Systems of equations and phase portraits

A system evolves a whole state

A first-order system is x′=F(t,x), where x(t) is a vector of state variables and F supplies every component’s rate at the same time and state. Initial data x(t₀)=x₀ give one value for each state component. Coupling means that one component’s derivative can depend on other components, not merely on its own value.

For a second-order oscillator q″+2δq′+ω₀²q=f(t), define x₁=q and x₂=q′. Then x₁′=x₂ and x₂′=−ω₀²x₁−2δx₂+f(t). The state contains position and velocity, which have different physical units. A Euclidean distance or circular plot using these raw components needs an explicit scaling before it acquires a physical meaning. The geometric examples below use dimensionless coordinates unless stated otherwise.

An autonomous system has no explicit time dependence: x′=F(x). In a two-component autonomous system, F assigns one arrow to each point of the state plane. A trajectory is a curve tangent to that field, traversed with time according to the differential equation. The vector field, one parametrised solution and the geometric trace of that solution are different objects.

A phase portrait shows equilibria, selected trajectories and directional information in state space. A time graph such as x₁(t) uses time as an axis and is a different drawing. A nonautonomous system can be made autonomous by adding a clock variable τ′=1 and replacing explicit t by τ, but that adds a state dimension. A two-dimensional projection should not be relabelled as the full autonomous state space.

Uniqueness constrains full trajectories, with time treated correctly

A useful sufficient local setting is F continuous on an open neighbourhood of (t₀,x₀) and locally Lipschitz in x, uniformly for nearby times. For example, continuous state partial derivatives on a small closed convex neighbourhood give a bounded Jacobian and hence such a Lipschitz estimate in a compatible norm. Then a unique local classical solution exists through the initial state. The conclusion is local unless stronger bounds justify a larger interval.

For an autonomous locally unique system, two solution curves meeting the same state can be aligned by a time shift. Uniqueness then makes their parametrised solutions agree locally after that shift and on their common continuation. Thus two distinct geometric orbits cannot cross and depart as different curves through one state. Periodic motion can revisit a state because it follows the same orbit again.

This statement does not mean that separate time graphs can never have the same first-component value. Their other state components may differ. Nor does it forbid intersections after projecting a higher-dimensional trajectory into a plane. The complete state, rather than its selected coordinate or projection, is what the uniqueness theorem compares.

For nonautonomous equations, the time is part of the initial point being compared. Two projected state paths can pass through the same state at different times while seeing different F(t,x). Equal states at unequal times do not force identical future histories in that setting. Likewise, where uniqueness hypotheses fail, intersecting or departing histories need separate analysis instead of an automatic phase-portrait rule.

Equilibria are solved before the portrait is classified

For x′=Ax with a constant real matrix, the origin is an equilibrium, and every vector in the kernel of A is also an equilibrium. If A is invertible, the origin is the only one. If A is singular, a line or higher-dimensional set of equilibria can occur. Classifying an isolated origin without first checking that kernel can misdescribe the actual system.

For the affine system x′=Ax+b, equilibria solve Ax_*+b=0. Invertibility gives the unique solution x_*=−A^(−1)b. With singular A, the equation can be inconsistent or have multiple solutions; neither outcome can be settled by a nonexistent inverse. Row reduction or an equivalent linear-system argument identifies the actual case.

If an equilibrium x_* is found, set z=x−x_*. Then z′=Az, so the homogeneous geometry is translated to x_*. Different equilibria in a nonisolated set still require the complete solution structure to interpret motion along and across that set. Constant forcing cannot always be absorbed into a unique shifted origin.

For example, A=diag(−1,−2) and b=(1,4) give x_*=(1,2). The translated components satisfy z₁′=−z₁ and z₂′=−2z₂, so all trajectories approach that equilibrium in future time. By contrast, x₁′=0,x₂′=−x₂ has an entire horizontal line of equilibria, while x₁′=1,x₂′=−x₂ has none.

