I have this problem
$v_t + xv_x = x v^2 $
With $v(t,1)=t^2$
My solution is
The initial curve is ($T,1,T^2)$ which hapoen when s = 0 and we have characteristic
$$dx/ds = x, dt/ds = 1, dv/ds = xv^2$$
Which i solved them to be
$$ x= e^s, t= s+T, u = \frac{ T^2+1- T^2 e^s }{T^2} $$
Now i just need to solve fot T and s and replace in u, correct¿ is my approach correct .?
I agree with your system of differential equations : $$dx/ds = x, dt/ds = 1, dv/ds = xv^2$$ or on an equivalent form (Charpit-Lagrange) : $$\frac{dt}{1}=\frac{dx}{x}=\frac{dv}{xv^2}=ds$$ A first characteristic equation comes from $\frac{dt}{1}=\frac{dx}{x}$ $$t-\ln|x|=c_1$$ A second characteristic equation comes from $\frac{dx}{x}=\frac{dv}{xv^2}$ $$\frac{1}{v}+x=c_2$$ An implicit form of the general solution of the PDE is : $$\frac{1}{v}+x=F(t-\ln|x|)$$ $F$ is an arbitrary function, to be determined according to the condition :
$v(t,1)=t^2\quad$ leading to $\quad\frac{1}{t^2}+1=F(t-\ln|1|)=F(t)$
So, the function $F(X)$ is determined : $$F(X)=\frac{1}{X^2}+1$$ We put it into the above general solution where $X=t-\ln|x|$ $$\frac{1}{v}+x=\frac{1}{ (t-\ln|x|)^2}+1$$ This is the particular solution (on implicit form) which satisfies the PDE and the boundary condition.
For the explicit solution, solve it for $v$ : $$v(x,t)= \frac{1}{\frac{1}{(t-\ln|x|)^2} +1-x}$$ $$v(x,t)= \frac{(t-\ln|x|)^2}{1+(1-x)(t-\ln|x|)^2 }$$