Consider the initial boundary value problem (IBVP)
\(\rm \left\{\begin{matrix}u_t+u_x=2u,&x>0; t>0\\\ u(0, t)=1+\sin t,&t>0\\\ u(x, 0)=e^x\cos x,&x>0\end{matrix}\right.\) If u is the solution of the IBVP, then the value of \(\rm \frac{u(2\pi, \pi)}{u(\pi, 2\pi)}\) is
1
eπ
2
e-π
3
-eπ
4
-e-π