Delta derivative distribution identity?
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It is easy to show that the Dirac $delta(x)$ distribution satisfies the distributional identity $$delta(x) = - x delta'(x).$$ Can we conclude that the following also holds $$delta'(x) = - frac{delta(x)}{x}?$$ In other words, are we allowed to multiply and divide distributions with ordinary functions?
functional-analysis distribution-theory dirac-delta
asked Nov 5 '18 at 11:50
FizikusFizikus
1591114
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add a comment |
$begingroup$
It is easy to show that the Dirac $delta(x)$ distribution satisfies the distributional identity $$delta(x) = - x delta'(x).$$ Can we conclude that the following also holds $$delta'(x) = - frac{delta(x)}{x}?$$ In other words, are we allowed to multiply and divide distributions with ordinary functions?
functional-analysis distribution-theory dirac-delta
asked Nov 5 '18 at 11:50
FizikusFizikus
1591114
$endgroup$
2
$begingroup$
I don't understand the downvote. This is a perfectly reasonable question. The problem of division is a standard one in distribution theory.
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– Giuseppe Negro
Nov 5 '18 at 13:25
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@GiuseppeNegro The question might be reasonable. The notation is not. What is the $x$ in the expressions $delta(x)$, $xdelta'(x)$? It does not make sense.
$endgroup$
– daw
Nov 6 '18 at 12:25
3
$begingroup$
@daw: It does. It is standard physicist's notation. The $delta$ is treated as if it were a function of $xinmathbb R$. I started to like this notation when I read the notes by Klainerman (who is a mathematician), part 1, chapter 3. He uses this notation and it works very well.
$endgroup$
– Giuseppe Negro
Nov 6 '18 at 13:58
$begingroup$
I've expanded my answer. My former answer wasn't incorrect, but it didn't say anything about what actually can be done.
$endgroup$
– md2perpe
Dec 22 '18 at 16:08
add a comment |
$begingroup$
It is easy to show that the Dirac $delta(x)$ distribution satisfies the distributional identity $$delta(x) = - x delta'(x).$$ Can we conclude that the following also holds $$delta'(x) = - frac{delta(x)}{x}?$$ In other words, are we allowed to multiply and divide distributions with ordinary functions?
functional-analysis distribution-theory dirac-delta
asked Nov 5 '18 at 11:50
FizikusFizikus
1591114
$endgroup$
It is easy to show that the Dirac $delta(x)$ distribution satisfies the distributional identity $$delta(x) = - x delta'(x).$$ Can we conclude that the following also holds $$delta'(x) = - frac{delta(x)}{x}?$$ In other words, are we allowed to multiply and divide distributions with ordinary functions?
functional-analysis distribution-theory dirac-delta
functional-analysis distribution-theory dirac-delta
asked Nov 5 '18 at 11:50
FizikusFizikus
1591114
asked Nov 5 '18 at 11:50
FizikusFizikus
1591114
asked Nov 5 '18 at 11:50
FizikusFizikus
1591114
asked Nov 5 '18 at 11:50
FizikusFizikus
1591114
asked Nov 5 '18 at 11:50
FizikusFizikus
1591114
1591114
2
$begingroup$
I don't understand the downvote. This is a perfectly reasonable question. The problem of division is a standard one in distribution theory.
$endgroup$
– Giuseppe Negro
Nov 5 '18 at 13:25
$begingroup$
@GiuseppeNegro The question might be reasonable. The notation is not. What is the $x$ in the expressions $delta(x)$, $xdelta'(x)$? It does not make sense.
$endgroup$
– daw
Nov 6 '18 at 12:25
3
$begingroup$
@daw: It does. It is standard physicist's notation. The $delta$ is treated as if it were a function of $xinmathbb R$. I started to like this notation when I read the notes by Klainerman (who is a mathematician), part 1, chapter 3. He uses this notation and it works very well.
$endgroup$
– Giuseppe Negro
Nov 6 '18 at 13:58
$begingroup$
I've expanded my answer. My former answer wasn't incorrect, but it didn't say anything about what actually can be done.
$endgroup$
– md2perpe
Dec 22 '18 at 16:08
add a comment |
2
$begingroup$
I don't understand the downvote. This is a perfectly reasonable question. The problem of division is a standard one in distribution theory.
$endgroup$
– Giuseppe Negro
Nov 5 '18 at 13:25
$begingroup$
@GiuseppeNegro The question might be reasonable. The notation is not. What is the $x$ in the expressions $delta(x)$, $xdelta'(x)$? It does not make sense.
