Solve $$\label{eq_0}\tag{1}\begin{cases} \partial_tu + (x-1) \partial_xu = 0\\ u(0,x) = x^3 \end{cases}$$ in the plane region $Q : = \{(t,x) \in \Bbb R^2 : x<1\}$
Using characteristic I found $$\label{eq_1}\tag{2}\begin{cases} \dot t = 1\\ \dot x = x-1 \end{cases}$$ so we have that for this ODE system the following characteristic curve: $\xi\in \Bbb R$ $$(x-1)\cdot e^{-t} = \xi\label{eq_2}\tag{3}$$ Using initial condition knowing from above that our solution has to be in the form $u(t,x) = f(\xi)$, I compute $$u(t,x) = [(x-1) e^{-t}+1]^3$$ I've checked via derivation that the solution and it looks alright. Here's the issue, in computing $\eqref{eq_2}$ via $\eqref{eq_1}$ in the calculation appears $\ln(x-1)$ that is defined for $x>1$ clearly in contrast with the request in $\eqref{eq_0}$. Is the solution unacceptable? The issues is that actually computing the integral should be $\ln|x-1|$? Thank you for the help!
Edit- Solving $\eqref{eq_1}$ : $$\frac{dx}{dt} = x-1\\ \frac{dx}{x-1} = dt \\ \int \frac{dx}{x-1} = \int dt \\ \ln(x-1) = t + c \\ x-1 = e^t\cdot \xi$$
If you can verify the answer does solve the PDE, and it is clearly well defined over the region Q, then the solution should be acceptable regardless on how you got there.