Eigenvectors give invariant directions and complete modes

If Av=λv, then x(t)=e^(λt)v solves x′=Ax by direct differentiation. If a real 2×2 matrix has distinct real eigenvalues λ₁,λ₂ with independent eigenvectors v₁,v₂, the general solution is C₁e^(λ₁t)v₁+C₂e^(λ₂t)v₂. In matrix form, A=PΛP^(−1), z=P^(−1)x gives z′=Λz, and the two decoupled scalar equations supply the formula.

An eigenvector line is invariant: a solution beginning on it stays on it. The eigenvectors need not be perpendicular for a general matrix, so diagonal coordinates are not automatically a rotation or an orthonormal basis. Distorting axes to make the picture convenient should not be confused with preserving ordinary lengths and angles.

For A=diag(1,−2), x₁=C₁e^t and x₂=C₂e^(−2t). The vertical axis is the stable line and the horizontal axis the unstable line. The origin is a saddle: points on one special line approach it, but arbitrarily nearby points with a nonzero unstable component leave a small neighbourhood in future time. Convergence along one selected trajectory is not stability of the equilibrium.

For two negative eigenvalues, all modes decay and the origin is a sink. A slower-decaying mode often determines the late-time direction, except for initial states that do not excite it. Two positive eigenvalues give a source in forward time. The eigenvalue signs give the available rates, while the eigenvectors and initial coefficients determine individual trajectories.

Trace and determinant classify the main planar linear cases

For a real 2×2 matrix, let T=tr A and D=det A. Its characteristic polynomial is λ²−Tλ+D, with discriminant Δ=T²−4D. These three numbers organise the possible eigenvalues. If D<0, the two real eigenvalues have opposite signs and the origin is an isolated saddle. If D>0 and T<0, both real parts are negative; T>0 gives positive real parts.

Within D>0 and nonzero T, Δ>0 gives distinct real modes and a node, while Δ<0 gives a complex conjugate pair and a spiral. The sign of T distinguishes sink from source. At Δ=0, repeated eigenvalues require the eigenvectors or Jordan structure to determine the detailed portrait. Trace and determinant alone do not distinguish a scalar matrix from a defective one.

If D>0 and T=0, the eigenvalues are purely imaginary and distinct, giving a centre for this exact planar linear system. Its trajectories are closed curves around the origin after an invertible real change of coordinates; they need not be circles in the original coordinates. The equilibrium is stable but does not attract nearby nonzero states.

If D=0, at least one eigenvalue is zero and equilibria may fail to be isolated. The zero matrix and a nonzero nilpotent Jordan matrix both have T=D=0 but very different motion. Inspect the matrix and solve the remaining case. Moreover, conclusions about an exact linear system must not be transferred without qualification to a nonlinear system whose Jacobian has zero or imaginary eigenvalues.

Complex eigenvalues determine rates, while arrows determine turning direction

For A=αI+βJ with J=[[0,−1],[1,0]], the system is x₁′=αx₁−βx₂, x₂′=βx₁+αx₂. Its solution is e^(αt) times a rotation through βt applied to the initial vector. Direct differentiation verifies this formula. The radius is multiplied by e^(αt), and the sign of β gives the rotation direction in the stated coordinate orientation.

For α<0 and β≠0, the origin is a spiral sink; α>0 gives a spiral source. At α=0 it is a linear centre with constant radius. A general real matrix with complex eigenvalues is similar to a rotation-and-scaling block, so the portrait is an invertible linear deformation of this picture. Original-coordinate trajectories can be elliptical spirals rather than circular ones.

Eigenvalues alone do not state clockwise versus counterclockwise orientation. The matrices J and −J have the same unordered eigenvalue pair ±i but opposite rotation directions. Evaluate an actual field vector, such as F(1,0), and compare it with the labelled axes. A change of coordinates with negative determinant can also reverse the apparent orientation of the drawing.

