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For a quasilinear partial differential equation, the characteristic equations for a quasilinear equation are the ordinary differential equationsA solution surface is locally a differentiable graph $z=u(x,y)$ whose tangent plane contains the vector $(P,Q,R)$ at every point. Its normal is $(-u_x,-u_y,1)$, so the tangency condition is precisely $P{u}_x+Q{u}_y=R$. Thus every characteristic lying on that graph follows the vector field without leaving the graph.

Parameterize the given initial curve by $\gamma (t)=(x_0(t),y_0(t),z_0(t))$. For each $t$, solve the characteristic equations for a quasilinear equation with initial value $\gamma (t)$ at $s=0$. The resulting map $(s,t)\mapsto (x(s,t),y(s,t),z(s,t))$ sweeps out a candidate solution surface. If the coefficients and initial data are continuously differentiable, the Picard-Lindelöf theorem supplies this local family. To make it a graph we also need the projection $(s,t)\mapsto (x,y)$ to be locally invertible. At $s=0$ its Jacobian determinant isThe Noncharacteristic Cauchy data condition is $J(0,t)\ne 0$. The inverse function theorem then gives $s=s(x,y)$ and $t=t(x,y)$, and hence $u(x,y)=z(s(x,y),t(x,y))$. Its derivatives obey the PDE by the chain rule. The same characteristic initial-value uniqueness gives local uniqueness of this solution wherever the projection remains invertible and the relevant characteristic meets the prescribed data.

The same variation of parameters construction works on $[0,\pi ]$. All solutions satisfying the boundary condition at zero have the formAt $x=\pi$, the homogeneous term vanishes for every $C$; it can no longer be chosen to correct the final boundary value. Since $\sin (\pi -t)=\sin t$, that boundary value is zero exactly whenThis condition is necessary by the displayed general solution, and sufficient by the same formula. When it holds, the full answer isFor complex-valued forcing, allow $C\in \mathbb{C}$ instead. The freedom is addition of the nonzero homogeneous Dirichlet solution $e^{-2x}\sin x$. At this endpoint separation, the homogeneous boundary problem has a nonzero solution. The resulting Fredholm solvability condition restricts the forcing to the compatible functions described above. In the language of the Fredholm alternative, the condition is orthogonality to the adjoint null solution $e^{2x}\sin x$ of $D^2-4D+5$ with zero endpoint values.

For this forcing the integral in part (a) simplifies toConsequently the unique boundary value problem solution isIt vanishes at zero and at $\pi /2$. To check the differential equation directly, set $y=e^{-2x}w$. Then $Ly=e^{-2x}(w^{\prime \prime }+w)$. For $w=1-\sin x-\cos x$, one has $w^{\prime \prime }+w=1$, as required.

For the linear differential operator $L=D^2+4D+5$, the characteristic equation is $m^2+4m+5=0$, with roots $-2\pm i$. Take the fundamental system of solutionsThe Wronskian is never zero, so variation of parameters is valid on the whole interval. Write a particular solution as $a(x)u(x)+b(x)v(x)$, and impose the auxiliary condition $a^{\prime }u+b^{\prime }v=0$. Substitution then gives $a^{\prime }{u}^{\prime }+b^{\prime }{v}^{\prime }=f$. These two equations giveChoosing both integrals to start at zero makes the resulting particular solution and its first derivative vanish there:A general solution is $Au+Cv+y_p$. The boundary condition at zero sets $A=0$. The condition at $\ell =\pi /2$ givesbecause $\sin \ell =1$ and $\sin (\ell -t)=\cos t$. Combining the two integrals yields the Green's function representationIndeed, for $t<x$ the kernel is $-e^{2(t-x)}\sin t\cos x$, since $\sin (x-t)-\sin x\cos t=-\cos x\sin t$; for $x<t$ it is $-e^{2(t-x)}\sin x\cos t$.

For fixed interior $t$, the Green's function vanishes at both endpoints, is continuous at $x=t$, solves the homogeneous equation away from $t$, and satisfiesThis derivative jump is the normalization required by the leading coefficient one in $L$. Finally, a homogeneous solution satisfying the first boundary condition is $C{e}^{-2x}\sin x$, and the second boundary condition at $\pi /2$ forces $C=0$. Thus the boundary value problem has exactly one solution for every continuous forcing, justifying the unique inverse represented by this kernel.

The Bendixson-Dulac theorem states that if $(P,Q)$ is a continuously differentiable planar vector field on a simply connected open region $D$, and a continuously differentiable Dulac function $B$ makeshave one weak sign throughout $D$, without vanishing identically on any nonempty open subset, then there is no nonconstant periodic orbit contained in $D$. To see why, a hypothetical simple periodic orbit bounds a region in $D$. The divergence theorem gives the integral of this expression as the outward flux of $B(P,Q)$ across the orbit. The flux is zero because the vector field is tangent to its orbit, whereas the one-sign area integral is nonzero, a contradiction. The usual stricter version assumes a strictly positive or strictly negative expression everywhere.

For the printed polynomial system, the origin is an equilibrium andThus its linearization is a centre equilibrium. For the full system take $B=1$. The divergence isOn the open disk $x^2+y^2<3$ this is nonpositive, and is negative wherever $x\ne 0$. Its zero set inside the disk is a line, so it cannot vanish on any open subset. The Bendixson-Dulac theorem therefore provesThis also exposes a terminology problem in the source: a nonlinear centre equilibrium has surrounding closed orbits, so the printed claim of a centre cannot hold in that sense. The imaginary eigenvalues establish only a centre of the linearized system. In fact the full equilibrium is an asymptotically stable weak focus, as the following direct check shows.

A quartic Lyapunov correction for a weak focus makes the nonlinear damping visible. Put $r^2=x^2+y^2$ and consider the Lyapunov functionIt is positive definite sufficiently close to zero. Differentiating along the polynomial vector field givesFor clarity, the quartic terms come from $-x^4$ in the derivative of $r^2/2$, and from applying $y{\partial }_x-x{\partial }_y$ to the quartic correction. Their sum is $-3x^4/8-3x^2{y}^2/4-3y^4/8$; all other terms have degree at least six. Therefore $\dot{V}<0$ in a sufficiently small punctured neighbourhood, proving asymptotic stability by Second Lyapunov theorem. Finallyso nonconstant nearby orbits turn clockwise as they converge. The equilibrium is a weak focus despite the centre in its linearization. This interpretation resolves both the local dynamics and the absence of closed orbits without changing the printed coefficients.

The first component of the vector field vanishes only when $y=0$ or $x=\pm 1$. The second vanishes only when $y=1$ or $x+y=3$. Intersecting these alternatives gives exactly five equilibria. Their Jacobian matrix isAt $(3,0)$,The real parts are negative and the imaginary parts are nonzero, so $(3,0)$ is a stable spiral.

At $(1,1)$ the Jacobian matrix is $\mathrm{diag}(2,-1)$. Its real eigenvalues have opposite signs, so $(1,1)$ is a saddle equilibrium.

At $(1,2)$,Both eigenvalues are positive, so $(1,2)$ is an unstable node.

At $(-1,1)$ the Jacobian matrix is $\mathrm{diag}(-2,-3)$, giving a stable node.

At $(-1,4)$,so $(-1,4)$ is a saddle equilibrium. All five equilibria are hyperbolic, so these linear classifications determine their nonlinear local stability as well.