Mixing
confusion about the laplacian in polar coordinates
$begingroup$
I have a confusion about the following. If we have a function $f:mathbb{R}^2 to mathbb{R}$ of class $C^2$ and we want to calculate the laplacian of this function polar coordinates.
I don't understand what it means. Does it mean we need to calculate :
$$frac{partial}{partial x} (frac{partial f(r cos theta, r sin theta)}{partial x}) + frac{partial}{partial y} (frac{partial f(r cos theta, r sin theta)}{partial y})$$
Or does it mean we need to calculate :
$$frac{partial ^2 f}{partial x^2}(r cos theta, r sin theta) + frac{partial ^2 f}{partial y^2}(r cos theta, r sin theta)$$
Moreover, I am really confused since I don't understand the fondamental difference between these two expressions.
Thank you !
real-analysis calculus multivariable-calculus partial-derivative
asked Jan 27 at 9:19
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
I have a confusion about the following. If we have a function $f:mathbb{R}^2 to mathbb{R}$ of class $C^2$ and we want to calculate the laplacian of this function polar coordinates.
I don't understand what it means. Does it mean we need to calculate :
$$frac{partial}{partial x} (frac{partial f(r cos theta, r sin theta)}{partial x}) + frac{partial}{partial y} (frac{partial f(r cos theta, r sin theta)}{partial y})$$
Or does it mean we need to calculate :
$$frac{partial ^2 f}{partial x^2}(r cos theta, r sin theta) + frac{partial ^2 f}{partial y^2}(r cos theta, r sin theta)$$
Moreover, I am really confused since I don't understand the fondamental difference between these two expressions.
Thank you !
real-analysis calculus multivariable-calculus partial-derivative
asked Jan 27 at 9:19
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
$begingroup$
A function $f(rcos(theta),rsin(theta))$ does not have a partial derivative with respect to $x$.
$endgroup$
– Good Morning Captain
Jan 27 at 9:24
add a comment |
$begingroup$
I have a confusion about the following. If we have a function $f:mathbb{R}^2 to mathbb{R}$ of class $C^2$ and we want to calculate the laplacian of this function polar coordinates.
I don't understand what it means. Does it mean we need to calculate :
$$frac{partial}{partial x} (frac{partial f(r cos theta, r sin theta)}{partial x}) + frac{partial}{partial y} (frac{partial f(r cos theta, r sin theta)}{partial y})$$
Or does it mean we need to calculate :
$$frac{partial ^2 f}{partial x^2}(r cos theta, r sin theta) + frac{partial ^2 f}{partial y^2}(r cos theta, r sin theta)$$
Moreover, I am really confused since I don't understand the fondamental difference between these two expressions.
Thank you !
real-analysis calculus multivariable-calculus partial-derivative
asked Jan 27 at 9:19
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
I have a confusion about the following. If we have a function $f:mathbb{R}^2 to mathbb{R}$ of class $C^2$ and we want to calculate the laplacian of this function polar coordinates.
I don't understand what it means. Does it mean we need to calculate :
$$frac{partial}{partial x} (frac{partial f(r cos theta, r sin theta)}{partial x}) + frac{partial}{partial y} (frac{partial f(r cos theta, r sin theta)}{partial y})$$
Or does it mean we need to calculate :
$$frac{partial ^2 f}{partial x^2}(r cos theta, r sin theta) + frac{partial ^2 f}{partial y^2}(r cos theta, r sin theta)$$
Moreover, I am really confused since I don't understand the fondamental difference between these two expressions.
Thank you !
real-analysis calculus multivariable-calculus partial-derivative
real-analysis calculus multivariable-calculus partial-derivative
asked Jan 27 at 9:19
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 9:19
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 9:19
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 9:19
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 9:19
dghkgfzyukzdghkgfzyukz
16612
16612
$begingroup$
A function $f(rcos(theta),rsin(theta))$ does not have a partial derivative with respect to $x$.