For A=[[-1,−2],[2,−1]], a solution starting at (1,0) is e^(−t)(cos(2t),sin(2t)). At the initial point its velocity is (−1,2), so it turns counterclockwise while spiralling inward. Its squared radius is e^(−2t), which provides a direct check of the envelope independently of the plotted arrows.

Repeated eigenvalues retain the Jordan information

If a repeated eigenvalue has two independent eigenvectors in the plane, the two exponential eigenmodes still supply a full basis. For a real 2×2 matrix with only the eigenvalue λ and two independent eigenvectors, A=λI. If there is only one independent eigenvector, the matrix is defective and needs a generalized eigenvector rather than a second copy of the same mode.

Choose v,w with (A−λI)v=0 and (A−λI)w=v. Then e^(λt)v and e^(λt)(tv+w) are solutions. For the second one, differentiation gives e^(λt)[λ(tv+w)+v], while applying A gives exactly the same expression. Their initial vectors v,w are independent, so they form the required two-dimensional solution basis.

For A=[[-1,1],[0,−1]], take v=(1,0),w=(0,1). The solution is x=e^(−t)(C₁+C₂t,C₂). Every state approaches zero in future time, but the polynomial shear distinguishes the portrait from A=−I. Thus equal trace, determinant and repeated eigenvalue do not make the two flow maps identical.

At λ=0 the distinction is sharper. The zero matrix makes every state constant. The nilpotent matrix N=[[0,1],[0,0]] gives x₁=C₁+C₂t,x₂=C₂. Nearby initial states with C₂≠0 drift arbitrarily far over time, so the origin is not stable. A list of zero eigenvalues without its eigenspace information is insufficient to decide boundedness or stability.

The matrix exponential solves constant-matrix evolution

For a constant finite matrix A, define e^(tA)=I+tA+t²A²/2!+⋯. In a submultiplicative matrix norm, each nonconstant term (n≥1) is bounded by |t|ⁿ‖A‖ⁿ/n!, while the identity matrix is a separate finite term. Thus the scalar exponential series proves absolute convergence and uniform convergence on bounded time intervals. The differentiated series has the same compact-interval control, giving (d/dt)e^(tA)=Ae^(tA)=e^(tA)A.

Consequently x(t)=e^((t−t₀)A)x₀ solves x′=Ax and has x(t₀)=x₀. Uniqueness makes it the whole initial-value solution. For this same fixed matrix, the convergent-series product gives e^(sA)e^(tA)=e^((s+t)A), with inverse e^(−tA). The flow map is invertible even when trajectories approach zero as t→∞; no distinct states merge at a finite time.

If A=PΛP^(−1), powers give e^(tA)=P e^(tΛ)P^(−1). A Jordan block supplies the polynomial factors just derived. For N²=0, the series terminates at e^(tN)=I+tN. This definition works without diagonalizability, while an efficient computation can exploit the matrix’s actual structure.

For two different matrices, e^(A+B)=e^Ae^B is guaranteed when they commute, and fails in general without that condition. Likewise, a time-varying A(t) cannot ordinarily be handled by substituting ∫A into the constant-matrix exponential formula. The order of successive evolutions matters. A later worked example compares two exact shear steps to make this limitation concrete.

A forcing vector is accumulated through the matrix response

For x′=Ax+f(t), with A constant and f continuous on the interval, multiply by e^(−(t−t₀)A). The product rule cancels the Ax terms and gives d[e^(−(t−t₀)A)x]/dt=e^(−(t−t₀)A)f(t). Integration yields x(t)=e^((t−t₀)A)x₀+∫_(t₀)^t e^((t−s)A)f(s) ds. Matrix multiplication order is retained.

The first term transports the initial state, and the second is the zero-initial-state response. Differentiating the latter gives f(t)+A times that integral, verifying the equation; its value at t₀ is zero. This is the vector counterpart of the first-order and oscillator response formulas in lessons 138–139. It does not require A to be invertible.

