I'm trying to solve rather formally one-dimensional transport equation: $$ u_{t}+cu_{x}=0\quad\text{in $(0,\infty)\times(-\infty,\infty)$} $$ with an initial data $u_{0}$, which is bounded and uniformly continuous, by using Laplace transform.
I now get $$ \mathcal{L}[u(t,x)](s)=\frac{1}{c}\int_{-\infty}^{x}\exp\left(-\frac{s}{c}(x-y)\right)u_{0}(y)dy $$ but I don't know how to get the function $u$. The solution (in some weak sense if necessary) should have the form of $u(t,x)=u_{0}(x-ct)$ and so I'm going toward this form.
Please give hints, comments or solutions. Thank you in advance.
Why not a substitution? $y=x-ct \implies dy=-cdt$, and $$ \mathcal{L}[u(t,x)](s) = \frac{1}{c} \int_{-\infty}^x \mathrm{exp}\left(-\frac{s}{c}(x-y)\right)u_0(y)dy = \int_0^\infty \mathrm{exp}(-st)\,u_0(x-ct)\,dt,$$ So $u(t,x) = u_0(x-ct)$.