Partial Derivative of Multivariate Indicator Function in the Distributional Sense
Clash Royale CLAN TAG#URR8PPP
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Define $f:mathbbR^2tomathbbR$ by $left(x,yright)mapsto1$ if $x^2+y^2leq1$ and $left(x,yright)mapsto0$ otherwise.
Is there a definition of $partial f/partial x$ in the distributional sense? If so, what is it? References are welcome.
I know that $partial f/partial x=0$ on $leftleft(x,yright)inmathbbR^2:x^2+y^2neq1right$, and my intuition tells me that $partial f/partial x$ behaves like the Dirac delta function on $leftleft(x,yright)inmathbbR^2:x^2+y^2=1right$, but I do not know how to express this rigorously.
distribution-theory
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Define $f:mathbbR^2tomathbbR$ by $left(x,yright)mapsto1$ if $x^2+y^2leq1$ and $left(x,yright)mapsto0$ otherwise.
Is there a definition of $partial f/partial x$ in the distributional sense? If so, what is it? References are welcome.
I know that $partial f/partial x=0$ on $leftleft(x,yright)inmathbbR^2:x^2+y^2neq1right$, and my intuition tells me that $partial f/partial x$ behaves like the Dirac delta function on $leftleft(x,yright)inmathbbR^2:x^2+y^2=1right$, but I do not know how to express this rigorously.
distribution-theory
Any locally integrable function has a derivative in the sense of distributions/ generalized functions because every distribution has a derivative.
â Kavi Rama Murthy
Sep 7 at 6:19
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Define $f:mathbbR^2tomathbbR$ by $left(x,yright)mapsto1$ if $x^2+y^2leq1$ and $left(x,yright)mapsto0$ otherwise.
Is there a definition of $partial f/partial x$ in the distributional sense? If so, what is it? References are welcome.
I know that $partial f/partial x=0$ on $leftleft(x,yright)inmathbbR^2:x^2+y^2neq1right$, and my intuition tells me that $partial f/partial x$ behaves like the Dirac delta function on $leftleft(x,yright)inmathbbR^2:x^2+y^2=1right$, but I do not know how to express this rigorously.
distribution-theory
Define $f:mathbbR^2tomathbbR$ by $left(x,yright)mapsto1$ if $x^2+y^2leq1$ and $left(x,yright)mapsto0$ otherwise.
Is there a definition of $partial f/partial x$ in the distributional sense? If so, what is it? References are welcome.
I know that $partial f/partial x=0$ on $leftleft(x,yright)inmathbbR^2:x^2+y^2neq1right$, and my intuition tells me that $partial f/partial x$ behaves like the Dirac delta function on $leftleft(x,yright)inmathbbR^2:x^2+y^2=1right$, but I do not know how to express this rigorously.
distribution-theory
distribution-theory
asked Sep 7 at 6:08
Cleric
3,08432464
3,08432464
Any locally integrable function has a derivative in the sense of distributions/ generalized functions because every distribution has a derivative.
â Kavi Rama Murthy
Sep 7 at 6:19
add a comment |Â
Any locally integrable function has a derivative in the sense of distributions/ generalized functions because every distribution has a derivative.
â Kavi Rama Murthy
Sep 7 at 6:19
Any locally integrable function has a derivative in the sense of distributions/ generalized functions because every distribution has a derivative.
â Kavi Rama Murthy
Sep 7 at 6:19
Any locally integrable function has a derivative in the sense of distributions/ generalized functions because every distribution has a derivative.
â Kavi Rama Murthy
Sep 7 at 6:19
add a comment |Â
1 Answer
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let $vec x = (x_1,dots,x_n)$, and let $vecphi:mathbb R^n to mathbb R^n $ be a vector valued test function, extend the pairing between smooth functions of compact support $langle vec f,vec grangle := int_mathbb R^n vec fcdotvec g$ to distributions analagously to the scalar case. By extending Gauss's theorem, the natural generalisation of integration by parts to higher dimensions, we have the natural generalisation of the distributional derivative,
$$langle nabla f, vecphirangle := -langle f,nablacdot vecphi rangle = -int_<1 nabla cdot vecphi(vec x) dvec x = -int_ vecphicdot vec n dl$$
so $nabla f$ is a vector valued distribution supported on $mathbb S^1 = $ which points in the direction of the outward normal from $mathbb S^1$. If you wanted to, you could define for any test function $varphi : mathbb R^n to mathbb R$,
$$langle delta_mathbb S^1,varphirangle := int_ varphi dl,$$
then (somewhat abusively since $vec n$ is only defined on $mathbb S^1$), $$nabla f = -vec n delta_mathbb S^1$$
The components $partial_i f$ are related as follows. If $varphi : mathbb R^n to mathbb R$ is a test function, set $vecphi = varphi vec e_i$, where $vec e_i$ is the $i$th standard basis vector of $vec R^n$. Then
$$langle partial_i f, varphirangle := - langle f , partial_i varphirangle = - langle f,nabla cdot vecphirangle = langle nabla f,vecphirangle = langle -vec n delta_mathbb S^1 , vecphirangle = langle -n_idelta_mathbb S^1, varphirangle$$
i.e. $$ fracpartial fpartial x_i = -n_idelta_mathbb S^1$$
in the sense of distributions.
