to exploit this property in more complicated situations, for example:
$$int limits _{Omega in mathcal{R}^4}f(x_1,x_2,x_3,x_4) delta(x_1-x_2+1) delta(x_2) mathrm{d}Omega rightarrow int limits _{Omega {x_2rightarrow 0,x_1rightarrow-1} in mathcal{R}^2}f(-1,0,x_3,x_4) mathrm{d}Omega$$
but it appears that this is not always possible (I have seen how certain examples work, but others do not$^{(1)}$). Can anyone explain to me why this is wrong? I was hoping to use this "trick" to numerically integrate functions that have a Dirac Delta inside.
$^{(1)}$ I have tried the following case using Mathematica v$11.0$:
1)The function is: $$f(x,y,z,zeta)=frac{z^4}{2 (y^2+z^2)^{5/2}}cdot delta (x+y-8) $$
2)The domain of integration is:$$[0leq xleq25]cup[-5leq yleq 5]cup[-0.5leq zleq0.5]cup[0leqzetaleq1]cup[0leq x+yleq25]cup[0leq z+zetaleq1] $$
I obtain different results when doing the domain reduction for $xrightarrow8-y$ or $yrightarrow8-x$
integration dirac-delta
asked Jan 15 at 18:25
RiccardoRiccardo
314
$endgroup$
$begingroup$
The purest way of understanding the Dirac delta function is using the language of distributions. Yoy may start having a look at Wiki page: en.wikipedia.org/wiki/Dirac_delta_function#As_a_distribution and en.wikipedia.org/wiki/Distribution_(mathematics)
$endgroup$
– Dog_69
Jan 15 at 18:48
2
$begingroup$
Can you give an example where it does not work?
$endgroup$
– Botond
Jan 15 at 18:49
$begingroup$
@Botond I have added one of the examples I am dealing with
$endgroup$
– Riccardo
Jan 16 at 11:18
add a comment |
$begingroup$
I would like to better understand how to use and manipulate the Dirac Delta function.
It seems to me that whenever the delta function appears in an integral, it reduces the dimension of the domain of integration by $1$, i.e.
$$int limits _{Omega in mathcal{R}^2}f(x_1,x_2) delta(x_1-y) mathrm{d}Omega rightarrow int limits _{Omega {x_1rightarrow y} in mathcal{R}}f(y,x_2) mathrm{d}x_2$$
For very easy functions and domains I was able to manually check this.
I would like to exploit this property in more complicated situations, for example:
$$int limits _{Omega in mathcal{R}^4}f(x_1,x_2,x_3,x_4) delta(x_1-x_2+1) delta(x_2) mathrm{d}Omega rightarrow int limits _{Omega {x_2rightarrow 0,x_1rightarrow-1} in mathcal{R}^2}f(-1,0,x_3,x_4) mathrm{d}Omega$$
but it appears that this is not always possible (I have seen how certain examples work, but others do not$^{(1)}$). Can anyone explain to me why this is wrong? I was hoping to use this "trick" to numerically integrate functions that have a Dirac Delta inside.
$^{(1)}$ I have tried the following case using Mathematica v$11.0$:
1)The function is: $$f(x,y,z,zeta)=frac{z^4}{2 (y^2+z^2)^{5/2}}cdot delta (x+y-8) $$
2)The domain of integration is:$$[0leq xleq25]cup[-5leq yleq 5]cup[-0.5leq zleq0.5]cup[0leqzetaleq1]cup[0leq x+yleq25]cup[0leq z+zetaleq1] $$
I obtain different results when doing the domain reduction for $xrightarrow8-y$ or $yrightarrow8-x$
integration dirac-delta
asked Jan 15 at 18:25
RiccardoRiccardo
314
$endgroup$
I would like to better understand how to use and manipulate the Dirac Delta function.
It seems to me that whenever the delta function appears in an integral, it reduces the dimension of the domain of integration by $1$, i.e.
$$int limits _{Omega in mathcal{R}^2}f(x_1,x_2) delta(x_1-y) mathrm{d}Omega rightarrow int limits _{Omega {x_1rightarrow y} in mathcal{R}}f(y,x_2) mathrm{d}x_2$$
For very easy functions and domains I was able to manually check this.
