find $\varphi \in C_{c}^{1}\left(\mathbb{R}_{*}^{+} \times \mathbb{R}\right)$ such that $\partial_{t} \varphi +c \partial_{x} \varphi=f$

Question

let $f \in C_{c}\left(\mathbb{R}_{*}^{+} \times \mathbb{R}\right)$, I am trying to find $\varphi \in C_{c}^{1}\left(\mathbb{R}_{*}^{+} \times \mathbb{R}\right)$ such that $\partial_{t} \varphi +c \partial_{x} \varphi=f$ where $c \in \mathbb{R}$

I don't really have an idea of how to attack this problem

Answer 1

Try the method of characteristics. We need this auxiliar system of ODEs,

$$\dfrac{dt}{1}=\dfrac{dx}{c}=\dfrac{d\varphi}{f(t,x)}$$

Integrating the first proportion we get, $x-ct=c_1$

We need a second equation. In order to get it we use the first and third ratios and the first equation for the $x$ dependence for $f$ ($x=ct+c_1$)

$$\dfrac{dt}{1}=\dfrac{d\varphi}{f(t,x)}$$

$$\int_{t_0}^t f(\tau,c\tau+c_1)d\tau+c_2=\varphi$$

or

$$\varphi=\int_{t_0}^t f(\tau,c\tau+x-ct)d\tau+c_2$$

Now, $c_1$ and $c_2$ must be related: $c_2=g(c_1)$, with $g$ a single variable, differentiable function. Then

$$\varphi(t,x)=\int_{t_0}^t f(\tau,c(\tau-t)+x)d\tau+g(x-ct)$$