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$varphi$ verify that $nabla cdotvarphi=0$ but doesn't exist $G:Bbb R^3to Bbb R^3$, $mathcal C^1$ such that $nabla times G=varphi$
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I have this problem:
Probe that $varphi:Bbb R^3-[0] to Bbb R^3, varphi=frac(x,y,z)vertvert(x,y,z)vertvert^3$ verify that $nabla cdotvarphi=0$ but does not exist $G:Bbb R^3to Bbb R^3$, of class $mathcal C^1$ such that $nabla times G=varphi$
It's easy to prove that $nabla cdotvarphi=0$ but I have a problem with the part of there is no $G$ such that $nabla times G=varphi$, if is $G=(G1,G2,G3)$ I have this conditions:
$fracxvertvert(x,y,z)vertvert^3= fracpartial G3partial y-fracpartial G2partial z$
$fracyvertvert(x,y,z)vertvert^3= fracpartial G3partial x-fracpartial G1partial z$
$fraczvertvert(x,y,z)vertvert^3= fracpartial G2partial x-fracpartial G1partial y$
And with $nabla cdotvarphi=0$ if the lateral derivatives of $G$ coincide I can find some $G$, so something is wrong. Has something to do with the fact that $varphi$ is not $mathcal C^1$?
vector-analysis divergence curl
asked Aug 6 at 2:32
Felipe Cignoli
734
add a comment |Â
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0
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I have this problem:
Probe that $varphi:Bbb R^3-[0] to Bbb R^3, varphi=frac(x,y,z)vertvert(x,y,z)vertvert^3$ verify that $nabla cdotvarphi=0$ but does not exist $G:Bbb R^3to Bbb R^3$, of class $mathcal C^1$ such that $nabla times G=varphi$
It's easy to prove that $nabla cdotvarphi=0$ but I have a problem with the part of there is no $G$ such that $nabla times G=varphi$, if is $G=(G1,G2,G3)$ I have this conditions:
$fracxvertvert(x,y,z)vertvert^3= fracpartial G3partial y-fracpartial G2partial z$
$fracyvertvert(x,y,z)vertvert^3= fracpartial G3partial x-fracpartial G1partial z$
$fraczvertvert(x,y,z)vertvert^3= fracpartial G2partial x-fracpartial G1partial y$
And with $nabla cdotvarphi=0$ if the lateral derivatives of $G$ coincide I can find some $G$, so something is wrong. Has something to do with the fact that $varphi$ is not $mathcal C^1$?
vector-analysis divergence curl
asked Aug 6 at 2:32
Felipe Cignoli
734
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have this problem:
Probe that $varphi:Bbb R^3-[0] to Bbb R^3, varphi=frac(x,y,z)vertvert(x,y,z)vertvert^3$ verify that $nabla cdotvarphi=0$ but does not exist $G:Bbb R^3to Bbb R^3$, of class $mathcal C^1$ such that $nabla times G=varphi$
It's easy to prove that $nabla cdotvarphi=0$ but I have a problem with the part of there is no $G$ such that $nabla times G=varphi$, if is $G=(G1,G2,G3)$ I have this conditions:
$fracxvertvert(x,y,z)vertvert^3= fracpartial G3partial y-fracpartial G2partial z$
$fracyvertvert(x,y,z)vertvert^3= fracpartial G3partial x-fracpartial G1partial z$
$fraczvertvert(x,y,z)vertvert^3= fracpartial G2partial x-fracpartial G1partial y$
And with $nabla cdotvarphi=0$ if the lateral derivatives of $G$ coincide I can find some $G$, so something is wrong. Has something to do with the fact that $varphi$ is not $mathcal C^1$?
