Dynamical systems and phase portraits

Start with two temperature differences

Consider two uniform bodies, each exchanging heat with its own maintained reservoir at T₀ = 300 K. Let their temperatures be T₁ and T₂. Define x = T₁ − T₀ and y = T₂ − T₀, measured in kelvins. In the first case, both bodies begin 4 K warmer than their reservoirs: the initial state is the ordered pair (x₀, y₀) = (4, 4) K.

Supply the rate laws dx/dt = −x/(2 s) and dy/dt = −y/(1 s) for t ≥ 0, restricting the physical histories here to nonnegative temperature excesses. A derivative here means the instantaneous rate of change of the named quantity as time changes; each rate has units K/s. The minus signs make a positive temperature excess decrease. The denominators are exact supplied time constants, not measured cooling times.

The model assumes each body has one uniform temperature, each reservoir stays at 300 K, the bodies do not influence one another, and the coefficients remain constant. Proportional cooling is a simplified physical model whose applicability must be checked for a real system. This lesson uses its stipulated equations to teach states and evolution, rather than to predict an unspecified experiment.

Prepare with Units, standards, and physical constants and Algebra, functions, graphs, and logarithms. First follow the rate signs, exact table and phase portrait. For a later return to the derivative checks, use Derivatives, Taylor series, and local approximation and Ordinary differential equations. We introduce the needed state and rate notation locally before using it.

Separate the state from its rate

A dynamical system specifies how a declared state evolves. In this model the state is z = (x, y), while the evolution rule gives its rate F(z) = (−x/(2 s), −y/(1 s)). The pair is ordered: the first component refers to body 1 and the second to body 2. A state is not the same quantity as its rate.

With the reservoirs and coefficients fixed, these rates depend on the current state without an explicit time input. Such a system is autonomous. At (x, y) = (2, 1) K, the rate is (−1, −1) K/s whether that state is supplied as an initial condition or reached later. The state-space vector field is this assignment of a rate vector to each allowed state.

The whole pair selects the whole history here. Cases A and B can have the same x while having different y. Because the bodies are uncoupled, x alone does satisfy a closed equation for its own evolution in this special model; it still does not specify body 2. Choosing variables for another system requires checking whether those variables and the supplied parameters close the intended evolution rule.

Attach times to exact states

The supplied laws have solutions x(t) = x₀ exp[−t/(2 s)] and y(t) = y₀ exp[−t/(1 s)]. The notation exp(a) means e to the power a, where e is the base of the natural logarithm. The arguments t/(2 s) and t/(1 s) are dimensionless. Differentiating the two exponentials gives precisely the supplied rates, and at t = 0 both factors are 1.

Choose Δ = 2 ln(2) s, where ln is the natural logarithm and exp[ln(2)] = 2. Over one interval Δ, x is multiplied by 1/2 and y by 1/4. Thus case A goes from (4, 4) K to (2, 1) K and then to (1, 1/4) K at times 0, Δ and 2Δ.

The other cases use the same rate laws with initial pairs B = (4, 2) K and C = (2, 4) K. Their exact values are listed below. The first column divides time by Δ, so its entries are dimensionless counts of that interval; each remaining heading divides a temperature difference by K. These are stipulated model values, not readings with an implied measurement precision.

Read state curves and rate arrows

Plot x horizontally and y vertically. A point now represents the two temperature differences at one instant. A solution history traces a curve through these states as time increases. A phase portrait shows a region of state space with representative solution curves, their direction of evolution and equilibrium states. These axes describe temperatures, not two physical directions in which a body moves.

The diagram draws only the three history segments from t = 0 to 4 ln(2) s. Circles mark case A on a solid curve, squares mark B on a dashed curve, and triangles mark C on a dash-dot curve. Each case has three markers corresponding to the time table. Both axes use the same kelvin scale; time is supplied by the table rather than by either phase axis.

The sixteen grey arrows show instantaneous rates at selected states. Each shaft displays F(z) multiplied by a common 0.1 s factor and the axis scale. This factor makes relative rates visible; an arrow tip is not asserted to be the exact state 0.1 s later. A solution curve is tangent to its rate direction where that rate is nonzero. The temperature excesses decrease toward the lower left, consistent with the arrow signs.

Check every rate for equilibrium

An equilibrium of this autonomous system is a state at which both rates vanish. Here that requires x = 0 K and y = 0 K: both bodies have temperature 300 K. The separate diamond marks that constant solution. It does not represent bodies at absolute zero.

At (0, 2) K the first rate is zero but the second is −2 K/s. The state is changing, so it is not an equilibrium. Likewise, a small arrow is not a test for an exact zero rate. The equations, including every component, supply that test.

For positive initial excesses the exponentials stay positive at every finite time and approach zero only as t grows without bound. The three drawn curves therefore do not reach the equilibrium diamond at their final markers. They approach the same limiting state without meeting there at a finite time.

Noncrossing requires a uniqueness condition. One sufficient local condition is that the autonomous rate components have continuous first partial derivatives in an open region around the state. Our linear field has this property on its mathematical extension to an open plane, and has an explicit unique solution. Two distinct integral orbits cannot cross there and then follow different continuations. Different clock origins can describe the same orbit; a projection that omits a state component can also conceal the distinction between histories.

Keep the clock separate from the curve

For x₀ > 0 K, put u = exp[−t/(2 s)]. Then x = x₀u and y = y₀u², so eliminating u gives y = y₀(x/x₀)². The ratio x/x₀ is dimensionless. This relation describes an orbit’s geometry, but does not say how quickly the state moves along it.

