Laplacian from cartesian to polar

Question

I am following this derivation or proof (which one would I call it?). Most of it is straight forward but there is a jump on a specific line that I do not understand. I am going to draw it out here.

Take the following as given: \begin{align} x &= r\cos\theta;\qquad y = r\sin\theta;\qquad u=u(x,y);\\\\ \frac{\partial u}{\partial r}&=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial r}=\cos\theta\frac{\partial u}{\partial x} + \sin\theta\frac{\partial u}{\partial y}. \tag{1}\label{main} \end{align}

I am attempting to convert the Laplace equation to polar. The proof that I am following says that $$ \frac{\partial^2 u}{\partial r^2}= \cos\theta\frac{\partial}{\partial r}\frac{\partial u}{\partial x} + \sin\theta\frac{\partial }{\partial r}\frac{\partial u}{\partial y}\label{alien} \tag{2} $$ This makes sense, just taking the partial derivative with respect to $r$ to all of the terms on both sides. The following, I do not understand and would appreciate explanation:

$$ \begin{align} \frac{\partial^2 u}{\partial r^2}= \cos\theta \left( \frac{\partial}{\partial x}\frac{\partial u}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial }{\partial y}\frac{\partial u}{\partial y}\frac{\partial y}{\partial r} \right) + \sin\theta\left( \frac{\partial}{\partial x}\frac{\partial u}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial }{\partial y}\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}\right ).\tag{3} \label{marsh} \end{align}$$

If I assumed that this could happen, then next line is a bit intuitive, but also odd: \begin{align} \frac{\partial^2 u}{\partial r^2}= \cos^2\theta\frac{\partial^2 u}{\partial x ^2} + \sin^2\theta\frac{\partial u^2}{\partial y^2} + 2\cos\theta\sin\theta\frac{\partial^2 u}{\partial x \partial y}.\label{dobop} \tag{4} \end{align} This seems a bit intuitive because of $\ref{main}$ has the equality: \begin{align} \frac{\partial u}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial r}=\cos\theta\frac{\partial u}{\partial x} + \sin\theta\frac{\partial u}{\partial y}. \end{align} Just looking at it I could be convinced that: \begin{align} \frac{\partial x}{\partial r} = \cos \theta \qquad \frac{\partial y}{\partial r} = \sin \theta. \end{align}

Here are my main questions:

• What is going on between $\ref{alien}$ and $\ref{marsh}$?
• What is going on between $\ref{marsh}$ and $\ref{dobop}$?

Side question:

• Any good books that would help me learn about partial derivatives and what is right and wrong to do?

Answer 1

Q1. I will assume you understand where Eq. (1) comes from, your first question is actually an application of the same idea (there is a typo in your Eq. (3)). In general if you have a function $g = g(x, y)$ and $x = r\cos \theta$, $y = r\sin \theta$ then

$$ \frac{\partial g}{\partial r} = \frac{\partial x}{\partial r}\frac{\partial g}{\partial x} + \frac{\partial y}{\partial r}\frac{\partial g}{\partial y} = \cos\theta\frac{\partial g}{\partial x} + \sin\theta\frac{\partial g}{\partial y}, $$

if $g = u$ then you recover your first equation. If $g = \partial u/\partial x$ then

$$ \frac{\partial}{\partial r}\left(\frac{\partial u}{\partial x} \right) = \frac{\partial x}{\partial r}\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}\right) + \frac{\partial y}{\partial r}\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial x}\right) = \cos\theta\frac{\partial^2 u}{\partial x^2} + \sin\theta\frac{\partial^2 u}{\partial y\partial x}, $$

and similarly if $g = \partial u/\partial y$

$$ \frac{\partial}{\partial r}\left(\frac{\partial u}{\partial y} \right) = \frac{\partial x}{\partial r}\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial y}\right) + \frac{\partial y}{\partial r}\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial y}\right) = \cos\theta\frac{\partial^2 u}{\partial x\partial y} + \sin\theta\frac{\partial^2 u}{\partial y^2}. $$

Eq. (3) just follows from adding these two results

Q2. Eq. (4) just follows from Eq. (3)

\begin{eqnarray*} \frac{\partial ^2 u}{\partial r^2} &=& \cos\theta\left( \cos\theta\frac{\partial^2 u}{\partial x^2} + \sin\theta\frac{\partial^2 u}{\partial y\partial x} \right) + \sin\theta \left( \cos\theta\frac{\partial^2 u}{\partial x\partial y} + \sin\theta\frac{\partial^2 u}{\partial y^2}\right) \\ &=& \cos^2\theta \frac{\partial^2 u}{\partial x^2} + \sin^2\theta \frac{\partial^2 u}{\partial y^2} + 2 \sin\theta\cos\theta \frac{\partial^2 u}{\partial y\partial x} \end{eqnarray*}

Q3. There are many good Multivariable Calculus books, it would be unfair to say which is best. This is a good reference

https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/