Solving a PDE with Giving Condition So I have come to one problem in Applied PDE Chapter 1, and the problem is give as following.
Solve the PDE $u_t+u_x=0$, with $u(\cos\theta,\sin\theta)=\theta$, $0\leq\theta<2\pi$.
I have a general idea that by solving $u_t+u_x=0$, we'll obtain that $C = x-t$, which indeed gives $u(x,t) = f(x-t)$.
But then there's no such initial value such as $u(x,0)=f(x)$ given. 
So what should I have done to deal with the giving condition associating with the given PDE to solve for $u$? 
Thanks for any suggestions!
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{\mrm{u}\pars{x,t} = \mrm{f}\pars{x - t}}$.

Then,
\begin{align}
\theta & =
\mrm{f}\pars{\cos\pars{\theta} - \sin\pars{\theta}} =
\mrm{f}\pars{\cos\pars{\theta + \pi/4} \over \cos\pars{\pi/4}} =
\mrm{f}\pars{\root{2}\cos\pars{\theta + {\pi \over 4}}}
\end{align}
Lets 

$\ds{\xi \equiv \root{2}\cos\pars{\theta + \pi/4}}$
  
  $\ds{\implies
\theta = \arccos\pars{\root{2}\xi/2 - \pi/4}\implies
\mrm{f}\pars{\xi} = \arccos\pars{\root{2}\xi/2 - \pi/4}}$

$$
\bbx{\mrm{u}\pars{x,t} =
\arccos\pars{{\root{2} \over 2}\,\bracks{x - t} - {\pi \over 4}}}
$$
