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How to prove $mathbfM/r$ is a continuously differentiable vector field?
Why does Green's Theorem require partial derivatives to be continuousQuestion for divergence theoremShowing volume and surface integration is unaffected by the singularity at $mathbfr'=mathbfr$Volume integral of graddivergence of a radial vector fieldDivergence theorem with nonorientable surfacesdivergence theorem; triple cross productIs it possible to evaluate this integral for general volume $V$?2D Divergence Theorem (example)What are the gradient, divergence and curl of the three-dimensional delta function?System of ODE's - Why does this vector function connect two constant vectors continuously?Prove a vector calculus identityShowing volume and surface integration is unaffected by the singularity at $mathbfr'=mathbfr$
$begingroup$
I had previously asked a question here regarding the applicability of divergence theorem.
$mathbfM'$ is a continuous vector field in volume $V'$ (which is compact and has a piecewise smooth boundary $S$). Consider the following integral:
$$int_V nabla . left( dfracmathbfMr right) dV$$
with $(r=0) in V$.
The answer says that, since the integrand is infinite at $(r=0)$, if we redefine our integral as:
$$lim limits_delta to 0 int_V-delta nabla . left( dfracmathbfMr right) dV$$
where $delta$ is a volume surrounding the point $(r=0)$
then we can apply $GDT$.
But that is only condition. How can we ensure $dfracmathbfMr$ is a continuously differentiable vector field?
Edit:
Comment : "Is $mathbfM$ continuously differentiable?"
Maybe... But break it down into two cases: $(i)$ $mathbfM$ is continuously differentiable $(ii)$ $mathbfM$ is continuous
Please show whether $dfracmathbfMr$ is a continuously differentiable vector field in the two cases.
functions continuity improper-integrals vector-fields divergence
edited Mar 16 at 9:34
Henning Makholm
242k17308551
asked Mar 16 at 5:00
JoeJoe
315214
$endgroup$
add a comment |
$begingroup$
I had previously asked a question here regarding the applicability of divergence theorem.
$mathbfM'$ is a continuous vector field in volume $V'$ (which is compact and has a piecewise smooth boundary $S$). Consider the following integral:
$$int_V nabla . left( dfracmathbfMr right) dV$$
with $(r=0) in V$.
The answer says that, since the integrand is infinite at $(r=0)$, if we redefine our integral as:
$$lim limits_delta to 0 int_V-delta nabla . left( dfracmathbfMr right) dV$$
where $delta$ is a volume surrounding the point $(r=0)$
then we can apply $GDT$.
But that is only condition. How can we ensure $dfracmathbfMr$ is a continuously differentiable vector field?
Edit:
Comment : "Is $mathbfM$ continuously differentiable?"
Maybe... But break it down into two cases: $(i)$ $mathbfM$ is continuously differentiable $(ii)$ $mathbfM$ is continuous
Please show whether $dfracmathbfMr$ is a continuously differentiable vector field in the two cases.
functions continuity improper-integrals vector-fields divergence
edited Mar 16 at 9:34
Henning Makholm
242k17308551
asked Mar 16 at 5:00
JoeJoe
315214
$endgroup$
$begingroup$
Is $mathbfM$ continuously differentiable?
$endgroup$
– RRL
Mar 16 at 5:56
$begingroup$
@RRL: Please have a look at the edit.
$endgroup$
– Joe
Mar 16 at 6:11
add a comment |
$begingroup$
I had previously asked a question here regarding the applicability of divergence theorem.
$mathbfM'$ is a continuous vector field in volume $V'$ (which is compact and has a piecewise smooth boundary $S$). Consider the following integral:
$$int_V nabla . left( dfracmathbfMr right) dV$$
with $(r=0) in V$.
The answer says that, since the integrand is infinite at $(r=0)$, if we redefine our integral as:
$$lim limits_delta to 0 int_V-delta nabla . left( dfracmathbfMr right) dV$$
where $delta$ is a volume surrounding the point $(r=0)$
then we can apply $GDT$.
