Nonlinear oscillators

A nonlinear oscillator has a restoring force that is not exactly proportional to displacement. Its frequency can depend on amplitude, waveforms need not be sinusoidal, and responses cannot generally be added by superposition. The same position-velocity state architecture still works, but the force law and interpretation become richer.

Linear motion is a local approximation

Near a stable equilibrium \(x_0\), a smooth potential can be expanded as

\[U(x_0+\eta)=U(x_0)+\frac12U''(x_0)\eta^2+\frac16U'''(x_0)\eta^3+\cdots\]

The first derivative vanishes at equilibrium. Keeping only the positive quadratic term produces simple harmonic motion. Larger displacement makes higher powers important. Thus “nonlinear” often means that the small-oscillation approximation is no longer accurate for the amplitude of interest.

The local small-oscillation angular frequency remains

\[\omega_0=\sqrt{\frac{U''(x_0)}{m}}\]

but it need not describe finite-amplitude periods.

A cubic-stiffness model

The Duffing-style unforced conservative equation is

\[m\ddot x+kx+\beta x^3=0\]

Its restoring force is

\[F(x)=-kx-\beta x^3\]

and its potential is

\[U(x)=\frac12kx^2+\frac14\beta x^4\]

because \(F=-dU/dx\). The cubic coefficient has units force per length cubed.

import numpy as np

linear_stiffness, cubic_stiffness, mass, position = 1.0, 0.5, 1.0, 1.0
force = -linear_stiffness * position - cubic_stiffness * position**3
print("Acceleration (m/s^2):", force / mass)

Setting \(\beta=0\) must recover the linear oscillator exactly.

Hardening and softening behavior

With positive linear stiffness and positive \(\beta\), the restoring force grows more rapidly than a linear spring at large displacement. This is a hardening oscillator, and its finite-amplitude frequency generally increases with amplitude.

A negative cubic coefficient initially softens the spring. However, the simple quartic potential then falls without bound at sufficiently large displacement, so it is only a local model unless stabilizing higher-order terms or physical travel limits are added. Never extrapolate a fitted negative quartic model to arbitrary amplitude.

For weak cubic nonlinearity and amplitude \(A\), an approximate hardening frequency is

\[\omega(A)\approx\omega_0\left(1+\frac{3\beta A^2}{8k}\right)\]

when the correction is small. This approximation shows the trend but should be checked against event-measured periods for the actual parameter range.

Energy and turning points

Without damping or drive,

\[E=\frac12mv^2+\frac12kx^2+\frac14\beta x^4\]

At a turning point, velocity is zero, so energy determines amplitude. Period can be written as an energy integral:

\[T=\sqrt{2m}\int_{x_-}^{x_+}\frac{dx}{\sqrt{E-U(x)}}\]

with appropriate doubling or full-cycle limits depending on symmetry. Unlike the harmonic case, the integral generally depends on energy and hence amplitude.

The phase-space trajectory remains on a constant-energy contour for an ideal conservative system, but it is not generally an ellipse. Comparing numerical energy with the analytic potential is a powerful check of the force sign and integrator.

The nonlinear pendulum

A simple pendulum obeys

\[\ddot\theta+\frac{g}{L}\sin\theta=0\]

The approximation

\(\sin\theta\approx\theta\)

produces harmonic motion with \(\omega_0=\sqrt{g/L}\). At larger amplitude, the exact restoring torque is weaker than the linear extrapolation near the extremes and the period grows.

For modest amplitude \(\theta_0\) in radians,

\[T\approx2\pi\sqrt{\frac{L}{g}}\left(1+\frac{\theta_0^2}{16}+\cdots\right)\]

The exact finite-amplitude period involves an elliptic integral and diverges as the release approaches the upright separatrix from below.

