Mixing
Is it true that $(Deltavarphi)phi=nabla cdot(nablaphi)varphi-nablaphicdotnablavarphi$ for complex-valued...
$begingroup$
I know from vector calculus that for a complex-valued functions $phi$ and $varphi$ say, the following identity hold: $$phiDeltavarphi=nabla cdot(varphinabla phi)-nabla phi cdot nabla varphi.$$
This is an obvious identity from
$$nabla cdot(varphinabla phi)=phiDeltavarphi+nabla phi cdot nabla varphi$$
Now, what about if I want simplify $(Delta varphi)phi$, would it be something like
$$(Deltavarphi) phi=nabla cdot(nabla phi)varphi-nabla phi cdot nabla varphi quad ?$$
If yes, why?
It might be easy to figure out, but it is not apparent to me.
vector-analysis
edited Feb 2 at 12:09
Blue
49.6k870158
asked Feb 2 at 12:04
M.D.M.D.
599
$endgroup$
|
show 5 more comments
$begingroup$
I know from vector calculus that for a complex-valued functions $phi$ and $varphi$ say, the following identity hold: $$phiDeltavarphi=nabla cdot(varphinabla phi)-nabla phi cdot nabla varphi.$$
This is an obvious identity from
$$nabla cdot(varphinabla phi)=phiDeltavarphi+nabla phi cdot nabla varphi$$
Now, what about if I want simplify $(Delta varphi)phi$, would it be something like
$$(Deltavarphi) phi=nabla cdot(nabla phi)varphi-nabla phi cdot nabla varphi quad ?$$
If yes, why?
It might be easy to figure out, but it is not apparent to me.
vector-analysis
edited Feb 2 at 12:09
Blue
49.6k870158
asked Feb 2 at 12:04
M.D.M.D.
599
$endgroup$
$begingroup$
isn't multiplication of complex valued functions commutative? Or what am I missing?
$endgroup$
– Calvin Khor
Feb 2 at 12:07
$begingroup$
What is $Delta varphi $ for a complex function ?
$endgroup$
– idm
Feb 2 at 12:16
$begingroup$
@Calvin. I guess so. But the product in the left-handside $phiDeltavarphi $ is not in general scalar-valued in $mathbb{R}^d$, right? Oh! Do you mean since the laplacian is scalar so the product in the left-handside is commutative and the same is extended to the right-handside?
$endgroup$
– M.D.
Feb 2 at 12:16
$begingroup$
@idm. $Delta varphi$ must be complex-valued too.
$endgroup$
– M.D.
Feb 2 at 12:18
1
$begingroup$
Oh, great. Thanks a lot guys @CalvinKhor and idm.
$endgroup$
– M.D.
Feb 2 at 12:43
|
show 5 more comments
$begingroup$
I know from vector calculus that for a complex-valued functions $phi$ and $varphi$ say, the following identity hold: $$phiDeltavarphi=nabla cdot(varphinabla phi)-nabla phi cdot nabla varphi.$$
This is an obvious identity from
$$nabla cdot(varphinabla phi)=phiDeltavarphi+nabla phi cdot nabla varphi$$
Now, what about if I want simplify $(Delta varphi)phi$, would it be something like
$$(Deltavarphi) phi=nabla cdot(nabla phi)varphi-nabla phi cdot nabla varphi quad ?$$
If yes, why?
It might be easy to figure out, but it is not apparent to me.
vector-analysis
edited Feb 2 at 12:09
Blue
49.6k870158
asked Feb 2 at 12:04
M.D.M.D.
599
$endgroup$
I know from vector calculus that for a complex-valued functions $phi$ and $varphi$ say, the following identity hold: $$phiDeltavarphi=nabla cdot(varphinabla phi)-nabla phi cdot nabla varphi.$$
This is an obvious identity from
$$nabla cdot(varphinabla phi)=phiDeltavarphi+nabla phi cdot nabla varphi$$
Now, what about if I want simplify $(Delta varphi)phi$, would it be something like
$$(Deltavarphi) phi=nabla cdot(nabla phi)varphi-nabla phi cdot nabla varphi quad ?$$
If yes, why?
It might be easy to figure out, but it is not apparent to me.
vector-analysis
vector-analysis
edited Feb 2 at 12:09
Blue
49.6k870158
asked Feb 2 at 12:04
M.D.M.D.
599
edited Feb 2 at 12:09
Blue
49.6k870158
asked Feb 2 at 12:04
M.D.M.D.
599
edited Feb 2 at 12:09
Blue
49.6k870158
edited Feb 2 at 12:09
Blue
49.6k870158
edited Feb 2 at 12:09
Blue
49.6k870158
49.6k870158
asked Feb 2 at 12:04
M.D.M.D.
