derivative of $f(x,y)$ w.r.t.$ u(x,y)$
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I have explicitly known expressions for:
$f(x,y), u(x,y), v(x,y), w(x,y)$
$fracpartial fpartial x(x,y), fracpartial fpartial y(x,y), fracpartial upartial x(x,y), fracpartial upartial y(x,y), fracpartial vpartial x(x,y), fracpartial vpartial y(x,y), fracpartial wpartial x(x,y), fracpartial wpartial y(x,y)$
, from which I want to obtain explicit expression for:
$fracpartial fpartial u(x,y)$ and $fracpartial fpartial v(x,y)$
At first, I thought I should go for chain-rule, but it seems I cannot get the answer (3 unknowns from 2 equation as below)
$fracpartial fpartial x=fracpartial fpartial ufracpartial upartial x+fracpartial fpartial vfracpartial vpartial x+fracpartial fpartial wfracpartial wpartial x$
$fracpartial fpartial y=fracpartial fpartial ufracpartial upartial y+fracpartial fpartial vfracpartial vpartial y+fracpartial fpartial wfracpartial wpartial y$
I couldn't find anything similar to my problem (searched for multivariate derivative chain-rule, functional derivative), and couldn't come up with better keywords for the questions.
Can anyone let me know how to solve my problem, or more relevant keywords for my problem? Thank you.
P.S. For better understanding of my problem, explicit expressions are:
$u(x,y) = sin(x)cos(y)$
$v(x,y) = sin(x)sin(y)$
$w(x,y) = cos(x)$
$f(x,y) = Y_l^m(x,y)$ is spherical harmonics
multivariable-calculus chain-rule
add a comment |Â
up vote
-1
down vote
favorite
I have explicitly known expressions for:
$f(x,y), u(x,y), v(x,y), w(x,y)$
$fracpartial fpartial x(x,y), fracpartial fpartial y(x,y), fracpartial upartial x(x,y), fracpartial upartial y(x,y), fracpartial vpartial x(x,y), fracpartial vpartial y(x,y), fracpartial wpartial x(x,y), fracpartial wpartial y(x,y)$
, from which I want to obtain explicit expression for:
$fracpartial fpartial u(x,y)$ and $fracpartial fpartial v(x,y)$
At first, I thought I should go for chain-rule, but it seems I cannot get the answer (3 unknowns from 2 equation as below)
$fracpartial fpartial x=fracpartial fpartial ufracpartial upartial x+fracpartial fpartial vfracpartial vpartial x+fracpartial fpartial wfracpartial wpartial x$
$fracpartial fpartial y=fracpartial fpartial ufracpartial upartial y+fracpartial fpartial vfracpartial vpartial y+fracpartial fpartial wfracpartial wpartial y$
I couldn't find anything similar to my problem (searched for multivariate derivative chain-rule, functional derivative), and couldn't come up with better keywords for the questions.
Can anyone let me know how to solve my problem, or more relevant keywords for my problem? Thank you.
P.S. For better understanding of my problem, explicit expressions are:
$u(x,y) = sin(x)cos(y)$
$v(x,y) = sin(x)sin(y)$
$w(x,y) = cos(x)$
$f(x,y) = Y_l^m(x,y)$ is spherical harmonics
multivariable-calculus chain-rule
I found a question that seems related link
â MJH
Sep 2 at 8:14
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have explicitly known expressions for:
$f(x,y), u(x,y), v(x,y), w(x,y)$
$fracpartial fpartial x(x,y), fracpartial fpartial y(x,y), fracpartial upartial x(x,y), fracpartial upartial y(x,y), fracpartial vpartial x(x,y), fracpartial vpartial y(x,y), fracpartial wpartial x(x,y), fracpartial wpartial y(x,y)$
, from which I want to obtain explicit expression for:
$fracpartial fpartial u(x,y)$ and $fracpartial fpartial v(x,y)$
At first, I thought I should go for chain-rule, but it seems I cannot get the answer (3 unknowns from 2 equation as below)
$fracpartial fpartial x=fracpartial fpartial ufracpartial upartial x+fracpartial fpartial vfracpartial vpartial x+fracpartial fpartial wfracpartial wpartial x$
$fracpartial fpartial y=fracpartial fpartial ufracpartial upartial y+fracpartial fpartial vfracpartial vpartial y+fracpartial fpartial wfracpartial wpartial y$
I couldn't find anything similar to my problem (searched for multivariate derivative chain-rule, functional derivative), and couldn't come up with better keywords for the questions.