vector-analysis divergence curl
asked Aug 6 at 2:32
Felipe Cignoli
734
I have this problem:
Probe that $varphi:Bbb R^3-[0] to Bbb R^3, varphi=frac(x,y,z)vertvert(x,y,z)vertvert^3$ verify that $nabla cdotvarphi=0$ but does not exist $G:Bbb R^3to Bbb R^3$, of class $mathcal C^1$ such that $nabla times G=varphi$
It's easy to prove that $nabla cdotvarphi=0$ but I have a problem with the part of there is no $G$ such that $nabla times G=varphi$, if is $G=(G1,G2,G3)$ I have this conditions:
$fracxvertvert(x,y,z)vertvert^3= fracpartial G3partial y-fracpartial G2partial z$
$fracyvertvert(x,y,z)vertvert^3= fracpartial G3partial x-fracpartial G1partial z$
$fraczvertvert(x,y,z)vertvert^3= fracpartial G2partial x-fracpartial G1partial y$
And with $nabla cdotvarphi=0$ if the lateral derivatives of $G$ coincide I can find some $G$, so something is wrong. Has something to do with the fact that $varphi$ is not $mathcal C^1$?
vector-analysis divergence curl
asked Aug 6 at 2:32
Felipe Cignoli
734
asked Aug 6 at 2:32
Felipe Cignoli
734
asked Aug 6 at 2:32
Felipe Cignoli
734
asked Aug 6 at 2:32
Felipe Cignoli
734
734
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
1
down vote
accepted
If $varphi = nabla times G$ on $mathbbR^3 - 0$ where $G in C^1(mathbbR^3)$, then by the divergence theorem, since $nabla cdot (nabla times G) = 0,$
$$int_partial B_R(0) varphi cdot mathbfn ,dS = int_partial B_R(0) nabla times G cdot mathbfn ,dS = int_B_R(0) nabla cdot (nabla times G), dV = 0$$
where $B_R(0)$ is the closed ball of radius $R$ centered at the origin.
However,
$$int_partial B_R(0) varphi cdot mathbfn ,dS = int_0^2piint_0^pifracRmathbfe_rR^3 cdot mathbfe_rR^2sin theta ,dtheta , dphi = 4pi$$
answered Aug 6 at 5:02
RRL
43.7k42260
I knew there was a simpler way to argue this than mine! Cheers!
– Robert Lewis
Aug 6 at 5:16
@RobertLewis: Although $varphi$ is clearly singular at the origin. So I think you show directly that it can't be the restriction of $nabla times G$ which is continuous everywhere if $G$ is $C^1$. Seems so obvious as to require no proof. Maybe we are missing something.
– RRL
Aug 6 at 5:24
It is pretty obvious. I was thinking of a proof like yours but ended up futzing around with all the junk in my answer--so I just posted it. A method like yours nagged at me the whole time, but I got on that tack and . . .
– Robert Lewis
Aug 6 at 5:29
add a comment |Â
up vote
2
down vote
We note that we are asked for a $mathcal C^1$ function
$G:Bbb R^3 to Bbb R^3 tag 1$
such that
$nabla times G = varphi, tag 2$
whilst $varphi$ is only defined on $Bbb R^3 setminus (0, 0, 0) $; indeed, with
$vec r = (x, y, z), tag 3$
we have
$r = Vert vec r Vert = (x^2 + y^2 + z^2)^1/2, tag 4$
whence
$varphi = dfrac(x, y, z)Vert (x, y, z) Vert^3 = r^-3 vec r; tag 5$
it follows that
$Vert varphi Vert^2 = r^-6 vec r cdot vec r = r^-6 r^2 = r^-4, tag 6$
so that
$Vert varphi Vert = r^-2 tag 7$
on $Bbb R^3 setminus (0, 0, 0) $; now suppose a $mathcal C^1$ function $G:Bbb R^3 to Bbb R^3$ exists which satisfies (2); then both $G$ and its derivatives are $mathcal C^0$; restricting $G$ to the closed ball $bar B(0, epsilon)$ of radius $epsilon$ centered at $(0, 0, 0)$, we have, by the compactness of $bar B(0, epsilon)$, that the derivatives of $G$ are bounded on $bar B(0, epsilon)$; but this clearly contradicts (7), which in concert with
$Vert nabla times G Vert = Vert varphi Vert tag 8$
(from (2)) shows the derivatives of $G$ must become arbitrarily large on $bar B(0, epsilon)$; hence no such $G$ exists.
The key here is that, while $varphi$ is only defined on $Bbb R^3 setminus (0, 0, 0) $, $G$ is hypothesized to be defined on all of $Bbb R^3$.
Appendix: Derivation of
$nabla cdot varphi = 0. tag9$
I worked this out whilst thinking about my solution; I though it might be worth including here.