But that is only condition. How can we ensure $dfracmathbfMr$ is a continuously differentiable vector field?
Edit:
Comment : "Is $mathbfM$ continuously differentiable?"
Maybe... But break it down into two cases: $(i)$ $mathbfM$ is continuously differentiable $(ii)$ $mathbfM$ is continuous
Please show whether $dfracmathbfMr$ is a continuously differentiable vector field in the two cases.
functions continuity improper-integrals vector-fields divergence
edited Mar 16 at 9:34
Henning Makholm
242k17308551
asked Mar 16 at 5:00
JoeJoe
315214
$endgroup$
I had previously asked a question here regarding the applicability of divergence theorem.
$mathbfM'$ is a continuous vector field in volume $V'$ (which is compact and has a piecewise smooth boundary $S$). Consider the following integral:
$$int_V nabla . left( dfracmathbfMr right) dV$$
with $(r=0) in V$.
The answer says that, since the integrand is infinite at $(r=0)$, if we redefine our integral as:
$$lim limits_delta to 0 int_V-delta nabla . left( dfracmathbfMr right) dV$$
where $delta$ is a volume surrounding the point $(r=0)$
then we can apply $GDT$.
But that is only condition. How can we ensure $dfracmathbfMr$ is a continuously differentiable vector field?
Edit:
Comment : "Is $mathbfM$ continuously differentiable?"
Maybe... But break it down into two cases: $(i)$ $mathbfM$ is continuously differentiable $(ii)$ $mathbfM$ is continuous
Please show whether $dfracmathbfMr$ is a continuously differentiable vector field in the two cases.
functions continuity improper-integrals vector-fields divergence
functions continuity improper-integrals vector-fields divergence
edited Mar 16 at 9:34
Henning Makholm
242k17308551
asked Mar 16 at 5:00
JoeJoe
315214
edited Mar 16 at 9:34
Henning Makholm
242k17308551
asked Mar 16 at 5:00
JoeJoe
315214
edited Mar 16 at 9:34
Henning Makholm
242k17308551
edited Mar 16 at 9:34
Henning Makholm
242k17308551
edited Mar 16 at 9:34
Henning Makholm
242k17308551
242k17308551
asked Mar 16 at 5:00
JoeJoe
315214
asked Mar 16 at 5:00
JoeJoe
315214
asked Mar 16 at 5:00
JoeJoe
315214
315214
$begingroup$
Is $mathbfM$ continuously differentiable?
$endgroup$
– RRL
Mar 16 at 5:56
$begingroup$
@RRL: Please have a look at the edit.
$endgroup$
– Joe
Mar 16 at 6:11
add a comment |
$begingroup$
Is $mathbfM$ continuously differentiable?
$endgroup$
– RRL
Mar 16 at 5:56
$begingroup$
@RRL: Please have a look at the edit.
$endgroup$
– Joe
Mar 16 at 6:11
$begingroup$
Is $mathbfM$ continuously differentiable?
$endgroup$
– RRL
Mar 16 at 5:56
$begingroup$
Is $mathbfM$ continuously differentiable?
$endgroup$
– RRL
Mar 16 at 5:56
$begingroup$
@RRL: Please have a look at the edit.
$endgroup$
– Joe
Mar 16 at 6:11
$begingroup$
@RRL: Please have a look at the edit.
$endgroup$
– Joe
Mar 16 at 6:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have $r = |mathbfx|$ and the Euclidean norm is a continuous function from $mathbbR^3$ into $mathbbR$ since by the reverse triangle inequality
$$|, |mathbfx| - |mathbfx_0|,| leqslant |mathbfx - mathbfx_0|.$$
Hence, $mathbfx mapsto frac1r$ is continuous in any region that excludes the origin. You can show this with an $epsilon - delta$ argument, or just quote the theorem that continuity of $f$ with domain in $mathbbR$ implies continuity of $1/f$ on any set where $f(x) neq 0$.
