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 Incoming links: Classical fourth-order Runge-Kutta method

Delay stability of proportional integrator control 2026-10-14

x^{'} (t) = - k x (t - τ)

with

k > 0

and

τ \geq 0

, the zero equilibrium point is asymptotically stable if and only if

k τ < π /2

. For positive

τ

, perturbations are measured by the supremum norm of the initial history of delayed feedback; in the stable regime, every continuous history produces a solution tending to zero. At equality, nondecaying sinusoids prevent asymptotic stability. For

k τ > π /2

, growing oscillatory solutions exist. These statements concern this ideal linear time-invariant system, rather than an arbitrary controlled plant. The corresponding linear equation and its threshold are developed in Cain and Reynolds, section 6.4, pages 174–177.

Substitute the exponential function

x (t) = e^{s t}

, allowing a complex number

s

. The characteristic equation is

s + k e^{- s τ} = 0.

(1)

For positive feedback delay this is not a characteristic polynomial: it has infinitely many roots. A root's real part determines growth or decay of its exponential mode, and its imaginary part determines the angular frequency. To find the first boundary, set

s = iω

with

ω > 0

and use Euler's formula:

k cos (ω τ) = 0, ω - k sin (ω τ) = 0.

(2)

The smallest positive crossing is

ω = k

and

ω τ = π /2

. Thus at

τ_{c} = π / (2 k)

the complex conjugate roots are

s = \pm ik

. In fact, with the matching initial history of delayed feedback,

x (t) = A cos (k t) + B sin (k t)

(3)

solves the delay differential equation exactly: a quarter-cycle feedback delay turns the past displacement into precisely the current derivative. Its cycle lasts

2 π / k = 4 τ_{c}

. The amplitude is arbitrary; this linear model does not select a finite-amplitude oscillation.

The roots cross toward growth as feedback delay increases. Differentiating the characteristic equation at fixed

k

gives

\frac{d s}{d τ} = - \frac{s ^{2}}{1 + s τ}, Re (\frac{d s}{d τ})_{s = ik, τ = τ_{c}} = \frac{k ^{2}}{1 + ( π /2 ) ^{2}} > 0.

(4)

This calculation identifies the direction of the first destabilization. The full threshold also requires ruling out earlier unstable roots: if

s = α + i β

with

α \geq 0

and

β > 0

, the characteristic equation implies

cos (β τ) \leq 0

and

0 < β \leq k

, hence

k τ \geq β τ \geq π /2

. A real nonnegative root is impossible because

s + k e^{- s τ} > 0

there. This supplies the geometric reason the threshold is a quarter-cycle rather than an arbitrary fitting constant.

Figure 1.
Feedback delay separates decay, persistent oscillation, and growth
. Left: numerical solutions with the same constant initial history of delayed feedback $x = 1$ on $[- τ, 0]$ , plotted against $q = t / τ$ for $k τ = 0.8$ , $π /2$ , and $2$ . The middle trajectory includes a transient because this history is not a pure sinusoid; the dashed curve is a separate exact solution with history $cos (π q /2)$ . Each row has its own vertical scale. Right: the exact boundary $k = π / (2 τ)$ for positive feedback gain and feedback delay. Toy responses are not measured data. The generator uses the classical fourth-order Runge-Kutta method with linear interpolation of delayed values and $Δ q = 1/2000$ , checked against a separately evaluated exact finite sum and two coarser steps. **Details**.

The tradeoff is concrete. Without feedback delay, increasing

k

shortens the exponential decay time constant

1/ k

. With known

τ > 0

, the same proportional control must satisfy

k < π / (2 τ)

. For

τ = 0.2 s

, the threshold is

k \approx 7.854 s^{- 1}

k = 5 s^{- 1}

is stable and

k = 8 s^{- 1}

is unstable. A design close to the boundary leaves little room for additional feedback delay; reducing

k

or reducing measurement latency increases that room. The zero-delay time constant is only a comparison scale, not the decay time of the delayed response. Frank discusses the gain–delay compromise in Control Theory Tutorial, section 13.4 and figure 13.3.

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Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 341 / 7 / Solution Created 2026-10-03 Updated 2026-10-20

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s

-stage Runge-Kutta method for

y^{'} = f (t, y)

is defined by

Y_{i} = y_{n} + h \sum_{j = 1}^{s} a_{ij} f (t_{n} + c_{j} h, Y_{j}), y_{n + 1} = y_{n} + h \sum_{i = 1}^{s} b_{i} f (t_{n} + c_{i} h, Y_{i}) .

