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I am stuck with an exercise that requires me to find the Laplacian $\Delta u=(D_x^2u+D_y^2u)$ of a 2d-function $u$ in polar coordinates (in the standard Euclidean plane).

I found the following article on the net, and tried to follow its logic, but I could not understand two steps: http://www.sci.brooklyn.cuny.edu/~mate/misc/laplacian_polarcoord_higherdim.pdf

at first, the representation of $D_x$ and $D_y$ in terms of $r, \theta$, at the bottom of page 2:

$D_y=\sin\theta D_r+\frac{\cos\theta}{r}D_\theta$ and

$D_x=\cos\theta D_r-\frac{\sin\theta}{r}D_\theta$ . When I draw a sketch of the plane with a circle and all the coordinates, I get that it should be $D_y=\sin\theta D_r+r\ \cos\theta D_\theta$, because the larger the radius is, the greater will be the impact of a change in $\theta$ on a change in $y$. What am I making wrong here?

And then, secondly, what looks like an easy multiplication, namely taking the square of the above terms (on the top of page 3 in the link):

$D_y^2=(\sin\theta D_r+\frac{\cos\theta}{r}D_\theta)(\sin\theta D_r+\frac{\cos\theta}{r}D_\theta)$

$=\sin^2\theta D_r^2+\frac{2\sin\theta \cos\theta}{r}D_\theta D_r+\frac{\sin^2\theta}{r^2}D_\theta^2 \\ -\frac{\cos\theta}{r^2}D_\theta + \frac{\cos^2\theta}{r}D_r - \frac{\cos\theta \sin\theta}{r^2}D_\theta$

and similarly for $D_x$.

I don't see from where the last three summands come, what I see is: $(a+b)(a+b)=a^2+2ab+b^2+c+d+e$ and I cannot see the context of $c,d,e.$

It would be great if you could explain those issues to me!

  • 1
    When you write something like `sin` in $\TeX$, it gets interpreted as juxtaposed variable names and is therefore italicized. To get proper formatting for function names, you need to use the corresponding command sequences, e.g. `\sin`. If you have a function name for which there's no command sequence, such as $\operatorname{tr}$, you can get it formatted properly using `\operatorname{tr}`.2012-04-15
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    thanks! I'll do that from now on.2012-04-15
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    Also note that you can use double dollar signs to get displayed equations, which look nicer and are easier to read; single dollar signs are for inline equations.2012-04-15

2 Answers 2

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$D_\theta$ and $D_y$ are not changes in $\theta$ and $y$, respectively, but changes with respect to these variables. Thus, precisely because, as you write, the greater the radius, the greater the impact of a change in $\theta$ on a change in $y$, it's also true that the greater the radius, the less $D_\theta$ contributes to $D_y$, since the changes in $\theta$ and $y$ are in the denominator of these derivatives, not the numerator.

On your second question, those terms arise because the differential operators in the first factor have to be applied to the entire result of applying the second factor to the function. The first three terms are what you get if you apply the differential operators in the first factor to the original function; the second three terms are what you get when you apply them to the functions in the second factor.

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Here is another approach.

Since $x=r\cos(\theta)$ and $y=r\sin(\theta)$ we get the Jacobian $$ \frac{\partial(x,y)}{\partial(r,\theta)}=\begin{bmatrix}\cos(\theta)&\sin(\theta)\\-r\sin(\theta)&r\cos(\theta) \end{bmatrix}\tag{1} $$ which inverted is $$ \begin{align} \frac{\partial(r,\theta)}{\partial(x,y)} &=\frac1r\begin{bmatrix}r\cos(\theta)&-\sin(\theta)\\r\sin(\theta)&\cos(\theta) \end{bmatrix}\\ &=\frac1r\begin{bmatrix}x&-y/r\\y&x/r \end{bmatrix}\tag{2} \end{align} $$ Thus, $$ \begin{align} \frac{\partial}{\partial x} &=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta}\\ &=\cos(\theta)\frac{\partial}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\tag{3} \end{align} $$ and $$ \begin{align} \frac{\partial}{\partial y} &=\frac{\partial r}{\partial y}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial y}\frac{\partial}{\partial \theta}\\ &=\sin(\theta)\frac{\partial}{\partial r}+\frac{\cos(\theta)}{r}\frac{\partial}{\partial \theta}\tag{4} \end{align} $$ Applying $(3)$ twice yields, $$ \begin{align} \frac{\partial^2}{\partial x^2}&= \cos^2(\theta)\frac{\partial^2}{\partial r^2} -\frac{2\sin(\theta)\cos(\theta)}{r}\frac{\partial^2}{\partial r\partial \theta} +\frac{\sin^2(\theta)}{r^2}\frac{\partial^2}{\partial \theta^2}\\ &+\frac{\sin^2(\theta)}{r}\frac{\partial}{\partial r} +2\frac{\sin(\theta)\cos(\theta)}{r^2}\frac{\partial}{\partial \theta}\tag{5} \end{align} $$ and applying $(4)$ twice gives $$ \begin{align} \frac{\partial^2}{\partial y^2}&= \sin^2(\theta)\frac{\partial^2}{\partial r^2} +\frac{2\sin(\theta)\cos(\theta)}{r}\frac{\partial^2}{\partial r\partial \theta} +\frac{\cos^2(\theta)}{r^2}\frac{\partial^2}{\partial \theta^2}\\ &+\frac{\cos^2(\theta)}{r}\frac{\partial}{\partial r} -2\frac{\sin(\theta)\cos(\theta)}{r^2}\frac{\partial}{\partial \theta}\tag{6} \end{align} $$ Adding $(5)$ and $(6)$ produces $$ \Delta=\frac{\partial^2}{\partial r^2}+\frac1r\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}\tag{7} $$


