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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?

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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 $\varphi$.

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  • $\begingroup$ Thank you for your answer! Two minor questions: 1.) So the fundamental problem is the fact that my initial data is prescribed exactly along one of the characteristic lines? 2.) Given any quasi-linear first oder equation with compatible initial data, is there a fast way to directly see whether it can be solved by method of characteristics? $\endgroup$
    – Joker123
    Jul 24, 2019 at 15:01
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    $\begingroup$ (1) yes. (2) locally: yes, if the data is compatible (in the sense that it is not characteristic) then a local solution exists and is unique. In the 2nd edition of Evans' PDE textbook, this is done in section 3.2.4. globally: no, at least if you require strong solutions. For example, consider the equation $x\partial_x u + y\partial_y u = 0$ with data prescribed on the unit circle. You have unique local $C^1$ solution for any $C^1$ data. But unless the data is constant, the solution cannot extend globally to a $C^1(R^2)$ function. $\endgroup$
    – Willie Wong
    Jul 24, 2019 at 17:13

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