Assume we would like to solve PDE by method of characteristics, and take one specific example to be the following, where $a$ is a constant.
$$ \partial_t u(x,t) + a \partial_x u(x,t) = 0 $$
Then, the idea is to parametrize both $x$ and $t$ with parameter $s$ to get the following.
$$ u'(x(s), t(s)) = t'(s) \partial_t u(x,t) + x'(s) \partial_x u(x,t) $$
This expression is zero if we choose $x'(s) = a$, $t'(s) = 1$. Then, we know that for arbitrary parameter $s^*$, we have $u(x(s), t(s)) = u(x(s^*), t(s^*))$.
Solving ODEs we get that $x(s) = as + C_1$ and $t(s) =s+ C_2$ where $C_1, C_2$ are some constants.
Question: I would like to show that general solution of such problem is $u(x,t) = u(x-at)$.
My attempt and confusion: The way I would proceed is the following. Intuitively, ($C_1, C_2)$ are initial coordinates from which characteristic curve goes with parameter $s$ until it hits $(x,t)$. In other words, if $(x,t)$ is given, then, for given $s$, I can compute what should be initial conditions of such curve. This can be done and the result is that $C_1(x,t,s) = x-as$ and $C_2(x,t,s) = t-s$. From here, we get the following.
$$ u(x,t) = u(C_1 (x,t,s), C_2 (x,t,s)) = u(x-as, t-s) $$
I have seen that in some cases people set $C_2 = 0$ but this is not clear to me why could that be allowed. For example, I imagine that some characteristic curves might be circles around $(1,1)$ with radius $<1$ and then none of the initial conditions would go through axis.
I appreciate your help!
The general solution is not $u(x,t)=u(x-at)$ but is : $$u(x,t)=F(x-at)$$ where $F$ is an arbitrary function (to be deterined according to some condition to be specified).
For example if the condition is $u(x,0)=x^3$ on $x^2+t^2=1$ $$u(x,0)=F(x-a(0))=F(x)=x^3$$ So the function $F$ is determined. We put it into the above general solution where $F(\chi)=\chi^3$ with $\chi=(x-at)$. Thus $F(x-at)=\chi^3=(x-at)^3$. The solution is : $$u(x,t)=(x-at)^3$$ The condition $u(x,0)=x^3$ is satisfied and the PDE is also satisfied.
Note : It seems a bit strange to write "I imagine that some characteristic curves might be circles around $(1,1)$ " because there is no $y$ in this problem. So, no cercle $x^2+y^2=1$. But this is OK if we consider $t$ as a geometrical ordinate but not "time", so that $x^2+t^2=1$ is the equation of a circle.