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A planar node whose Jacobian matrix has a repeated nonzero real eigenvalue and only one independent eigenvector is an improper node. In suitable real coordinates its linearization of a dynamical system has the displayed form, with solutionsIt is a stable node for $\lambda <0$ and an unstable node for $\lambda >0$. The factor $t$ makes generic trajectories tangent to the unique eigendirection as they approach the equilibrium in the attracting time direction. For a nonlinear autonomous system with this hyperbolic equilibrium, the sign of $\lambda$ still determines local stability; the local phase portrait has the same topological type as its linearization.

A node is a planar hyperbolic equilibrium whose two linearized eigenvalues are real and have the same sign. Negative eigenvalues give a stable node; positive eigenvalues give an unstable node. Nearby motion approaches or recedes without the repeated rotation characteristic of a focus.

For $\dot{x}=-ax$, $\dot{y}=-by$ with $a,b>0$, solutions are $x=x_0{e}^{-at}$ and $y=y_0{e}^{-bt}$. Each coordinate decays, while unequal decay rates usually make a trajectory align with the slower direction as it nears the origin. Reversing both signs gives outward motion. Repeated eigenvalues require attention to the number of independent eigenvectors and can add polynomial factors to the exponential response. The node classification describes local behavior near an equilibrium; it does not determine the geometry of remote orbits.

Intersect the two nullcline families. The axes yield $(0,0)$ and $(1,0)$, and $y=1$ with $xy=2$ yields $(2,1)$. The possible remaining intersection would have $x+y=1$ and $xy=2$, requiring $x^2-x+2=0$, whose discriminant is $-7$. Hence there are no further real equilibria.

The Jacobian matrix isEvaluate it at each equilibrium:

All three are hyperbolic equilibria, so their classifications persist for the nonlinear system. Immediately to the right of $(2,1)$ on $y=1$, the velocity is vertically downward, showing that the attracting spiral turns clockwise.

The second equilibrium point equation gives $y=-(x+1)/2$. Substituting into the first gives $x+e^{(x-1)/2}=0$, so $x<0$. Put $x=-a$, with $a>0$. ThenThe function $h(a)=a{e}^{(a+1)/2}$ is strictly increasing on $a\ge 0$, since $h^{\prime }(a)=e^{(a+1)/2}(1+a/2)>0$. It starts at zero and tends to infinity, so it takes the value one exactly once. Moreover $h(1)=e>1$, so $0<a<1$. This proves uniqueness of the stated equilibrium point.

At that point $e^{x+y}=a$, and the Jacobian matrix isIts trace is $3+a$ and determinant $2+a$. The characteristic discriminant is $(3+a{)}^2-4(2+a)=(a+1{)}^2$, so its two eigenvalues areThey are distinct and positive, with eigenvectors $(1,-1)$ and $(a,1)$ respectively. Thus the linearized critical point is an unstable node. It is a hyperbolic equilibrium, so the original nonlinear equilibrium point is also locally repelling; this last observation is stronger than the requested linearized classification.

Both eigenvalues are real and positive: $\boxed{{\lambda }_1>0,{\lambda }_2>0}$. The real eigendirections carry exponentially growing factors, giving an unstable node.

Both eigenvalues are real and positive: $\boxed{{\lambda }_1>0,{\lambda }_2>0}$. The real eigendirections carry exponentially growing factors, giving an unstable node.

An unstable node hasBoth linear modes grow forward in time. Nearby nonconstant orbits move away from the equilibrium, and approach it under reversed time; their local directions come from the two real eigenvectors.

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.

Define the auxiliary variable $y=ϵ\dot{x}-x+x^3/3$. Differentiating gives $\dot{y}=ϵ\ddot{x}+(x^2-1)\dot{x}=-x$, while rearrangement gives $ϵ\dot{x}=x-x^3/3+y$. Conversely eliminating $y$ recovers the second-order equation, so this phase plane system is equivalent. At an equilibrium $\dot{y}=0$ forces $x=0$, and the first equation then forces $y=0$. Its Jacobian matrix isThus $ϵ{\lambda }^2-\lambda +1=0$. Expanding $\sqrt{1-4ϵ}=1-2ϵ-2{ϵ}^2-4{ϵ}^3+O({ϵ}^4)$ givesFor sufficiently small positive $ϵ$ both are positive: the equilibrium is an unstable node. For negative $ϵ$ the fast root is negative and the slow root positive, so it becomes a saddle equilibrium.

