Using polar co-ordinates to solve first order linear PDE

Question

Let $u = u(x,y)$ be a scalar fields on $\mathbb R^2$. Consider the linear first order PDE for $u$,$$(x\partial_y - y\partial_x)u = 0$$ By introducing $\partial_{\phi}u = 0$planar polar co-ordinates $r$ and $\phi$, show that the PDE is equivalent to $\partial_{\phi}u = 0$, and thus find its general solution.

Ok so far I have tried to find its general solution first:

First, rewrite the above PDE as \begin{align} \partial_x u+\frac{y}{x}\partial_y u=0 \ \ (1) \end{align} Using the method of characteristic, further rewrite $(1)$ as follows \begin{align} \frac{d}{dx}u(x, y(x))= \partial_x u+ y'(x)\partial_y u = \partial_x u+\frac{y}{x}\partial_y u=0, \end{align} we have the ode \begin{align} y' = \frac{y}{x} \ \ \Rightarrow \ \ y = Cx. \end{align} Let us impose the artifical initial condition $y(1) = y_0$ and $u(1, y_0) = f(y_0)$ . Hence it follows \begin{align} y = y_0x \ \ \Rightarrow \ \ y_0 = \frac{y}{x}. \end{align} which means \begin{align} u(x, y) = u\left(x, y_0x\right) = f(y_0) = f\left(\frac{y}{x} \right). \end{align}

I am having trouble using planar polar co-ordinates $r$ and $\phi$ to show that the PDE is equivalent to $\partial_{\phi} u = 0$. Any help to solve that bit of the problem will be appreciated.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{x = r\cos\pars{\theta}\,,\quad y = r\sin\pars{\theta}}$.

\begin{align} \partiald{}{r} & = \partiald{x}{r}\,\partiald{}{x} + \partiald{y}{r}\,\partiald{}{y} = \cos\pars{\theta}\,\partiald{}{x} + \sin\pars{\theta}\,\partiald{}{y} \\[5mm] \partiald{}{\theta} & = \partiald{x}{\theta}\,\partiald{}{x} + \partiald{y}{\theta}\,\partiald{}{y} = -r\sin\pars{\theta}\,\partiald{}{x} + r\cos\pars{\theta}\,\partiald{}{y} \end{align}

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$$ \left\{\begin{array}{rcrcl} \ds{\cos\pars{\theta}\,\partiald{}{x}} & \ds{+} & \ds{\sin\pars{\theta}\,\partiald{}{y}} & \ds{=} & \ds{\partiald{}{r}} \\ \ds{-r\sin\pars{\theta}\,\partiald{}{x}} & \ds{+} & \ds{r\cos\pars{\theta}\,\partiald{}{y}} & \ds{=} & \ds{\partiald{}{\theta}} \end{array}\right. $$

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$$ \left\{\begin{array}{rcrcl} \ds{\partiald{}{x}} & \ds{=} & \ds{\cos\pars{\theta}\,\partiald{}{r}} & \ds{-} & \ds{{\sin\pars{\theta} \over r}\,\partiald{}{\theta}} \\[2mm] \ds{\partiald{}{y}} & \ds{=} & \ds{\sin\pars{\theta}\,\partiald{}{r}} & \ds{+} & \ds{{\cos\pars{\theta} \over r}\,\partiald{}{\theta}} \end{array}\right. $$

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\begin{align} &\left\{\begin{array}{rcrcl} \ds{y\,\partiald{}{x}} & \ds{=} & \ds{{1 \over 2}\,r\sin\pars{2\theta}\,\partiald{}{r}} & \ds{-} & \ds{\sin^{2}\pars{\theta}\,\partiald{}{\theta}} \\[2mm] \ds{x\,\partiald{}{y}} & \ds{=} & \ds{{1 \over 2}\,r\sin\pars{2\theta}\,\partiald{}{r}} & \ds{+} & \ds{\cos^{2}\pars{\theta}\,\partiald{}{\theta}} \end{array}\right. \\[5mm] \implies &\ \bbox[10px,#ffe,border:1px dotted navy]{\ds{% x\,\partiald{}{y} - y\,\partiald{}{x} = \partiald{}{\theta}}} \quad \implies \quad \bbox[10px,#ffe,border:1px dotted navy]{\ds{% \partiald{u}{\theta} = 0}} \end{align}