Measuring period reproducibly

Visual peak selection is imprecise. Define an event such as consecutive positive- direction equilibrium crossings. The time between such crossings is one period. Locate the crossing within a time-step bracket rather than assigning it to the nearest sample.

x, v, dt, crossings = 1.0, 0.0, 0.001, []
for step in range(20_000):
    previous = x
    v += (-x - 0.5 * x**3) * dt
    x += v * dt
    if previous < 0 <= x and v > 0:
        fraction = -previous / (x - previous)
        crossings.append((step + fraction) * dt)
print("Periods from upward crossings (s):", np.diff(crossings))

Measure several cycles after any transient and report mean and variation. Repeat at half the time step. Change amplitude without changing solver settings so physical amplitude dependence is not confused with numerical phase error.

Superposition no longer works

For a linear equation, if two functions are solutions, their sum is also a solution under corresponding combined initial data or forcing. The cubic term breaks this property:

\[(x_1+x_2)^3\ne x_1^3+x_2^3\]

A multi-frequency drive can therefore create harmonics and combination frequencies. Fourier analysis remains useful for describing output, but output frequencies need not match only the input frequencies.

Damping and drive

A driven damped nonlinear oscillator can be written

\[m\ddot x+c\dot x+kx+\beta x^3=F_0\cos(\omega t)\]

Its instantaneous mechanical-energy rate is

\[\frac{dE}{dt}=-cv^2+F(t)v\]

Drive can replace energy removed by damping. The settled response may contain harmonics, multiple coexisting periodic states, jumps, or chaos. Initial condition and sweep direction can select different attractors at identical parameters.

position, velocity, time = 0.2, 0.1, 0.5
spring = -position - 0.5 * position**3
damping = -0.2 * velocity
drive = 0.3 * np.cos(1.2 * time)
print("Net force (N):", spring + damping + drive)

Do not compare a nonlinear frequency sweep with the linear amplitude formula except as a small-amplitude limiting test.

Multiple equilibria

Changing signs can create a double-well potential. For example,

\[U(x)=-\frac12ax^2+\frac14\beta x^4\qquad a>0,\ \beta>0\]

has an unstable equilibrium at the origin and stable minima at

\[x=\pm\sqrt{\frac{a}{\beta}}\]

Low-energy motion stays in one well; energy above the central barrier can cross between wells. Damping, drive, and noise can produce transitions whose interpretation depends on the full energy and transfer model.

Numerical care

Nonlinear force can become extremely large, making a previously adequate time step unstable. Estimate the highest local curvature over the visited amplitude range and refine adaptively or choose a robust method. Reject nonfinite state rather than drawing it.

Run the zero-nonlinearity regression, force odd-symmetry tests where appropriate, finite-difference checks of \(F=-dU/dx\), and conservative energy tests. Compare period, turning points, phase portrait, harmonics, and energy transfer under time-step and duration refinement.

A surprising nonlinear response becomes credible only after simpler limiting cases pass and the same qualitative result persists under numerical refinement and plausible parameter uncertainty.

Nondimensional parameters

Scaling makes it easier to compare different oscillators and identify which terms are important. Let displacement be measured relative to a characteristic amplitude \(X\) and time relative to \(1/\omega_0\). Write \(x=Xq\) and \(\tau=\omega_0t\). The conservative cubic equation becomes

\[\frac{d^2q}{d\tau^2}+q+\alpha q^3=0\]

with

\[\alpha=\frac{\beta X^2}{k}\]

If magnitude of \(\alpha\) is much less than one over the motion range, the linear approximation should be accurate. If it is order one, nonlinear effects are expected. This comparison is more informative than calling a dimensional cubic coefficient “large” without a displacement scale.

Nondimensionalization also improves numerical conditioning when physical values span many orders of magnitude. Convert output back to physical units only after solving and retain the scale definitions with the result.

Connecting a model to measurements

Nonlinear coefficients should be inferred from force-displacement data, period-versus- amplitude data, or another explicit measurement. Fit the simplest range that supports the intended prediction. A cubic approximation that matches moderate motion may fail near contact, yielding, or geometric limits.

Validate with data not used for fitting. For a conservative candidate, integrate the measured force to obtain potential and check whether observed turning-point speeds match energy conversion. For a damped model, compare decay across several amplitudes. For a driven model, compare both upward and downward frequency sweeps when multiple responses may exist.