$endgroup$
– daw
Nov 6 '18 at 12:25
3
$begingroup$
@daw: It does. It is standard physicist's notation. The $delta$ is treated as if it were a function of $xinmathbb R$. I started to like this notation when I read the notes by Klainerman (who is a mathematician), part 1, chapter 3. He uses this notation and it works very well.
$endgroup$
– Giuseppe Negro
Nov 6 '18 at 13:58
$begingroup$
I've expanded my answer. My former answer wasn't incorrect, but it didn't say anything about what actually can be done.
$endgroup$
– md2perpe
Dec 22 '18 at 16:08
2
2
$begingroup$
I don't understand the downvote. This is a perfectly reasonable question. The problem of division is a standard one in distribution theory.
$endgroup$
– Giuseppe Negro
Nov 5 '18 at 13:25
$begingroup$
I don't understand the downvote. This is a perfectly reasonable question. The problem of division is a standard one in distribution theory.
$endgroup$
– Giuseppe Negro
Nov 5 '18 at 13:25
$begingroup$
@GiuseppeNegro The question might be reasonable. The notation is not. What is the $x$ in the expressions $delta(x)$, $xdelta'(x)$? It does not make sense.
$endgroup$
– daw
Nov 6 '18 at 12:25
$begingroup$
@GiuseppeNegro The question might be reasonable. The notation is not. What is the $x$ in the expressions $delta(x)$, $xdelta'(x)$? It does not make sense.
$endgroup$
– daw
Nov 6 '18 at 12:25
3
3
$begingroup$
@daw: It does. It is standard physicist's notation. The $delta$ is treated as if it were a function of $xinmathbb R$. I started to like this notation when I read the notes by Klainerman (who is a mathematician), part 1, chapter 3. He uses this notation and it works very well.
$endgroup$
– Giuseppe Negro
Nov 6 '18 at 13:58
$begingroup$
@daw: It does. It is standard physicist's notation. The $delta$ is treated as if it were a function of $xinmathbb R$. I started to like this notation when I read the notes by Klainerman (who is a mathematician), part 1, chapter 3. He uses this notation and it works very well.
$endgroup$
– Giuseppe Negro
Nov 6 '18 at 13:58
$begingroup$
I've expanded my answer. My former answer wasn't incorrect, but it didn't say anything about what actually can be done.
$endgroup$
– md2perpe
Dec 22 '18 at 16:08
$begingroup$
I've expanded my answer. My former answer wasn't incorrect, but it didn't say anything about what actually can be done.
$endgroup$
– md2perpe
Dec 22 '18 at 16:08
add a comment |
1 Answer
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Multiplication of a distribution $u$ with a function $f$ is defined by $langle fu, phi rangle := langle u, fphi rangle.$ Since $C_c^infty$ is closed under multiplication with $C^infty$ functions, we are allowed to multiply distributions with $C^infty$ functions. We are, using this definition, not allowed to multiply distributions with non-$C^infty$ functions like $frac{1}{x}.$
However, we can try to solve distribution equations like $fu = v,$ where $f$ is a $C^infty$ function and $v$ is a given distribution. This equation often has solutions, but there might be an infinitude of solutions. For example, the solutions to $xu = delta$ are $u = -delta' + C delta,$ where $C$ is a constant.
More generally, if $f$ has zeros of finite order then $fu = v$ has a family of solutions which differ by $delta$:s and derivatives thereof placed at the zeros.
In the specific case $xu = delta$ we can say that $u = -delta'$ is the canonical solution, since $x$ being odd and $delta$ being even should result in $u$ being odd. But in the general case we can not do this.
answered Nov 5 '18 at 13:19
md2perpemd2perpe
8,61411129
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add a comment |
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$begingroup$
Multiplication of a distribution $u$ with a function $f$ is defined by $langle fu, phi rangle := langle u, fphi rangle.$ Since $C_c^infty$ is closed under multiplication with $C^infty$ functions, we are allowed to multiply distributions with $C^infty$ functions. We are, using this definition, not allowed to multiply distributions with non-$C^infty$ functions like $frac{1}{x}.$
However, we can try to solve distribution equations like $fu = v,$ where $f$ is a $C^infty$ function and $v$ is a given distribution. This equation often has solutions, but there might be an infinitude of solutions. For example, the solutions to $xu = delta$ are $u = -delta' + C delta,$ where $C$ is a constant.
More generally, if $f$ has zeros of finite order then $fu = v$ has a family of solutions which differ by $delta$:s and derivatives thereof placed at the zeros.