With
$varphi = dfrac(x, y, z)Vert (x, y, z) Vert^3 tag10$
and
$vec r = (x, y, z), tag11$
so that
$r = (x^2 + y^2 + z^2)^1/2, tag12$
we may write
$varphi = r^-3 vec r; tag13$
then, in accord with the vector identity
$nabla cdot f vec X = nabla f cdot X + f nabla cdot X, tag14$
which holds for scalar and vector fields $f$ and $X$, respectively, we have
$nabla cdot varphi = nabla r^-3 cdot vec r + r^-3 nabla cdot vec r; tag15$
now,
$nabla cdot vec r = 3; tag16$
thus,
$nabla cdot varphi = nabla r^-3 cdot vec r + 3 r^-3 ; tag17$
$nabla r^-3 = -3r^-4 nabla r; tag18$
from (11) and (12),
$nabla r = dfrac12(x^2 + y^2 + z^2)^-1/2 (2x, 2y, 2z) = r^-1(x, y, z) = r^-1 vec r; tag19$
$nabla r cdot vec r = r^-1 vec r cdot vec r = r^-1r^2 = r; tag20$
$nabla cdot varphi = -3r^-4 r + 3r^-3 = 0. tag21$
End: Appendix
answered Aug 6 at 4:31
Robert Lewis
37.2k22356
1
Go Beavers .....
– RRL
Aug 6 at 5:08
@RRL oh really? When were you "institutionalized"?
– Robert Lewis
Aug 6 at 5:09
1
PhD long ago ... but not called Throop
– RRL
Aug 6 at 5:09
1
@RRL: Ancient history for me as well! Cheers!
– Robert Lewis
Aug 6 at 5:10
@RRL: But not Throop ancient, though the buildiing was still there in my undergrad sojourn..
– Robert Lewis
Aug 6 at 5:17
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $varphi = nabla times G$ on $mathbbR^3 - 0$ where $G in C^1(mathbbR^3)$, then by the divergence theorem, since $nabla cdot (nabla times G) = 0,$
$$int_partial B_R(0) varphi cdot mathbfn ,dS = int_partial B_R(0) nabla times G cdot mathbfn ,dS = int_B_R(0) nabla cdot (nabla times G), dV = 0$$
where $B_R(0)$ is the closed ball of radius $R$ centered at the origin.
However,
$$int_partial B_R(0) varphi cdot mathbfn ,dS = int_0^2piint_0^pifracRmathbfe_rR^3 cdot mathbfe_rR^2sin theta ,dtheta , dphi = 4pi$$
answered Aug 6 at 5:02
RRL
43.7k42260
I knew there was a simpler way to argue this than mine! Cheers!
– Robert Lewis
Aug 6 at 5:16
@RobertLewis: Although $varphi$ is clearly singular at the origin. So I think you show directly that it can't be the restriction of $nabla times G$ which is continuous everywhere if $G$ is $C^1$. Seems so obvious as to require no proof. Maybe we are missing something.
– RRL
Aug 6 at 5:24
It is pretty obvious. I was thinking of a proof like yours but ended up futzing around with all the junk in my answer--so I just posted it. A method like yours nagged at me the whole time, but I got on that tack and . . .
– Robert Lewis
Aug 6 at 5:29
add a comment |Â
up vote
1
down vote
accepted
If $varphi = nabla times G$ on $mathbbR^3 - 0$ where $G in C^1(mathbbR^3)$, then by the divergence theorem, since $nabla cdot (nabla times G) = 0,$
$$int_partial B_R(0) varphi cdot mathbfn ,dS = int_partial B_R(0) nabla times G cdot mathbfn ,dS = int_B_R(0) nabla cdot (nabla times G), dV = 0$$
where $B_R(0)$ is the closed ball of radius $R$ centered at the origin.
However,
$$int_partial B_R(0) varphi cdot mathbfn ,dS = int_0^2piint_0^pifracRmathbfe_rR^3 cdot mathbfe_rR^2sin theta ,dtheta , dphi = 4pi$$
answered Aug 6 at 5:02
RRL
43.7k42260
I knew there was a simpler way to argue this than mine! Cheers!
