I am trying to solve this particular problem:

Suppose that $u(x, y)$ is a continuously differentiable, circularly symmetric function, so that when expressed in polar coordinates, $x = r \cos \theta$, $y = r \sin \theta$, it depends solely on the radius $r$; that is $u = f(r)$. Show that $u_x(x, y) = f′(r) \cos \theta$ and hence deduce that $f′(0) = 0$, which implies the Neumann boundary condition $u_r = 0$ when $r = 0$.

I didn't understand how to do this problem, so I wanted to attempt a similar problem from my textbook, in the hope that it will give me the experience needed to understand how to complete the other problem:

Consider the change of variable to polar coordinates: $x = r \cos(\theta)$, $y = r \sin \theta$. Use the chain rule to obtain $u_r$ and $u_\theta$ in terms of $u_x$ and $u_y$ and hence show that

$$\partial_x = \cos(\theta) \partial_r - \frac{1}{r} \sin(\theta) \partial_\theta$$

$$\partial_y = \sin(\theta) \partial_r + \frac{1}{r} \cos(\theta) \partial_\theta$$

Hence, by considering

$$\partial_x^2 u = (\cos(\theta) \partial_r - \frac{1}{r} \sin(\theta) \partial_\theta)(\cos(\theta) \partial_r - \frac{1}{r} \sin(\theta) \partial_\theta)u $$

or otherwise, show that

$$u_{xx} + u_{yy} = u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta \theta}$$

So the first thing I did was attempt to use change of variables:

$$\frac{\partial{u}}{\partial{x}} = \frac{\partial{u}}{\partial{x}} \frac{\partial{x}}{\partial{r}} + \frac{\partial{u}}{\partial{x}} \frac{\partial{x}}{\partial{\theta}} = \frac{\partial{u}}{\partial{x}} \cos(\theta) + \frac{\partial{u}}{\partial{x}} (-r \sin(\theta))$$

$$\frac{\partial{u}}{\partial{y}} = \frac{\partial{u}}{\partial{y}} \frac{\partial{y}}{\partial{r}} + \frac{\partial{u}}{\partial{y}} \frac{\partial{y}}{\partial{\theta}} = \frac{\partial{u}}{\partial{y}} \sin(\theta) + \frac{\partial{u}}{\partial{y}} (r \cos(\theta))$$

Why do I think the derivatives are done this way? Because we have $u(x(r, \theta), y(r, \theta))$, so I think this is the only way it makes sense.

But this is obviously different to the two equations that the author mentioned:

$$\partial_x = \cos(\theta) \partial_r - \frac{1}{r} \sin(\theta) \partial_\theta$$

$$\partial_y = \sin(\theta) \partial_r + \frac{1}{r} \cos(\theta) \partial_\theta$$

Did I do the change of variables wrong? I'm unsure of how to proceed from here.

Also, the original problem seems different from this: It asks us to show that $u_x(x, y) = f'(r) \cos(\theta)$, where $u = f(r)$. How is the change of variables different in this case? The way $f(r)$ is used here is especially confusing for me, since I do not have a lot of practice with the chain rule.

I apologise for doing (or not doing, haha) change of variables so badly. I was only taught how to do it for integration, but never have I done it like this. I want to understand these problems well, because that'll help me understand change of variables in this context (rather than just integration). Thank you.


2 Answers 2


I'll show you my method, which I prefer over @Niki Di Giano's.

You can obtain the partial derivatives without solving for $r$ and $\theta$ in terms of $x,y$.

First, obtain the $r$ and $\theta$ partial derivatives first. Using the multivariable chain rule

\begin{align} \frac{\partial u}{\partial r} &= \frac{\partial u}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial y} = \cos\theta\frac{\partial u}{\partial x} + \sin\theta \frac{\partial u}{\partial y} \\ \frac{\partial u}{\partial \theta} &= \frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta} = -r\sin\theta\frac{\partial u}{\partial x} + r\cos\theta\frac{\partial u}{\partial y} \end{align}

You can treat this as a system of linear equations

\begin{align} \cos\theta\frac{\partial u}{\partial x} + \sin\theta \frac{\partial u}{\partial y} &= \frac{\partial u}{\partial r} \\ -\sin\theta\frac{\partial u}{\partial x} + \cos\theta\frac{\partial u}{\partial y} &= \frac{1}{r}\frac{\partial u}{\partial \theta} \end{align}

where $u_x$ and $u_y$ are the unknowns. Use whatever method you want to solve for the $x$ and $y$ partials, and obtain

\begin{align} \frac{\partial u}{\partial x} &= \cos\theta\frac{\partial u}{\partial r} - \frac{\sin\theta}{r}\frac{\partial u}{\partial \theta} \\ \frac{\partial u}{\partial y} &= \sin\theta\frac{\partial u}{\partial r} + \frac{\cos\theta}{r}\frac{\partial u}{\partial \theta} \end{align}

I prefer to do this when the inverse coordinate change is less straightforward and it's more convenient to solve the linear system. Otherwise, you should get the same result.