I would like to exploit this property in more complicated situations, for example:
$$int limits _{Omega in mathcal{R}^4}f(x_1,x_2,x_3,x_4) delta(x_1-x_2+1) delta(x_2) mathrm{d}Omega rightarrow int limits _{Omega {x_2rightarrow 0,x_1rightarrow-1} in mathcal{R}^2}f(-1,0,x_3,x_4) mathrm{d}Omega$$
but it appears that this is not always possible (I have seen how certain examples work, but others do not$^{(1)}$). Can anyone explain to me why this is wrong? I was hoping to use this "trick" to numerically integrate functions that have a Dirac Delta inside.
$^{(1)}$ I have tried the following case using Mathematica v$11.0$:
1)The function is: $$f(x,y,z,zeta)=frac{z^4}{2 (y^2+z^2)^{5/2}}cdot delta (x+y-8) $$
2)The domain of integration is:$$[0leq xleq25]cup[-5leq yleq 5]cup[-0.5leq zleq0.5]cup[0leqzetaleq1]cup[0leq x+yleq25]cup[0leq z+zetaleq1] $$
I obtain different results when doing the domain reduction for $xrightarrow8-y$ or $yrightarrow8-x$
integration dirac-delta
integration dirac-delta
asked Jan 15 at 18:25
RiccardoRiccardo
314
asked Jan 15 at 18:25
RiccardoRiccardo
314
edited Jan 16 at 11:17
Riccardo
asked Jan 15 at 18:25
RiccardoRiccardo
314
asked Jan 15 at 18:25
RiccardoRiccardo
314
asked Jan 15 at 18:25
RiccardoRiccardo
314
314
$begingroup$
The purest way of understanding the Dirac delta function is using the language of distributions. Yoy may start having a look at Wiki page: en.wikipedia.org/wiki/Dirac_delta_function#As_a_distribution and en.wikipedia.org/wiki/Distribution_(mathematics)
$endgroup$
– Dog_69
Jan 15 at 18:48
2
$begingroup$
Can you give an example where it does not work?
$endgroup$
– Botond
Jan 15 at 18:49
$begingroup$
@Botond I have added one of the examples I am dealing with
$endgroup$
– Riccardo
Jan 16 at 11:18
add a comment |
$begingroup$
The purest way of understanding the Dirac delta function is using the language of distributions. Yoy may start having a look at Wiki page: en.wikipedia.org/wiki/Dirac_delta_function#As_a_distribution and en.wikipedia.org/wiki/Distribution_(mathematics)
$endgroup$
– Dog_69
Jan 15 at 18:48
2
$begingroup$
Can you give an example where it does not work?
$endgroup$
– Botond
Jan 15 at 18:49
$begingroup$
@Botond I have added one of the examples I am dealing with
$endgroup$
– Riccardo
Jan 16 at 11:18
$begingroup$
The purest way of understanding the Dirac delta function is using the language of distributions. Yoy may start having a look at Wiki page: en.wikipedia.org/wiki/Dirac_delta_function#As_a_distribution and en.wikipedia.org/wiki/Distribution_(mathematics)
$endgroup$
– Dog_69
Jan 15 at 18:48
$begingroup$
The purest way of understanding the Dirac delta function is using the language of distributions. Yoy may start having a look at Wiki page: en.wikipedia.org/wiki/Dirac_delta_function#As_a_distribution and en.wikipedia.org/wiki/Distribution_(mathematics)
$endgroup$
– Dog_69
Jan 15 at 18:48
2
2
$begingroup$
Can you give an example where it does not work?
$endgroup$
– Botond
Jan 15 at 18:49
$begingroup$
Can you give an example where it does not work?
$endgroup$
– Botond
Jan 15 at 18:49
$begingroup$
@Botond I have added one of the examples I am dealing with
$endgroup$
– Riccardo
Jan 16 at 11:18
$begingroup$
@Botond I have added one of the examples I am dealing with
$endgroup$
– Riccardo
Jan 16 at 11:18
add a comment |
1 Answer
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oldest
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$begingroup$
As stated in another answer I gave, integrals with the "delta-function" in them should (or at least may) be handled by considering the dirac measure
$$
delta_b(A)=begin{cases}1&bin A\0&bnotin Aend{cases}
$$
for some $binmathbb R$ as integrating over this measure $delta_b(dx)$ then corresponds to the "function" / "integrand" $delta(x-b),dx$ as it is used in the physics literature.
