# 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? $$$$t\frac{\partial u}{\partial t} (t,x) + x\frac{\partial u}{\partial x}(t,x) = 1, \\ u(0,x) = \varphi(x).$$$$

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?

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 $$\varphi$$.
• (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. Jul 24, 2019 at 17:13