How can we prove the uniqueness in $[0, +\infty)\times (0,1)$ of the solution of a PDE as the following:
$\frac{\partial}{\partial t}v(t,x)=(\frac{1}{2}x-\frac{1}{4})\frac{\partial}{\partial x}v(t,x)+\frac{1}{2}v(t,x)$
$v(0,x)=g(x)$,
where $g\in C^\infty(0, 1)$? Thanks for your help
You need a value for $u$ along each of the characteristic curves (passing by the domain) to form the surface solution in order for the solution to be unique. The intial conditions give such values if the curve along the values of $u$ are given cuts every one of the characteristic curves. A sufficient condition is that the curve for the boundary conditions ($t=t_0(s);x=x_0(s)$ with $v(t_0(s),x_0(s))=u_0(s)$) is nowhere tangent to any of the characteristic curves, so is considering $a(t,x,v)v_t+b(t,x,v)v_x=c(t,x,v)$, that $b(t_0,x_0,v_0)\dfrac{dt_0}{ds}-a(t_0,x_0,v_0)\dfrac{dx_0}{ds}\neq 0$
In your problem, $t_0=0,\,x_0=s,\,u_0=g(s)$ and $a=1,\,b=1/4-x/2$. Clearly the no tangency condition is fullfilled for the given domain.
Drawing the curve for the initial conditions and the characteristics helps to understand the problem. In this problem, the characteristics are determined by the intersection of:
$$\begin{array}\ x=c_1e^{-t/2}+1/2\\ v=c_2e^{t/2} \end{array}$$
and suitable values for $c_1$ and $c_2$. The curve along the values for $v$ are given is $t=0$ that clearly cuts every of the projected characteristics at $x=c_1+1/2$