1
One Euler step multiplies by one number
For y′ = −a·y the step is yₙ₊₁ = yₙ + h·(−a·yₙ) = (1 − a·h)·yₙ. The factor 1 − a·h is the same at every step, so yₙ = y₀·(1 − a·h)ⁿ exactly: a geometric sequence of fractions, whatever the size of a or h.
Lesson · First-Order Linear Equations · Stiffness
Stiffness: When the Step Is Too Big
On y′ = −a·y one Euler step multiplies the value by the exact factor 1 − a·h, and the whole column is a geometric sequence in that factor. The steps decay only when the factor lies between −1 and 1, which is h < 2/a. A larger step makes the exact fractions grow while the true solution falls to nearly zero. The backward step divides by 1 + a·h instead, and that factor lies between 0 and 1 for every step size.
Euler on y′ = −a·y: yₙ = y₀·(1 − a·h)ⁿ
decays smoothly when h < 1/a
wobbles but decays when 1/a < h < 2/a
grows when h > 2/a
backward: yₙ₊₁ = yₙ/(1 + a·h), always decays
The factor that decides whether Euler decays
Core ideas
What this lesson establishes
Three ideas: the factor, the bound it gives, and the different step that has no bound.
2
The steps decay only if the factor's size is below one
A geometric sequence goes to zero exactly when the size of its factor is less than 1. The condition |1 − a·h| < 1 means −1 < 1 − a·h < 1, which is 0 < a·h < 2, and so h < 2/a. The true solution decays for every h, so the bound belongs to the method alone.
3
The backward step divides instead
Backward Euler evaluates the rate at the new point: yₙ₊₁ = yₙ + h·(−a·yₙ₊₁). Solving for yₙ₊₁ gives yₙ₊₁ = yₙ/(1 + a·h). For a > 0 and h > 0 the denominator exceeds 1, so the factor lies strictly between 0 and 1 at every step size.
The material
The factor, its bound, and a step that has none
Euler's factor, what each range of the step does, the bound with its proof, the lab's three steps on one decay, and the backward alternative with its limits.
Take y′ = −5y with y(0) = 1. The true solution is e^(−5t), which falls toward zero quickly. The question for this lesson is whether Euler's polygon falls with it, and the answer depends on the step alone.
yₙ₊₁ = yₙ + h·(−5·yₙ) = (1 − 5h)·yₙ
yₙ = (1 − 5h)ⁿ
Everything is carried by the factor ρ = 1 − a·h, an exact fraction. Its sign says whether successive values change sign, and its size says whether they shrink.
Theorem · The step bound for decay
Forward Euler on y′ = −a·y with a > 0 goes to zero for every starting value exactly when 0 < h < 2/a.
Euler gives yₙ = y₀·ρⁿ with ρ = 1 − a·h. A geometric sequence goes to zero for every start when |ρ| < 1, and does not when |ρ| ≥ 1.
Now |1 − a·h| < 1 means −1 < 1 − a·h < 1. Subtracting 1 and changing signs gives 0 < a·h < 2, which is 0 < h < 2/a. At h = 2/a the factor is exactly −1 and the values alternate forever without shrinking.
□
Three steps on one decay
For a = 5 the bound is 2/5, and a second marker, 1/a = 1/5, is where the factor changes sign. Below 1/5 the factor is positive and the steps decay smoothly. Between 1/5 and 2/5 it is negative: the values alternate in sign and still shrink. Above 2/5 they alternate and grow.
h factor what the steps do y₈ (exact)
1/10 1/2 decays 1/256
1/4 −1/4 oscillates and decays 1/65536
1/2 −3/2 oscillates and grows 6561/256
The last row is the preset the lab opens with. With h = 1/2 and eight steps the polygon reaches t = 4, where it holds (3/2)⁸ = 6561/256, which is about 25.6289, rounded. The true value is e^(−20) ≈ 2.06115·10⁻⁹, also rounded, and the tile writes it as 2.06115e-9. Every fraction in the column is correct arithmetic; the method has turned a decay into a growth.
Small is relative to 2/a
A step of 1/2 looks small, and for a slow decay such as y′ = −y it is: the bound there is 2. The bound shrinks as a grows, so a quickly decaying component forces a small step even after it has stopped mattering to the answer. That is what the word stiff names here: the step is capped by the fastest decay in the problem, not by the accuracy one wants.
