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










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  • I found a question that seems related link
    – MJH
    Sep 2 at 8:14














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










share|cite|improve this question























  • I found a question that seems related link
    – MJH
    Sep 2 at 8:14












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










share|cite|improve this question















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






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edited Sep 2 at 9:38









José Carlos Santos

121k16101186




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asked Sep 2 at 8:11









MJH

1




1











  • I found a question that seems related link
    – MJH
    Sep 2 at 8:14
















  • I found a question that seems related link
    – MJH
    Sep 2 at 8:14















I found a question that seems related link
– MJH
Sep 2 at 8:14




I found a question that seems related link
– MJH
Sep 2 at 8:14










1 Answer
1






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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
$$fracpartial ypartial v=+fracuu^2+v^2=+fraccos(y)sin(x),.$$






share|cite|improve this answer






















  • 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).
    – 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.
    – MJH
    Sep 2 at 13:05











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1 Answer
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up vote
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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
$$fracpartial ypartial v=+fracuu^2+v^2=+fraccos(y)sin(x),.$$






share|cite|improve this answer






















  • 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).
    – 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.
    – MJH
    Sep 2 at 13:05















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
$$fracpartial ypartial v=+fracuu^2+v^2=+fraccos(y)sin(x),.$$






share|cite|improve this answer






















  • 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).
    – 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.
    – MJH
    Sep 2 at 13:05













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
$$fracpartial ypartial v=+fracuu^2+v^2=+fraccos(y)sin(x),.$$






share|cite|improve this answer














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
$$fracpartial ypartial v=+fracuu^2+v^2=+fraccos(y)sin(x),.$$







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share|cite|improve this answer








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).
    – 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.
    – 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).
    – 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.
    – 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).
– 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).
– 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.
– 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.
– MJH
Sep 2 at 13:05


















 

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