For a separate model, double both rates: dx/dt = −x/(1 s) and dy/dt = −2y/(1 s). Starting from (4, 4) K gives the same geometric relation y = (4 K)[x/(4 K)]². It reaches (2, 1) K at ln(2) s, instead of 2 ln(2) s. The first model’s arrows and time table do not describe this faster clock.

Temperature-rate arrows have units K/s before display scaling. They are not spatial velocities in m/s, and they do not show forces. For phase variables of different quantity kinds, such as position and momentum, each axis needs its own declared unit and scale; equal screen distances alone would not define a physical speed.

Recognize an explicit time input

Now supply a separate driven contrast: dx/dt = −x/(2 s) + (1 K/s) sin[t/(1 s)], with dy/dt = −y/(1 s). The sine uses the dimensionless numerical radian phase t/(1 s). This equation includes an explicit time input as well as the state; it is nonautonomous. It is not the model used in the portrait.

At the supplied state (2, 1) K and t = 0, the rates are (−1, −1) K/s. At the same supplied state and t = (π/2) s, the sine is 1 and the rates are (0, −1) K/s. These are two instantaneous probes, not a claim that one solution visits that state at both times. A single static state-only arrow assignment would omit the changing drive.

One can include the drive phase or a clock among the variables when constructing a larger evolution description, but its evolution and initial value must also be specified. Merely adding a label does not establish a closed physical model. Here it is enough to retain the explicit time input.

Distinguish sampling from continuous evolution

Observe the original model at tₙ = nΔ, where Δ = 2 ln(2) s and n is a nonnegative integer. Write xₙ = x(tₙ) and yₙ = y(tₙ). Its exact observation map is xₙ₊₁ = xₙ/2 and yₙ₊₁ = yₙ/4. This discrete map connects selected observations of a continuously changing state; it is not an Euler approximation.

The table and markers show only n = 0, 1 and 2, while the supplied mathematical laws extend to t ≥ 0. Sampling does not mean that nothing happens between observations. Conversely, a discrete dynamical model can be specified as an update rule in its own right; it need not have been derived from a continuous equation.

A useful portrait must retain the meaning of its state variables, the evolution rule, its parameters and domain, and any calibration of its arrows. The finite curves here establish no general stability classification or chaotic behavior. The next lesson, Fixed points and stability, examines how to describe behavior near equilibria more systematically.

Worked examples

Try it yourself

1. In the original model, is (0, 2) K an equilibrium? Give its rate and explain the test.

   Show the answer and reasoning

   No. Its rate is (0, −2) K/s.

   Only body 1 has zero excess and zero rate. Body 2 is still changing. Equilibrium requires all component rates to vanish.

2. At t = 2 ln(2) s, cases A and B both have x = 2 K. Are their whole states the same, and does x still close its own equation?

   Show the answer and reasoning

   The states differ: A is (2, 1) K and B is (2, 1/2) K. Here x still closes its own equation.

   Their initial x values and x rate laws agree, but their initial y values differ. The uncoupled x equation predicts x alone; that does not determine the two-body state.

3. For case A, why is y = (4 K)[x/(4 K)]² unit-consistent? Could its geometric relation alone distinguish the original and doubled-rate clocks?

   Show the answer and reasoning

   The squared ratio is dimensionless, leaving units K for y. The relation alone does not distinguish the two clocks.

   Both models follow this same orbit from the same initial pair, but reach (2, 1) K at different times. Timing needs the rate rule or a calibrated time table.

4. Case A reaches (1, 1/4) K at t = 4 ln(2) s. Under the supplied model, what is its next observation one Δ later? Is that state drawn in the figure?

   Show the answer and reasoning

   (1/2, 1/16) K at t = 6 ln(2) s. That observation lies beyond the drawn time segment.

   The exact sampling map divides x by 2 and y by 4. The model’s t ≥ 0 domain permits this calculation, while the diagram ends at 4 ln(2) s.

5. Does case A reach the equilibrium diamond at its final drawn marker? What does that diamond mean for the physical temperatures?

   Show the answer and reasoning

   No: its final drawn state is (1, 1/4) K. The diamond is the separate state T₁ = T₂ = 300 K.

   Positive exponential factors never become exactly zero at a finite time. Zero temperature excess means equality with the reservoir temperature, not absolute zero.

6. Starting from (4, 4) K, double both original rate coefficients. When does the state reach (2, 1) K, and may the original time table be reused unchanged?

   Show the answer and reasoning

   At ln(2) s. The original time table may not be reused unchanged.

   Doubling the rates halves the time to each corresponding state. It preserves this orbit’s geometry, while changing its time parametrization and rate-arrow magnitudes.

What to carry forward

• A state names the model’s variables; an evolution rule gives their rates or their updates, with parameters and domains specified.
• A phase portrait displays states and representative histories. Its axes need not be physical spatial directions.
• Instantaneous rate arrows, finite solution curves and sampled points convey different information. Arrow calibration must be stated.
• Equilibrium requires all component rates to vanish. Positive excesses in this model approach equilibrium only asymptotically.
• Orbit geometry alone does not specify a clock. Explicit-time forcing and omitted state variables can change what a portrait can represent.
• The exact half-and-quarter observation map samples the supplied continuous solution; it is not a numerical approximation or an empirical cooling result.