This measure is what is called $sigma$-finite (it even is finite), as is the usual Lebesgue-Borel measure on $mathbb R$. Therefore we can apply Tonelli's theorem which for states that if $f:Xtimes Yto [0,infty)$ with $X,Ysubseteqmathbb R$ is measurable then
$$
int_XBig(int_Y f(x,y),dyBig),delta_b(dx)=int_YBig(int_X f(x,y),delta_b(dx)Big),dytag{1}
$$
so we can interchange the integrals. This interchanging also works if we use the dirac measure twice, i.e. for the $x$ and the y integral. Also I chose this version of Fubini-Tonelli because the mapping from your example is obviously non-negative.
Note that here, if $f$ is continuous on $Xtimes Y$ then it for sure is measurable and the right-hand side of (1) evaluates to
$$
int_YBig(int_X f(x,y),delta_b(dx)Big),dy=begin{cases} int_Y f(b,y),dy&text{if }bin X\0&text{if }bnotin X end{cases},.
$$
However, if one of the delta "displacements" (the $b$) depends on the other integration variable, things start breaking down. Related to the second equation in your original post: for any continuous $f:mathbb R^2tomathbb R$ one gets
$$
int_{mathbb R}int_{mathbb R} f(x,y)delta(x-y)delta(y),dx,dy=int_{mathbb R}Big(int_{mathbb R} f(x,y),delta_y(dx)Big) ,delta_0(dy)=int_{mathbb R} f(y,y),delta_0(dy)=f(0,0)
$$
but
$$
int_{mathbb R}Big(int_{mathbb R} f(x,y)delta(x-y)delta(y),dyBig),dx
$$
involves the product of two "delta functions" in a single integral which is problematic for multiple reasons.
All of this of course is a rather general answer / comment but just goes to show that once a dirac measure in some sense depends on two integration variables, (a) one should be quite careful on what to do precisely and (b) interchanging integrals is not a given anymore (although it might still be possible, varies case by case).
answered Jan 16 at 22:56
Frederik vom EndeFrederik vom Ende
7521321
$endgroup$
add a comment |
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$begingroup$
As stated in another answer I gave, integrals with the "delta-function" in them should (or at least may) be handled by considering the dirac measure
$$
delta_b(A)=begin{cases}1&bin A\0&bnotin Aend{cases}
$$
for some $binmathbb R$ as integrating over this measure $delta_b(dx)$ then corresponds to the "function" / "integrand" $delta(x-b),dx$ as it is used in the physics literature.
This measure is what is called $sigma$-finite (it even is finite), as is the usual Lebesgue-Borel measure on $mathbb R$. Therefore we can apply Tonelli's theorem which for states that if $f:Xtimes Yto [0,infty)$ with $X,Ysubseteqmathbb R$ is measurable then
$$
int_XBig(int_Y f(x,y),dyBig),delta_b(dx)=int_YBig(int_X f(x,y),delta_b(dx)Big),dytag{1}
$$
so we can interchange the integrals. This interchanging also works if we use the dirac measure twice, i.e. for the $x$ and the y integral. Also I chose this version of Fubini-Tonelli because the mapping from your example is obviously non-negative.
Note that here, if $f$ is continuous on $Xtimes Y$ then it for sure is measurable and the right-hand side of (1) evaluates to
$$
int_YBig(int_X f(x,y),delta_b(dx)Big),dy=begin{cases} int_Y f(b,y),dy&text{if }bin X\0&text{if }bnotin X end{cases},.
$$
However, if one of the delta "displacements" (the $b$) depends on the other integration variable, things start breaking down. Related to the second equation in your original post: for any continuous $f:mathbb R^2tomathbb R$ one gets
$$
int_{mathbb R}int_{mathbb R} f(x,y)delta(x-y)delta(y),dx,dy=int_{mathbb R}Big(int_{mathbb R} f(x,y),delta_y(dx)Big) ,delta_0(dy)=int_{mathbb R} f(y,y),delta_0(dy)=f(0,0)
$$
but
$$
int_{mathbb R}Big(int_{mathbb R} f(x,y)delta(x-y)delta(y),dyBig),dx
$$
involves the product of two "delta functions" in a single integral which is problematic for multiple reasons.