$endgroup$
– Good Morning Captain
Jan 27 at 9:24
add a comment |
$begingroup$
A function $f(rcos(theta),rsin(theta))$ does not have a partial derivative with respect to $x$.
$endgroup$
– Good Morning Captain
Jan 27 at 9:24
$begingroup$
A function $f(rcos(theta),rsin(theta))$ does not have a partial derivative with respect to $x$.
$endgroup$
– Good Morning Captain
Jan 27 at 9:24
$begingroup$
A function $f(rcos(theta),rsin(theta))$ does not have a partial derivative with respect to $x$.
$endgroup$
– Good Morning Captain
Jan 27 at 9:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The basic interpretation is the second one. The Laplacian $Delta f(x,y)$ is a function of $x$ and $y$, which is defined as $frac{partial^2 f}{partial x^2}(x,y) + frac{partial^2 f}{partial y^2}(x,y)$, and expressing a function in polar coordinates just means substituting $x=r cos theta$ and $y = r sin theta$. So first compute the second partials with respect to $x$ and $y$, then substitute.
However, the chain rule gives you another way of obtaining the same result, by performing a different sequence of operations. This involves first expressing $f$ in polar coordinates, and then taking partial derivatives of that expression with respect to $r$ and $theta$. (Not with respect to $x$ and $y$ – that's sort of pointless, because if you wanted to do that, you should have done it directly on $f(x,y)$ instead of changing to polar coordinates to begin with.) How this works is explained in several other answers on this site, for example here or here.
answered Jan 27 at 9:52
Hans LundmarkHans Lundmark
35.9k564115
$endgroup$
$begingroup$
Thank you ! So it means that we have : $Delta f(x,y) = frac{partial ^2 f}{partial x}(x,y) + frac{partial^2f}{partial y}(x,y)$, and this can be expressed in polar coordinates by substituing so we have : $Delta f(r cos theta, r sin theta) = frac{partial ^2 f}{partial x}(r cos theta,r sin theta) + frac{partial^2f}{partial y}(r cos theta,r sin theta)$ which is the laplacian in polar coordinates. But an other way to expressed this laplacian in polar coordinates is :
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
$begingroup$
$frac{partial}{partial r} (frac{partial f(r cos theta, r sin theta)}{partial r}) + frac{partial}{partial theta} (frac{partial f(r cos theta, r sin theta)}{partial theta})$ which can be calculated using the chain rule.
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
1
$begingroup$
@dghkgfzyukz: No, that's not the correct expression for the Laplacian in polar coordinates! See the questions that I linked to.
$endgroup$
– Hans Lundmark
Jan 27 at 12:57
1
$begingroup$
@dghkgfzyukz: The first comment was OK, but the formula in the second comment was wrong.
$endgroup$
– Hans Lundmark
Jan 27 at 14:18
1
$begingroup$
Thanks to the link you provided I think I understand now, just need to do some calculations to see is evrything is effectively clear. Once again thank you for your patience and help.
$endgroup$
– dghkgfzyukz
Jan 27 at 14:32
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
The basic interpretation is the second one. The Laplacian $Delta f(x,y)$ is a function of $x$ and $y$, which is defined as $frac{partial^2 f}{partial x^2}(x,y) + frac{partial^2 f}{partial y^2}(x,y)$, and expressing a function in polar coordinates just means substituting $x=r cos theta$ and $y = r sin theta$. So first compute the second partials with respect to $x$ and $y$, then substitute.
However, the chain rule gives you another way of obtaining the same result, by performing a different sequence of operations. This involves first expressing $f$ in polar coordinates, and then taking partial derivatives of that expression with respect to $r$ and $theta$. (Not with respect to $x$ and $y$ – that's sort of pointless, because if you wanted to do that, you should have done it directly on $f(x,y)$ instead of changing to polar coordinates to begin with.) How this works is explained in several other answers on this site, for example here or here.
