First extract the polynomial content from $f$:The primitive quartic is Eisenstein at the prime $2$: every nonleading coefficient is divisible by $2$, while the constant term is not divisible by $4$. Eisenstein's criterion makes it irreducible over $\mathbb{Q}$, and Gauss's lemma makes it irreducible in $\mathbb{Z}[X]$. The constant factor $2$ is itself irreducible in $\mathbb{Z}[X]$.

The polynomial $g$ is already irreducible over $\mathbb{Q}$: it is Eisenstein at $3$. Its nonleading coefficients, including the missing zero coefficients, are divisible by $3$, and $-21$ is not divisible by $9$. Thus $g$ is its own sole nonconstant irreducible factor over $\mathbb{Q}$.

Over the field ${\mathbb{Z}}_3$, grouping terms givesThe first two factors are linear and therefore irreducible. The quadratic takes the values $1,2,2$ at $0,1,2$ respectively, so it has no root in ${\mathbb{Z}}_3$. A degree-two polynomial over a field is reducible exactly when it has a root, proving its irreducibility.

All three factorizations are unique up to order and multiplication by units. The ring $\mathbb{Z}$ is a unique factorization domain, hence so is $\mathbb{Z}[X]$ by Gauss's lemma. A polynomial ring over a field is a Euclidean domain, so both $\mathbb{Q}[X]$ and ${\mathbb{Z}}_3[X]$ have unique factorization. The units are $\pm 1$ in the first ring, nonzero rational constants in the second, and $1,2$ in the third.

An equilibrium has $y=-x/2$ from the first equation. Substitution in the second gives $-x(x^2+3)/2=0$, so the origin is the only real equilibrium. Its Jacobian matrix isPurely imaginary linear eigenvalues alone do not classify the nonlinear equilibrium. Use insteadIts derivatives are $H_x=2x+y+x^3/2$ and $H_y=x+2y$, so the vector field is $(H_y,-H_x)$. The chain rule gives $\dot{H}=H_x{H}_y-H_y{H}_x=0$. Therefore the nonstationary trajectories in the phase portrait lie on the level curvesThe quadratic part is positive definite, with eigenvalues $1/2$ and $3/2$, so $H$ has a strict local minimum at zero. Small positive level sets are smooth closed ovals surrounding it, and the vector field never vanishes on them. They are periodic orbits. Thus the origin is a nonlinear centre, Lyapunov stable but not asymptotically stable. The near-origin ellipses have their long axis along $y=-x$. At $(x,0)$ with $x>0$, the velocity points downward, fixing clockwise orientation.

The characteristic equations for a transport equation are $\dot{x}=1$, $\dot{y}=2$, $\dot{z}=-3z$. Label an initial point by $s\in [0,1]$ and choose characteristic time zero on the data segment. Integration givesThus $s=x-y/2$ and $t=y/2$, givingThis is the full infinite strip between the parallel characteristic lines $y=2x$ and $y=2x-2$. Both positive and negative characteristic times are allowed because the problem specifies no forward-time restriction. Each characteristic in this strip meets the data segment exactly once. Outside it, the general solution $z=e^{-3y/2}F(x-y/2)$ contains unrestricted values of $F$, so the data do not determine a unique solution there.

If two solutions exist, their difference satisfies $d^{\prime \prime }+d=0$ with $d(0)=d(L)=0$, where $L=\pi /4$. The general solution is $d=a\sin x+b\cos x$. The first endpoint gives $b=0$ and the second gives $a\sin L=0$, hence $a=0$. Thus the solution is unique if it exists.

For the operator $-D^2-1$, choose the homogeneous solutions $\sin x$ and $\sin (L-x)$ satisfying the left and right endpoint conditions. Continuity at $x=s$ and derivative jump $G_x(s^{+},s)-G_x(s^{-},s)=-1$ yield the Dirichlet Green functionIndeed it vanishes at both endpoints and satisfies $-G_{xx}-G={\delta }_s$. Splitting the integral at $s=x$ and differentiating verifies that $y={\int }_0^{L}G(x,s)f(s)ds$ is twice continuously differentiable for continuous $f$ and solves the boundary-value problem, establishing existence as well.

At zero, $G_x(0,s)=\sin (L-s)/\sin L=\cos s-\sin s$, since $\cot L=1$. Thus