– Robert Lewis
Aug 6 at 5:16
@RobertLewis: Although $varphi$ is clearly singular at the origin. So I think you show directly that it can't be the restriction of $nabla times G$ which is continuous everywhere if $G$ is $C^1$. Seems so obvious as to require no proof. Maybe we are missing something.
– RRL
Aug 6 at 5:24
It is pretty obvious. I was thinking of a proof like yours but ended up futzing around with all the junk in my answer--so I just posted it. A method like yours nagged at me the whole time, but I got on that tack and . . .
– Robert Lewis
Aug 6 at 5:29
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $varphi = nabla times G$ on $mathbbR^3 - 0$ where $G in C^1(mathbbR^3)$, then by the divergence theorem, since $nabla cdot (nabla times G) = 0,$
$$int_partial B_R(0) varphi cdot mathbfn ,dS = int_partial B_R(0) nabla times G cdot mathbfn ,dS = int_B_R(0) nabla cdot (nabla times G), dV = 0$$
where $B_R(0)$ is the closed ball of radius $R$ centered at the origin.
However,
$$int_partial B_R(0) varphi cdot mathbfn ,dS = int_0^2piint_0^pifracRmathbfe_rR^3 cdot mathbfe_rR^2sin theta ,dtheta , dphi = 4pi$$
answered Aug 6 at 5:02
RRL
43.7k42260
If $varphi = nabla times G$ on $mathbbR^3 - 0$ where $G in C^1(mathbbR^3)$, then by the divergence theorem, since $nabla cdot (nabla times G) = 0,$
$$int_partial B_R(0) varphi cdot mathbfn ,dS = int_partial B_R(0) nabla times G cdot mathbfn ,dS = int_B_R(0) nabla cdot (nabla times G), dV = 0$$
where $B_R(0)$ is the closed ball of radius $R$ centered at the origin.
However,
$$int_partial B_R(0) varphi cdot mathbfn ,dS = int_0^2piint_0^pifracRmathbfe_rR^3 cdot mathbfe_rR^2sin theta ,dtheta , dphi = 4pi$$
answered Aug 6 at 5:02
RRL
43.7k42260
answered Aug 6 at 5:02
RRL
43.7k42260
answered Aug 6 at 5:02
RRL
43.7k42260
answered Aug 6 at 5:02
RRL
43.7k42260
43.7k42260
I knew there was a simpler way to argue this than mine! Cheers!
– Robert Lewis
Aug 6 at 5:16
@RobertLewis: Although $varphi$ is clearly singular at the origin. So I think you show directly that it can't be the restriction of $nabla times G$ which is continuous everywhere if $G$ is $C^1$. Seems so obvious as to require no proof. Maybe we are missing something.
– RRL
Aug 6 at 5:24
It is pretty obvious. I was thinking of a proof like yours but ended up futzing around with all the junk in my answer--so I just posted it. A method like yours nagged at me the whole time, but I got on that tack and . . .
– Robert Lewis
Aug 6 at 5:29
add a comment |Â
I knew there was a simpler way to argue this than mine! Cheers!
– Robert Lewis
Aug 6 at 5:16
@RobertLewis: Although $varphi$ is clearly singular at the origin. So I think you show directly that it can't be the restriction of $nabla times G$ which is continuous everywhere if $G$ is $C^1$. Seems so obvious as to require no proof. Maybe we are missing something.
– RRL
Aug 6 at 5:24
It is pretty obvious. I was thinking of a proof like yours but ended up futzing around with all the junk in my answer--so I just posted it. A method like yours nagged at me the whole time, but I got on that tack and . . .
– Robert Lewis
Aug 6 at 5:29
I knew there was a simpler way to argue this than mine! Cheers!
– Robert Lewis
Aug 6 at 5:16
I knew there was a simpler way to argue this than mine! Cheers!
– Robert Lewis
Aug 6 at 5:16
@RobertLewis: Although $varphi$ is clearly singular at the origin. So I think you show directly that it can't be the restriction of $nabla times G$ which is continuous everywhere if $G$ is $C^1$. Seems so obvious as to require no proof. Maybe we are missing something.