(1)

The Butcher tableau records

A = (a_{ij})

b

and

c

; internal consistency usually sets

c = A e

, and first-order consistency requires

b^{T} e = 1

. A strictly lower triangular

A

gives an explicit method, while other stage dependencies generally require an implicit Runge-Kutta method. The stage solvability of an implicit Runge-Kutta method must be checked separately: if

f

is globally Lipschitz continuous in

y

with constant

L

, the stage fixed-point map is a contraction in the maximum stage norm whenever

h L max_{i} \sum_{j} ∣ a_{ij} ∣ < 1

. This is a sufficient small-step condition, not a restriction intrinsic to the definitions of A-stability or B-stability.

Linear stability begins with the Dahlquist test equation

y^{'} = λ y

. Solving the stages gives

y_{n + 1} = R (z) y_{n}, R (z) = 1 + z b^{T} (I - z A)^{- 1} e, z = hλ .

(2)

The linear stability domain consists of test values for which the stage system is well defined and

∣ R (z) ∣ \leq 1

. A-stability means this includes the closed left half-plane. For

Re λ < 0

, this gives stability with no scalar decay-mode step restriction. For a normal matrix

L

y^{'} = L y

, unitary diagonalization of a normal matrix reduces the norm estimate to these scalar factors. For a non-normal matrix, eigenvalues alone do not establish a uniform norm bound: eigenvector conditioning and transient amplification matter. Mesh-uniform estimates for discretized PDEs must control operator powers, not only their spectra.

Every consistent explicit Runge-Kutta method has a nonconstant polynomial stability function and therefore cannot be A-stable, since that polynomial is unbounded on the negative real axis. For example, forward Euler method has

R (z) = 1 + z

, stability disk

∣1 + z ∣ \leq 1

, and restriction

0 \leq h q \leq 2

for

λ = - q < 0

. For a pure imaginary mode

λ = iω \neq = 0

∣1 + ihω ∣ > 1

, so it is unstable for every positive step. The classical fourth-order Runge-Kutta method instead has

R (z) = 1 + z + \frac{1}{2} z^{2} + \frac{1}{6} z^{3} + \frac{1}{24} z^{4}, ∣ R (i q) ∣^{2} = 1 - \frac{q ^{6}}{72} + \frac{q ^{8}}{576} .

(3)

It is stable on the imaginary axis exactly for

∣ q ∣ \leq 22

, an illustrative conditional stability range for oscillatory evolution.

Backward Euler method has

R (z) = (1 - z)^{- 1}

and is A-stable. L-stability adds

R (z) \to 0

∣ z ∣ \to \infty

in the left half-plane; backward Euler satisfies this and strongly damps unresolved rapidly decaying modes in a stiff differential equation. The implicit midpoint rule and the trapezoidal rule both have

R (z) = (1 + z /2) / (1 - z /2)

, so they are A-stable but not L-stable. They preserve the modulus of a pure imaginary test mode, but their stiff-decay limit is

- 1

, leaving oscillatory numerical remnants. The three-stage Lobatto IIIA method has the fourth-order rational function found in Question 2 and likewise lacks L-stability. Thus high order and A-stability do not themselves imply efficient stiff damping.

Nonlinear stability measures differences between solutions of the same equation. A one-sided Lipschitz condition is

Re ⟨ f (t, y) - f (t, y), y - y ⟩ \leq ν ∥ y - y ∥^{2} .

(4)

Differentiating the squared difference and applying the Gronwall inequality gives

∥ y (t) - y (t) ∥ \leq e^{ν (t - t_{0})} ∥ y (t_{0}) - y (t_{0}) ∥

. A dissipative vector field has

ν \leq 0

and is contractive. A B-stable method preserves this property for every positive step size for which the stages are well defined:

∥ y_{n + 1} - y_{n + 1} ∥ \leq ∥ y_{n} - y_{n} ∥.

(5)

This includes time-dependent vector fields when dissipativity holds at each common time argument. Applying it to scalar linear dissipative fields shows that B-stability implies A-stability when the linear stages are well defined; the converse fails.

The standard sufficient theorem is algebraic stability implies B-stability, subject to stage solvability. Define

B = diag (b)

and

M = B A + A^{T} B - b b^{T}

. Algebraic stability of a Runge-Kutta method means

b_{i} \geq 0

and

M

is a positive semidefinite matrix. To prove the theorem, set

d = y_{n} - y_{n}

D_{i} = Y_{i} - Y_{i}

and

F_{i} = f (t_{n} + c_{i} h, Y_{i}) - f (t_{n} + c_{i} h, Y_{i})

. Expanding the output difference and using

D_{i} = d + h \sum_{j} a_{ij} F_{j}

yields the Runge-Kutta contractivity identity

∥ d_{n + 1} ∥^{2} = ∥ d ∥^{2} + 2 h \sum_{i} b_{i} Re ⟨ D_{i}, F_{i} ⟩ - h^{2} \sum_{i, j} m_{ij} Re ⟨ F_{i}, F_{j} ⟩ .