All of this of course is a rather general answer / comment but just goes to show that once a dirac measure in some sense depends on two integration variables, (a) one should be quite careful on what to do precisely and (b) interchanging integrals is not a given anymore (although it might still be possible, varies case by case).
answered Jan 16 at 22:56
Frederik vom EndeFrederik vom Ende
7521321
$endgroup$
add a comment |
$begingroup$
As stated in another answer I gave, integrals with the "delta-function" in them should (or at least may) be handled by considering the dirac measure
$$
delta_b(A)=begin{cases}1&bin A\0&bnotin Aend{cases}
$$
for some $binmathbb R$ as integrating over this measure $delta_b(dx)$ then corresponds to the "function" / "integrand" $delta(x-b),dx$ as it is used in the physics literature.
This measure is what is called $sigma$-finite (it even is finite), as is the usual Lebesgue-Borel measure on $mathbb R$. Therefore we can apply Tonelli's theorem which for states that if $f:Xtimes Yto [0,infty)$ with $X,Ysubseteqmathbb R$ is measurable then
$$
int_XBig(int_Y f(x,y),dyBig),delta_b(dx)=int_YBig(int_X f(x,y),delta_b(dx)Big),dytag{1}
$$
so we can interchange the integrals. This interchanging also works if we use the dirac measure twice, i.e. for the $x$ and the y integral. Also I chose this version of Fubini-Tonelli because the mapping from your example is obviously non-negative.
Note that here, if $f$ is continuous on $Xtimes Y$ then it for sure is measurable and the right-hand side of (1) evaluates to
$$
int_YBig(int_X f(x,y),delta_b(dx)Big),dy=begin{cases} int_Y f(b,y),dy&text{if }bin X\0&text{if }bnotin X end{cases},.
$$
However, if one of the delta "displacements" (the $b$) depends on the other integration variable, things start breaking down. Related to the second equation in your original post: for any continuous $f:mathbb R^2tomathbb R$ one gets
$$
int_{mathbb R}int_{mathbb R} f(x,y)delta(x-y)delta(y),dx,dy=int_{mathbb R}Big(int_{mathbb R} f(x,y),delta_y(dx)Big) ,delta_0(dy)=int_{mathbb R} f(y,y),delta_0(dy)=f(0,0)
$$
but
$$
int_{mathbb R}Big(int_{mathbb R} f(x,y)delta(x-y)delta(y),dyBig),dx
$$
involves the product of two "delta functions" in a single integral which is problematic for multiple reasons.
All of this of course is a rather general answer / comment but just goes to show that once a dirac measure in some sense depends on two integration variables, (a) one should be quite careful on what to do precisely and (b) interchanging integrals is not a given anymore (although it might still be possible, varies case by case).
answered Jan 16 at 22:56
Frederik vom EndeFrederik vom Ende
7521321
$endgroup$
add a comment |
$begingroup$
As stated in another answer I gave, integrals with the "delta-function" in them should (or at least may) be handled by considering the dirac measure
$$
delta_b(A)=begin{cases}1&bin A\0&bnotin Aend{cases}
$$
for some $binmathbb R$ as integrating over this measure $delta_b(dx)$ then corresponds to the "function" / "integrand" $delta(x-b),dx$ as it is used in the physics literature.
This measure is what is called $sigma$-finite (it even is finite), as is the usual Lebesgue-Borel measure on $mathbb R$. Therefore we can apply Tonelli's theorem which for states that if $f:Xtimes Yto [0,infty)$ with $X,Ysubseteqmathbb R$ is measurable then
$$
int_XBig(int_Y f(x,y),dyBig),delta_b(dx)=int_YBig(int_X f(x,y),delta_b(dx)Big),dytag{1}
$$
so we can interchange the integrals. This interchanging also works if we use the dirac measure twice, i.e. for the $x$ and the y integral. Also I chose this version of Fubini-Tonelli because the mapping from your example is obviously non-negative.
