 Solution

With $A$ invertible, write the system as ${\mathbf{u}}_x+M{\mathbf{u}}_y=\mathbf{d}$, $M=A^{-1}B$. Real distinct eigenvalues of $M$ give strict hyperbolicity; a real complete eigenbasis gives the usual diagonalizable hyperbolic form. The characteristic curves satisfy $dy/dx={\lambda }_i$. Left eigenvectors form characteristic dependent combinations $w_i={\ell }_i\mathbf{u}$; for constant coefficients they obey separate transport equations. Characteristic coordinates are functions constant on those curves, such as $y-{\lambda }_{i}x$ in the constant case. For variable coefficients extra derivatives of the eigenvectors can couple the combinations; local characteristic directions still follow the real matrix pencil. If $A$ is singular one first chooses a noncharacteristic coordinate direction with an invertible coefficient combination.

For the specified constants,Its speeds are $2$ and $-1$, with left eigenvectors $(2,1)$ and $(1,1)$. HenceThe transported Riemann invariants are $\boxed{R_{+}=2u+v-4x,R_{-}=u+v-x}$, constant on $y-2x=\text{constant}$ and $y+x=\text{constant}$, respectively. Initial data give $w_{+}(0,s)=s-\sin (s/2)$ and $w_{-}(0,s)=-e^s$, soInvert these two combinations:Both characteristic feet must lie on the supplied positive initial half-line. ThusOutside this sector at least one invariant needs unsupplied initial data. The closed expressions give one possible extension there but are not uniquely forced by the question's initial half-line.