Can anyone let me know how to solve my problem, or more relevant keywords for my problem? Thank you.
P.S. For better understanding of my problem, explicit expressions are:
$u(x,y) = sin(x)cos(y)$
$v(x,y) = sin(x)sin(y)$
$w(x,y) = cos(x)$
$f(x,y) = Y_l^m(x,y)$ is spherical harmonics
multivariable-calculus chain-rule
I have explicitly known expressions for:
$f(x,y), u(x,y), v(x,y), w(x,y)$
$fracpartial fpartial x(x,y), fracpartial fpartial y(x,y), fracpartial upartial x(x,y), fracpartial upartial y(x,y), fracpartial vpartial x(x,y), fracpartial vpartial y(x,y), fracpartial wpartial x(x,y), fracpartial wpartial y(x,y)$
, from which I want to obtain explicit expression for:
$fracpartial fpartial u(x,y)$ and $fracpartial fpartial v(x,y)$
At first, I thought I should go for chain-rule, but it seems I cannot get the answer (3 unknowns from 2 equation as below)
$fracpartial fpartial x=fracpartial fpartial ufracpartial upartial x+fracpartial fpartial vfracpartial vpartial x+fracpartial fpartial wfracpartial wpartial x$
$fracpartial fpartial y=fracpartial fpartial ufracpartial upartial y+fracpartial fpartial vfracpartial vpartial y+fracpartial fpartial wfracpartial wpartial y$
I couldn't find anything similar to my problem (searched for multivariate derivative chain-rule, functional derivative), and couldn't come up with better keywords for the questions.
Can anyone let me know how to solve my problem, or more relevant keywords for my problem? Thank you.
P.S. For better understanding of my problem, explicit expressions are:
$u(x,y) = sin(x)cos(y)$
$v(x,y) = sin(x)sin(y)$
$w(x,y) = cos(x)$
$f(x,y) = Y_l^m(x,y)$ is spherical harmonics
multivariable-calculus chain-rule
multivariable-calculus chain-rule
edited Sep 2 at 9:38
José Carlos Santos
121k16101186
121k16101186
asked Sep 2 at 8:11
MJH
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1
I found a question that seems related link
â MJH
Sep 2 at 8:14
add a comment |Â
I found a question that seems related link
â MJH
Sep 2 at 8:14
I found a question that seems related link
â MJH
Sep 2 at 8:14
I found a question that seems related link
â MJH
Sep 2 at 8:14
add a comment |Â
1 Answer
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You have to pick two of $u,v,w$ to be free variables (noting that $u^2+v^2+w^2=1$, whence all three of them cannot be treated as free variables). I am guessing that $u$ and $v$ are free. However, the assignment $(x,y)mapsto (u,v)$ is an almost everywhere $4$-to-$1$ map from $(mathbbR/2pimathbbZ)times(mathbbR/2pimathbbZ)$ to $[-1,+1]times[-1,+1]$. In other words, $$u(x,y)=u(-x,y+pi)=u(x+pi,y+pi)=u(-x+pi,y)$$ and $$v(x,y)=v(-x,y+pi)=v(x+pi,y+pi)=v(-x+pi,y),,$$ where the operations $zmapsto -z$ and $zmapsto z+pi$ are considered modulo $2pi$. Consequently, it is quite problematic to go from $(u,v)$ back to $(x,y)$. However, if $x$ and $y$ are both known, then you can see that
$$x=s,textarcsinbig(sqrtu^2+v^2big)+ttext and y=textarctanleft(fracvuright)+tau,,$$
where $sin-1,+1$ and $t,tauin0,pi$ (indeed, $s$ is the sign of $sin(x)$). The values of $s$, $t$, and $tau$ are fixed in a small neighborhood of each noncritical point. That is,
$$fracpartial fpartial u(x,y)=left(fracpartial fpartial x(x,y)right),left(fracpartial xpartial uright)+left(fracpartial fpartial y(x,y)right),left(fracpartial ypartial uright)$$
and
$$fracpartial fpartial v(x,y)=left(fracpartial fpartial x(x,y)right),left(fracpartial xpartial vright)+left(fracpartial fpartial y(x,y)right),left(fracpartial ypartial vright)$$
can be determined from
$$fracpartial xpartial u=fracsusqrt(1-u^2-v^2)(u^2+v^2)=fraccos(y),,$$
$$fracpartial xpartial v=fracsvsqrt(1-u^2-v^2)(u^2+v^2)=fracsin(y),,$$
$$fracpartial ypartial u=-fracvu^2+v^2=-fracsin(y)sin(x)$$
and
$$fracpartial ypartial v=+fracuu^2+v^2=+fraccos(y)sin(x),.$$
Thank you for tips - It worked for some test conditions! However, it did not work for other conditions (possibly due to numerical error because I tested using some extreme values where reference answers exist).
