I have what I imagine is the simplest partial differential equation one could ask for but I'm uncertain how its solved. We are initially told $u(x,t) = f(\chi,\tau)$ where $\chi = x +bt$ and $\tau = t$. We also have an initial condition where $u(x,0) = x +1$.
I was able to rewrite the initial complicated PDE of $u(x,t)$ as:
$$ \frac {\partial f}{\partial \tau} + bf = 0 $$
I want to solve for $f(\chi,\tau)$ and then get the general solution for $u(x,t)$. My initial idea was to solve with simple integration, which gave me $$f(\chi,\tau) = e^{-b\tau} + C(\chi) $$
I don't know how to solve for the constant term with $\chi$, but I would think it would require an initial condition. Thus, I have to get $u(x,t)$ and then use the initial condition given. But how does one write $u(x,t)$ without knowing what $C(\chi)$ is?
Any guidance would be appreciated.
You have: $f(\chi,\tau)=e^{-b\tau}+C(\chi)$
You know
$\chi=x+bt,\tau=t,u(x,t)=f(\chi,\tau)$
Substitute things now, $$u(x,t)=e^{-b[t]}+C([x+bt])$$
Your initial condition is: $u(x,0)=x+1$, so $$u(x,0)=e^{-b\cdot0}+C(x+b\cdot0)=1+C(x)=x+1\rightarrow C(x)=x$$
Therefore, $u(x,t)=e^{-bt}+x$.
Is that it?