Parameter uncertainty can move bifurcation or jump locations substantially. Report a range of responses over plausible coefficients instead of extra digits in one curve. If two different force laws fit the available data equally well but predict different large-amplitude behavior, the limitation is model identifiability, not a problem that a smaller time step can solve.

Symmetry as a diagnostic

An odd restoring force satisfies \(F(-x)=-F(x)\) and produces a symmetric potential. With symmetric initial conditions, positive and negative turning points should have equal magnitude. A persistent asymmetry can indicate an even force term, shifted equilibrium, gravity bias, contact, or an implementation error.

Test symmetry directly at force-function level before integrating. Then check trajectory symmetry under refined time steps. This separates a wrong force evaluation from an otherwise correct solver and makes the nonlinear behavior easier to explain.

A worked Python notebook

Add a cubic stiffness term to a linear spring and compare local slopes. The question for this calculation is precise: how does restoring force change when we change displacement, while keeping the other choices fixed? The earlier sections explain the physical relationship. Here we turn one part of that relationship into a calculation that you can run, inspect, and change in Google Colab.

This hardening-spring model is an approximation to a specified force law, not every nonlinear oscillator. Keep that sentence beside the results. A formula can return a number outside its useful range, and a plotting library will usually draw that number without asking whether the physical assumptions still hold. The selected interval is from -2 to 2 m. It gives us a specific domain to discuss instead of leaving the graph's range to a software default.

Use the notebook link above the lesson, run the setup cell, and then run the cells in order. The code below is also included there. NumPy handles the arrays of numerical values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these libraries or notebook cells are unfamiliar, start with Your first physics notebook.

State the relationship before choosing a graph

In this small example, q stands for displacement, expressed in m. The output is restoring force, expressed in N. The equation below uses the numerical values in those units. Any fixed coefficients therefore carry the units needed to make the relation consistent. A value supplied in another unit must be converted before it enters this expression.

\[f(q) = - 0.5 q^{3} - q\]

This is a function of one changing input. It is not a statement that the physical system has only one variable. We deliberately hold the other inputs fixed so that one cause of variation is visible. Changing several inputs at once would make it harder to tell which change produced the output difference. Write down the held-fixed choices before trying a different experiment.

At 1 m, the expression gives -1.5 N. This is our reference calculation. It is a prediction of the stated example, not a new measurement. Its displayed digits help us compare implementations. They should not be read as a claim that physical inputs are known to six significant figures.

import numpy as np
import sympy as sp
import matplotlib.pyplot as plt

q = sp.Symbol("q", real=True)
expression = sp.sympify("-q-0.5*q**3", locals={"q": q})
reference_input = 1.0
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "m")
print("Predicted restoring force:", reference_output, "N")

sp.Symbol declares a mathematical symbol. It does not assign a numerical measurement to that symbol. subs then replaces the symbol with our chosen input, and float converts the resulting symbolic number into a standard Python numerical value. Keeping these steps separate lets us inspect the equation before numerical substitution. It also prevents a later plot from becoming the only surviving record of the model.

Compare a few cases you can explain

Before drawing hundreds of points, calculate three. The lower endpoint gives 6 N. The midpoint, at 0 m, gives 0 N. The upper endpoint gives -6 N. Read these together with the assumptions: they describe different choices of displacement within the same example, with the other inputs held fixed.

These three values are a useful first check, but they do not establish the shape between them. Two endpoints can be equal even when the curve bends or oscillates in between. Likewise, a large change between endpoints does not tell us whether the relationship is linear. We need the equation and intermediate samples to answer those questions.

evaluate = sp.lambdify(q, expression, modules="numpy")
inputs = np.array([-2.0, 0.0, 2.0])
outputs = np.asarray(evaluate(inputs), dtype=float)
outputs = np.broadcast_to(outputs, inputs.shape)
for input_value, output_value in zip(inputs, outputs):
    print(f"{input_value:.6g} m -> {output_value:.6g} N")
assert np.all(np.isfinite(outputs))

lambdify turns the symbolic expression into a numerical function. We explicitly select NumPy as its numerical backend so it can accept an array. The same expression now has two useful forms: one for algebra and one for calculating many cases. The assertion checks that the chosen inputs produce finite outputs. It would catch an infinity or an undefined numerical result, but a finite value can still be physically inappropriate.