I don't have any references for the above computation; it is not difficult, but I have never seen it before. It would be interesting to have a theory of generalised $n$-forms that would allow both the above and the usual computations with generalised functions, but I don't know of it. So I would also greatly appreciate some references.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
let $vec x = (x_1,dots,x_n)$, and let $vecphi:mathbb R^n to mathbb R^n $ be a vector valued test function, extend the pairing between smooth functions of compact support $langle vec f,vec grangle := int_mathbb R^n vec fcdotvec g$ to distributions analagously to the scalar case. By extending Gauss's theorem, the natural generalisation of integration by parts to higher dimensions, we have the natural generalisation of the distributional derivative,
$$langle nabla f, vecphirangle := -langle f,nablacdot vecphi rangle = -int_<1 nabla cdot vecphi(vec x) dvec x = -int_ vecphicdot vec n dl$$
so $nabla f$ is a vector valued distribution supported on $mathbb S^1 = $ which points in the direction of the outward normal from $mathbb S^1$. If you wanted to, you could define for any test function $varphi : mathbb R^n to mathbb R$,
$$langle delta_mathbb S^1,varphirangle := int_ varphi dl,$$
then (somewhat abusively since $vec n$ is only defined on $mathbb S^1$), $$nabla f = -vec n delta_mathbb S^1$$
The components $partial_i f$ are related as follows. If $varphi : mathbb R^n to mathbb R$ is a test function, set $vecphi = varphi vec e_i$, where $vec e_i$ is the $i$th standard basis vector of $vec R^n$. Then
$$langle partial_i f, varphirangle := - langle f , partial_i varphirangle = - langle f,nabla cdot vecphirangle = langle nabla f,vecphirangle = langle -vec n delta_mathbb S^1 , vecphirangle = langle -n_idelta_mathbb S^1, varphirangle$$
i.e. $$ fracpartial fpartial x_i = -n_idelta_mathbb S^1$$
in the sense of distributions.
I don't have any references for the above computation; it is not difficult, but I have never seen it before. It would be interesting to have a theory of generalised $n$-forms that would allow both the above and the usual computations with generalised functions, but I don't know of it. So I would also greatly appreciate some references.
add a comment |Â
up vote
2
down vote
accepted
let $vec x = (x_1,dots,x_n)$, and let $vecphi:mathbb R^n to mathbb R^n $ be a vector valued test function, extend the pairing between smooth functions of compact support $langle vec f,vec grangle := int_mathbb R^n vec fcdotvec g$ to distributions analagously to the scalar case. By extending Gauss's theorem, the natural generalisation of integration by parts to higher dimensions, we have the natural generalisation of the distributional derivative,
$$langle nabla f, vecphirangle := -langle f,nablacdot vecphi rangle = -int_<1 nabla cdot vecphi(vec x) dvec x = -int_ vecphicdot vec n dl$$
so $nabla f$ is a vector valued distribution supported on $mathbb S^1 = $ which points in the direction of the outward normal from $mathbb S^1$. If you wanted to, you could define for any test function $varphi : mathbb R^n to mathbb R$,
$$langle delta_mathbb S^1,varphirangle := int_ varphi dl,$$
then (somewhat abusively since $vec n$ is only defined on $mathbb S^1$), $$nabla f = -vec n delta_mathbb S^1$$
The components $partial_i f$ are related as follows. If $varphi : mathbb R^n to mathbb R$ is a test function, set $vecphi = varphi vec e_i$, where $vec e_i$ is the $i$th standard basis vector of $vec R^n$. Then
$$langle partial_i f, varphirangle := - langle f , partial_i varphirangle = - langle f,nabla cdot vecphirangle = langle nabla f,vecphirangle = langle -vec n delta_mathbb S^1 , vecphirangle = langle -n_idelta_mathbb S^1, varphirangle$$
i.e. $$ fracpartial fpartial x_i = -n_idelta_mathbb S^1$$
in the sense of distributions.