– RRL
Aug 6 at 5:24
@RobertLewis: Although $varphi$ is clearly singular at the origin. So I think you show directly that it can't be the restriction of $nabla times G$ which is continuous everywhere if $G$ is $C^1$. Seems so obvious as to require no proof. Maybe we are missing something.
– RRL
Aug 6 at 5:24
It is pretty obvious. I was thinking of a proof like yours but ended up futzing around with all the junk in my answer--so I just posted it. A method like yours nagged at me the whole time, but I got on that tack and . . .
– Robert Lewis
Aug 6 at 5:29
It is pretty obvious. I was thinking of a proof like yours but ended up futzing around with all the junk in my answer--so I just posted it. A method like yours nagged at me the whole time, but I got on that tack and . . .
– Robert Lewis
Aug 6 at 5:29
add a comment |Â
up vote
2
down vote
We note that we are asked for a $mathcal C^1$ function
$G:Bbb R^3 to Bbb R^3 tag 1$
such that
$nabla times G = varphi, tag 2$
whilst $varphi$ is only defined on $Bbb R^3 setminus (0, 0, 0) $; indeed, with
$vec r = (x, y, z), tag 3$
we have
$r = Vert vec r Vert = (x^2 + y^2 + z^2)^1/2, tag 4$
whence
$varphi = dfrac(x, y, z)Vert (x, y, z) Vert^3 = r^-3 vec r; tag 5$
it follows that
$Vert varphi Vert^2 = r^-6 vec r cdot vec r = r^-6 r^2 = r^-4, tag 6$
so that
$Vert varphi Vert = r^-2 tag 7$
on $Bbb R^3 setminus (0, 0, 0) $; now suppose a $mathcal C^1$ function $G:Bbb R^3 to Bbb R^3$ exists which satisfies (2); then both $G$ and its derivatives are $mathcal C^0$; restricting $G$ to the closed ball $bar B(0, epsilon)$ of radius $epsilon$ centered at $(0, 0, 0)$, we have, by the compactness of $bar B(0, epsilon)$, that the derivatives of $G$ are bounded on $bar B(0, epsilon)$; but this clearly contradicts (7), which in concert with
$Vert nabla times G Vert = Vert varphi Vert tag 8$
(from (2)) shows the derivatives of $G$ must become arbitrarily large on $bar B(0, epsilon)$; hence no such $G$ exists.
The key here is that, while $varphi$ is only defined on $Bbb R^3 setminus (0, 0, 0) $, $G$ is hypothesized to be defined on all of $Bbb R^3$.
Appendix: Derivation of
$nabla cdot varphi = 0. tag9$
I worked this out whilst thinking about my solution; I though it might be worth including here.
With
$varphi = dfrac(x, y, z)Vert (x, y, z) Vert^3 tag10$
and
$vec r = (x, y, z), tag11$
so that
$r = (x^2 + y^2 + z^2)^1/2, tag12$
we may write
$varphi = r^-3 vec r; tag13$
then, in accord with the vector identity
$nabla cdot f vec X = nabla f cdot X + f nabla cdot X, tag14$
which holds for scalar and vector fields $f$ and $X$, respectively, we have
$nabla cdot varphi = nabla r^-3 cdot vec r + r^-3 nabla cdot vec r; tag15$
now,
$nabla cdot vec r = 3; tag16$
thus,
$nabla cdot varphi = nabla r^-3 cdot vec r + 3 r^-3 ; tag17$
$nabla r^-3 = -3r^-4 nabla r; tag18$
from (11) and (12),
$nabla r = dfrac12(x^2 + y^2 + z^2)^-1/2 (2x, 2y, 2z) = r^-1(x, y, z) = r^-1 vec r; tag19$
$nabla r cdot vec r = r^-1 vec r cdot vec r = r^-1r^2 = r; tag20$
$nabla cdot varphi = -3r^-4 r + 3r^-3 = 0. tag21$
End: Appendix
answered Aug 6 at 4:31
Robert Lewis
37.2k22356
1
Go Beavers .....
– RRL
Aug 6 at 5:08
@RRL oh really? When were you "institutionalized"?
– Robert Lewis
Aug 6 at 5:09
1
PhD long ago ... but not called Throop
– RRL
Aug 6 at 5:09
1
@RRL: Ancient history for me as well! Cheers!
