Could someone check my reasoning on this problem?
I'm supposed to solve this IVP by finding $u$.
$\left\{ \begin{array}{lr} u_t+b\cdot Du +c u=0, & \mbox{ in } \mathbb{R}^n \times (0, \infty)\\ u=g & \mbox{ on }\mathbb{R}^n \times \{ t=0\}.\end{array} \right.$
Where $c\in \mathbb{R}$ and $b\in \mathbb{R}^n$ are constants.
So, using the method of characteristics,
I let $v(x,t)=e^{ct}u(x,t)$. Which gives $Dv(x,t)=e^{ct}Du(x,t)$ and $v_t(x,t)=ce^{ct}u(x,t)+e^{ct}u_t(x,t)$.
So substituting $v_t$ and $Dv$, I get
$v_t(x,t)+b\cdot Dv=e^{ct}[u_t(x,t)+bDu(x,t)+cu(x,t)]=0.$
and for $t=0$, I also have $v=u(x,0)=g.$
So, I have a new PDE to solve.
$\left\{ \begin{array}{lr} v_t+b\cdot Dv=0, & \mbox{ in } \mathbb{R}^n \times (0, \infty)\\ v=g & \mbox{ on }\mathbb{R}^n \times \{ t=0\}.\end{array} \right.$
But this is just the usual homogenous transport equation with solution $v(x,t)=g(x-tb)$
Thus $v(x,t)=e^{ct}u(x,t)$ implies that $u(x,t)=e^{-ct}v(x,t)=e^{-ct}g(x-tb).$
The second part of the problem asks to find a physical interpretation of $c$ and $cu$ as it relates to the derivation of the equation.
I'm clueless when it comes to physics so I might be way off base.
But given that the transport equation uses conservation laws, it seems for this equation something is either "growing" or "decaying" depending on whether c is positive or negative. In which case, $cu$ is the concentration of whatever is being transported that is either shrinking or growing exponentially. Is this correct?