+ \lambda   u =  f\\
  u(0,x,v) =  h_0(x,v)
\end{gathered} \right.$$ Let $u_-(t,x,v)= \max \{-(  u(t,x,v)),0 \}$. In what follows, we will use the notation ${\ensuremath{\left\langle \cdot , \cdot  \right\rangle}}$ to denote the $L^2$ inner product in $(t,x,v)$ and ${\ensuremath{\left\| \cdot \right\|}}_2$ to denote the corresponding induced norm. It follows immediately from [@Degond; @Tartar] that $$\label{Tartar}
{\ensuremath{\left\langle \frac{{\partial}u}{{\partial}t} + \hat{v}_1 {\partial}_x   u,   \ u_- \right\rangle}}= \frac{1}{2} \left( \iint {\ensuremath{\left|  u_-(0,x,v)\right|}}^2 \, dx \, dv - \iint {\ensuremath{\left|  u_-(t,x,v)\right|}}^2 \, dx \, dv \right) .$$ Using this, we find $$\begin{aligned}
{\ensuremath{\left\langle  f,   u_- \right\rangle}} & = & {\ensuremath{\left\langle \mcL   u + \lambda   u,   u_- \right\rangle}}\\
& = & {\ensuremath{\left\langle \frac{{\partial}u}{{\partial}t} + \hat{v}_1 {\partial}_x   u,   u_- \right\rangle}} 
+ {\ensuremath{\left\langle F \cdot \nabla_v  u,   u_- \right\rangle}} - {\ensuremath{\left\langle {\nabla}_v \cdot (D {\nabla}_v \cdot u),   u_- \right\rangle}} 
+ \lambda {\ensuremath{\left\langle   u,   u_- \right\rangle}}  \end{aligned}$$ For the last term we split the integral into two portions, namely $$\begin{aligned}
{\ensuremath{\left\langle   u,   u_- \right\rangle}} &= \int_0^T\iint u(t,x,v) u_{-}(t,x,v)\,dv\,dx\,dt\\ & = \int_{A} u(t,x,v) u_{-}(t,x,v)\,dv\,dx\,dt + \int_{A^c} u(t,x,v) u_{-}(t,x,v)\,dv\,dx\,dt\end{aligned}$$ where $A = {\{(t,x,v) : u(t,x,v) \geq 0\}}$. On the set $A$, we have $u_{-}(t,x,v) = 0$, and the corresponding integrals vanish. On $A^c$ we have $u_{-}(t,x,v) = -u(t,x,v)$ and hence 