– Robert Lewis
Aug 6 at 5:10
@RRL: But not Throop ancient, though the buildiing was still there in my undergrad sojourn..
– Robert Lewis
Aug 6 at 5:17
add a comment |Â
up vote
2
down vote
We note that we are asked for a $mathcal C^1$ function
$G:Bbb R^3 to Bbb R^3 tag 1$
such that
$nabla times G = varphi, tag 2$
whilst $varphi$ is only defined on $Bbb R^3 setminus (0, 0, 0) $; indeed, with
$vec r = (x, y, z), tag 3$
we have
$r = Vert vec r Vert = (x^2 + y^2 + z^2)^1/2, tag 4$
whence
$varphi = dfrac(x, y, z)Vert (x, y, z) Vert^3 = r^-3 vec r; tag 5$
it follows that
$Vert varphi Vert^2 = r^-6 vec r cdot vec r = r^-6 r^2 = r^-4, tag 6$
so that
$Vert varphi Vert = r^-2 tag 7$
on $Bbb R^3 setminus (0, 0, 0) $; now suppose a $mathcal C^1$ function $G:Bbb R^3 to Bbb R^3$ exists which satisfies (2); then both $G$ and its derivatives are $mathcal C^0$; restricting $G$ to the closed ball $bar B(0, epsilon)$ of radius $epsilon$ centered at $(0, 0, 0)$, we have, by the compactness of $bar B(0, epsilon)$, that the derivatives of $G$ are bounded on $bar B(0, epsilon)$; but this clearly contradicts (7), which in concert with
$Vert nabla times G Vert = Vert varphi Vert tag 8$
(from (2)) shows the derivatives of $G$ must become arbitrarily large on $bar B(0, epsilon)$; hence no such $G$ exists.
The key here is that, while $varphi$ is only defined on $Bbb R^3 setminus (0, 0, 0) $, $G$ is hypothesized to be defined on all of $Bbb R^3$.
Appendix: Derivation of
$nabla cdot varphi = 0. tag9$
I worked this out whilst thinking about my solution; I though it might be worth including here.
With
$varphi = dfrac(x, y, z)Vert (x, y, z) Vert^3 tag10$
and
$vec r = (x, y, z), tag11$
so that
$r = (x^2 + y^2 + z^2)^1/2, tag12$
we may write
$varphi = r^-3 vec r; tag13$
then, in accord with the vector identity
$nabla cdot f vec X = nabla f cdot X + f nabla cdot X, tag14$
which holds for scalar and vector fields $f$ and $X$, respectively, we have
$nabla cdot varphi = nabla r^-3 cdot vec r + r^-3 nabla cdot vec r; tag15$
now,
$nabla cdot vec r = 3; tag16$
thus,
$nabla cdot varphi = nabla r^-3 cdot vec r + 3 r^-3 ; tag17$
$nabla r^-3 = -3r^-4 nabla r; tag18$
from (11) and (12),
$nabla r = dfrac12(x^2 + y^2 + z^2)^-1/2 (2x, 2y, 2z) = r^-1(x, y, z) = r^-1 vec r; tag19$
$nabla r cdot vec r = r^-1 vec r cdot vec r = r^-1r^2 = r; tag20$
$nabla cdot varphi = -3r^-4 r + 3r^-3 = 0. tag21$
End: Appendix
answered Aug 6 at 4:31
Robert Lewis
37.2k22356
1
Go Beavers .....
– RRL
Aug 6 at 5:08
@RRL oh really? When were you "institutionalized"?
– Robert Lewis
Aug 6 at 5:09
1
PhD long ago ... but not called Throop
– RRL
Aug 6 at 5:09
1
@RRL: Ancient history for me as well! Cheers!
– Robert Lewis
Aug 6 at 5:10
@RRL: But not Throop ancient, though the buildiing was still there in my undergrad sojourn..