Since $mathbfM$ is continuous, so too is $mathbfx mapsto fracmathbfM(mathbfx)r$, since with $r_0 = mathbfx_0$ we have
$$left|fracmathbfM(mathbfx)r - fracmathbfM(mathbfx_0)r_0right| leqslant frac1r|mathbfM(mathbfx)-mathbfM(mathbfx_0)|+ |mathbfM(mathbfx_0)|left|frac1r - frac1r_0 right|,$$
and $mathbfM(mathbfx_0)$ and $1/r$ are bounded in any compact neighborhood of $mathbfx_0$.
For continuous differentiability you need $mathbfM$ to be continuously differentiable unless some removable discontinuity arises. It remains to show that $mathbfx mapsto frac1r$ is continuously differentiable which amounts to showing that the partial derivatives are continuous by an argument similar to that given above.
answered Mar 16 at 6:56
RRLRRL
53k42573
$endgroup$
$begingroup$
Thanks for the answer, sir.... However what bothers me is that by just stating electric field $mathbfE$ is continuous (rather than continuously differentiable), the divergence theorem is applied to vector field $mathbfE$. How is it justifible?
$endgroup$
– Joe
Mar 16 at 15:19
$begingroup$
@Joe: You're welcome. I don't know where you are seeing this. Perhaps "they" are not stating the hypotheses carefully, which is not uncommon in a physics setting. It is usually safe to assume that a physical entity like the electric field is specified by a function sufficiently smooth for the divergence theorem to apply. Furthermore it is sufficient but not necessary that the vector field be continuously differentiable for the theorem to apply. See, for example, here and here.
$endgroup$
– RRL
Mar 16 at 21:39
$begingroup$
In your link it is said "Continuity of the partial derivatives is strong sufficient condition". Almost all of the physics textbook (which gives a brief introduction to mathematical physics) like Griffiths (page 31) just writes down the equation without a statement. Other elementary mathematical physics textbooks uses the "strong sufficient condition" discussed above. Late 19th century physics textbooks do make the statement about $GDT$ (along with elementary proofs) but doesn't mention anything about "continuity of partial derivatives".
$endgroup$
– Joe
Mar 17 at 6:06
$begingroup$
Please see here as well as here. Are these statements sufficient in order to apply $GDT$ for electric field $mathbfE$ and $fracmathbfMr$?
$endgroup$
– Joe
Mar 17 at 6:06
$begingroup$
@Joe: I would not expect 19th century books and papers to specify these conditions carefully. They are about the (important) physics content and assume tacitly that there is enough regularity for the theorem to hold. To be honest I'm not entirely sure what the question is anymore.
$endgroup$
– RRL
Mar 18 at 0:13
|
show 3 more comments
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1 Answer
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1 Answer
1
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oldest
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active
oldest
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active
oldest
votes
$begingroup$
We have $r = |mathbfx|$ and the Euclidean norm is a continuous function from $mathbbR^3$ into $mathbbR$ since by the reverse triangle inequality
$$|, |mathbfx| - |mathbfx_0|,| leqslant |mathbfx - mathbfx_0|.$$
Hence, $mathbfx mapsto frac1r$ is continuous in any region that excludes the origin. You can show this with an $epsilon - delta$ argument, or just quote the theorem that continuity of $f$ with domain in $mathbbR$ implies continuity of $1/f$ on any set where $f(x) neq 0$.
Since $mathbfM$ is continuous, so too is $mathbfx mapsto fracmathbfM(mathbfx)r$, since with $r_0 = mathbfx_0$ we have
$$left|fracmathbfM(mathbfx)r - fracmathbfM(mathbfx_0)r_0right| leqslant frac1r|mathbfM(mathbfx)-mathbfM(mathbfx_0)|+ |mathbfM(mathbfx_0)|left|frac1r - frac1r_0 right|,$$
and $mathbfM(mathbfx_0)$ and $1/r$ are bounded in any compact neighborhood of $mathbfx_0$.