The equilibrium equations are $x(-1+x^2+\beta y^2)=y(-1+\beta x^2+y^2)=0$. For $\beta \ne 1$ the fixed points are the origin, four axis points $(\pm 1,0),(0,\pm 1)$, and four diagonal points $(\pm s,\pm s)$ with independent signs and $s=(1+\beta {)}^{-1/2}$. The hypothesis $\beta >-1$ ensures that $s$ is real. The Jacobian matrix has diagonal entries $-1+3x^2+\beta y^2$, $-1+\beta x^2+3y^2$ and off-diagonal entries $2\beta xy$.

At the origin both eigenvalues are $-1$, so it is a stable node. At each axis point the eigenvalues are $2,\beta -1$. At each diagonal point they are $2,2(1-\beta )/(1+\beta )$, along the radial and transverse diagonal directions respectively. ThusThe axes and diagonals are invariant lines. For $\beta =1/2$ the axis saddle points have stable directions transverse to their axes and unstable directions along them; the diagonal points are unstable nodes. For $\beta =2$ these roles switch: the diagonal saddle points have stable transverse directions and unstable radial directions, and the axis points are unstable nodes. Their stable manifolds separate trajectories attracted to the origin from outward-escaping trajectories. Symmetry gives the other quadrants.

For $\beta =1$, the entire unit circle consists of equilibria. In polar coordinates $\dot{r}=r(r^2-1)$ and $\dot{\theta }=0$: every ray inside the circle moves toward the origin, and every ray outside moves outward. The circle has neutral tangential and unstable radial directions; its points are not isolated saddle points or unstable nodes.

All three phase portraits are shown below. The potential $\Phi =-(x^2+y^2)/2+(x^4+y^4)/4+\beta x^2{y}^2/2$ satisfies $(\dot{x},\dot{y})=\nabla \Phi$ and $\dot{\Phi }=|\nabla \Phi {|}^2$, ruling out nonconstant periodic orbits. Blue curves show the saddle stable separatrices, integrated backwards from their local stable directions. The flow arrows and invariant lines, together with the equilibrium classification, specify the topology of the sketches.

At an equilibrium point, $x^2-1=0$ and $kxy=0$. If $k\ne 0$, the only equilibria areTheir Jacobian matrix isAt $(1,0)$ its eigenvalues are $2,k$, and at $(-1,0)$ they are $-2,-k$. Hence for $k>0$, $(1,0)$ is an unstable node and $(-1,0)$ is a stable node; for $k<0$, both points are saddles.

If $k=0$, every point on each line $x=1$ and $x=-1$ is an equilibrium. The line $x=1$ is transversely repelling because the nonzero eigenvalue is $2$, while $x=-1$ is transversely attracting because the nonzero eigenvalue is $-2$; the tangential eigenvalue is zero on both nonisolated equilibrium lines.

For $\nu =1$, the equilibria are $(0,0)$ and $(\pm 1,0)$. At the origin the eigenvalues are $1,-1$, so it is a saddle for every $\kappa$.

At either nonzero point the Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -2& \kappa \end{array}\right)$, with eigenvaluesBoth points are attracting for $\kappa <0$ and repelling for $\kappa >0$. For $-2\sqrt{2}<\kappa <0$ they are stable foci; for $\kappa <-2\sqrt{2}$ stable nodes. For $0<\kappa <2\sqrt{2}$ they are unstable foci; for $\kappa >2\sqrt{2}$ unstable nodes. At the two repeated-root boundaries they are degenerate nodes. At $\kappa =0$, linearization has purely imaginary eigenvalues; the conserved energy in the next part shows that they are actual nonlinear centres. Thus the zero-real-part case is resolved by an additional first integral, rather than by the linear test alone.

The nullclines meet at the origin and $P=(2-k,k(2-k))$. At the origin, the Jacobian matrix has trace three, determinant $2-k>0$ and discriminant $1+4k>0$, giving an unstable node. At $P$, its determinant is $k-2<0$, giving a saddle equilibrium. The horizontal component is positive below the parabolic nullcline and negative above it; the vertical component is positive above the straight nullcline and negative below it. These signs determine the directional regions of the phase portrait.

The quadrant is not positively invariant: nonzero points on either axis have an outward-pointing component. Restricting a drawing to that quadrant does not keep its trajectories there. For $k>2$, only the saddle at the origin remains in the quadrant, and an exponential Dulac multiplier for a quadratic planar system excludes nonconstant periodic orbits there.