– Robert Lewis
Aug 6 at 5:17
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We note that we are asked for a $mathcal C^1$ function
$G:Bbb R^3 to Bbb R^3 tag 1$
such that
$nabla times G = varphi, tag 2$
whilst $varphi$ is only defined on $Bbb R^3 setminus (0, 0, 0) $; indeed, with
$vec r = (x, y, z), tag 3$
we have
$r = Vert vec r Vert = (x^2 + y^2 + z^2)^1/2, tag 4$
whence
$varphi = dfrac(x, y, z)Vert (x, y, z) Vert^3 = r^-3 vec r; tag 5$
it follows that
$Vert varphi Vert^2 = r^-6 vec r cdot vec r = r^-6 r^2 = r^-4, tag 6$
so that
$Vert varphi Vert = r^-2 tag 7$
on $Bbb R^3 setminus (0, 0, 0) $; now suppose a $mathcal C^1$ function $G:Bbb R^3 to Bbb R^3$ exists which satisfies (2); then both $G$ and its derivatives are $mathcal C^0$; restricting $G$ to the closed ball $bar B(0, epsilon)$ of radius $epsilon$ centered at $(0, 0, 0)$, we have, by the compactness of $bar B(0, epsilon)$, that the derivatives of $G$ are bounded on $bar B(0, epsilon)$; but this clearly contradicts (7), which in concert with
$Vert nabla times G Vert = Vert varphi Vert tag 8$
(from (2)) shows the derivatives of $G$ must become arbitrarily large on $bar B(0, epsilon)$; hence no such $G$ exists.
The key here is that, while $varphi$ is only defined on $Bbb R^3 setminus (0, 0, 0) $, $G$ is hypothesized to be defined on all of $Bbb R^3$.
Appendix: Derivation of
$nabla cdot varphi = 0. tag9$
I worked this out whilst thinking about my solution; I though it might be worth including here.
With
$varphi = dfrac(x, y, z)Vert (x, y, z) Vert^3 tag10$
and
$vec r = (x, y, z), tag11$
so that
$r = (x^2 + y^2 + z^2)^1/2, tag12$
we may write
$varphi = r^-3 vec r; tag13$
then, in accord with the vector identity
$nabla cdot f vec X = nabla f cdot X + f nabla cdot X, tag14$
which holds for scalar and vector fields $f$ and $X$, respectively, we have
$nabla cdot varphi = nabla r^-3 cdot vec r + r^-3 nabla cdot vec r; tag15$
now,
$nabla cdot vec r = 3; tag16$
thus,
$nabla cdot varphi = nabla r^-3 cdot vec r + 3 r^-3 ; tag17$
$nabla r^-3 = -3r^-4 nabla r; tag18$
from (11) and (12),
$nabla r = dfrac12(x^2 + y^2 + z^2)^-1/2 (2x, 2y, 2z) = r^-1(x, y, z) = r^-1 vec r; tag19$
$nabla r cdot vec r = r^-1 vec r cdot vec r = r^-1r^2 = r; tag20$
$nabla cdot varphi = -3r^-4 r + 3r^-3 = 0. tag21$
End: Appendix
answered Aug 6 at 4:31
Robert Lewis
37.2k22356
We note that we are asked for a $mathcal C^1$ function
$G:Bbb R^3 to Bbb R^3 tag 1$
such that
$nabla times G = varphi, tag 2$
whilst $varphi$ is only defined on $Bbb R^3 setminus (0, 0, 0) $; indeed, with
$vec r = (x, y, z), tag 3$
we have
$r = Vert vec r Vert = (x^2 + y^2 + z^2)^1/2, tag 4$
whence
$varphi = dfrac(x, y, z)Vert (x, y, z) Vert^3 = r^-3 vec r; tag 5$
it follows that
$Vert varphi Vert^2 = r^-6 vec r cdot vec r = r^-6 r^2 = r^-4, tag 6$
so that
$Vert varphi Vert = r^-2 tag 7$
on $Bbb R^3 setminus (0, 0, 0) $; now suppose a $mathcal C^1$ function $G:Bbb R^3 to Bbb R^3$ exists which satisfies (2); then both $G$ and its derivatives are $mathcal C^0$; restricting $G$ to the closed ball $bar B(0, epsilon)$ of radius $epsilon$ centered at $(0, 0, 0)$, we have, by the compactness of $bar B(0, epsilon)$, that the derivatives of $G$ are bounded on $bar B(0, epsilon)$; but this clearly contradicts (7), which in concert with
$Vert nabla times G Vert = Vert varphi Vert tag 8$
(from (2)) shows the derivatives of $G$ must become arbitrarily large on $bar B(0, epsilon)$; hence no such $G$ exists.
