Given the equation y'=(ax+by+f(x,y))/(cx+dy+varphi(x,y))…

Given the equation $$y'=\frac{ax+by+f(x,y)}{cx+dy+\varphi(x,y)},$$ where $f$ and $\varphi$ are defined and continuous in a neighbourhood of $(0,0)$ and vanish together with their first partial derivatives at $(0,0)$, prove that if the equation is unchanged under the substitution $y\mapsto-y$, and the roots of the characteristic equation $$\begin{vmatrix}c-\lambda & d\\ a & b-\lambda\end{vmatrix}=0$$ are purely imaginary, then the singular point $(0,0)$ is a centre.

Step 1 · Structure imposed by the symmetry condition

The equation is unchanged when $y$ is replaced by $-y$, which forces the right-hand side $R(x,y)$ to satisfy $R(x,-y)=-R(x,y)$: $R$ is an odd function of $y$. For the linear part this requires $b(-y)=-by$ and $d(-y)=-dy$ automatically, while the nonlinear remainders $f,\varphi$ must combine so the quotient remains odd in $y$.

Step 2 · Symmetry of the phase portrait with respect to the $x$-axis

Since $R(x,-y)=-R(x,y)$, if $y=\Phi(x)$ is an integral curve then so is $\tilde\Phi(x)=-\Phi(x)$: $$\tilde\Phi'(x)=-\Phi'(x)=-R(x,\Phi(x))=-R(x,-\tilde\Phi(x))=R(x,\tilde\Phi(x)).$$ Hence the phase portrait is symmetric with respect to the $x$-axis: every integral curve above it has a mirror image below it.

Step 3 · Purely imaginary eigenvalues — linearised behaviour

The characteristic equation expands to $\lambda^2-(b+c)\lambda+(bc-ad)=0$. Purely imaginary eigenvalues $\lambda_{1,2}=\pm\beta i$ ($\beta>0$) require $b+c=0$ (zero trace) and $bc-ad>0$ (positive determinant), so $\beta=\sqrt{bc-ad}$.

Step 4 · Local behaviour near the origin

The associated planar system is $\dot x=cx+dy+\varphi(x,y)$, $\dot y=ax+by+f(x,y)$. Since $f,\varphi$ vanish to first order at $(0,0)$, they are $o(r)$ as $r\to0$. The linearised system has purely imaginary eigenvalues, so the linear part alone has a centre at the origin; for the full nonlinear system this is, by itself, inconclusive (the nonlinear terms could turn the centre into a focus) — the symmetry resolves the ambiguity.

Step 5 · Closure of trajectories via the Poincar\'e return map

Let $\Sigma=\{(x,0):x>0\}$ be a cross-section near the origin, and let $\Pi:\Sigma\to\Sigma$ be the Poincar\'e first-return map. A trajectory starting at $P_0\in\Sigma$ traces an upper arc $\Gamma^+$ to some return point $P_1\in\Sigma$. By the $x$-axis symmetry of Section~2, the reflected lower arc $\Gamma^-$ is also an integral curve joining $P_0$ to $P_1$. Traversing $\Gamma^+$ forward and $\Gamma^-$ backward forms a closed curve through $P_0$; by uniqueness of solutions (the right-hand side is Lipschitz in a punctured neighbourhood of the origin), the trajectory through $P_0$ coincides with this closed curve. Hence $\Pi(P_0)=P_0$ for all sufficiently small $P_0\in\Sigma$.

Step 6 · The role of each hypothesis

• Purely imaginary eigenvalues: guarantee the linearised flow rotates around the origin rather than forming a node or saddle.
• Symmetry $y\mapsto-y$: forces the upper and lower arcs of each trajectory to be exact reflections, making the Poincar\'e map the identity.
• Vanishing of $f,\varphi$ to first order: ensures the nonlinear terms do not disrupt the rotation and that uniqueness of solutions holds near the origin.

Step 7 · Conclusion

$$\boxed{ \text{If the eigenvalues are purely imaginary and the equation is symmetric under } y\mapsto-y,\text{ then }(0,0)\text{ is a \textbf{centre}.} }$$ The symmetry forces the Poincar\'e return map along the $x$-axis to be the identity: every orbit starting on the positive $x$-axis is closed, since its upper and lower halves are exact reflections of each other. Combined with the rotational behaviour guaranteed by the purely imaginary eigenvalues, this proves that the singular point is a centre.

Summary Table

Additional Notes

• The argument hinges on relating the scalar equation to its associated planar system and translating the geometric property (centre, exactness, symmetry) into an algebraic condition on the trace and determinant of the corresponding matrix.