Solving a simple $2$-dimensional, linear, first order PDE with method of characteristics. How can I solve the following problem using the method of characteristics?
\begin{equation}
t\frac{\partial u}{\partial t} (t,x) + x\frac{\partial u}{\partial x}(t,x) =  1,
\\
u(0,x) = \varphi(x).
\end{equation}
The initial function $\varphi$ is $\mathcal{C}^1$ and defined on the whole real line. 
I tried using the method of characteristics, but all characteristic integral curves seem to lie in the $xu$-plane. 
Is it even possible to solve this equation with the method of characteristics? If not, why not?
 A: With the given "initial condition" on $u(0,x)$, the answer is no: you can't guarantee the existence of a unique solution. This is because you've prescribed characteristic initial data. 
The vector field $ t\partial_t + x\partial_x$ is the "radial" vector field pointing out from the origin $(0,0)$. Your "data" is prescribed along one (two, if you count the parts with positive and negative $x$ differently) integral curves. Hence you data is only compatible with the equation if you know $x \varphi'(x) = 1$, which incidentally requires $\varphi(x) = \ln(x) + C$. 
Even if $\varphi$ is compatible, the solution is non-unique: re-writing in terms of the polar coordinates $(r,\theta)$ for $\mathbb{R}^2 \setminus \{0\}$, your differential equation can be rewritten as $r \partial_r u(r,\theta) = 1$. 
This means that any function of the form $u(r,\theta) = C(\theta) + \ln(r)$ is a solution, and the function $C(\theta)$ can be chosen arbitrarily. And there are infinitely many different functions $C(\theta)$ compatible with the prescribed $\varphi$. 