â MJH
Sep 2 at 12:58
Actually, the question above was reformulated by myself. The original problem involves gradient in reciprocal k-space: $veck times fracddveck Y_lm(hatk)$, $fracddveck=k_x hatx+k_y haty+k_z hatz$, which I have never met before. I expressed the k-vectors as $u,v,w$ in the original question. I wonder if my approach was acceptable.
â MJH
Sep 2 at 13:05
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You have to pick two of $u,v,w$ to be free variables (noting that $u^2+v^2+w^2=1$, whence all three of them cannot be treated as free variables). I am guessing that $u$ and $v$ are free. However, the assignment $(x,y)mapsto (u,v)$ is an almost everywhere $4$-to-$1$ map from $(mathbbR/2pimathbbZ)times(mathbbR/2pimathbbZ)$ to $[-1,+1]times[-1,+1]$. In other words, $$u(x,y)=u(-x,y+pi)=u(x+pi,y+pi)=u(-x+pi,y)$$ and $$v(x,y)=v(-x,y+pi)=v(x+pi,y+pi)=v(-x+pi,y),,$$ where the operations $zmapsto -z$ and $zmapsto z+pi$ are considered modulo $2pi$. Consequently, it is quite problematic to go from $(u,v)$ back to $(x,y)$. However, if $x$ and $y$ are both known, then you can see that
$$x=s,textarcsinbig(sqrtu^2+v^2big)+ttext and y=textarctanleft(fracvuright)+tau,,$$
where $sin-1,+1$ and $t,tauin0,pi$ (indeed, $s$ is the sign of $sin(x)$). The values of $s$, $t$, and $tau$ are fixed in a small neighborhood of each noncritical point. That is,
$$fracpartial fpartial u(x,y)=left(fracpartial fpartial x(x,y)right),left(fracpartial xpartial uright)+left(fracpartial fpartial y(x,y)right),left(fracpartial ypartial uright)$$
and
$$fracpartial fpartial v(x,y)=left(fracpartial fpartial x(x,y)right),left(fracpartial xpartial vright)+left(fracpartial fpartial y(x,y)right),left(fracpartial ypartial vright)$$
can be determined from
$$fracpartial xpartial u=fracsusqrt(1-u^2-v^2)(u^2+v^2)=fraccos(y),,$$
$$fracpartial xpartial v=fracsvsqrt(1-u^2-v^2)(u^2+v^2)=fracsin(y),,$$
$$fracpartial ypartial u=-fracvu^2+v^2=-fracsin(y)sin(x)$$
and
$$fracpartial ypartial v=+fracuu^2+v^2=+fraccos(y)sin(x),.$$
Thank you for tips - It worked for some test conditions! However, it did not work for other conditions (possibly due to numerical error because I tested using some extreme values where reference answers exist).
â MJH
Sep 2 at 12:58
Actually, the question above was reformulated by myself. The original problem involves gradient in reciprocal k-space: $veck times fracddveck Y_lm(hatk)$, $fracddveck=k_x hatx+k_y haty+k_z hatz$, which I have never met before. I expressed the k-vectors as $u,v,w$ in the original question. I wonder if my approach was acceptable.