(6)

Dissipativity makes the first sum nonpositive, while positive semidefiniteness makes the last quadratic form nonnegative. The latter follows by factoring

M = Q^{T} Q

and writing it as a sum of squared norms of linear combinations of the

F_{i}

. This proves the claimed contraction. It is a sufficient theorem; absence of algebraic stability is not, by itself, a proof of failure of B-stability for every representation.

Several examples make the distinctions concrete. For backward Euler,

b = 1

A = (1)

, so

M = (1)

: it is both L-stable and B-stable. For implicit midpoint,

b = 1

A = (1/2)

, so

M = (0)

: it is B-stable and preserves squared norms for a skew-Hermitian linear equation, but lacks stiff damping. For the two-stage Gauss--Legendre Runge-Kutta method,

A = (1/4 1/4 + 3 /6 1/4 - 3 /6 1/4), b = (1/2, 1/2)^{T},

(7)

the matrix

M

vanishes as well. The collocation theorem gives order four, and its stability function is exactly the same as the three-stage Lobatto IIIA function. Yet the Gauss method is algebraically stable, whereas that Lobatto method is not. Identical scalar linear stability functions need not imply identical nonlinear stability properties.

An example combining stiff damping and nonlinear contraction is the two-stage Radau IIA method, with

A = (5/12 3/4 - 1/12 1/4), b = (3/4, 1/4)^{T}, M = \frac{1}{16} (1 - 1 - 1 1), R (z) = \frac{1 + z /3}{1 - 2 z /3 + z ^{2} /6} .

(8)

Its positive weights and positive semidefinite

M

prove B-stability; its rational function has denominator roots

2 \pm i 2

and, for

z = x + i y

, satisfies

∣1 - 2 z /3 + z^{2} /6 ∣^{2} - ∣1 + z /3 ∣^{2} = \frac{∣ z ∣ ^{4} - 8 x ∣ z ∣ ^{2} + 24 x ^{2} - 72 x}{36} \geq 0 (x \leq 0) .

(9)

Thus it is A-stable; the limit

R (z) \to 0

then proves L-stability. The standard Radau IIA collocation theorem gives order

2 s - 1

, here three. These properties explain its usefulness for stiff nonlinear equations.

Finally, A-stability alone does not imply B-stability. The trapezoidal rule fails B-stability even for the scalar dissipative equation

y^{'} = - y^{3}

. With

h = 1

, its step map

T

is uniquely defined by

T (y) + \frac{1}{2} T (y)^{3} = y - \frac{1}{2} y^{3} .

(10)

y = 2

T (y) = 0

. Implicit differentiation gives

T^{'} (y) = \frac{1 - \frac{3}{2} y ^{2}}{1 + \frac{3}{2} T ( y ) ^{2}}, T^{'} (2) = - 2.

(11)

Thus nearby starting values are separated by approximately twice their original distance after one step, although the exact flow is contractive. This example and the contractivity theorem show why scalar linear analysis, stiff damping and genuinely nonlinear norm estimates answer different stability questions.

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Past exam of the systems biology course of the University of Cambridge / 2024 / Paper 3 / 3 / c / Solution 2026-10-20

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At a fixed input the production terms are constant, so both linear ordinary differential equations can also be solved exactly:

Y (t) = Y^{*} + [Y (0) - Y^{*}] e^{- t}, Z (t) = Z_{direct}^{*} + [Z (0) - Z_{direct}^{*}] e^{- t} .

(1)

Each displacement from its equilibrium concentration decays with time constant one. The Jacobian matrix is

- I

, with two eigenvalues equal to

- 1

, so the fixed point is globally asymptotically stable for the nonnegative state space.