The backward step has no bound
The backward factor 1/(1 + a·h) is between 0 and 1 whatever h is. For a = 5 and h = 1/2 it is 1/(1 + 5/2) = 2/7, so the steps decay, where the forward factor was −3/2. Stable does not mean accurate: the true factor over one step is e^(−5/2) ≈ 0.082085, rounded, and 2/7 ≈ 0.285714 is more than three times as large. After eight steps the backward column holds (2/7)⁸ = 256/5764801, still far above the true value, but it has not blown up.
Example · What the backward method costs
For y′ = −a·y the backward step is solved by one division, because the equation is linear in y. For a nonlinear right-hand side the new value appears on both sides inside the function, and finding it takes an equation to be solved at every step.
That cost is the price of the bound's disappearing. This course shows the one linear case; the general theory is not part of it.
Interactive
Stiffness
Euler's factor 1 − a·h, exactly
Method
Deciding whether a step size is safe
Compute the factor, read its size and sign, compare the step with the bound, and only then trust the column.
1
Write the equation as y′ = −a·y
Read a, which is positive for a decay. For an equation with a forcing, the gap to the steady state obeys this equation, so a is the same.
2
Compute the forward factor 1 − a·h
An exact fraction. Then yₙ = y₀·(1 − a·h)ⁿ, and every value in the Euler column is a power of the factor.
3
Read its sign and size
Between 0 and 1: smooth decay. Between −1 and 0: alternating and decaying. Exactly −1: alternating without shrinking. Below −1: alternating and growing.
4
Compare h with 2/a
The steps decay exactly when h < 2/a. If the step is larger, shrink it, or change the method.
5
Compute the backward factor if the bound is too tight
1/(1 + a·h), between 0 and 1 for every h. It decays, but how closely it follows the true decay is a separate question.
Factors, bounds and the backward step
Score 0 / 0Choose the answer you can justify.
Euler on y′ = −5y with a step of 1/2
The decay rate is a = 5 and the step is h = 1/2, so the bound is 2/5 and the step exceeds it. The forward factor and the backward factor are both exact.
forward factor: 1 − 5·(1/2) = −3/2
bound: h < 2/5, but h = 1/2
y₈ = (−3/2)⁸ = 6561/256 ≈ 25.6289
true y(4) = e^(−20) ≈ 2.06115·10⁻⁹
backward factor: 1/(1 + 5/2) = 2/7
The polygon does not merely lose accuracy. It reverses its behaviour: a quantity that decays to nearly zero is described as one that grows and changes sign at every step. A smaller error per step would not have prevented this, because the factor amplifies whatever error there is.
For a rehearsal, take the second preset. The step 1/4 is above 1/5 and below 2/5. Predict the factor, the verdict and y₈ before pressing the button, then read the backward factor as well.
Common mistakes
1
Thinking any small step will doThe step h = 1/2 is small next to 1, and it makes Euler fail on y′ = −5y: the factor is −3/2, and the column grows to 6561/256 while the true value falls to ≈ 2.06115·10⁻⁹. What counts as small is h < 2/a, which depends on the equation. Lowering the error made in a single step does not help when the factor multiplies the error that is already there.
2
Treating the backward method as the accurate oneThe backward factor 2/7 decays for h = 1/2, but the true factor for that step is ≈ 0.082085. After eight steps the backward column holds 256/5764801, still far above the true value. Backward Euler removes the bound; it does not remove the error.
3
Taking any sign change as the method failingWith h = 1/4 the factor is −1/4: the values alternate in sign, and they decay all the same, to 1/65536 after eight steps. The failure is the factor's size passing 1, which happens at 2/a, not the sign changing, which happens at 1/a. The wobble is inaccurate, but it is not unstable.
Check your understanding
✓
Finish when you can compute Euler's factor for a decay, say what the steps do, and state the largest step that decays.
You should be able to write the forward factor 1 − a·h as an exact fraction, classify the behaviour from its sign and size, derive the bound h < 2/a, compute the backward factor 1/(1 + a·h), and say why a small step size alone does not guarantee a faithful answer.
Note: That is the end of the first-order equations. Everything so far has been one unknown and its rate; the next course adds a second derivative, and the single exponential factor becomes a pair of them whose exponents are roots of a quadratic equation from Algebra.