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For the normalized equations $\dot{x}=x(1-y)$ and $\dot{y}=ry(x-1)$ on the positive quadrant,is a first integral. It is nonnegative, strictly convex, and tends to infinity at the boundary and at infinity. Every positive regular level is therefore a compact simple closed curve traversed by one periodic orbit.

For the particular equation, write $D_x=x{\partial }_x$ and $D_y=y{\partial }_y$. Its left side is $(D_x^2-D_y^2)u$. Choose $\varphi =xy$ and $\psi =x/y$; these are smooth invertible coordinates in the positive quadrant, with inverse $x=\sqrt{\varphi \psi }$, $y=\sqrt{\varphi /\psi }$. In them $D_x=\varphi {\partial }_{\varphi }+\psi {\partial }_{\psi }$ and $D_y=\varphi {\partial }_{\varphi }-\psi {\partial }_{\psi }$. Subtraction of their squares leaves $4\varphi \psi u_{\varphi \psi }$, so the canonical equation is $u_{\varphi \psi }=0$. Integrating it gives $u=F(\varphi )+G(\psi )$.

Parameterize the data curve by $(x,y)=(s,s^2)$, $s\ge 1$. Its coordinate values are $(\varphi ,\psi )=(s^3,s^{-1})$, soDifferentiate the first identity along the curve to get $3s^2{F}^{\prime }(s^3)-s^{-2}{G}^{\prime }(s^{-1})=6s^5-s^{-2}$. Adding gives $F^{\prime }(s^3)=2s^3$, and then $G^{\prime }(s^{-1})=1$. Thus $F(\varphi )={\varphi }^2+C$ for $\varphi \ge 1$ and $G(\psi )=\psi -C$ for $0<\psi \le 1$. ThereforeThe coordinate rectangle gives exactly the regionIts boundary consists of $xy=1$ up to $(1,1)$ and the ray $y=x$ from that point onward. It is unbounded upward. The original data curve lies inside this region. The sketch displays both characteristic boundaries and that data arc.

To establish the exact extent of uniqueness, outside this region either $\varphi <1$ or $\psi >1$. A smooth function supported in that unprescribed coordinate range can be added to $F$ or $G$ without changing either datum on the arc. Hence the displayed global formula is one extension, but data alone do not uniquely determine values outside the shaded region. Boundary values are fixed by continuity. The following classification criteria are pointwise and assume a nonzero principal coefficient matrix.

The equation is hyperbolic at a point when $\boxed{b^2-ac>0}$. The principal form has one positive and one negative eigenvalue and two distinct real characteristic directions. In the given equation $a=x^2$, $b=0$, $c=-y^2$, so $b^2-ac=x^2{y}^2>0$ throughout the positive quadrant, consistent with the two canonical characteristic families used above.

Let $X,Y$ be prey and predator abundances. With constant positive demographic and predation coefficients, the Lotka-Volterra equations areTheir positive equilibrium is $X_{\ast }=\gamma /\delta$, $Y_{\ast }=\alpha /\beta$. Its Jacobian matrix and eigenvalues areThe linearized populations oscillate with period $2\pi /\sqrt{\alpha \gamma }$, with the predator peak following the prey peak. This is a center equilibrium, not an attracting equilibrium. The conclusion is supported beyond linearization by the Logarithmic first integral of the Lotka-Volterra equations: in these units one may useDirect differentiation gives $\dot{V}=0$. Its Hessian matrix is a positive-definite matrix in the positive quadrant, and its minimum is at $(X_{\ast },Y_{\ast })$. Nearby level curves are closed periodic orbits, so there is no damping toward a preferred oscillation amplitude.

For $a=b=0$, take the Logarithmic first integral of the Lotka-Volterra equationsAlong a trajectory,Thus $V$ is a first integral. The scalar function $h(s)=s-\log s-1$ is nonnegative for $s>0$, vanishes only at $s=1$, and tends to infinity as $s\downarrow 0$ or $s\to \infty$. Moreover,is positive definite, so $V$ is strictly convex. It follows that $(1,1)$ is its unique critical point and every level set $V=c>0$ is a compact smooth simple closed curve in the positive quadrant. The vector field is nonzero on such a level curve, so uniqueness of solutions makes each connected level curve one periodic orbit. The zero level consists of the fixed point $(1,1)$. Hence every trajectory in $\Lambda$ is either periodic or the fixed point $(1,1)$.

Use the Dulac theorem on the simply connected positive quadrant with $\varphi (x,y)=1/(xy)$. If $F=(\dot{x},\dot{y})$, thenand henceFor $ab<0$, both terms have the same strict sign throughout $\Lambda$. The Bendixson-Dulac criterion therefore gives no periodic orbit in $\Lambda$. The same computation also excludes the case $ab=0$ unless $a=b=0$.

SetBecause the region lies in the positive quadrant, its four inequalities becomeThus this change of variables sends $D$ to a rectangle in the $uv$-plane. Its Jacobian determinant issoSince $x^2=uv$, the double integral is

The two coordinate axes divide a Cartesian coordinate system in the plane into four open quadrants, distinguished by the signs of their coordinates. The positive quadrant is the first quadrant.