In the specific case $xu = delta$ we can say that $u = -delta'$ is the canonical solution, since $x$ being odd and $delta$ being even should result in $u$ being odd. But in the general case we can not do this.
answered Nov 5 '18 at 13:19
md2perpemd2perpe
8,61411129
$endgroup$
add a comment |
$begingroup$
Multiplication of a distribution $u$ with a function $f$ is defined by $langle fu, phi rangle := langle u, fphi rangle.$ Since $C_c^infty$ is closed under multiplication with $C^infty$ functions, we are allowed to multiply distributions with $C^infty$ functions. We are, using this definition, not allowed to multiply distributions with non-$C^infty$ functions like $frac{1}{x}.$
However, we can try to solve distribution equations like $fu = v,$ where $f$ is a $C^infty$ function and $v$ is a given distribution. This equation often has solutions, but there might be an infinitude of solutions. For example, the solutions to $xu = delta$ are $u = -delta' + C delta,$ where $C$ is a constant.
More generally, if $f$ has zeros of finite order then $fu = v$ has a family of solutions which differ by $delta$:s and derivatives thereof placed at the zeros.
In the specific case $xu = delta$ we can say that $u = -delta'$ is the canonical solution, since $x$ being odd and $delta$ being even should result in $u$ being odd. But in the general case we can not do this.
answered Nov 5 '18 at 13:19
md2perpemd2perpe
8,61411129
$endgroup$
add a comment |
$begingroup$
Multiplication of a distribution $u$ with a function $f$ is defined by $langle fu, phi rangle := langle u, fphi rangle.$ Since $C_c^infty$ is closed under multiplication with $C^infty$ functions, we are allowed to multiply distributions with $C^infty$ functions. We are, using this definition, not allowed to multiply distributions with non-$C^infty$ functions like $frac{1}{x}.$
However, we can try to solve distribution equations like $fu = v,$ where $f$ is a $C^infty$ function and $v$ is a given distribution. This equation often has solutions, but there might be an infinitude of solutions. For example, the solutions to $xu = delta$ are $u = -delta' + C delta,$ where $C$ is a constant.
More generally, if $f$ has zeros of finite order then $fu = v$ has a family of solutions which differ by $delta$:s and derivatives thereof placed at the zeros.
In the specific case $xu = delta$ we can say that $u = -delta'$ is the canonical solution, since $x$ being odd and $delta$ being even should result in $u$ being odd. But in the general case we can not do this.
answered Nov 5 '18 at 13:19
md2perpemd2perpe
8,61411129
$endgroup$
Multiplication of a distribution $u$ with a function $f$ is defined by $langle fu, phi rangle := langle u, fphi rangle.$ Since $C_c^infty$ is closed under multiplication with $C^infty$ functions, we are allowed to multiply distributions with $C^infty$ functions. We are, using this definition, not allowed to multiply distributions with non-$C^infty$ functions like $frac{1}{x}.$
However, we can try to solve distribution equations like $fu = v,$ where $f$ is a $C^infty$ function and $v$ is a given distribution. This equation often has solutions, but there might be an infinitude of solutions. For example, the solutions to $xu = delta$ are $u = -delta' + C delta,$ where $C$ is a constant.
More generally, if $f$ has zeros of finite order then $fu = v$ has a family of solutions which differ by $delta$:s and derivatives thereof placed at the zeros.
In the specific case $xu = delta$ we can say that $u = -delta'$ is the canonical solution, since $x$ being odd and $delta$ being even should result in $u$ being odd. But in the general case we can not do this.
answered Nov 5 '18 at 13:19
md2perpemd2perpe
8,61411129
edited Dec 22 '18 at 16:06
answered Nov 5 '18 at 13:19
md2perpemd2perpe
8,61411129
answered Nov 5 '18 at 13:19
md2perpemd2perpe
8,61411129
answered Nov 5 '18 at 13:19
md2perpemd2perpe
8,61411129
8,61411129
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$begingroup$
I don't understand the downvote. This is a perfectly reasonable question. The problem of division is a standard one in distribution theory.
$endgroup$
– Giuseppe Negro
Nov 5 '18 at 13:25
$begingroup$
@GiuseppeNegro The question might be reasonable. The notation is not. What is the $x$ in the expressions $delta(x)$, $xdelta'(x)$? It does not make sense.
$endgroup$
– daw
Nov 6 '18 at 12:25
3
$begingroup$
@daw: It does. It is standard physicist's notation. The $delta$ is treated as if it were a function of $xinmathbb R$. I started to like this notation when I read the notes by Klainerman (who is a mathematician), part 1, chapter 3. He uses this notation and it works very well.
$endgroup$
– Giuseppe Negro
Nov 6 '18 at 13:58
$begingroup$
I've expanded my answer. My former answer wasn't incorrect, but it didn't say anything about what actually can be done.
$endgroup$
– md2perpe
Dec 22 '18 at 16:08