I have to find the general solution for the following initial value problem:
$
\left.
\begin{eqnarray}
u_t+div(a\cdot u)=0, (t,x)\in(0,T)\times\mathbb{R}^d\atop
u(0,x)=u_0(x), x\in\mathbb{R}^d, a\in\mathbb{R}^d
\end{eqnarray}
\right\}
$
I've working on the one dimensional case and I have found by the method of characteristics that the solution is $u(t,x)=u_0(x-at)$. How can I generalize this?
I am guessing what you meant to write is the higher dimensional advection equation with a constant velocity $\textbf{a}$: $$ u_t(\textbf{x},t) = - \nabla \cdot \textbf{a} u(\textbf{x},t).$$ You can separate variables: $$u = \prod_i U_i(x_i,t),$$ giving $$\sum_i \partial_t U_i(x_i,t)\prod_{j\neq i}U_j(x_j,t) = -\sum_i \partial_{x_i} a_i U_i(x_i,t)\prod_{j\neq i}U_j(x_j,t),$$ or rearranging, $$ \sum_i \Big[\partial_t U_i(x_i,t) + \partial_{x_i}a_i U_i(x_i,t)\Big] = 0.$$ This means each separated component satisfies the 1D problem: $$ \partial_t U_i(x_i,t) + \partial_{x_i}a_i U_i(x_i,t) = 0.$$ That is, $U_i(x_i,t) = U_{i,0}(x_i-a_i t)$. Therefore the general solution to your problem is $$u(\textbf{x},t) = \prod_i U_{i,0}(x_i-a_i t),$$ i.e. $$ u = u_0(\textbf{x}-\textbf{a} t).$$