Lyapunov exponents and predictability

Start with a small initial change

Suppose a dimensionless state x changes continuously according to dx/dt = x(1-x)/T, with T = 2 s. Choose an initial state x₀ between 0 and 1 and follow it for t ≥ 0. This is an exact mathematical model supplied for the lesson; its state is not a measured position or a population estimate.

For the reference initial state x₀ = 1/8, the state at t = (2 ln 2) s is 2/9. At that same time, a very small initial change is amplified, to first order, by 128/81. The factor is greater than 1, so the initial change grows over this interval. We will find that the same reference trajectory eventually contracts tangent changes.

Prepare with Algebra, functions, graphs, and logarithms for fractions, exponentials and natural logarithms, and Units, standards, and physical constants for rates per second. The first reading can follow the exact table and the three comparisons. For the derivative arguments, return after Derivatives, Taylor series, and local approximation; Ordinary differential equations explains the initial-value notation used here.

Define the fixed-time tangent

Write φₜ(x₀) for the state reached at time t from initial state x₀. Put a = exp(t/T), a dimensionless number. The supplied equation has the exact solution phi_t(x0) = x0*a/(1-x0+x0*a). The denominator is positive for 0 ≤ x₀ ≤ 1 and t ≥ 0, so these forward trajectories are defined throughout that domain.

The tangent gain is G(t;x0) = d phi_t(x0)/dx0 = a/(1-x0+x0*a)^2. Here we change the initial state while holding the elapsed time fixed. For a small nonzero offset ε, the state difference is approximately Gε; G itself is the limiting ratio as ε tends to zero. A nonzero finite offset generally retains higher-order corrections.

At x₀ = 0 or 1, the derivative uses the smooth equation extended to nearby real initial states at each fixed finite horizon. Such a local extension does not assert that every negative initial state has a trajectory for all forward times. In this lesson G is positive, and both its input and output changes are dimensionless.

Read gain and logarithmic rate

The upper panel plots G for x₀ = 1/8. The level G = 1 means that a tangent change has recovered its initial magnitude. Above it there is amplification; below it there is contraction relative to the starting magnitude. The circles mark the exact times and values in the table.

For t > 0 define the finite-time rate lambda_F(t;x0) = ln(abs(G(t;x0)))/t. Natural logarithms turn multiplicative gain into an additive amount, and division by elapsed time gives a rate. Its unit is s⁻¹. A positive value means net tangent amplification over that entire interval, even if the change was contracting during the interval’s final part.

The lower panel uses its own vertical scale for λ_F. Diamonds mark positive-time values. At t = 0 the quotient is undefined: ln G and t are both zero. The open circle shows only the right-hand limit, +3/8 s⁻¹ for this reference. It is not a value of λ_F at zero.

Both panels use the same time axis. Their vertical quantities and units differ. Lines join analytically calculated samples; the exact expressions in the table supply the selected values.

Locate growth and return

For the reference, the expressions simplify to x(t) = a/(7+a) and G(t;1/8) = 64*a/(7+a)^2. Gain starts at 1. Differentiating with respect to a gives a numerator proportional to 7 − a, so gain increases until a = 7 and decreases afterward.

At t = (2 ln 7) s the state is 1/2 and the maximum gain is 16/7. The instantaneous tangent rate changes sign there: d ln(G)/dt = (1-2*x(t))/T. Once the state exceeds 1/2, this local rate is negative, although accumulated gain can still exceed 1.

Solving G = 1 gives (a-1)*(a-49) = 0. Besides the initial time, the return occurs at t = (4 ln 7) s, where x = 7/8. Only after this return does the finite-time rate become negative. A local negative rate and a negative elapsed-time average answer different questions.

Take a trajectory-specific limit

An asymptotic Lyapunov exponent asks for the long-time tangent rate: lambda(x0) = limit(t->infinity) ln(abs(G(t;x0)))/t, when this limit exists. The initial-state derivative is taken first, at each finite t; the time limit follows. It is not the infinite-time rate of a fixed nonzero separation.

For any fixed 0 < x₀ ≤ 1, factor a from the denominator to obtain G = a^(-1)/(x0+(1-x0)/a)^2. The second factor tends to the positive constant 1/x₀². Its logarithm divided by t tends to zero, while ln(a)/t = 1/T. Therefore λ = −1/T = −1/2 s⁻¹ along every one of these supplied trajectories.

In particular, our x₀ = 1/8 reference has a negative exponent despite its earlier growth. The finite graph ends at t = (12 ln 2) s, about 8.32 s; its final rate is only about −0.025 s⁻¹. The graph does not claim that a short interval has already reached the asymptotic rate.

Compare two stationary trajectories

At x₀ = 0 the state stays zero, but the tangent gain is a. Thus λ_F = +1/T and λ = +1/T. At x₀ = 1 the state stays one, but G = 1/a, giving λ_F = −1/T and λ = −1/T. A stationary state can amplify or contract a nearby change; the exponent is not the rate of motion of the reference state.

