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Orient the circle counterclockwise and define its Cauchy transformThe inside and outside limits satisfy ${\Phi }_{+}-{\Phi }_{-}=\varphi$ and ${\Phi }_{+}+{\Phi }_{-}=(\pi i{)}^{-1}\mathrm{PV}{\int }_C\varphi (\tau )/(\tau -t)d\tau$. Also ${\Phi }_{-}(z)=O(z^{-1})$ at infinity. Applying these formulas to the two polynomial coefficients and dividing by $t^2$ gives the Riemann–Hilbert conditionThe homogeneous jump is $G=t/(t^2-1)$. A canonical factorization of a rational scalar Riemann-Hilbert jump isThe inside factor is analytic and nonvanishing; the exterior factor has no zeros or poles outside the circle and grows like $z$. The jump index is $1-2=-1$, consistently with this growth.

Put $F_{+}={\Phi }_{+}$ and $F_{-}={\Phi }_{-}/X_{-}$. Their additive jump is $R(t)=((A-1)t+1)/(t^2-1)$. Since $R$ is analytic outside the circle and decays there, the functions $F_{+}$ inside and $F_{-}+R$ outside glue to an entire function. This function tends to zero at infinity, so Liouville's theorem gives $F_{+}=0$, $F_{-}=-R$. ConsequentlyThe required decay of a Cauchy transform forces $A=1$. Then ${\Phi }_{-}=-1/z$, and the jump givesFor the homogeneous problem, ${\Phi }_{+}/X_{+}$ and ${\Phi }_{-}/X_{-}$ glue to an entire function vanishing at infinity, so only the zero solution exists; this proves uniqueness. Direct verification uses the principal-value integral $-\pi i/t$: the two terms on the left combine to $(t^4+t^3-t^2-t^4+t^3+t^2)/(2t)=t^2$, exactly the right side at $A=1$.