For continuous differentiability you need $mathbfM$ to be continuously differentiable unless some removable discontinuity arises. It remains to show that $mathbfx mapsto frac1r$ is continuously differentiable which amounts to showing that the partial derivatives are continuous by an argument similar to that given above.
answered Mar 16 at 6:56
RRLRRL
53k42573
$endgroup$
$begingroup$
Thanks for the answer, sir.... However what bothers me is that by just stating electric field $mathbfE$ is continuous (rather than continuously differentiable), the divergence theorem is applied to vector field $mathbfE$. How is it justifible?
$endgroup$
– Joe
Mar 16 at 15:19
$begingroup$
@Joe: You're welcome. I don't know where you are seeing this. Perhaps "they" are not stating the hypotheses carefully, which is not uncommon in a physics setting. It is usually safe to assume that a physical entity like the electric field is specified by a function sufficiently smooth for the divergence theorem to apply. Furthermore it is sufficient but not necessary that the vector field be continuously differentiable for the theorem to apply. See, for example, here and here.
$endgroup$
– RRL
Mar 16 at 21:39
$begingroup$
In your link it is said "Continuity of the partial derivatives is strong sufficient condition". Almost all of the physics textbook (which gives a brief introduction to mathematical physics) like Griffiths (page 31) just writes down the equation without a statement. Other elementary mathematical physics textbooks uses the "strong sufficient condition" discussed above. Late 19th century physics textbooks do make the statement about $GDT$ (along with elementary proofs) but doesn't mention anything about "continuity of partial derivatives".
$endgroup$
– Joe
Mar 17 at 6:06
$begingroup$
Please see here as well as here. Are these statements sufficient in order to apply $GDT$ for electric field $mathbfE$ and $fracmathbfMr$?
$endgroup$
– Joe
Mar 17 at 6:06
$begingroup$
@Joe: I would not expect 19th century books and papers to specify these conditions carefully. They are about the (important) physics content and assume tacitly that there is enough regularity for the theorem to hold. To be honest I'm not entirely sure what the question is anymore.
$endgroup$
– RRL
Mar 18 at 0:13
|
show 3 more comments
$begingroup$
We have $r = |mathbfx|$ and the Euclidean norm is a continuous function from $mathbbR^3$ into $mathbbR$ since by the reverse triangle inequality
$$|, |mathbfx| - |mathbfx_0|,| leqslant |mathbfx - mathbfx_0|.$$
Hence, $mathbfx mapsto frac1r$ is continuous in any region that excludes the origin. You can show this with an $epsilon - delta$ argument, or just quote the theorem that continuity of $f$ with domain in $mathbbR$ implies continuity of $1/f$ on any set where $f(x) neq 0$.
Since $mathbfM$ is continuous, so too is $mathbfx mapsto fracmathbfM(mathbfx)r$, since with $r_0 = mathbfx_0$ we have
$$left|fracmathbfM(mathbfx)r - fracmathbfM(mathbfx_0)r_0right| leqslant frac1r|mathbfM(mathbfx)-mathbfM(mathbfx_0)|+ |mathbfM(mathbfx_0)|left|frac1r - frac1r_0 right|,$$
and $mathbfM(mathbfx_0)$ and $1/r$ are bounded in any compact neighborhood of $mathbfx_0$.
For continuous differentiability you need $mathbfM$ to be continuously differentiable unless some removable discontinuity arises. It remains to show that $mathbfx mapsto frac1r$ is continuously differentiable which amounts to showing that the partial derivatives are continuous by an argument similar to that given above.
answered Mar 16 at 6:56
RRLRRL
53k42573
$endgroup$
$begingroup$
Thanks for the answer, sir.... However what bothers me is that by just stating electric field $mathbfE$ is continuous (rather than continuously differentiable), the divergence theorem is applied to vector field $mathbfE$. How is it justifible?