Here is a numerical solver using the classical fourth-order Runge-Kutta method; it uses only the repository's existing NumPy and matplotlib dependencies. Continue the definitions from part (b):

def integrate(rhs, initial, times):
    out = np.empty((len(times), len(initial)))
    out[0] = initial
    for j, dt in enumerate(np.diff(times)):
        state = out[j]
        k1 = rhs(state)
        k2 = rhs(state + dt*k1/2)
        k3 = rhs(state + dt*k2/2)
        k4 = rhs(state + dt*k3)
        out[j+1] = state + dt*(k1 + 2*k2 + 2*k3 + k4)/6
    return out

def direct_rhs(state, X):
    Y, Z = state
    return np.array([2*hill(X, 1, 2) - Y,
                     3*hill(X, 2, 4) - Z])

times = np.linspace(0, 8, 801)
plt.figure(figsize=(7, 4))
for initial, style in [([0, 0], "-"), ([3, 4], "--")]:
    trajectory = integrate(lambda s: direct_rhs(s, 2), initial, times)
    exact = np.array([1.6, 1.5]) + (np.array(initial) - [1.6, 1.5]) \
            * np.exp(-times[:, None])
    assert np.max(np.abs(trajectory - exact)) < 1e-8
    plt.plot(times, trajectory[:, 0], style, label=f"Y, initial {initial}")
    plt.plot(times, trajectory[:, 1], style, label=f"Z, initial {initial}")
plt.xlabel("Time")
plt.ylabel("Concentration")
plt.legend()
plt.tight_layout()
plt.show()

For

X = 2

, $Y^{*} = 1.6, Z_{direct}^{*} = 1.5$ . The initial state

(0, 0)

approaches those values from below, while

(3, 4)

approaches from above. Over eight time units the initial displacement is reduced by

e^{- 8} \approx 0.000335

, explaining why both trajectories nearly coincide with the equilibria from part (b). The two-panel figure above displays these simulations.

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Paul trap secular-motion example 2026-10-18

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Choose these dimensionless parameters for an original example, with no collisions or damping. They are teaching inputs, not measurements from either lecture. With

s = Ω t

and slow initial amplitude

X_{0}

, the radial equation and its leading approximation are

\frac{d ^{2} x}{d s ^{2}} = \frac{q _{x}}{2} cos s x, X = X_{0} cos (\frac{q _{x} s}{2 2}), x_{app} = X (1 - \frac{q _{x}}{2} cos s) .

(1)

Set

x (0) = 0.95 X_{0}

and

d x / d s (0) = 0

to match the approximate initial conditions. The leading secular angular frequency is

ω_{x} /Ω = 0.0353553

. The velocity Verlet method integrates the full RF equation over 80 drive periods; a separate fourth-order Runge-Kutta method checks the computed trajectory. Halving the timestep tests numerical error independently of the averaging error.

Figure 1.
Full RF motion and the leading secular approximation
. Original chosen small-parameter calculation in units of $X_{0}$ . The lower panel resolves six drive cycles and includes the first micromotion correction. Agreement over this finite interval does not establish a stability diagram or an exact secular frequency. **Details**.

Using 400, 800, 1600 and 3200 steps per RF period gives successive maximum position differences of approximately

1.40 \times 1 0^{- 4}

3.50 \times 1 0^{- 5}

and

8.74 \times 1 0^{- 6}

in units of

X_{0}

, consistent with second-order convergence. The finest trajectory differs from the independent classical fourth-order Runge-Kutta method result by at most

2.92 \times 1 0^{- 6} X_{0}

, whereas

x_{app}

differs by as much as

0.0378 X_{0}

. Over a sufficiently long time, the omitted correction to the secular angular frequency produces a growing phase discrepancy even if the numerical trajectory has converged. For the chosen point only, integration of a fundamental solution matrix over one RF period checks that both the radial and axial Floquet multipliers lie on the unit circle, away from its endpoints. This tests simultaneous ideal confinement without extrapolating to larger

∣ q_{i} ∣

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Proportional-integral control 2026-10-14

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Proportional-integral control is the proportional and integral part of a PID controller, with the derivative term omitted. Proportional-integral control combines proportional control with integral action. For the ideal controlled plant, a constant reference signal, exact measurement and a constant control disturbance, the ordinary differential equations give

\overset{e}{˙} = - k_{p} e - k_{i} z - w, \overset{z}{˙} = e, \overset{e}{¨} + k_{p} \overset{e}{˙} + k_{i} e = 0.

(1)

The characteristic polynomial is

λ^{2} + k_{p} λ + k_{i}

. Positive

k_{p}

and

k_{i}

imply asymptotic stability of the equilibrium point: real roots of a polynomial have negative sum and positive product, hence are both negative; a nonreal complex conjugate pair has real part

- k_{p} /2 < 0

. This is a short stability check for this ideal controlled plant, rather than a rule for every dynamical system.

At the equilibrium point,

e_{*} = 0

x_{*} = r

, and

z_{*} = - w / k_{i}

. For

w = 0.2

and

k_{i} = 1

, the stored value

z_{*} = - 0.2

supplies

u_{*} = - 0.2

even though the tracking error is zero. This separates the eventual accuracy from the transient response: integral action cancels the lasting offset but can make the plant output pass beyond the goal. The authors discuss practical limits on integral action in Åström and Murray, Feedback Systems, Chapter 11.

