PDE-Charactestic Method I have the following problem :
Let $a,b,c$ be real numbers, and $ (t, x) ∈ [0, +∞[×[1, 2[→ u(t, x) ∈ \Bbb R $ such that :
$ u_t + cu_x - au - bu^2 = 0 $ and  $ u(0,x) = u_0(x) $

i) Show that $ v(t)=u(t,x+ct) $ is solution of an ODE, and solve it, assuming $u$ verify the PDE above.

I'm not entirely sure how to deal with this, but that's what I got so far :
$v'(t) = u_t(t, X(t, 0, x)) + (d/dt)X(t, 0, x)u_x(t, X(t, 0, x)) $ where $X(t, 0, x)=x+ct$
$v'(t) = u_t(t, X(t, 0, x)) + cu_x(t, X(t, 0, x)) $
since $u$ is solution of the PDE : $v'(t) = av(t) + bv(t)^2 $
$v(t) = \dfrac{a}{a.C.e^{-at} - b}$

ii) Use this result to solve the PDE when :

*

*

*

*$b=0$



*


*$a=-b>0$ and $u_0(x)∈[1,2]$



*


*$a=0, b=1$  and  $u_0(x)<0$ bounded



*


*$a=0, b=1$ and $u_0=-e^{-x^2}$

there I don't really see how to proceed while using $v$,
 A: This isn't a direct answer to the question where some ambiguities appear in the notations as pointed out in comments. Hopping that the final resut below will help you, at least to compare and check your own result.
$$u_t+c\,u_x=a\,u+b\,u^2$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dt}{1}=\frac{dx}{c}=\frac{du}{u+bu^2}$$
A first characteristic equation comes from solving $\frac{dt}{1}=\frac{dx}{c}$ :
$$x-ct=c_1$$
A second characteristic equation comes from solving $\frac{dt}{1}=\frac{du}{u+bu^2}$ :
$$(b+\frac{a}{u})e^{at}=c_2$$
The general solution of the PDE on implicit form $c_2=F(c_1)$ is:
$$(b+\frac{a}{u})e^{at}=F(x-ct)$$
$F$ is an arbitrary function until no condition is taken into account.
$$\boxed{u(t,x)=\frac{a}{-b+e^{-at}F(x-ct)}}$$
CONDITION : $u(0,x)=u_0(x)$
$$u(0,x)=u_0(x)=\frac{a}{-b+e^{-a*0}F(x-c*0)}=\frac{a}{-b+F(x)}$$
$$F(x)=b+\frac{a}{u_0(x)}$$
Now the function $F$ is determined. We put it into the above general solution with argument $(x-ct)$ thus $F(x-ct)=b+\frac{a}{u_0(x-ct)}$
$$\boxed{u(t,x)=\frac{a}{-b+e^{-at}\left(b+\frac{a}{u_0(x-ct)} \right)}}$$
This is the particular solution satisfying both the PDE and the condition.
Note : I guess that there is a typo in the wording of the question : It should be $v=u(t,x-ct)$ instead of $v=u(t,x+ct)$.
A: Following your solution for $v$, which is indeed the computation along a characteristic curve, you get
$$
av^{[x]}(t)^{-1}+b=(au_0(x)^{-1}+b)e^{-at}\\
v^{[x]}(t)=\frac{au_0(x)}{(a+bu_0(x))e^{-at}-bu_0(x)}
$$
Now in reverse you get the PDE solution as
$$
u(t,x)=u(t,(x-ct)+ct)=v^{[x-ct]}(t)=\dfrac{a}{a·C(x-ct)·e^{-at}-b}
\\=\frac{au_0(x-ct)}{(a+bu_0(x-ct))e^{-at}-bu_0(x-ct)}.
$$
You might have to take care of boundary conditions if the PDE is restricted to the strip $x\in[1,2]$.