The key here is that, while $varphi$ is only defined on $Bbb R^3 setminus (0, 0, 0) $, $G$ is hypothesized to be defined on all of $Bbb R^3$.
Appendix: Derivation of
$nabla cdot varphi = 0. tag9$
I worked this out whilst thinking about my solution; I though it might be worth including here.
With
$varphi = dfrac(x, y, z)Vert (x, y, z) Vert^3 tag10$
and
$vec r = (x, y, z), tag11$
so that
$r = (x^2 + y^2 + z^2)^1/2, tag12$
we may write
$varphi = r^-3 vec r; tag13$
then, in accord with the vector identity
$nabla cdot f vec X = nabla f cdot X + f nabla cdot X, tag14$
which holds for scalar and vector fields $f$ and $X$, respectively, we have
$nabla cdot varphi = nabla r^-3 cdot vec r + r^-3 nabla cdot vec r; tag15$
now,
$nabla cdot vec r = 3; tag16$
thus,
$nabla cdot varphi = nabla r^-3 cdot vec r + 3 r^-3 ; tag17$
$nabla r^-3 = -3r^-4 nabla r; tag18$
from (11) and (12),
$nabla r = dfrac12(x^2 + y^2 + z^2)^-1/2 (2x, 2y, 2z) = r^-1(x, y, z) = r^-1 vec r; tag19$
$nabla r cdot vec r = r^-1 vec r cdot vec r = r^-1r^2 = r; tag20$
$nabla cdot varphi = -3r^-4 r + 3r^-3 = 0. tag21$
End: Appendix
answered Aug 6 at 4:31
Robert Lewis
37.2k22356
edited Aug 6 at 5:02
answered Aug 6 at 4:31
Robert Lewis
37.2k22356
answered Aug 6 at 4:31
Robert Lewis
37.2k22356
answered Aug 6 at 4:31
Robert Lewis
37.2k22356
37.2k22356
1
Go Beavers .....
– RRL
Aug 6 at 5:08
@RRL oh really? When were you "institutionalized"?
– Robert Lewis
Aug 6 at 5:09
1
PhD long ago ... but not called Throop
– RRL
Aug 6 at 5:09
1
@RRL: Ancient history for me as well! Cheers!
– Robert Lewis
Aug 6 at 5:10
@RRL: But not Throop ancient, though the buildiing was still there in my undergrad sojourn..
– Robert Lewis
Aug 6 at 5:17
add a comment |Â
1
Go Beavers .....
– RRL
Aug 6 at 5:08
@RRL oh really? When were you "institutionalized"?
– Robert Lewis
Aug 6 at 5:09
1
PhD long ago ... but not called Throop
– RRL
Aug 6 at 5:09
1
@RRL: Ancient history for me as well! Cheers!
– Robert Lewis
Aug 6 at 5:10
@RRL: But not Throop ancient, though the buildiing was still there in my undergrad sojourn..
– Robert Lewis
Aug 6 at 5:17
1
1
Go Beavers .....
– RRL
Aug 6 at 5:08
Go Beavers .....
– RRL
Aug 6 at 5:08
@RRL oh really? When were you "institutionalized"?
– Robert Lewis
Aug 6 at 5:09
@RRL oh really? When were you "institutionalized"?
– Robert Lewis
Aug 6 at 5:09
1
1
PhD long ago ... but not called Throop
– RRL
Aug 6 at 5:09
PhD long ago ... but not called Throop
– RRL
Aug 6 at 5:09
1
1
@RRL: Ancient history for me as well! Cheers!
– Robert Lewis
Aug 6 at 5:10
@RRL: Ancient history for me as well! Cheers!
– Robert Lewis
Aug 6 at 5:10
@RRL: But not Throop ancient, though the buildiing was still there in my undergrad sojourn..
– Robert Lewis
Aug 6 at 5:17
@RRL: But not Throop ancient, though the buildiing was still there in my undergrad sojourn..
– Robert Lewis
Aug 6 at 5:17
add a comment |Â
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