answered Jan 27 at 9:52
Hans LundmarkHans Lundmark
35.9k564115
$endgroup$
$begingroup$
Thank you ! So it means that we have : $Delta f(x,y) = frac{partial ^2 f}{partial x}(x,y) + frac{partial^2f}{partial y}(x,y)$, and this can be expressed in polar coordinates by substituing so we have : $Delta f(r cos theta, r sin theta) = frac{partial ^2 f}{partial x}(r cos theta,r sin theta) + frac{partial^2f}{partial y}(r cos theta,r sin theta)$ which is the laplacian in polar coordinates. But an other way to expressed this laplacian in polar coordinates is :
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
$begingroup$
$frac{partial}{partial r} (frac{partial f(r cos theta, r sin theta)}{partial r}) + frac{partial}{partial theta} (frac{partial f(r cos theta, r sin theta)}{partial theta})$ which can be calculated using the chain rule.
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
1
$begingroup$
@dghkgfzyukz: No, that's not the correct expression for the Laplacian in polar coordinates! See the questions that I linked to.
$endgroup$
– Hans Lundmark
Jan 27 at 12:57
1
$begingroup$
@dghkgfzyukz: The first comment was OK, but the formula in the second comment was wrong.
$endgroup$
– Hans Lundmark
Jan 27 at 14:18
1
$begingroup$
Thanks to the link you provided I think I understand now, just need to do some calculations to see is evrything is effectively clear. Once again thank you for your patience and help.
$endgroup$
– dghkgfzyukz
Jan 27 at 14:32
|
show 1 more comment
$begingroup$
The basic interpretation is the second one. The Laplacian $Delta f(x,y)$ is a function of $x$ and $y$, which is defined as $frac{partial^2 f}{partial x^2}(x,y) + frac{partial^2 f}{partial y^2}(x,y)$, and expressing a function in polar coordinates just means substituting $x=r cos theta$ and $y = r sin theta$. So first compute the second partials with respect to $x$ and $y$, then substitute.
However, the chain rule gives you another way of obtaining the same result, by performing a different sequence of operations. This involves first expressing $f$ in polar coordinates, and then taking partial derivatives of that expression with respect to $r$ and $theta$. (Not with respect to $x$ and $y$ – that's sort of pointless, because if you wanted to do that, you should have done it directly on $f(x,y)$ instead of changing to polar coordinates to begin with.) How this works is explained in several other answers on this site, for example here or here.
answered Jan 27 at 9:52
Hans LundmarkHans Lundmark
35.9k564115
$endgroup$
$begingroup$
Thank you ! So it means that we have : $Delta f(x,y) = frac{partial ^2 f}{partial x}(x,y) + frac{partial^2f}{partial y}(x,y)$, and this can be expressed in polar coordinates by substituing so we have : $Delta f(r cos theta, r sin theta) = frac{partial ^2 f}{partial x}(r cos theta,r sin theta) + frac{partial^2f}{partial y}(r cos theta,r sin theta)$ which is the laplacian in polar coordinates. But an other way to expressed this laplacian in polar coordinates is :
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
$begingroup$
$frac{partial}{partial r} (frac{partial f(r cos theta, r sin theta)}{partial r}) + frac{partial}{partial theta} (frac{partial f(r cos theta, r sin theta)}{partial theta})$ which can be calculated using the chain rule.
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
1
$begingroup$
@dghkgfzyukz: No, that's not the correct expression for the Laplacian in polar coordinates! See the questions that I linked to.
$endgroup$
– Hans Lundmark
Jan 27 at 12:57
1
$begingroup$
@dghkgfzyukz: The first comment was OK, but the formula in the second comment was wrong.
$endgroup$
– Hans Lundmark
Jan 27 at 14:18
1
$begingroup$
Thanks to the link you provided I think I understand now, just need to do some calculations to see is evrything is effectively clear. Once again thank you for your patience and help.
$endgroup$
– dghkgfzyukz
Jan 27 at 14:32
|
show 1 more comment
$begingroup$
The basic interpretation is the second one. The Laplacian $Delta f(x,y)$ is a function of $x$ and $y$, which is defined as $frac{partial^2 f}{partial x^2}(x,y) + frac{partial^2 f}{partial y^2}(x,y)$, and expressing a function in polar coordinates just means substituting $x=r cos theta$ and $y = r sin theta$. So first compute the second partials with respect to $x$ and $y$, then substitute.