With constant f=b and invertible A, an equilibrium shift can simplify the formula to x=x_*+e^((t−t₀)A)(x₀−x_*), x_*=−A^(−1)b. If A is singular, retain the integral or solve components instead of inserting an invalid inverse. The example x₁′=1,x₂′=−x₂ illustrates why a constant input may cause drift rather than a shifted equilibrium.

For a continuous variable matrix A(t), a known invertible fundamental matrix Φ(t) gives x(t)=Φ(t)[Φ(t₀)^(−1)x₀+∫_(t₀)^t Φ(s)^(−1)f(s) ds]. Substitution checks the formula using Φ′=A(t)Φ. The noncommuting products cannot be reordered arbitrarily. Obtaining Φ is an additional task, so this identity is not a promise that every variable-coefficient system has an elementary solution.

Nullclines locate directional changes, while trajectories follow the whole field

In x′=P(x,y), y′=Q(x,y), an x-nullcline is the set P=0 and a y-nullcline is Q=0. Their intersections are equilibria. At a point with P=0,Q≠0 the vector is vertical; with Q=0,P≠0 it is horizontal. These sets help organise directional signs but are not automatically trajectories or invariant curves.

For x′=y−x,y′=−x−y, the nullclines are y=x and y=−x. At (1,1) the vector is (0,−2), vertical and not tangent to y=x. The trajectory crosses that nullcline. Declaring every zero-rate line a solution path would therefore give the wrong phase portrait.

For a differentiable candidate invariant curve H(x,y)=C, check dH/dt=H_xP+H_yQ along its points. If that expression vanishes on the curve, and local uniqueness and regularity permit the corresponding tangent evolution, the curve is preserved. An eigenvector line or a conserved-energy level is supported by such a calculation. An arrow sampled at one point does not establish invariance of an entire curve.

For a planar linear flow Φ=e^(tA), expanding the derivative of its 2×2 determinant using Φ′=AΦ gives (det Φ)′=(tr A)det Φ and det Φ(0)=1. Hence det Φ=e^((tr A)t)>0. Oriented area scales by this determinant. This connects divergence of the linear field, which is tr A, with local area expansion or contraction, while the full eigenstructure still controls stretching and turning.

Linearization gives local information under a precise spectral test

An equilibrium x_* is stable if every sufficiently close initial state remains as close as prescribed for all future time, with existence on those future intervals understood. It is asymptotically stable if it is stable and all states in some neighbourhood approach x_* as t→∞. It is unstable if it is not stable. One approaching orbit does not meet these neighbourhood conditions, and attraction alone should not be silently substituted for the definition of asymptotic stability.

For an autonomous C¹ field near x_*, write x=x_*+z. Differentiability gives z′=Az+r(z), where A=DF(x_*), r(0)=0 and ‖r(z)‖/‖z‖→0 as z→0. The Jacobian therefore gives a first-order local approximation. Computing its eigenvalues is useful, but the smallness of the remainder is a neighbourhood statement, not a uniform guarantee along an arbitrary distant or very long trajectory.

A standard local linearization result for such a C¹ field states that if every eigenvalue of A has negative real part, the equilibrium is locally asymptotically stable; if at least one has positive real part, it is unstable. In a planar hyperbolic saddle, one eigenvalue is negative and one positive, so instability follows. Hyperbolic means that no eigenvalue has zero real part. These are local nonlinear conclusions under the stated regularity, not claims that the original nonlinear trajectories equal straight eigenvector lines.

If an eigenvalue has zero real part and none has positive real part, this test is inconclusive. Nonlinear terms can decide behaviour, particularly for a linear centre or a zero eigenvalue. The theorem itself is stated here rather than fully proved; the next two sections derive explicit examples showing both useful nonlinear information and the limits of relying only on a Jacobian.

