Problem: $$ \begin{cases} u_t = u_{xx} +5 u_{x} &\text{ in } R_+\times(0,1)\\ u(0,x)=g(x) &\text{ for } x\in(0,1)\\ u_x(t,0)=\alpha(t) &\text{ for } t>1\\ u_x(t,1)=\beta(t) &\text{ for } t>1 \end{cases}$$
To show that the problem has at most one solution I assumed that there are two solutions $u,v$. Now let $w=u-v$. It's easy to see that $w$ satisfies equation with zero boundary conditions.
I started with considering energy function $E(t)=\int_0^1 w^2(t,x)dx$
$$E'(t)=\int_0^12ww_tdx=\int_0^12w(w_{xx}+5w_x)dx=\int_0^12ww_{xx}dx+\int_0^110ww_xdx = 2(ww_x)|_0^1-\int_0^12w^2_xdx+\int_0^110ww_xdx=0-\int_0^12w^2_xdx+5(w^2(t,1)-w^2(t,0))$$ And that's the moment when I got stuck. What can I do about this difference, I think that it's probably just zero, but I don't know how to show that.
To show the difference is zero you can first note that at $t=0$ it is zero and then differentiate it to get:
$2(w(t,1)w_t(t,1)-w(t,0)w_t(t,0))=0$
from $w_t=w_{xx}+5w_x$ and your boundary conditions give you $w_x(t,0)=w_{xx}(t,0)=0$, $w_x(t,1)=w_{xx}(t,1)=0$. So it is zero as you thought!