$endgroup$
– Joe
Mar 16 at 15:19
$begingroup$
@Joe: You're welcome. I don't know where you are seeing this. Perhaps "they" are not stating the hypotheses carefully, which is not uncommon in a physics setting. It is usually safe to assume that a physical entity like the electric field is specified by a function sufficiently smooth for the divergence theorem to apply. Furthermore it is sufficient but not necessary that the vector field be continuously differentiable for the theorem to apply. See, for example, here and here.
$endgroup$
– RRL
Mar 16 at 21:39
$begingroup$
In your link it is said "Continuity of the partial derivatives is strong sufficient condition". Almost all of the physics textbook (which gives a brief introduction to mathematical physics) like Griffiths (page 31) just writes down the equation without a statement. Other elementary mathematical physics textbooks uses the "strong sufficient condition" discussed above. Late 19th century physics textbooks do make the statement about $GDT$ (along with elementary proofs) but doesn't mention anything about "continuity of partial derivatives".
$endgroup$
– Joe
Mar 17 at 6:06
$begingroup$
Please see here as well as here. Are these statements sufficient in order to apply $GDT$ for electric field $mathbfE$ and $fracmathbfMr$?
$endgroup$
– Joe
Mar 17 at 6:06
$begingroup$
@Joe: I would not expect 19th century books and papers to specify these conditions carefully. They are about the (important) physics content and assume tacitly that there is enough regularity for the theorem to hold. To be honest I'm not entirely sure what the question is anymore.
$endgroup$
– RRL
Mar 18 at 0:13
|
show 3 more comments
$begingroup$
We have $r = |mathbfx|$ and the Euclidean norm is a continuous function from $mathbbR^3$ into $mathbbR$ since by the reverse triangle inequality
$$|, |mathbfx| - |mathbfx_0|,| leqslant |mathbfx - mathbfx_0|.$$
Hence, $mathbfx mapsto frac1r$ is continuous in any region that excludes the origin. You can show this with an $epsilon - delta$ argument, or just quote the theorem that continuity of $f$ with domain in $mathbbR$ implies continuity of $1/f$ on any set where $f(x) neq 0$.
Since $mathbfM$ is continuous, so too is $mathbfx mapsto fracmathbfM(mathbfx)r$, since with $r_0 = mathbfx_0$ we have
$$left|fracmathbfM(mathbfx)r - fracmathbfM(mathbfx_0)r_0right| leqslant frac1r|mathbfM(mathbfx)-mathbfM(mathbfx_0)|+ |mathbfM(mathbfx_0)|left|frac1r - frac1r_0 right|,$$
and $mathbfM(mathbfx_0)$ and $1/r$ are bounded in any compact neighborhood of $mathbfx_0$.
For continuous differentiability you need $mathbfM$ to be continuously differentiable unless some removable discontinuity arises. It remains to show that $mathbfx mapsto frac1r$ is continuously differentiable which amounts to showing that the partial derivatives are continuous by an argument similar to that given above.
answered Mar 16 at 6:56
RRLRRL
53k42573
$endgroup$
We have $r = |mathbfx|$ and the Euclidean norm is a continuous function from $mathbbR^3$ into $mathbbR$ since by the reverse triangle inequality
$$|, |mathbfx| - |mathbfx_0|,| leqslant |mathbfx - mathbfx_0|.$$
Hence, $mathbfx mapsto frac1r$ is continuous in any region that excludes the origin. You can show this with an $epsilon - delta$ argument, or just quote the theorem that continuity of $f$ with domain in $mathbbR$ implies continuity of $1/f$ on any set where $f(x) neq 0$.
Since $mathbfM$ is continuous, so too is $mathbfx mapsto fracmathbfM(mathbfx)r$, since with $r_0 = mathbfx_0$ we have
$$left|fracmathbfM(mathbfx)r - fracmathbfM(mathbfx_0)r_0right| leqslant frac1r|mathbfM(mathbfx)-mathbfM(mathbfx_0)|+ |mathbfM(mathbfx_0)|left|frac1r - frac1r_0 right|,$$
and $mathbfM(mathbfx_0)$ and $1/r$ are bounded in any compact neighborhood of $mathbfx_0$.
