For an introductory partial differential equations class, I'm trying to prove that the solution $u=u(t,x)$ for the initial value problem $$ \begin{cases}au_t+bu_x = 0\\ u(0,x) = u_0(x)\end{cases} , \quad a,b \in\mathbb{R}.$$ exists.
I know that using the Characteristic's method, we find the solution $$u(t,x) = f(ax-bt) $$ where $f$ is an arbitrary differentiable function.
Does the function $f$ always exists? What if $f\not\in\mathcal{C}^1$?
Should $u_0(x)\in\mathcal{C}^1$?
PS: I've tried to use Cauchy-Lipschitz (or Picard–Lindelöf) theorem, for the ODE in terms of $\xi = at+bx$
$$\dfrac{\partial u}{\partial\xi} = 0 $$
but i don't know how to proceed, cause in this case the initial condition $u_0(x)$ is a function, and the theorem statement is for $u_0(x)$ constant.
Thanks in advance.
$$ \begin{cases}au_t+bu_x = 0\\ u(0,x) = u_0(x)\end{cases}$$
You general solution is correct : $$u(t,x) = f(ax-bt) $$
Initial condition : $$u(x,0)=f(ax)=u_0(x)$$ Let $X=ax$ . The function $f(X)$ is determined : $$f(X)=u_0\left(\frac{X}{a}\right)$$ We put this function into the above general solution where $X=ax-bt$ $$u(x,t)=u_0\left(\frac{ax-bt}{a}\right)=u_0\left(x-\frac{b}{a}t\right)$$ In so far the function $u_0(x)$ is a given function thus supposed to be an unique existing function, I don't see the necessity to give more prove that $u_0\left(x-\frac{b}{a}t\right)$ is an unique existing function.