Runge–Kutta order conditions
Derive low-order conditions by matching solution derivatives and distinguish quadrature accuracy from nonlinear stage coupling.
On this page
The question for today
What you’ll understand
- Read a Runge–Kutta tableau as a stage computation.
- Verify consistency and the conditions through order three.
- Explain why a scalar linear test does not certify general nonlinear order.
Runge–Kutta methods evaluate the vector field at several predicted states before combining the results. More evaluations do not automatically give higher order: the stage positions and final weights must cooperate to reproduce the correct derivative expansion. Low-order conditions expose that cooperation. They also separate two questions often confused in practice: how accurately the weights integrate a time-dependent forcing, and how accurately the stages follow a solution whose state changes during the step.
Stages and their internal time coordinates
An s-stage method is specified by a matrix A, weights b, and time coordinates c. Its stage derivatives evaluate f at a shifted time and a state assembled from other stage derivatives. A strictly lower-triangular A makes the method explicit, because stage i uses only earlier stages. An implicit tableau couples unknown stages and requires an algebraic solve. Here we assume the usual internal consistency condition , so the time coordinates agree with the stage row sums.
The final weights must sum to one to reproduce a constant vector field. Matching the next derivative introduces the weighted average of the stage times. This is already enough to rule out many plausible-looking formulas. A two-stage predictor-corrector that averages the starting slope and an Euler-predicted endpoint slope satisfies the first two conditions. Its order is a statement about smooth general differential equations with those coefficients, not merely about the number of samples in the average.
Nonlinear derivatives create different conditions
For an autonomous equation, Taylor expansion first produces the vector field, then its Jacobian applied to itself, then two distinct third-order expressions: a second derivative acting on two copies of the field, and a Jacobian applied twice in a chain. A method must reproduce both third-order coefficients. The first is governed by squared stage row sums, while the second depends on the matrix A acting on c. Treating all third-order information as a single scalar coefficient misses that distinction.
With , the conditions displayed below are necessary and sufficient for classical order at least three on sufficiently smooth general systems, including nonautonomous systems represented by adjoining the time variable. The square is componentwise. These are not the complete conditions for fourth or higher order, where additional derivative patterns occur. Rooted trees organize those patterns systematically, but the low-order equations can be checked directly by expanding the stages and matching derivatives.
What a test equation can and cannot verify
The scalar linear equation reproduces the chain-shaped derivative terms but has no second derivative of the vector field. A method whose stability polynomial agrees with the exponential through a certain power can therefore still fail a branching nonlinear order condition at that order. Conversely a time-only equation tests moment conditions on the quadrature weights but does not test the full stage coupling. A meaningful verification suite needs problems that expose both time dependence and nonlinear state dependence.
Order conditions are algebraic, while usable accuracy also depends on error constants, stage cost, stability, and the smoothness of the actual trajectory. Two second-order methods can agree on every scalar linear problem and differ on a nonlinear problem. This is a reason to inspect a method's full tableau rather than identify it only by its scalar amplification factor. At discontinuities or near singularities, even a correctly designed high-order tableau may not exhibit its nominal convergence rate until the integration is appropriately segmented.
Work it through
Checking Heun's two stages
Verify the order of Heun's method with A₂₁=1 and b=(1/2,1/2), then take one h=0.2 step for y′=t+y from y(0)=1.
The row sums give c=(0,1). The weights sum to one and bᵀc=1/2, so the method has order at least two. But bᵀc²=1/2 differs from 1/3, and bᵀAc=0 differs from 1/6, so it is not order three.
The first slope is k₁=1. The Euler-predicted endpoint is 1+0.2=1.2 at time 0.2, so the second slope is k₂=0.2+1.2=1.4.
Average the slopes: Y₁=1+0.2(1+1.4)/2=1.24. The exact solution is 2exp(t)−t−1, whose value at 0.2 is about 1.2428055163.
The tableau is genuinely second order, and this step has exact-minus-numerical error approximately 0.0028055163.
The living notebook
Explore the model
Interactive lab · function
A curvature term that linear tests miss
Pause and Reset, then set coefficient a=1 and position x=1. Predict the tangent slope of y=x² and compare it with the readout. For a fresh comparison, Reset and set a=1 and x=0.5. Use the nonzero curvature to explain why evaluating a field at a predicted state produces more than a linear chain term.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe tangent slopes are two and one, while the second derivative is two in both cases. This plot illustrates derivative information in a nonlinear field; it does not execute RK stages or verify all order conditions.
Pause & reason
Check your understanding
Your turn
Try it yourself
Use the explicit midpoint method on y′=y², y(0)=1, with h=0.1. Compute its two slopes and endpoint, then compare with the exact value.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The second stage uses time h/2 and the state 1+(h/2)k₁. The exact solution is 1/(1−t) before t=1.
Compare with the solution
The first slope is k₁=1. The midpoint stage state is 1.05, giving k₂=1.05²=1.1025. The update is Y₁=1+0.1·1.1025=1.11025. The exact endpoint is 1/0.9≈1.1111111111, so exact-minus-numerical error is about 0.0008611111. Its tableau has c₂=1/2 and b=(0,1), satisfying the two order-two conditions but not the order-three moment condition.
Keep the idea
Take it with you
- A tableau specifies stage states, stage times, and final derivative weights.
- General third order needs both a quadrature-moment condition and a stage-coupling condition.
- Scalar linear accuracy does not test every nonlinear derivative structure.
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