For continuous differentiability you need $mathbfM$ to be continuously differentiable unless some removable discontinuity arises. It remains to show that $mathbfx mapsto frac1r$ is continuously differentiable which amounts to showing that the partial derivatives are continuous by an argument similar to that given above.
answered Mar 16 at 6:56
RRLRRL
53k42573
edited Mar 16 at 7:01
answered Mar 16 at 6:56
RRLRRL
53k42573
answered Mar 16 at 6:56
RRLRRL
53k42573
answered Mar 16 at 6:56
RRLRRL
53k42573
53k42573
$begingroup$
Thanks for the answer, sir.... However what bothers me is that by just stating electric field $mathbfE$ is continuous (rather than continuously differentiable), the divergence theorem is applied to vector field $mathbfE$. How is it justifible?
$endgroup$
– Joe
Mar 16 at 15:19
$begingroup$
@Joe: You're welcome. I don't know where you are seeing this. Perhaps "they" are not stating the hypotheses carefully, which is not uncommon in a physics setting. It is usually safe to assume that a physical entity like the electric field is specified by a function sufficiently smooth for the divergence theorem to apply. Furthermore it is sufficient but not necessary that the vector field be continuously differentiable for the theorem to apply. See, for example, here and here.
$endgroup$
– RRL
Mar 16 at 21:39
$begingroup$
In your link it is said "Continuity of the partial derivatives is strong sufficient condition". Almost all of the physics textbook (which gives a brief introduction to mathematical physics) like Griffiths (page 31) just writes down the equation without a statement. Other elementary mathematical physics textbooks uses the "strong sufficient condition" discussed above. Late 19th century physics textbooks do make the statement about $GDT$ (along with elementary proofs) but doesn't mention anything about "continuity of partial derivatives".
$endgroup$
– Joe
Mar 17 at 6:06
$begingroup$
Please see here as well as here. Are these statements sufficient in order to apply $GDT$ for electric field $mathbfE$ and $fracmathbfMr$?
$endgroup$
– Joe
Mar 17 at 6:06
$begingroup$
@Joe: I would not expect 19th century books and papers to specify these conditions carefully. They are about the (important) physics content and assume tacitly that there is enough regularity for the theorem to hold. To be honest I'm not entirely sure what the question is anymore.
$endgroup$
– RRL
Mar 18 at 0:13
|
show 3 more comments
$begingroup$
Thanks for the answer, sir.... However what bothers me is that by just stating electric field $mathbfE$ is continuous (rather than continuously differentiable), the divergence theorem is applied to vector field $mathbfE$. How is it justifible?
$endgroup$
– Joe
Mar 16 at 15:19
$begingroup$
@Joe: You're welcome. I don't know where you are seeing this. Perhaps "they" are not stating the hypotheses carefully, which is not uncommon in a physics setting. It is usually safe to assume that a physical entity like the electric field is specified by a function sufficiently smooth for the divergence theorem to apply. Furthermore it is sufficient but not necessary that the vector field be continuously differentiable for the theorem to apply. See, for example, here and here.
$endgroup$
– RRL
Mar 16 at 21:39
$begingroup$
In your link it is said "Continuity of the partial derivatives is strong sufficient condition". Almost all of the physics textbook (which gives a brief introduction to mathematical physics) like Griffiths (page 31) just writes down the equation without a statement. Other elementary mathematical physics textbooks uses the "strong sufficient condition" discussed above. Late 19th century physics textbooks do make the statement about $GDT$ (along with elementary proofs) but doesn't mention anything about "continuity of partial derivatives".
$endgroup$
– Joe
Mar 17 at 6:06
$begingroup$
Please see here as well as here. Are these statements sufficient in order to apply $GDT$ for electric field $mathbfE$ and $fracmathbfMr$?