Make a graph that answers the question

The horizontal axis shows displacement; the vertical axis shows restoring force. Every plotted point is a prediction of the same expression. We mark the reference case separately so that a reader can connect the earlier printed value to its place on the curve. The axis labels include units because a shape without a scale cannot tell us the size of a physical effect.

input_grid = np.linspace(-2.0, 2.0, 301)
output_grid = np.asarray(evaluate(input_grid), dtype=float)
output_grid = np.broadcast_to(output_grid, input_grid.shape)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(input_grid, output_grid, label="Stated model")
ax.scatter([reference_input], [reference_output], label="Reference case", zorder=3)
ax.set_xlabel("Displacement (m)")
ax.set_ylabel("Restoring force (N)")
ax.legend()
ax.grid(alpha=0.25)
fig.tight_layout()
plt.show()

Look for a sign change, a turning point, or a region where the curve becomes steep. Connect each feature to the expression rather than to the colour of the line. If the curve has a minimum or maximum, changing the input in opposite directions may have similar effects on the output near that point. If it is nearly straight over a short interval, a local linear approximation may be useful even when the complete expression is nonlinear.

Increasing the number of samples makes the line smoother and may reveal features that a coarse grid missed. It does not improve the physical inputs or enlarge the model's validity range. For quantities that can take only discrete values, use the line as a guide to the relationship and interpret only the allowed input values. A plotting function does not enforce such restrictions for you.

Measure the local sensitivity

A derivative asks a more specific question than the whole graph: how quickly does restoring force change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 1 m is -2.5 in those ratio units. Locally, the output decreases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = - 1.5 q^{2} - 1\]

For a small input change, multiply the derivative by that change to estimate the output change. Then compare that estimate with a fresh evaluation of the full expression. The distinction between these two calculations matters: the derivative gives a tangent approximation, while substitution evaluates the chosen expression at the new point.

slope_expression = sp.diff(expression, q)
slope_at_reference = float(slope_expression.subs(q, reference_input))
changed_input = 1.02
linear_prediction = reference_output + slope_at_reference * (
    changed_input - reference_input
)
full_prediction = float(evaluate(changed_input))
print("Symbolic derivative:", slope_expression)
print("Tangent prediction:", linear_prediction, "N")
print("Full expression:", full_prediction, "N")
print("Approximation difference:", full_prediction - linear_prediction, "N")

Here the changed input is 1.02 m. The tangent estimate is -1.55 N, while the full expression gives -1.5506 N. If the two values are close, it means the chosen input change is small enough for this local approximation at this point. It does not mean the entire model is linear. If the slope is zero, a quadratic or higher-order term may supply the leading change instead.

This is also the beginning of uncertainty propagation. If the input has a small stated standard uncertainty, the magnitude of the slope times that uncertainty estimates one contribution to output uncertainty. Do not automatically use the same rule for a wide interval, a discontinuity, a turning point with zero first derivative, or several correlated uncertain inputs. Those situations need more of the model than one slope.

Check the derivative with numerical differences

We can estimate the same slope from values on either side of the reference point. Subtract the lower output from the upper output and divide by twice the input step. Parentheses are essential: the entire output difference must be divided by the entire input interval. Without them, Python can execute a different mathematical expression while still producing a plausible number.