You can treat $\frac{\partial }{\partial x}$ as an operator and apply it twice, i.e

\begin{align} \frac{\partial^2 u}{\partial x^2} &= \left(\cos\theta\frac{\partial }{\partial r} - \frac{\sin\theta}{r}\frac{\partial }{\partial \theta} \right)\left(\cos\theta\frac{\partial u}{\partial r} - \frac{\sin\theta}{r}\frac{\partial u}{\partial \theta} \right) \\ &= \cos\theta\frac{\partial }{\partial r}\left(\cos\theta\frac{\partial u}{\partial r}\right) - \frac{\sin\theta}{r}\frac{\partial }{\partial \theta}\left(\cos\theta\frac{\partial u}{\partial r}\right) \\ &\quad - \cos\theta\frac{\partial }{\partial r}\left(\frac{\sin\theta}{r}\frac{\partial u}{\partial \theta}\right) + \frac{\sin\theta}{r}\frac{\partial }{\partial \theta}\left(\frac{\sin\theta}{r}\frac{\partial u}{\partial \theta}\right) \\ &= \dots \end{align}

and likewise for $u_{yy}$ to prove the Laplacian identity.


Your way of applying the chain rule to the partial differentiation is almost correct. However, if you're trying to obtain the partial derivative with respect to $x$, and $x$ is a function of other variables (in this case, $r$ and $\theta$), you have to consider the sum of how much the function varies over $r$, $\theta$ and then consider how much they vary over $x$. Without further ado: $$ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial x} $$ With the expressions for $r$ and $\theta$ given as functions of both $x$ and $y$. Can you take it from here?

HINT: Given $x(r, \theta)$ and $y(r, \theta)$, we determine: $$r^2 = x^2 + y^2 \quad \tan \theta ={ y\over{x}} \\ r = \sqrt{x^2 + y^2} \quad \theta =\arctan{ y\over{x}}$$ The arctan function is a bit finicky. For example, it gives you the wrong $\theta$ when both $y$ and $x$ are negative. You will have to add $\pi$ to the value of the function to get the correct angle when that happens. However, since we're talking about derivatives, a constant is not going to change much.$\\$ With that adjustment in mind, we can now find the correct values for the partial derivatives: $${\partial{r}\over{\partial{x}}} =\frac{1}{2} \frac{2x}{\sqrt{x^2 + y^2}} = \frac{r \cos \theta}{r} = \cos \theta \\ {\partial{\theta}\over{\partial{x}}} =\frac{-\frac{y}{x^2}}{1 + \frac{y^2}{x^2}} = - \frac{\sin\theta}{r}$$

  • $\begingroup$ Hmm, but how do we get the values for the derivatives in $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial x}$ when our equations are in the form $x = r \cos \theta$ and $y = r \sin \theta$? That's what made me do it the other way. $\endgroup$
    – user423167
    Commented Aug 3, 2018 at 9:49
  • $\begingroup$ Can you determine the expression of $r$ and $\theta$ as functions of $x$ and $y$? I'll add in some hints so you can check. $\endgroup$ Commented Aug 3, 2018 at 10:20
  • $\begingroup$ Thanks for the help! I got it now. So was that change of variables? Or is there something more we must do? What was the point of getting $\partial_x = \cos(\theta) \partial_r - \frac{1}{r} \sin(\theta) \partial_\theta$ and $\partial_y = \sin(\theta) \partial_r + \frac{1}{r} \cos(\theta) \partial_\theta$ when it comes to change of variables? $\endgroup$
    – user423167
    Commented Aug 3, 2018 at 11:55
  • $\begingroup$ If you're familiar with change of variables and their Jacobian matrices, you may consider what you're doing here akin to transforming the space using the Jacobian for this particular change of variables. It is also connected to change of variables while integrating a multivariate function. The question is always the same: what linear function does my function look like, locally, in the space described by this pair (or more) of coordinates? $\endgroup$ Commented Aug 3, 2018 at 12:02
  • 1
    $\begingroup$ Always glad I could help. $\endgroup$ Commented Aug 3, 2018 at 12:24

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