â MJH
Sep 2 at 13:05
add a comment |Â
up vote
0
down vote
You have to pick two of $u,v,w$ to be free variables (noting that $u^2+v^2+w^2=1$, whence all three of them cannot be treated as free variables). I am guessing that $u$ and $v$ are free. However, the assignment $(x,y)mapsto (u,v)$ is an almost everywhere $4$-to-$1$ map from $(mathbbR/2pimathbbZ)times(mathbbR/2pimathbbZ)$ to $[-1,+1]times[-1,+1]$. In other words, $$u(x,y)=u(-x,y+pi)=u(x+pi,y+pi)=u(-x+pi,y)$$ and $$v(x,y)=v(-x,y+pi)=v(x+pi,y+pi)=v(-x+pi,y),,$$ where the operations $zmapsto -z$ and $zmapsto z+pi$ are considered modulo $2pi$. Consequently, it is quite problematic to go from $(u,v)$ back to $(x,y)$. However, if $x$ and $y$ are both known, then you can see that
$$x=s,textarcsinbig(sqrtu^2+v^2big)+ttext and y=textarctanleft(fracvuright)+tau,,$$
where $sin-1,+1$ and $t,tauin0,pi$ (indeed, $s$ is the sign of $sin(x)$). The values of $s$, $t$, and $tau$ are fixed in a small neighborhood of each noncritical point. That is,
$$fracpartial fpartial u(x,y)=left(fracpartial fpartial x(x,y)right),left(fracpartial xpartial uright)+left(fracpartial fpartial y(x,y)right),left(fracpartial ypartial uright)$$
and
$$fracpartial fpartial v(x,y)=left(fracpartial fpartial x(x,y)right),left(fracpartial xpartial vright)+left(fracpartial fpartial y(x,y)right),left(fracpartial ypartial vright)$$
can be determined from
$$fracpartial xpartial u=fracsusqrt(1-u^2-v^2)(u^2+v^2)=fraccos(y),,$$
$$fracpartial xpartial v=fracsvsqrt(1-u^2-v^2)(u^2+v^2)=fracsin(y),,$$
$$fracpartial ypartial u=-fracvu^2+v^2=-fracsin(y)sin(x)$$
and
$$fracpartial ypartial v=+fracuu^2+v^2=+fraccos(y)sin(x),.$$
Thank you for tips - It worked for some test conditions! However, it did not work for other conditions (possibly due to numerical error because I tested using some extreme values where reference answers exist).
â MJH
Sep 2 at 12:58
Actually, the question above was reformulated by myself. The original problem involves gradient in reciprocal k-space: $veck times fracddveck Y_lm(hatk)$, $fracddveck=k_x hatx+k_y haty+k_z hatz$, which I have never met before. I expressed the k-vectors as $u,v,w$ in the original question. I wonder if my approach was acceptable.
â MJH
Sep 2 at 13:05
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You have to pick two of $u,v,w$ to be free variables (noting that $u^2+v^2+w^2=1$, whence all three of them cannot be treated as free variables). I am guessing that $u$ and $v$ are free. However, the assignment $(x,y)mapsto (u,v)$ is an almost everywhere $4$-to-$1$ map from $(mathbbR/2pimathbbZ)times(mathbbR/2pimathbbZ)$ to $[-1,+1]times[-1,+1]$. In other words, $$u(x,y)=u(-x,y+pi)=u(x+pi,y+pi)=u(-x+pi,y)$$ and $$v(x,y)=v(-x,y+pi)=v(x+pi,y+pi)=v(-x+pi,y),,$$ where the operations $zmapsto -z$ and $zmapsto z+pi$ are considered modulo $2pi$. Consequently, it is quite problematic to go from $(u,v)$ back to $(x,y)$. However, if $x$ and $y$ are both known, then you can see that
$$x=s,textarcsinbig(sqrtu^2+v^2big)+ttext and y=textarctanleft(fracvuright)+tau,,$$
where $sin-1,+1$ and $t,tauin0,pi$ (indeed, $s$ is the sign of $sin(x)$). The values of $s$, $t$, and $tau$ are fixed in a small neighborhood of each noncritical point. That is,
$$fracpartial fpartial u(x,y)=left(fracpartial fpartial x(x,y)right),left(fracpartial xpartial uright)+left(fracpartial fpartial y(x,y)right),left(fracpartial ypartial uright)$$
and
$$fracpartial fpartial v(x,y)=left(fracpartial fpartial x(x,y)right),left(fracpartial xpartial vright)+left(fracpartial fpartial y(x,y)right),left(fracpartial ypartial vright)$$
can be determined from
$$fracpartial xpartial u=fracsusqrt(1-u^2-v^2)(u^2+v^2)=fraccos(y),,$$
$$fracpartial xpartial v=fracsvsqrt(1-u^2-v^2)(u^2+v^2)=fracsin(y),,$$
$$fracpartial ypartial u=-fracvu^2+v^2=-fracsin(y)sin(x)$$
and
$$fracpartial ypartial v=+fracuu^2+v^2=+fraccos(y)sin(x),.$$