The calculation below uses three decreasing step sizes. Each estimate is compared with the symbolic derivative, so the reference is independent of the numerical difference formula. Near an endpoint, it is acceptable here to evaluate a small extension only because the displayed mathematical expression is smooth there; such a check does not assert that extrapolated values describe additional physical cases.

input_span = 4.0
for fraction in [1e-2, 1e-3, 1e-4]:
    step = input_span * fraction
    upper = float(evaluate(reference_input + step))
    lower = float(evaluate(reference_input - step))
    numerical_slope = (upper - lower) / (2 * step)
    difference = numerical_slope - slope_at_reference
    print(f"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}")

For a smooth function, reducing the step often reduces the approximation error at first. Very small steps can eventually make subtraction of nearly equal floating-point values less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare the observed behaviour over a useful range of steps instead of choosing the smallest number the computer can represent.

An absolute slope difference is useful when the expected slope is near zero. A relative difference would divide by a very small reference and could look enormous even when the absolute discrepancy is tiny. Away from zero, a relative difference can help compare quantities with very different scales. The tolerance should reflect the question and the numerical method, rather than a universal decimal threshold copied into every test.

Change the example and preserve the reasoning

Make one copy of the reference calculation before editing it. Choose a second value of displacement inside the stated interval, predict whether restoring force will rise or fall, and then evaluate the expression. Place the prediction and the result in a text cell. If they disagree, identify whether the mistake was in the sign, the unit conversion, the algebra, or the assumption about how the curve behaves.

Next choose a shorter plotting interval around the reference point. Compare the local curve with its tangent approximation. Then widen the interval again. The difference shows why a useful local estimate need not describe the whole range. Keep the axis labels and numerical scale visible when comparing the plots; automatic axis limits can make two different effects appear equally large.

Finally, return to the physical limit stated at the start: This hardening-spring model is an approximation to a specified force law, not every nonlinear oscillator. Describe one change to the setup that would violate that limit. State what new input, equation, or observation would be needed before trusting an answer in the changed situation. This last step connects the Python experiment to the physics instead of treating successful execution as the only criterion for a useful result.

Before saving, restart the runtime and run all cells in order. A result that depends on an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep unrounded values for later calculations, round only the reported output, and label any new data as measured, assumed, or generated. The graph and printed values should remain traceable to the same expression and the same choices of inputs.

Recover the output change by integrating the slope

There is another way to check that the expression and its derivative describe the same relationship. Add up the small output changes predicted by the slope across the chosen input interval. In the continuous limit, this integral equals the output at the upper endpoint minus the output at the lower endpoint. The result has units of N: the input unit in the integration cancels the input unit in the denominator of the slope.

This statement is about the mathematics of this particular relationship. It does not assign a new physical meaning to an arbitrary area under a graph. For example, an area under force versus distance represents work, while an area under a different pair of quantities may represent something else or have no standard name. Always carry the axis units through the calculation before naming an integral.

slope_function = sp.lambdify(q, slope_expression, modules="numpy")
endpoint_change = float(evaluate(2.0) - evaluate(-2.0))
for points in [101, 201, 401]:
    grid = np.linspace(-2.0, 2.0, points)
    slopes = np.asarray(slope_function(grid), dtype=float)
    slopes = np.broadcast_to(slopes, grid.shape)
    integrated_change = np.trapezoid(slopes, grid)
    print(
        points,
        "integrated change:",
        integrated_change,
        "endpoint change:",
        endpoint_change,
    )

The trapezoidal rule joins neighbouring slope samples with straight segments and adds their signed areas. A negative contribution remains negative; replacing every slope by its magnitude would calculate a different quantity. Increasing the sample count provides a practical refinement check. If the values do not settle toward the endpoint difference, first inspect the interval, singularities, and array ordering.

Do not conclude that two agreeing calculations prove the model is physically complete. Both use the same selected relationship, so an omitted effect would be omitted from both. Their agreement is evidence that the numerical tools implement that relationship consistently. Comparing with an independent experiment or a more appropriate physical model is a separate test.

Keep this comparison in the notebook when you change the expression. It is particularly useful after adding a coefficient, changing a sign convention, or converting an input unit. A code cell that produces a smooth graph can conceal these mistakes; a second calculation of the same output change gives you another place to notice them.

Sources

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