However, the chain rule gives you another way of obtaining the same result, by performing a different sequence of operations. This involves first expressing $f$ in polar coordinates, and then taking partial derivatives of that expression with respect to $r$ and $theta$. (Not with respect to $x$ and $y$ – that's sort of pointless, because if you wanted to do that, you should have done it directly on $f(x,y)$ instead of changing to polar coordinates to begin with.) How this works is explained in several other answers on this site, for example here or here.
answered Jan 27 at 9:52
Hans LundmarkHans Lundmark
35.9k564115
$endgroup$
The basic interpretation is the second one. The Laplacian $Delta f(x,y)$ is a function of $x$ and $y$, which is defined as $frac{partial^2 f}{partial x^2}(x,y) + frac{partial^2 f}{partial y^2}(x,y)$, and expressing a function in polar coordinates just means substituting $x=r cos theta$ and $y = r sin theta$. So first compute the second partials with respect to $x$ and $y$, then substitute.
However, the chain rule gives you another way of obtaining the same result, by performing a different sequence of operations. This involves first expressing $f$ in polar coordinates, and then taking partial derivatives of that expression with respect to $r$ and $theta$. (Not with respect to $x$ and $y$ – that's sort of pointless, because if you wanted to do that, you should have done it directly on $f(x,y)$ instead of changing to polar coordinates to begin with.) How this works is explained in several other answers on this site, for example here or here.
answered Jan 27 at 9:52
Hans LundmarkHans Lundmark
35.9k564115
answered Jan 27 at 9:52
Hans LundmarkHans Lundmark
35.9k564115
answered Jan 27 at 9:52
Hans LundmarkHans Lundmark
35.9k564115
answered Jan 27 at 9:52
Hans LundmarkHans Lundmark
35.9k564115
35.9k564115
$begingroup$
Thank you ! So it means that we have : $Delta f(x,y) = frac{partial ^2 f}{partial x}(x,y) + frac{partial^2f}{partial y}(x,y)$, and this can be expressed in polar coordinates by substituing so we have : $Delta f(r cos theta, r sin theta) = frac{partial ^2 f}{partial x}(r cos theta,r sin theta) + frac{partial^2f}{partial y}(r cos theta,r sin theta)$ which is the laplacian in polar coordinates. But an other way to expressed this laplacian in polar coordinates is :
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
$begingroup$
$frac{partial}{partial r} (frac{partial f(r cos theta, r sin theta)}{partial r}) + frac{partial}{partial theta} (frac{partial f(r cos theta, r sin theta)}{partial theta})$ which can be calculated using the chain rule.
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
1
$begingroup$
@dghkgfzyukz: No, that's not the correct expression for the Laplacian in polar coordinates! See the questions that I linked to.
$endgroup$
– Hans Lundmark
Jan 27 at 12:57
1
$begingroup$
@dghkgfzyukz: The first comment was OK, but the formula in the second comment was wrong.
$endgroup$
– Hans Lundmark
Jan 27 at 14:18
1
$begingroup$
Thanks to the link you provided I think I understand now, just need to do some calculations to see is evrything is effectively clear. Once again thank you for your patience and help.
$endgroup$
– dghkgfzyukz
Jan 27 at 14:32
|
show 1 more comment
$begingroup$
Thank you ! So it means that we have : $Delta f(x,y) = frac{partial ^2 f}{partial x}(x,y) + frac{partial^2f}{partial y}(x,y)$, and this can be expressed in polar coordinates by substituing so we have : $Delta f(r cos theta, r sin theta) = frac{partial ^2 f}{partial x}(r cos theta,r sin theta) + frac{partial^2f}{partial y}(r cos theta,r sin theta)$ which is the laplacian in polar coordinates. But an other way to expressed this laplacian in polar coordinates is :
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
$begingroup$
$frac{partial}{partial r} (frac{partial f(r cos theta, r sin theta)}{partial r}) + frac{partial}{partial theta} (frac{partial f(r cos theta, r sin theta)}{partial theta})$ which can be calculated using the chain rule.
