Burgers' equation with boundary conditions Consider the signaling problem
$u_t + c(u)u_x= 0, t> 0, x> 0$
$u(x, 0) = u_0, x> 0,$
$u(0, t) = g(t), t> 0,$
where $c$ and $g$ are given functions and $u_0$ is a positive constant. If $c
(u) > 0$,
under what conditions on the signal $g$ will no shocks form? Determine the
solution in this case in the domain $x > 0, t> 0$.
Here's what I have:
Characteristic lines are given by:
$x = ut + x_o$
Solving for $du/dt$ gives:
$u = k(x_o)$
The initial condition gives:
$u(x_o, 0) = u_o = k(x_o) = k(ut-x)$
$u(x-ut, 0) = k(ut-x)$
The boundary condition gives:
$u(0, t) = g(t)$
$u(0, \frac{x-x_o}{u}) = g(\frac{x-x_o}{u})$
$u(x,t) = g(\frac{x-x_o}{u})$
But from here I am quite confused... I don't see how I could solve for $u$ or determine conditions on $g$ where a shock will not develop?
 A: $$u_t+C(u)u_x=0\quad\text{where }C(u)\text{ is a given function}$$
GENERAL SOLUTION :
The system of characteristic differential equations is :
$$\frac{dt}{1}=\frac{dx}{C(u)}=\frac{du}{0}$$
A first equation of characteristic cuves comes from $du=0\quad\to\quad u=c_1$ .
A second equation of characteristic cuves comes from $\frac{dt}{1}=\frac{dx}{C(c_1)}\quad\to\quad x-C(c_1)t=c_2$
The general solution of the PDE is expressed on the form of implicit equation $\Phi\left(c_1,c_2\right)=0$ where $\Phi$ is any differentiable function of two variables.
$$\Phi\left(u\:,\:x-C(u)t\right)=0$$
This is a manner to express any relationship between the two variables. This is equivalent to express the relationship by any function $F$ :
$$x-C(u)t=F(u)$$
where $F$ is any derivable function. 
In general, this implicit equation cannot be solved for $u$ on closed form.
DETERMINATION OF THE FUNCTION $F$ according to the condition $u(0,t)=g(t)$ :
$0-C\left(g(t)\right)t=F\left(g(t)\right)$ 
Let $g(t)=X \quad\to\quad t=g^{-1}(X)\quad$ where $g^{-1}$ is the inverse function of $g$.
$$-C(X)g^{-1}(X)=F(X)$$
Thus the function $F$ is now determined, given the functions $C$ and $g$.
PARTICULAR SOLUTION FITTED WITH THE GIVEN CONDITION :
With the particular function $F$ found above :
$x-C(u)t=F(u)$ with $F(u)=-C(u)g^{-1}(u)$
$$x-C(u)t=-C(u)g^{-1}(u)$$
$$x+C(u)\left(g^{-1}(u)-t \right)=0$$
$g^{-1}(u)=t-\frac{x}{C(u)}$
$$u=g\left(t-\frac{x}{C(u)}\right)$$
The result is on the form of implicit equation. Solving for $u$ in order to obtain an explicit form $u(x,t)$ is generally not possible, except in case of particular functions $C$ and $g$.