You have to pick two of $u,v,w$ to be free variables (noting that $u^2+v^2+w^2=1$, whence all three of them cannot be treated as free variables). I am guessing that $u$ and $v$ are free. However, the assignment $(x,y)mapsto (u,v)$ is an almost everywhere $4$-to-$1$ map from $(mathbbR/2pimathbbZ)times(mathbbR/2pimathbbZ)$ to $[-1,+1]times[-1,+1]$. In other words, $$u(x,y)=u(-x,y+pi)=u(x+pi,y+pi)=u(-x+pi,y)$$ and $$v(x,y)=v(-x,y+pi)=v(x+pi,y+pi)=v(-x+pi,y),,$$ where the operations $zmapsto -z$ and $zmapsto z+pi$ are considered modulo $2pi$. Consequently, it is quite problematic to go from $(u,v)$ back to $(x,y)$. However, if $x$ and $y$ are both known, then you can see that
$$x=s,textarcsinbig(sqrtu^2+v^2big)+ttext and y=textarctanleft(fracvuright)+tau,,$$
where $sin-1,+1$ and $t,tauin0,pi$ (indeed, $s$ is the sign of $sin(x)$). The values of $s$, $t$, and $tau$ are fixed in a small neighborhood of each noncritical point. That is,
$$fracpartial fpartial u(x,y)=left(fracpartial fpartial x(x,y)right),left(fracpartial xpartial uright)+left(fracpartial fpartial y(x,y)right),left(fracpartial ypartial uright)$$
and
$$fracpartial fpartial v(x,y)=left(fracpartial fpartial x(x,y)right),left(fracpartial xpartial vright)+left(fracpartial fpartial y(x,y)right),left(fracpartial ypartial vright)$$
can be determined from
$$fracpartial xpartial u=fracsusqrt(1-u^2-v^2)(u^2+v^2)=fraccos(y),,$$
$$fracpartial xpartial v=fracsvsqrt(1-u^2-v^2)(u^2+v^2)=fracsin(y),,$$
$$fracpartial ypartial u=-fracvu^2+v^2=-fracsin(y)sin(x)$$
and
$$fracpartial ypartial v=+fracuu^2+v^2=+fraccos(y)sin(x),.$$
edited Sep 2 at 13:21
answered Sep 2 at 10:10
Batominovski
25.7k22881
25.7k22881
Thank you for tips - It worked for some test conditions! However, it did not work for other conditions (possibly due to numerical error because I tested using some extreme values where reference answers exist).
â MJH
Sep 2 at 12:58
Actually, the question above was reformulated by myself. The original problem involves gradient in reciprocal k-space: $veck times fracddveck Y_lm(hatk)$, $fracddveck=k_x hatx+k_y haty+k_z hatz$, which I have never met before. I expressed the k-vectors as $u,v,w$ in the original question. I wonder if my approach was acceptable.
â MJH
Sep 2 at 13:05
add a comment |Â
Thank you for tips - It worked for some test conditions! However, it did not work for other conditions (possibly due to numerical error because I tested using some extreme values where reference answers exist).
â MJH
Sep 2 at 12:58
Actually, the question above was reformulated by myself. The original problem involves gradient in reciprocal k-space: $veck times fracddveck Y_lm(hatk)$, $fracddveck=k_x hatx+k_y haty+k_z hatz$, which I have never met before. I expressed the k-vectors as $u,v,w$ in the original question. I wonder if my approach was acceptable.
â MJH
Sep 2 at 13:05
Thank you for tips - It worked for some test conditions! However, it did not work for other conditions (possibly due to numerical error because I tested using some extreme values where reference answers exist).
â MJH
Sep 2 at 12:58
Thank you for tips - It worked for some test conditions! However, it did not work for other conditions (possibly due to numerical error because I tested using some extreme values where reference answers exist).
â MJH
Sep 2 at 12:58
Actually, the question above was reformulated by myself. The original problem involves gradient in reciprocal k-space: $veck times fracddveck Y_lm(hatk)$, $fracddveck=k_x hatx+k_y haty+k_z hatz$, which I have never met before. I expressed the k-vectors as $u,v,w$ in the original question. I wonder if my approach was acceptable.
â MJH
Sep 2 at 13:05
Actually, the question above was reformulated by myself. The original problem involves gradient in reciprocal k-space: $veck times fracddveck Y_lm(hatk)$, $fracddveck=k_x hatx+k_y haty+k_z hatz$, which I have never met before. I expressed the k-vectors as $u,v,w$ in the original question. I wonder if my approach was acceptable.
â MJH
Sep 2 at 13:05
add a comment |Â
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I found a question that seems related link
â MJH
Sep 2 at 8:14