Note that here, if $f$ is continuous on $Xtimes Y$ then it for sure is measurable and the right-hand side of (1) evaluates to
$$
int_YBig(int_X f(x,y),delta_b(dx)Big),dy=begin{cases} int_Y f(b,y),dy&text{if }bin X\0&text{if }bnotin X end{cases},.
$$
However, if one of the delta "displacements" (the $b$) depends on the other integration variable, things start breaking down. Related to the second equation in your original post: for any continuous $f:mathbb R^2tomathbb R$ one gets
$$
int_{mathbb R}int_{mathbb R} f(x,y)delta(x-y)delta(y),dx,dy=int_{mathbb R}Big(int_{mathbb R} f(x,y),delta_y(dx)Big) ,delta_0(dy)=int_{mathbb R} f(y,y),delta_0(dy)=f(0,0)
$$
but
$$
int_{mathbb R}Big(int_{mathbb R} f(x,y)delta(x-y)delta(y),dyBig),dx
$$
involves the product of two "delta functions" in a single integral which is problematic for multiple reasons.
All of this of course is a rather general answer / comment but just goes to show that once a dirac measure in some sense depends on two integration variables, (a) one should be quite careful on what to do precisely and (b) interchanging integrals is not a given anymore (although it might still be possible, varies case by case).
answered Jan 16 at 22:56
Frederik vom EndeFrederik vom Ende
7521321
$endgroup$
As stated in another answer I gave, integrals with the "delta-function" in them should (or at least may) be handled by considering the dirac measure
$$
delta_b(A)=begin{cases}1&bin A\0&bnotin Aend{cases}
$$
for some $binmathbb R$ as integrating over this measure $delta_b(dx)$ then corresponds to the "function" / "integrand" $delta(x-b),dx$ as it is used in the physics literature.
This measure is what is called $sigma$-finite (it even is finite), as is the usual Lebesgue-Borel measure on $mathbb R$. Therefore we can apply Tonelli's theorem which for states that if $f:Xtimes Yto [0,infty)$ with $X,Ysubseteqmathbb R$ is measurable then
$$
int_XBig(int_Y f(x,y),dyBig),delta_b(dx)=int_YBig(int_X f(x,y),delta_b(dx)Big),dytag{1}
$$
so we can interchange the integrals. This interchanging also works if we use the dirac measure twice, i.e. for the $x$ and the y integral. Also I chose this version of Fubini-Tonelli because the mapping from your example is obviously non-negative.
Note that here, if $f$ is continuous on $Xtimes Y$ then it for sure is measurable and the right-hand side of (1) evaluates to
$$
int_YBig(int_X f(x,y),delta_b(dx)Big),dy=begin{cases} int_Y f(b,y),dy&text{if }bin X\0&text{if }bnotin X end{cases},.
$$
However, if one of the delta "displacements" (the $b$) depends on the other integration variable, things start breaking down. Related to the second equation in your original post: for any continuous $f:mathbb R^2tomathbb R$ one gets
$$
int_{mathbb R}int_{mathbb R} f(x,y)delta(x-y)delta(y),dx,dy=int_{mathbb R}Big(int_{mathbb R} f(x,y),delta_y(dx)Big) ,delta_0(dy)=int_{mathbb R} f(y,y),delta_0(dy)=f(0,0)
$$
but
$$
int_{mathbb R}Big(int_{mathbb R} f(x,y)delta(x-y)delta(y),dyBig),dx
$$
involves the product of two "delta functions" in a single integral which is problematic for multiple reasons.
All of this of course is a rather general answer / comment but just goes to show that once a dirac measure in some sense depends on two integration variables, (a) one should be quite careful on what to do precisely and (b) interchanging integrals is not a given anymore (although it might still be possible, varies case by case).
answered Jan 16 at 22:56
Frederik vom EndeFrederik vom Ende
7521321
edited Jan 16 at 23:08
answered Jan 16 at 22:56
Frederik vom EndeFrederik vom Ende
7521321
answered Jan 16 at 22:56
Frederik vom EndeFrederik vom Ende
7521321
answered Jan 16 at 22:56
Frederik vom EndeFrederik vom Ende
7521321
7521321
add a comment |
add a comment |
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The purest way of understanding the Dirac delta function is using the language of distributions. Yoy may start having a look at Wiki page: en.wikipedia.org/wiki/Dirac_delta_function#As_a_distribution and en.wikipedia.org/wiki/Distribution_(mathematics)
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– Dog_69
Jan 15 at 18:48
2
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Can you give an example where it does not work?
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– Botond
Jan 15 at 18:49
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@Botond I have added one of the examples I am dealing with
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– Riccardo
Jan 16 at 11:18