I don't have any references for the above computation; it is not difficult, but I have never seen it before. It would be interesting to have a theory of generalised $n$-forms that would allow both the above and the usual computations with generalised functions, but I don't know of it. So I would also greatly appreciate some references.
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
let $vec x = (x_1,dots,x_n)$, and let $vecphi:mathbb R^n to mathbb R^n $ be a vector valued test function, extend the pairing between smooth functions of compact support $langle vec f,vec grangle := int_mathbb R^n vec fcdotvec g$ to distributions analagously to the scalar case. By extending Gauss's theorem, the natural generalisation of integration by parts to higher dimensions, we have the natural generalisation of the distributional derivative,
$$langle nabla f, vecphirangle := -langle f,nablacdot vecphi rangle = -int_<1 nabla cdot vecphi(vec x) dvec x = -int_ vecphicdot vec n dl$$
so $nabla f$ is a vector valued distribution supported on $mathbb S^1 = $ which points in the direction of the outward normal from $mathbb S^1$. If you wanted to, you could define for any test function $varphi : mathbb R^n to mathbb R$,
$$langle delta_mathbb S^1,varphirangle := int_ varphi dl,$$
then (somewhat abusively since $vec n$ is only defined on $mathbb S^1$), $$nabla f = -vec n delta_mathbb S^1$$
The components $partial_i f$ are related as follows. If $varphi : mathbb R^n to mathbb R$ is a test function, set $vecphi = varphi vec e_i$, where $vec e_i$ is the $i$th standard basis vector of $vec R^n$. Then
$$langle partial_i f, varphirangle := - langle f , partial_i varphirangle = - langle f,nabla cdot vecphirangle = langle nabla f,vecphirangle = langle -vec n delta_mathbb S^1 , vecphirangle = langle -n_idelta_mathbb S^1, varphirangle$$
i.e. $$ fracpartial fpartial x_i = -n_idelta_mathbb S^1$$
in the sense of distributions.
I don't have any references for the above computation; it is not difficult, but I have never seen it before. It would be interesting to have a theory of generalised $n$-forms that would allow both the above and the usual computations with generalised functions, but I don't know of it. So I would also greatly appreciate some references.
let $vec x = (x_1,dots,x_n)$, and let $vecphi:mathbb R^n to mathbb R^n $ be a vector valued test function, extend the pairing between smooth functions of compact support $langle vec f,vec grangle := int_mathbb R^n vec fcdotvec g$ to distributions analagously to the scalar case. By extending Gauss's theorem, the natural generalisation of integration by parts to higher dimensions, we have the natural generalisation of the distributional derivative,
$$langle nabla f, vecphirangle := -langle f,nablacdot vecphi rangle = -int_<1 nabla cdot vecphi(vec x) dvec x = -int_ vecphicdot vec n dl$$
so $nabla f$ is a vector valued distribution supported on $mathbb S^1 = $ which points in the direction of the outward normal from $mathbb S^1$. If you wanted to, you could define for any test function $varphi : mathbb R^n to mathbb R$,
$$langle delta_mathbb S^1,varphirangle := int_ varphi dl,$$
then (somewhat abusively since $vec n$ is only defined on $mathbb S^1$), $$nabla f = -vec n delta_mathbb S^1$$
The components $partial_i f$ are related as follows. If $varphi : mathbb R^n to mathbb R$ is a test function, set $vecphi = varphi vec e_i$, where $vec e_i$ is the $i$th standard basis vector of $vec R^n$. Then
$$langle partial_i f, varphirangle := - langle f , partial_i varphirangle = - langle f,nabla cdot vecphirangle = langle nabla f,vecphirangle = langle -vec n delta_mathbb S^1 , vecphirangle = langle -n_idelta_mathbb S^1, varphirangle$$
i.e. $$ fracpartial fpartial x_i = -n_idelta_mathbb S^1$$
in the sense of distributions.
I don't have any references for the above computation; it is not difficult, but I have never seen it before. It would be interesting to have a theory of generalised $n$-forms that would allow both the above and the usual computations with generalised functions, but I don't know of it. So I would also greatly appreciate some references.
edited Sep 7 at 14:25
answered Sep 7 at 6:49
Calvin Khor
8,85621133
8,85621133
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Any locally integrable function has a derivative in the sense of distributions/ generalized functions because every distribution has a derivative.
â Kavi Rama Murthy
Sep 7 at 6:19