The dashed and dotted horizontal lines compare these two stationary initial states. The dotted value also equals the reference trajectory’s eventual exponent, but it does not replace that reference’s finite-time curve.

This example also shows why an exponent must name its reference trajectory. The positive exponent at the unstable equilibrium x = 0 does not make this scalar flow chaotic. Every interior state here increases monotonically toward 1; the stationary trajectories stay constant. Exponential tangent growth by itself does not establish recurrent chaotic dynamics.

Set a finite-error tolerance

Now compare the stationary reference x = 0 with the same exact model started at ε = 1/1000. Declare that a prediction is acceptable while its absolute state error is at most η = 1/10. There is no noise, forcing or parameter error in this comparison; the only difference is the specified initial state.

The tangent estimate gives ε exp(t/T). Its formal tolerance crossing is t_linear = T*ln(eta/epsilon) = (2 ln 100) s. This is a linear estimate, not an exact finite-offset forecast horizon.

The exact error is delta(t) = epsilon*a/(1-epsilon+epsilon*a). At a = 100 it is 100/1099, still less than 1/10. Setting the exact expression equal to η gives a = eta*(1-epsilon)/(epsilon*(1-eta)) = 111, hence the true crossing in this comparison is t = (2 ln 111) s. Equality is acceptable, so the allowed forward interval includes that endpoint.

For this fixed positive ε, the error eventually approaches 1 rather than growing without bound. Its logarithmic amplification divided by t tends to zero. The tangent exponent at zero is still +1/2 s⁻¹ because it takes the infinitesimal limit first. A tolerance, initial offset and declared error model are essential parts of a predictability question.

Use the result with its conditions

Keep three questions separate: how a tangent change behaves over a chosen finite interval, its asymptotic rate along a specified trajectory, and when a specified finite error exceeds a tolerance. This lesson solves each within one exact scalar equation. It does not estimate an exponent from measurements or validate a real forecasting system.

For a continuous model, changing the time unit changes the numerical rate. With τ = t/T, the stationary-zero exponent is +1 per unit τ, equivalent to +1/T per second. Logarithms act on the dimensionless gain; a dimensional raw error cannot be inserted into a logarithm without a declared reference ratio.

In more than one state dimension, the perturbation direction and the norm used to compare it must also be declared. A changing tangent map cannot generally be replaced by the eigenvalues of its instantaneous matrix. No general ergodic or trajectory-independent theorem is assumed here: the limits used above were obtained directly from the supplied solution.

Worked examples

Try it yourself

1. For T = 2 s and x₀ = 1/8, calculate the initial state rate dx/dt. Is it the Lyapunov exponent?

   Show the answer and reasoning

   dx/dt = 7/128 s⁻¹; it is not the Lyapunov exponent.

   The state rate is (1/8)(7/8)/(2 s). The asymptotic tangent exponent along this initial state is −1/2 s⁻¹; it concerns sensitivity, not the state’s speed.

2. At t = (4 ln 2) s for x₀ = 1/8, find the state, tangent gain and sign of λ_F.

   Show the answer and reasoning

   The state is 4/11, the gain is 256/121, and λ_F is positive.

   Here a = 4. Gain exceeds 1, so its natural logarithm is positive and division by positive time preserves the sign.

3. When does the x₀ = 1/8 tangent gain first return to 1 after t = 0, and is the reference state stationary then?

   Show the answer and reasoning

   At t = (4 ln 7) s, with state 7/8; the state is not stationary.

   The positive-time root is a = 49. The state rate remains positive at 7/8, even though the accumulated tangent gain has returned to its initial magnitude.

4. Does the open lower-panel circle at t = 0 mean λ_F(0;1/8) = 3/8 s⁻¹? Explain.

   Show the answer and reasoning

   No. It marks the right-hand limit; λ_F at t = 0 is undefined.

   The finite-time definition divides ln G by a positive elapsed time. Its continuous right-hand limit exists here, but substituting zero would give 0/0.

5. For the finite-offset comparison ε = 1/1000, is the exact error acceptable at t = (2 ln 100) s under the inclusive tolerance η = 1/10?

   Show the answer and reasoning

   Yes: the exact error is 100/1099, less than 1/10.

   Cross multiplication gives 1000 < 1099. The exact error reaches equality only at a = 111; the tangent estimate reaches equality at a = 100.

6. In a separate supplied model with the same dimensionless equation but T = 1 s, what is the stationary-zero tangent exponent per second? What is it per unit τ = t/T?

   Show the answer and reasoning

   It is +1 s⁻¹ and +1 per unit τ.

   The tangent gain at zero is exp(t/T). The exponent is 1/T in physical time and 1 in normalized time. This separate clock scale does not change the T = 2 s figure.

What to carry forward

• A tangent gain differentiates the flow with respect to its initial state at a fixed finite time.
• Finite-time logarithmic rates and asymptotic exponents concern different horizons and must name the reference trajectory.
• Temporary amplification can occur along a trajectory with a negative asymptotic exponent.
• A finite-error prediction horizon also needs an initial error, tolerance and declared error model; an exponent alone does not supply it.