Edit: Judging from the second part of the question, a bit more detail regarding the composition of operators giving $(5)$ and $(6)$ might be useful.

As usual, there are four terms resulting from the distribution over addition: $$ \begin{align} &\left(\cos(\theta)\frac{\partial}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\right) \left(\cos(\theta)\frac{\partial}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\right)\\ &=\left(\cos(\theta)\frac{\partial}{\partial r}\right)\left(\cos(\theta)\frac{\partial}{\partial r}\right) -\left(\cos(\theta)\frac{\partial}{\partial r}\right)\left(\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\right)\\ &-\left(\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\right)\left(\cos(\theta)\frac{\partial}{\partial r}\right) +\left(\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\right) \left(\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\right) \end{align} $$ In the first term, there is no interaction between the first $\dfrac{\partial}{\partial r}$ and the second $\cos(\theta)$. Thus, the first term becomes $$ \cos^2(\theta)\dfrac{\partial^2}{\partial r^2} $$ However, in the second term, the $\dfrac{\partial}{\partial r}$ does interact with the $\dfrac{\sin(\theta)}{r}$. The product rule says the second term becomes $$ -\dfrac{\cos(\theta)\sin(\theta)}{r^2}\dfrac{\partial}{\partial \theta}+\dfrac{\cos(\theta)\sin(\theta)}{r}\dfrac{\partial^2}{\partial r\partial \theta} $$ Similar things happen in the remaining two terms since $\dfrac{\partial}{\partial \theta}$ interacts with both $\cos(\theta)$ and $\dfrac{\sin(\theta)}{r}$.

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    Why did you use the matrix at the start? It looks like after that you applied the chain rule.2016-11-29
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    I think the first jacobian is not correct... in the first row you differentiated respect to $r$ both cells instead of just the first, and the second respect $\theta$.2017-09-12
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    @Masacroso: The first row is $(x,y)$ differentiated wrt $r$ and the second row is $(x,y)$ differentiated with respect to $\theta$.2017-09-12
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    yes, but this is not the standard definition of the jacobian.2017-09-12
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    @Masacroso: to which "standard" definition do you refer? It all depends on what you consider the "standard" vectors. If you consider row vectors as the vectors for $(x,y)$, as I have written it here, then my matrix is the "standard" definition. However, if you consider column vectors as the vectors for $\begin{pmatrix}x\\y\end{pmatrix}$, then the transpose of my matrix is the "standard" definition. In the end, it's the determinant that is important, and then, transpose or not, it doesn't much matter.2017-09-12
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    the unique definition that I know coincides with that on the [wikipedia article](https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant). Following it the jacobian is the transpose of what you did. This is not about transposing the vectors, the first coordinate is $x$ and the second is $y$ regardless of the transposition, right?2017-09-12
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    @Masacroso: If we use row vectors then the Jacobian is given by $$\begin{bmatrix}\mathrm{d}x&\mathrm{d}y\end{bmatrix} =\begin{bmatrix}\mathrm{d}r&\mathrm{d}\theta\end{bmatrix} \begin{bmatrix}\frac{\partial x}{\partial r}&\frac{\partial y}{\partial r}\\ \frac{\partial x}{\partial \theta}&\frac{\partial y}{\partial \theta}\end{bmatrix}$$ If we use column vectors, then the Jacobian is as given in the Wikipedia article. There is no *unique* definition. This is why I wrote it as $\frac{\partial(x,y)}{\partial(r,\theta)}$ (with row vectors).2017-09-12