Mixing
Mixed operation of divergence and gradient
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When I calculus the term $ucdot nabla u$, where $u$ is a vector, I get two different results by calculus in different ways i.e.
$((ucdot nabla) u)_i=sumlimits_k=1^n(u_knabla_k)u_i$ and $(ucdot (nabla u))_i=sumlimits_k=1^nu_k(nabla u)_ki=sumlimits_k=1^nu_k(nabla_iu_k)$.
Which one is right?
differential-equations multivariable-calculus vector-analysis
asked Jul 15 at 8:51
UAne
102
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up vote
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When I calculus the term $ucdot nabla u$, where $u$ is a vector, I get two different results by calculus in different ways i.e.
$((ucdot nabla) u)_i=sumlimits_k=1^n(u_knabla_k)u_i$ and $(ucdot (nabla u))_i=sumlimits_k=1^nu_k(nabla u)_ki=sumlimits_k=1^nu_k(nabla_iu_k)$.
Which one is right?
differential-equations multivariable-calculus vector-analysis
asked Jul 15 at 8:51
UAne
102
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
When I calculus the term $ucdot nabla u$, where $u$ is a vector, I get two different results by calculus in different ways i.e.
$((ucdot nabla) u)_i=sumlimits_k=1^n(u_knabla_k)u_i$ and $(ucdot (nabla u))_i=sumlimits_k=1^nu_k(nabla u)_ki=sumlimits_k=1^nu_k(nabla_iu_k)$.
Which one is right?
differential-equations multivariable-calculus vector-analysis
asked Jul 15 at 8:51
UAne
102
When I calculus the term $ucdot nabla u$, where $u$ is a vector, I get two different results by calculus in different ways i.e.
$((ucdot nabla) u)_i=sumlimits_k=1^n(u_knabla_k)u_i$ and $(ucdot (nabla u))_i=sumlimits_k=1^nu_k(nabla u)_ki=sumlimits_k=1^nu_k(nabla_iu_k)$.
Which one is right?
differential-equations multivariable-calculus vector-analysis
asked Jul 15 at 8:51
UAne
102
asked Jul 15 at 8:51
UAne
102
asked Jul 15 at 8:51
UAne
102
asked Jul 15 at 8:51
UAne
102
102
add a comment |Â
add a comment |Â
1 Answer
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up vote
1
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That's for you to decide in defining the notation you use, or to clarify by using parentheses.
I personally would tend to take $ucdotnabla u$ to mean $(ucdotnabla)u$, because I studied physics and that construction is more common there than the other one. Also, a dot is much more usual for the scalar product of two vectors (the first interpretation) than for the product of a vector and a matrix (the second interpretation). But still it's somewhat ambiguous, and if you can't expect your readers to know which one you mean, parentheses seem advisable.
Anyway, $nabla u$ is a bad (though common) notation for the gradient of a vector field, since it doesn't fit with the usual rules for matrices and vectors. The “right†way to write it would be $(nabla u^top)^top$, and then there would be no confusion, since $(ucdotnabla)uneq ucdot(nabla u^top)^top$ and the only consistent way to interptet $ucdotnabla u$ would be $(ucdotnabla)u$.
answered Jul 15 at 9:01
joriki
165k10180328
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
That's for you to decide in defining the notation you use, or to clarify by using parentheses.
I personally would tend to take $ucdotnabla u$ to mean $(ucdotnabla)u$, because I studied physics and that construction is more common there than the other one. Also, a dot is much more usual for the scalar product of two vectors (the first interpretation) than for the product of a vector and a matrix (the second interpretation). But still it's somewhat ambiguous, and if you can't expect your readers to know which one you mean, parentheses seem advisable.
Anyway, $nabla u$ is a bad (though common) notation for the gradient of a vector field, since it doesn't fit with the usual rules for matrices and vectors. The “right†way to write it would be $(nabla u^top)^top$, and then there would be no confusion, since $(ucdotnabla)uneq ucdot(nabla u^top)^top$ and the only consistent way to interptet $ucdotnabla u$ would be $(ucdotnabla)u$.
answered Jul 15 at 9:01
joriki
165k10180328
add a comment |Â
up vote
1
down vote
accepted
That's for you to decide in defining the notation you use, or to clarify by using parentheses.
I personally would tend to take $ucdotnabla u$ to mean $(ucdotnabla)u$, because I studied physics and that construction is more common there than the other one. Also, a dot is much more usual for the scalar product of two vectors (the first interpretation) than for the product of a vector and a matrix (the second interpretation). But still it's somewhat ambiguous, and if you can't expect your readers to know which one you mean, parentheses seem advisable.
Anyway, $nabla u$ is a bad (though common) notation for the gradient of a vector field, since it doesn't fit with the usual rules for matrices and vectors. The “right†way to write it would be $(nabla u^top)^top$, and then there would be no confusion, since $(ucdotnabla)uneq ucdot(nabla u^top)^top$ and the only consistent way to interptet $ucdotnabla u$ would be $(ucdotnabla)u$.
answered Jul 15 at 9:01
joriki
165k10180328
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
That's for you to decide in defining the notation you use, or to clarify by using parentheses.
I personally would tend to take $ucdotnabla u$ to mean $(ucdotnabla)u$, because I studied physics and that construction is more common there than the other one. Also, a dot is much more usual for the scalar product of two vectors (the first interpretation) than for the product of a vector and a matrix (the second interpretation). But still it's somewhat ambiguous, and if you can't expect your readers to know which one you mean, parentheses seem advisable.
Anyway, $nabla u$ is a bad (though common) notation for the gradient of a vector field, since it doesn't fit with the usual rules for matrices and vectors. The “right†way to write it would be $(nabla u^top)^top$, and then there would be no confusion, since $(ucdotnabla)uneq ucdot(nabla u^top)^top$ and the only consistent way to interptet $ucdotnabla u$ would be $(ucdotnabla)u$.
answered Jul 15 at 9:01
joriki
165k10180328
That's for you to decide in defining the notation you use, or to clarify by using parentheses.
I personally would tend to take $ucdotnabla u$ to mean $(ucdotnabla)u$, because I studied physics and that construction is more common there than the other one. Also, a dot is much more usual for the scalar product of two vectors (the first interpretation) than for the product of a vector and a matrix (the second interpretation). But still it's somewhat ambiguous, and if you can't expect your readers to know which one you mean, parentheses seem advisable.
Anyway, $nabla u$ is a bad (though common) notation for the gradient of a vector field, since it doesn't fit with the usual rules for matrices and vectors. The “right†way to write it would be $(nabla u^top)^top$, and then there would be no confusion, since $(ucdotnabla)uneq ucdot(nabla u^top)^top$ and the only consistent way to interptet $ucdotnabla u$ would be $(ucdotnabla)u$.
answered Jul 15 at 9:01
joriki
165k10180328
edited Jul 15 at 9:17
answered Jul 15 at 9:01
joriki
165k10180328
answered Jul 15 at 9:01
joriki
165k10180328
answered Jul 15 at 9:01
joriki
165k10180328
165k10180328
add a comment |Â
add a comment |Â
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