Context - Method of characteristics to integrate PDE given initial conditions:
$$ u_t + a u_x + b u = f(t, x) \tag{1a.}$$ $$ u(0, x) = u_0(x) \tag{1b.}$$
I am quoting some intermediate results here for brevity:
The substitution that has been done: $$ t \rightarrow \tau \tag{2}$$ $$ x \rightarrow \xi + a \tau \tag{3}$$ $$ \widetilde{u}(\tau, \xi) = u(t,x) \tag{4}$$
a, b are arbitrary constants.
After performing the variable change, the solution to the original PDE (1a.) becomes:
$$(e^{bs} \widetilde{u}(s, \xi))|_{s = 0}^{s = \tau} = \int_{0}^{\tau} f(s,\xi + a s) e^{bs} \;ds \tag{5}$$
$$e^{b\tau} \widetilde{u}(\tau, \xi) - \widetilde{u}(0, \xi) = \int_{0}^{\tau} f(s,\xi + a s) e^{bs} \;ds \tag{6}$$ $$e^{b\tau} \widetilde{u}(\tau, \xi) = \widetilde{u}(0, \xi) + \int_{0}^{\tau} f(s,\xi + a s) e^{bs} \;ds \tag{7}$$
The book I am referring to now, after going back to the original variables, substitutes:
$$\widetilde{u}(0, \xi) = u_0(x - a t) $$
I am not convinced with that substitution as is $$\widetilde{u}(0, \xi)$$ not evaluated at $$ \tau = 0 $$ and does it not mean (?) from (2) that $$t = 0$$
Final expression in the book:
$$e^{bt} u(t, x) = u_{0}(x - a t) + \int_{0}^{t} f(s, x - a(t - s)) e^{bs} \;ds \tag{8}$$
Could someone please point out what rule of substitution I am breaking? I also have some problem accepting the final expression involving "f"
Thanks!