For

k_{p} = 2

k_{i} = 1

x (0) = z (0) = 0

and

w = 0

, the characteristic polynomial is

(λ + 1)^{2}

. The initial conditions are

e (0) = 1

and

\overset{e}{˙} (0) = - 2

, giving

e (t) = (1 - t) e^{- t}, x (t) = 1 - (1 - t) e^{- t}, z (t) = t e^{- t} .

(2)

The tracking error changes sign at

t = 1

. Since

\overset{x}{˙} = (2 - t) e^{- t}

, the largest plant output occurs at

t = 2

and is

1 + e^{- 2} ≃ 1.135335

, about

13.53%

above the reference signal.

For a separate comparison, start the control disturbance

w = 0.2

t = 4

. For

t \geq 4

, write

s = t - 4

. The exact responses are

x_{P} (t) = 1 - e^{- 2 t} + 0.1 (1 - e^{- 2 s}), x_{P I} (t) = 1 - (1 - t) e^{- t} + 0.2 s e^{- s} .

(3)

The proportional control response tends to

1.1

, while the proportional-integral control response returns to

1

. At

t = 8

, the values are approximately

1.099966

and

1.017001

. The added term in the proportional-integral control response eventually vanishes, exposing how memory removes a constant control disturbance's lasting effect.

Figure 1.
Proportional control and integral action: arrival and disturbance rejection
. Left: the ideal controlled plant has reference signal $r = 1$ and no control disturbance; proportional-integral control reaches a peak $1.135335$ at $t = 2$ . Right: a control disturbance $w = 0.2$ begins at $t = 4$ ; proportional control settles at $1.1$ , while proportional-integral control returns to $1$ . The generator uses the classical fourth-order Runge-Kutta method with step $0.002$ , splits the computation at $t = 4$ , and checks both full responses and the stored memory against their exact formulas. All coordinates are dimensionless quantities. **Details**.

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Runge-Kutta method Created 2026-09-24 Updated 2026-10-14

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A Runge-Kutta method advances an initial value problem

y^{'} = f (t, y)

y (t_{0}) = y_{0}

, by sampling its vector field at several trial states within one time step. These samples are Runge-Kutta stages; a weighted combination of their slopes produces the next state. The construction improves on the single tangent line used by the Euler method while requiring evaluations of

f

, rather than explicit higher derivatives of the solution. It applies to scalar equations and systems of ordinary differential equations alike. The coefficient choices distinguish methods such as Heun's method, the classical fourth-order Runge-Kutta method and implicit Runge-Kutta methods.

For

s

stages, write

A = (a_{ij})

for the stage matrix,

c_{i}

for relative time nodes and

b_{i}

for output weights. A time step of length

h

computes

Y_{i} = y_{n} + h \sum_{j = 1}^{s} a_{ij} k_{j}, k_{i} = f (t_{n} + c_{i} h, Y_{i}), y_{n + 1} = y_{n} + h \sum_{i = 1}^{s} b_{i} k_{i} .

(1)

Here

k_{i}

is a slope, with units of state per time;

h k_{i}

is a state increment. Another common convention writes

K_{i} = h k_{i}

, so its stage states are

Y_{i} = y_{n} + \sum_{j} a_{ij} K_{j}

and its output is

y_{n} + \sum_{i} b_{i} K_{i}

. These are equivalent conventions: including

h

both inside

K_{i}

and again in the output would count the time step twice. The textbook formulation uses the increment convention.^[ref]

Usually the Butcher tableau obeys

c_{i} = \sum_{j} a_{ij}

, called internal consistency. Stage

i

then combines slopes over an effective time

c_{i} h

. The second-order conditions with independent Runge-Kutta abscissae explain what changes when time nodes are specified independently. A Runge-Kutta stage is generally an auxiliary approximation, not a separate accepted solution value at its node. Several stages can even have the same time node but different state arguments, as in the classical fourth-order Runge-Kutta method.

The useful idea is to approximate the integral

y (t_{n} + h) - y (t_{n}) = \int_{t_{n}}^{t_{n} + h} f (t, y (t)) d t

without knowing the intermediate exact trajectory. Trial states provide places to sample

f

, and Butcher order conditions make the resulting combination agree with the exact Taylor expansion. More stages alone do not guarantee a higher order of a numerical method: their coefficients must satisfy the appropriate conditions. Embedded Runge-Kutta pairs reuse these same samples for a second output and a practical local error estimator.

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