$endgroup$
– Joe
Mar 17 at 6:06
$begingroup$
@Joe: I would not expect 19th century books and papers to specify these conditions carefully. They are about the (important) physics content and assume tacitly that there is enough regularity for the theorem to hold. To be honest I'm not entirely sure what the question is anymore.
$endgroup$
– RRL
Mar 18 at 0:13
$begingroup$
Thanks for the answer, sir.... However what bothers me is that by just stating electric field $mathbfE$ is continuous (rather than continuously differentiable), the divergence theorem is applied to vector field $mathbfE$. How is it justifible?
$endgroup$
– Joe
Mar 16 at 15:19
$begingroup$
Thanks for the answer, sir.... However what bothers me is that by just stating electric field $mathbfE$ is continuous (rather than continuously differentiable), the divergence theorem is applied to vector field $mathbfE$. How is it justifible?
$endgroup$
– Joe
Mar 16 at 15:19
$begingroup$
@Joe: You're welcome. I don't know where you are seeing this. Perhaps "they" are not stating the hypotheses carefully, which is not uncommon in a physics setting. It is usually safe to assume that a physical entity like the electric field is specified by a function sufficiently smooth for the divergence theorem to apply. Furthermore it is sufficient but not necessary that the vector field be continuously differentiable for the theorem to apply. See, for example, here and here.
$endgroup$
– RRL
Mar 16 at 21:39
$begingroup$
@Joe: You're welcome. I don't know where you are seeing this. Perhaps "they" are not stating the hypotheses carefully, which is not uncommon in a physics setting. It is usually safe to assume that a physical entity like the electric field is specified by a function sufficiently smooth for the divergence theorem to apply. Furthermore it is sufficient but not necessary that the vector field be continuously differentiable for the theorem to apply. See, for example, here and here.
$endgroup$
– RRL
Mar 16 at 21:39
$begingroup$
In your link it is said "Continuity of the partial derivatives is strong sufficient condition". Almost all of the physics textbook (which gives a brief introduction to mathematical physics) like Griffiths (page 31) just writes down the equation without a statement. Other elementary mathematical physics textbooks uses the "strong sufficient condition" discussed above. Late 19th century physics textbooks do make the statement about $GDT$ (along with elementary proofs) but doesn't mention anything about "continuity of partial derivatives".
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– Joe
Mar 17 at 6:06
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In your link it is said "Continuity of the partial derivatives is strong sufficient condition". Almost all of the physics textbook (which gives a brief introduction to mathematical physics) like Griffiths (page 31) just writes down the equation without a statement. Other elementary mathematical physics textbooks uses the "strong sufficient condition" discussed above. Late 19th century physics textbooks do make the statement about $GDT$ (along with elementary proofs) but doesn't mention anything about "continuity of partial derivatives".
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– Joe
Mar 17 at 6:06
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Please see here as well as here. Are these statements sufficient in order to apply $GDT$ for electric field $mathbfE$ and $fracmathbfMr$?
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– Joe
Mar 17 at 6:06
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Please see here as well as here. Are these statements sufficient in order to apply $GDT$ for electric field $mathbfE$ and $fracmathbfMr$?
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– Joe
Mar 17 at 6:06
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@Joe: I would not expect 19th century books and papers to specify these conditions carefully. They are about the (important) physics content and assume tacitly that there is enough regularity for the theorem to hold. To be honest I'm not entirely sure what the question is anymore.
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– RRL
Mar 18 at 0:13
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@Joe: I would not expect 19th century books and papers to specify these conditions carefully. They are about the (important) physics content and assume tacitly that there is enough regularity for the theorem to hold. To be honest I'm not entirely sure what the question is anymore.
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– RRL
Mar 18 at 0:13
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Is $mathbfM$ continuously differentiable?
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– RRL
Mar 16 at 5:56
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@RRL: Please have a look at the edit.
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– Joe
Mar 16 at 6:11