The same linear centre can hide decay, growth or closed orbits

Consider three dimensionless planar systems with r²=x²+y². The first is x′=−y,y′=x. The second is x′=−y−xr²,y′=x−yr². The third is x′=−y+xr²,y′=x+yr². All have the same Jacobian J=[[0,−1],[1,0]] at the origin and hence the same purely imaginary eigenvalues ±i. Their nonlinear radial behaviour is nevertheless different.

For r>0, differentiate r²: 2rr′=2xx′+2yy′. The first system has r′=0. The second has r′=−r³ and the third r′=r³. The angular rate is θ′=(xy′−yx′)/r²=1 in all three cases. Thus the first follows circles, the second spirals inward and the third spirals outward, even though every linearization gives the same centre.

For initial radius r₀>0 at t=0, separation gives r=r₀/√(1+2r₀²t) in the inward system for future time. It never exceeds its starting radius and approaches zero, proving asymptotic stability of the origin directly. The outward system has r=r₀/√(1−2r₀²t) up to its pole at 1/(2r₀²). Arbitrarily small nonzero initial radii eventually leave any fixed small neighbourhood, so the origin is unstable.

The exact linear centre is stable without attraction: every nonzero radius remains constant. A Jacobian with imaginary eigenvalues therefore cannot choose among these three conclusions. The nonlinear rate calculation, not a finely sampled linear portrait, supplies the missing information. The formulas also distinguish slow algebraic decay from exponential decay and local instability from an assertion about every possible global nonlinear model.

Direct scalar quantities can reveal behaviour beyond eigenvectors

For a differentiable scalar V(x), its derivative along a system is V′=∇V·F. This connects multivariable derivatives to dynamics. In the inward radial example choose V=(x²+y²)/2. Then V′=−(x²+y²)², negative away from the origin. The explicit radial formula above establishes the convergence claim; V provides a second concise check of its sign and confinement.

A conserved scalar V′=0 need not imply attraction. In the linear centre V is constant on circles, so nonzero states do not converge to the origin. Likewise, merely finding V′≤0 in an arbitrary system does not by itself identify all long-time limits without further hypotheses. A Lyapunov or invariance theorem must be stated precisely before its stronger conclusions are used.

For a nonlinear saddle example, take x′=x and y′=−y+x². Its Jacobian at the origin is diag(1,−1), but its unstable invariant curve is y=x²/3 rather than the horizontal eigenvector line itself. Indeed, put H=y−x²/3. Then H′=−H, so H=0 stays zero. Along that curve, x grows as e^t and y as e^(2t), showing departure in forward time.

The vertical line x=0 is a stable invariant set in this example: x remains zero and y decays as e^(−t). Generic nearby states have a growing x component. The explicit curved invariant calculation illustrates why the linear eigendirections can be tangent information rather than exact nonlinear trajectories. No general stable-manifold construction is inferred from this one solved example.

A numerical spiral can be created by the step rule

For the exact rotation x′=Jx with J=[[0,−1],[1,0]], ‖x(t)‖² is constant. Explicit Euler instead gives wₙ₊₁=(I+hJ)wₙ. Since Jᵀ=−J and JᵀJ=I, direct multiplication gives (I+hJ)ᵀ(I+hJ)=(1+h²)I. Thus each nonzero step multiplies the squared radius by 1+h², producing an artificial outward spiral.

For any fixed h>0 the Euler iterates grow over sufficiently many steps, although the true solution is a stable centre. Reducing h improves a fixed finite-time approximation under appropriate analysis, but it does not make that fixed-step recurrence conserve radius for arbitrarily long time. A numerical portrait should disclose its method, step size, time window and any exact reference available.

Axes and units also matter. For the physical oscillator state (q,q′), a natural dimensionless pair can be u=q/Q and v=q′/(ω₀Q), with Q>0 and ω₀>0 fixed. Undamped energy contours become circles in these scaled variables. Equal-looking arrows on unscaled position and velocity axes do not imply equal physical rates, and changing plot aspect ratio can visually alter angles and ellipses.