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
1
$begingroup$
@dghkgfzyukz: No, that's not the correct expression for the Laplacian in polar coordinates! See the questions that I linked to.
$endgroup$
– Hans Lundmark
Jan 27 at 12:57
1
$begingroup$
@dghkgfzyukz: The first comment was OK, but the formula in the second comment was wrong.
$endgroup$
– Hans Lundmark
Jan 27 at 14:18
1
$begingroup$
Thanks to the link you provided I think I understand now, just need to do some calculations to see is evrything is effectively clear. Once again thank you for your patience and help.
$endgroup$
– dghkgfzyukz
Jan 27 at 14:32
$begingroup$
Thank you ! So it means that we have : $Delta f(x,y) = frac{partial ^2 f}{partial x}(x,y) + frac{partial^2f}{partial y}(x,y)$, and this can be expressed in polar coordinates by substituing so we have : $Delta f(r cos theta, r sin theta) = frac{partial ^2 f}{partial x}(r cos theta,r sin theta) + frac{partial^2f}{partial y}(r cos theta,r sin theta)$ which is the laplacian in polar coordinates. But an other way to expressed this laplacian in polar coordinates is :
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
$begingroup$
Thank you ! So it means that we have : $Delta f(x,y) = frac{partial ^2 f}{partial x}(x,y) + frac{partial^2f}{partial y}(x,y)$, and this can be expressed in polar coordinates by substituing so we have : $Delta f(r cos theta, r sin theta) = frac{partial ^2 f}{partial x}(r cos theta,r sin theta) + frac{partial^2f}{partial y}(r cos theta,r sin theta)$ which is the laplacian in polar coordinates. But an other way to expressed this laplacian in polar coordinates is :
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
$begingroup$
$frac{partial}{partial r} (frac{partial f(r cos theta, r sin theta)}{partial r}) + frac{partial}{partial theta} (frac{partial f(r cos theta, r sin theta)}{partial theta})$ which can be calculated using the chain rule.
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
$begingroup$
$frac{partial}{partial r} (frac{partial f(r cos theta, r sin theta)}{partial r}) + frac{partial}{partial theta} (frac{partial f(r cos theta, r sin theta)}{partial theta})$ which can be calculated using the chain rule.
$endgroup$
– dghkgfzyukz
Jan 27 at 12:13
1
1
$begingroup$
@dghkgfzyukz: No, that's not the correct expression for the Laplacian in polar coordinates! See the questions that I linked to.
$endgroup$
– Hans Lundmark
Jan 27 at 12:57
$begingroup$
@dghkgfzyukz: No, that's not the correct expression for the Laplacian in polar coordinates! See the questions that I linked to.
$endgroup$
– Hans Lundmark
Jan 27 at 12:57
1
1
$begingroup$
@dghkgfzyukz: The first comment was OK, but the formula in the second comment was wrong.
$endgroup$
– Hans Lundmark
Jan 27 at 14:18
$begingroup$
@dghkgfzyukz: The first comment was OK, but the formula in the second comment was wrong.
$endgroup$
– Hans Lundmark
Jan 27 at 14:18
1
1
$begingroup$
Thanks to the link you provided I think I understand now, just need to do some calculations to see is evrything is effectively clear. Once again thank you for your patience and help.
$endgroup$
– dghkgfzyukz
Jan 27 at 14:32
$begingroup$
Thanks to the link you provided I think I understand now, just need to do some calculations to see is evrything is effectively clear. Once again thank you for your patience and help.
$endgroup$
– dghkgfzyukz
Jan 27 at 14:32
|
show 1 more comment
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$begingroup$
A function $f(rcos(theta),rsin(theta))$ does not have a partial derivative with respect to $x$.
$endgroup$
– Good Morning Captain
Jan 27 at 9:24