General solution of Transport equation (homogeneous): Method of Characteristics
$$u_t+cu_x=0 (\star)$$
We know that if $f: \mathbb{R} \to \mathbb{R}$ is differentiable then $u(x,t)=f(x-ct)$ is a solution of $(\star)$.
We will show that each solution of $(\star)$ is of the form $u(x,t)=f(x-ct)$, where $f$ is an arbitrary differentiable function, $f: \mathbb{R} \to \mathbb{R}$.
We consider the lines $\epsilon_a: x-ct=a, a \in \mathbb{R}$.
We define the function $z(s)=u(x(s),t(s))$, where $x(s)=cs+a$, $t(s)=s$.
Then $(x(s), t(s)) \in \epsilon_a \forall s \in \mathbb{R}$. Let $u$ be a solution of $(\star)$ and for this $u$ I define $z(s)$ as previously.
Then $z'(s)=c u_x+ u_t=0$, i.e. $z(s)=\beta \forall s \in \mathbb{R}$.
$u(x(s),t(s))=z(s)=z(0)=u(a,0)=u(x(s)-ct(s),0) \forall s$
Thus $u(x,t)=u(x-ct,0)=:f(x-ct)$, where $f: \mathbb{R} \to \mathbb{R}$ is differentiable.
I am looking at the following exercise:
Find the general solution of the non-homogeneous Transport equation $u_t+cu_x=g(x,t)$ where $g$ is "smooth" (how smooth?) and using the above find the general solution of the wave equation.
Hint: We work as previously : $z'(s)=g(x(s),t(s))$.
So in this case do we know that $u(x,t)=f(x-ct)$ is again a solution of the given equation?
Or do we have to consider an other solution and show that all the solutions are of this form?
here is one way to do $$u_t + cu_x = g(x, t), u(x,0) = f(x).\tag 1 $$
define the characteristic lines parametrically by $x = a+cs, t = s.$ you can verify that $$\frac{d}{ds}u(a+cs, s)=u_t+cu_x=g(a+cs, s) \tag 2$$ integrating $(2)$ from $0$ to $t,$ gives you $$u(a+ct, t)=f(a)+\int_0^tg(a+cs, s)\, ds $$ which can be rewritten as $$u(x, t) = f(x-ct)+\int_0^t g(x-ct + cs,s)ds$$ is a solution of $(1).$