A phase portrait sampled from a small finite set of initial states cannot prove a basin boundary, a global absence of other equilibria or an asymptotic theorem. Use it to inspect a stated mathematical model alongside exact calculations and controlled numerical evidence. A plot that crosses an invariant boundary or creates energy should prompt a check of the method and domain before a new dynamical interpretation.

Verify the vector equation and the qualitative conclusion separately

For an explicit vector candidate, differentiate every component, multiply by the stated matrix or evaluate the original nonlinear field, and compare all components. Check the full initial vector and its connected time interval. A match in one component or in a projected curve does not verify a system solution.

For a linear portrait, solve the equilibrium equations and retain the matrix, trace, determinant, discriminant and eigenspace information required by the selected case. Check direction along a labelled axis rather than inferring rotation from eigenvalues alone. At repeated or zero eigenvalues, compute the actual modes or exponential instead of forcing the case into a distinct-root picture.

For nonlinear claims, state regularity and the neighbourhood of the equilibrium, distinguish a proved direct calculation from a cited local theorem, and mark inconclusive linearizations honestly. Nullclines need tangency checks before they are called invariant curves. Stable directions of a saddle and stability of the equilibrium are different assertions.

The explorer should keep its field, selected initial state, time direction, scaling and exact-versus-numerical layers visible. Independently test eigenvector residuals, derivative residuals, equilibrium conditions, invariant quantities and finite-step effects. These checks qualify the displayed cases; they do not replace the theorem hypotheses or establish a full portrait from a handful of arrows.

Worked examples

Try it yourself

1. Rewrite q″+3q′+2q=0 as a first-order state system.

   Show the answer and reasoning

   x₁=q,x₂=q′ gives x₁′=x₂ and x₂′=−2x₁−3x₂.

   The state retains the initial position and velocity. Their physical units differ until an explicit scaling is introduced.

2. Can distinct geometric orbits of a locally unique autonomous system cross at one full state and then depart differently?

   Show the answer and reasoning

   No.

   Align their meeting times by a shift and apply uniqueness. Repeated visits on the same periodic orbit and crossings of projections are different situations.

3. Does x₁′=1,x₂′=−x₂ have an equilibrium?

   Show the answer and reasoning

   No.

   The first component rate is always 1. A singular matrix does not guarantee a line of equilibria when a constant input is present.

4. Find the equilibrium set of x₁′=0,x₂′=−x₂.

   Show the answer and reasoning

   Every point with x₂=0 is an equilibrium.

   Both rates vanish on that line. The origin is not an isolated equilibrium of this system.

5. For A=diag(2,−3), which coordinate line approaches the origin in forward time? Is the origin stable?

   Show the answer and reasoning

   The vertical line approaches; the origin is an unstable saddle.

   A nonzero horizontal component grows as e^(2t). One special converging line does not satisfy the neighbourhood definition of stability.

6. Classify a real planar linear system with T=−4,D=5.

   Show the answer and reasoning

   A spiral sink.

   Δ=T²−4D=−4 gives a complex pair; the negative trace gives negative real part, and D>0 makes the equilibrium isolated.

7. Do J=[[0,−1],[1,0]] and −J turn states in the same direction because their eigenvalue sets match?

   Show the answer and reasoning

   No; they turn in opposite directions.

   At (1,0), J gives (0,1) and −J gives (0,−1). The unordered eigenvalues ±i do not encode that orientation.

8. If (A−λI)v=0 and (A−λI)w=v, what second generalized solution accompanies e^(λt)v?

   Show the answer and reasoning

   e^(λt)(tv+w).

   Its derivative agrees with A times itself. The independent initial vector w supplies the missing mode.

9. If N²=0, what is e^(tN)? Does it follow that every trajectory is bounded?

   Show the answer and reasoning

   e^(tN)=I+tN; boundedness does not follow.

   A nonzero N can produce a component growing linearly in time. The zero-eigenvalue Jordan structure matters.