599
asked Feb 2 at 12:04
M.D.M.D.
599
asked Feb 2 at 12:04
M.D.M.D.
599
599
$begingroup$
isn't multiplication of complex valued functions commutative? Or what am I missing?
$endgroup$
– Calvin Khor
Feb 2 at 12:07
$begingroup$
What is $Delta varphi $ for a complex function ?
$endgroup$
– idm
Feb 2 at 12:16
$begingroup$
@Calvin. I guess so. But the product in the left-handside $phiDeltavarphi $ is not in general scalar-valued in $mathbb{R}^d$, right? Oh! Do you mean since the laplacian is scalar so the product in the left-handside is commutative and the same is extended to the right-handside?
$endgroup$
– M.D.
Feb 2 at 12:16
$begingroup$
@idm. $Delta varphi$ must be complex-valued too.
$endgroup$
– M.D.
Feb 2 at 12:18
1
$begingroup$
Oh, great. Thanks a lot guys @CalvinKhor and idm.
$endgroup$
– M.D.
Feb 2 at 12:43
|
show 5 more comments
$begingroup$
isn't multiplication of complex valued functions commutative? Or what am I missing?
$endgroup$
– Calvin Khor
Feb 2 at 12:07
$begingroup$
What is $Delta varphi $ for a complex function ?
$endgroup$
– idm
Feb 2 at 12:16
$begingroup$
@Calvin. I guess so. But the product in the left-handside $phiDeltavarphi $ is not in general scalar-valued in $mathbb{R}^d$, right? Oh! Do you mean since the laplacian is scalar so the product in the left-handside is commutative and the same is extended to the right-handside?
$endgroup$
– M.D.
Feb 2 at 12:16
$begingroup$
@idm. $Delta varphi$ must be complex-valued too.
$endgroup$
– M.D.
Feb 2 at 12:18
1
$begingroup$
Oh, great. Thanks a lot guys @CalvinKhor and idm.
$endgroup$
– M.D.
Feb 2 at 12:43
$begingroup$
isn't multiplication of complex valued functions commutative? Or what am I missing?
$endgroup$
– Calvin Khor
Feb 2 at 12:07
$begingroup$
isn't multiplication of complex valued functions commutative? Or what am I missing?
$endgroup$
– Calvin Khor
Feb 2 at 12:07
$begingroup$
What is $Delta varphi $ for a complex function ?
$endgroup$
– idm
Feb 2 at 12:16
$begingroup$
What is $Delta varphi $ for a complex function ?
$endgroup$
– idm
Feb 2 at 12:16
$begingroup$
@Calvin. I guess so. But the product in the left-handside $phiDeltavarphi $ is not in general scalar-valued in $mathbb{R}^d$, right? Oh! Do you mean since the laplacian is scalar so the product in the left-handside is commutative and the same is extended to the right-handside?
$endgroup$
– M.D.
Feb 2 at 12:16
$begingroup$
@Calvin. I guess so. But the product in the left-handside $phiDeltavarphi $ is not in general scalar-valued in $mathbb{R}^d$, right? Oh! Do you mean since the laplacian is scalar so the product in the left-handside is commutative and the same is extended to the right-handside?
$endgroup$
– M.D.
Feb 2 at 12:16
$begingroup$
@idm. $Delta varphi$ must be complex-valued too.
$endgroup$
– M.D.
Feb 2 at 12:18
$begingroup$
@idm. $Delta varphi$ must be complex-valued too.
$endgroup$
– M.D.
Feb 2 at 12:18
1
1
$begingroup$
Oh, great. Thanks a lot guys @CalvinKhor and idm.
$endgroup$
– M.D.
Feb 2 at 12:43
$begingroup$
Oh, great. Thanks a lot guys @CalvinKhor and idm.
$endgroup$
– M.D.
Feb 2 at 12:43
|
show 5 more comments
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$begingroup$
isn't multiplication of complex valued functions commutative? Or what am I missing?
$endgroup$
– Calvin Khor
Feb 2 at 12:07
$begingroup$
What is $Delta varphi $ for a complex function ?
$endgroup$
– idm
Feb 2 at 12:16
$begingroup$
@Calvin. I guess so. But the product in the left-handside $phiDeltavarphi $ is not in general scalar-valued in $mathbb{R}^d$, right? Oh! Do you mean since the laplacian is scalar so the product in the left-handside is commutative and the same is extended to the right-handside?
$endgroup$
– M.D.
Feb 2 at 12:16
$begingroup$
@idm. $Delta varphi$ must be complex-valued too.
$endgroup$
– M.D.
Feb 2 at 12:18
1
$begingroup$
Oh, great. Thanks a lot guys @CalvinKhor and idm.
$endgroup$
– M.D.
Feb 2 at 12:43