10. For constant A, what is the inverse of e^(tA)? Can two distinct initial states merge at a finite time under x′=Ax?

    Show the answer and reasoning

    The inverse is e^(−tA); distinct states do not merge at a finite time.

    The same-matrix exponential group law gives identity. Asymptotic convergence as t→∞ does not contradict finite-time invertibility.

11. In general, may e^Ae^B be replaced by e^(A+B)? Give a useful sufficient condition.

    Show the answer and reasoning

    Not in general; AB=BA is a sufficient condition.

    Noncommuting evolutions retain order. The exact two-shear example shows different products and a different single exponential.

12. For x′=y−x,y′=−x−y, why is the x-nullcline y=x not itself an invariant line?

    Show the answer and reasoning

    At (1,1) the field is (0,−2), which is not tangent to y=x.

    A zero x rate makes the arrow vertical there. Invariance requires tangency along the proposed curve, not just a vanishing component.

13. For a planar constant-matrix flow with trace −3, what is the area scaling after time 2?

    Show the answer and reasoning

    e^(−6).

    The flow determinant is e^((tr A)t)>0. It scales oriented and ordinary area by that positive factor.

14. What does hyperbolic mean for an equilibrium’s Jacobian?

    Show the answer and reasoning

    No eigenvalue has zero real part.

    Nonzero determinant alone is insufficient: a matrix with purely imaginary eigenvalues is invertible but not hyperbolic.

15. For a C¹ autonomous field near an equilibrium, what follows if every Jacobian eigenvalue has negative real part?

    Show the answer and reasoning

    The equilibrium is locally asymptotically stable.

    This is a local linearization theorem with regularity assumptions. It does not make the nonlinear solution identical to its linear approximation everywhere.

16. Can a Jacobian pair ±i alone distinguish the three radial examples’ nonlinear stability?

    Show the answer and reasoning

    No.

    The same Jacobian accompanies constant radius, inward cubic radial decay and outward cubic growth. Direct nonlinear terms determine their different conclusions.

17. For x′=x,y′=−y+x², compute the derivative of H=y−x²/3.

    Show the answer and reasoning

    H′=−H.

    Differentiate using both state equations. The level H=0 is invariant, giving an actual parabola tangent to the linear unstable direction.

18. For explicit Euler on x′=Jx, what is the squared-radius factor per step of size h?

    Show the answer and reasoning

    1+h².

    (I+hJ)ᵀ(I+hJ)=(1+h²)I. The artificial growth differs from exact rotation, which preserves radius.

What to carry forward

• A system evolves a complete state, and a phase portrait is different from a time graph or projection.
• Local uniqueness restricts full autonomous orbits after time alignment, not arbitrary projected crossings.
• Solve equilibrium equations before classifying the portrait; singular affine systems can have many equilibria or none.
• Eigenvectors supply invariant linear directions and the initial coefficients select the actual motion.
• Trace, determinant and discriminant organise strict planar linear cases, with repeated and zero cases inspected separately.
• Eigenvalues alone do not determine rotation direction in the labelled coordinates.
• A defective repeated root needs generalized modes and polynomial factors.
• The matrix exponential defines constant-matrix evolution without requiring diagonalizability.
• Different matrix evolutions retain their order unless suitable commutation conditions allow simplification.
• Forced-response integrals remain valid for singular constant matrices without an invalid inverse.
• Nullclines mark component rates; invariant curves require a tangency or conservation calculation.
• The trace of a planar linear field controls its flow’s area factor, while the full matrix controls its shape.
• Nonlinear linearization results are local and require regularity plus an applicable spectral condition.
• Zero or imaginary linearized real parts can leave stability undecided; direct nonlinear calculations can settle particular examples.
• Nonlinear invariant curves can be tangent to eigendirections without being those straight lines.
• Numerical stepping can create a false spiral, so exact invariants, method, step and time window must remain visible.
• Physical state components may need explicit nondimensionalisation before plotted lengths and circles carry a common meaning.