Why Step Size Affects Decay
For a differential equation describing exponential decay, the exact solution moves smoothly toward zero. A numerical method can behave differently: its values may shrink, remain the same size, or alternate in sign. The forward Euler method makes these behaviors especially clear because each step multiplies the current value by one fixed factor.
We will analyze that factor for the initial-value problem \(y'=-ky\), where \(k>0\). The goal is to determine exactly which positive step sizes produce numerical decay without sign alternation, and how those conditions differ from decay that alternates in sign.
The Euler Recurrence and Amplification Factor
Let \(h>0\) be a fixed step size, and let \(t_n=t_0+nh\). Write \(y_n\) for the Euler approximation at \(t_n\). Forward Euler advances by adding the step size times the slope at the current point. Since the slope is \(-ky_n\), the update is
The number multiplying \(y_n\) is called the amplification factor for this recurrence. Set \(q=1-kh\). Then each step has the same simple form, \(y_{n+1}=qy_n\), and repeated substitution gives \(y_n=q^n y_0\) for every integer \(n\ge 0\).
Here \(k\) has units of inverse time and \(h\) has units of time, so \(kh\) and \(q\) are dimensionless. Step-size conditions should therefore be expressed using the dimensionless product \(kh\), not by comparing \(h\) alone to a pure number.
Deriving the Decay Conditions
Assume first that \(y_0\ne 0\). From \(y_n=q^n y_0\), the magnitude of each value is \(|y_n|=|q|^n|y_0|\). Thus the size behavior is determined by \(|q|\), while the sign behavior is determined by whether \(q\) is positive, zero, or negative.
The geometric factor must satisfy \(|q|<1\). Then \(|y_n|=|q|^n|y_0|\) tends to zero.
The multiplier must be nonnegative, \(q\ge 0\). If \(q<0\), multiplication reverses the sign at each step.
Nonnegative \(q\) with \(|q|<1\) means \(0\le q<1\). Substituting \(q=1-kh\) gives \(0<kh\le 1\).
There is an important distinction at the endpoint \(q=0\). The values satisfy \(y_1=0\) and remain zero thereafter. Their magnitudes are nonincreasing and tend to zero, but they are not strictly decreasing at every step after reaching zero. Strictly decreasing magnitudes with no sign reversal require \(0<q<1\), or equivalently \(0<kh<1\).
To see why the strict decay condition follows, if \(0<q<1\), then \(|y_{n+1}|=q|y_n|<|y_n|\) whenever \(y_n\ne0\). Also \(q^n\) tends to zero, so the magnitudes tend to zero. At \(q=0\), \(|y_1|=0\), after which all magnitudes are zero; this is nonincreasing decay to zero, not strict decrease at every later step.
For comparison, allow sign alternation but still require magnitudes to tend to zero. Then \(|q|<1\), which becomes \(0<kh<2\). The part \(1<kh<2\) has \(q<0\): the values alternate signs while shrinking. At \(kh=2\), \(q=-1\), so the values alternate with constant magnitude and do not decay. For \(kh>2\), \(q<-1\), so the magnitudes grow while the signs alternate.
| Condition on \(kh\) | Multiplier \(q=1-kh\) | Behavior for \(y_0\ne0\) |
|---|---|---|
| \(0<kh<1\) | \(0<q<1\) | Same sign; magnitudes strictly decrease to zero. |
| \(kh=1\) | \(q=0\) | Zero after one step; thereafter the values stay zero. |
| \(1<kh<2\) | \(-1<q<0\) | Signs alternate; magnitudes strictly decrease to zero. |
| \(kh=2\) | \(q=-1\) | Signs alternate; magnitude is constant. |
| \(kh>2\) | \(q<-1\) | Signs alternate; magnitudes grow. |
Worked Examples
Worked Example: Decay Without Sign Alternation
Consider \(y'=-3y\), with \(y_0=10\) and step size \(h=0.2\). Here \(k=3\), so \(kh=3(0.2)=0.6\). The amplification factor is
Thus the Euler recurrence is \(y_{n+1}=0.4y_n\). The first values are
Each value is positive and is \(0.4\) times the preceding value, so the values strictly decrease. The general expression is \(y_n=10(0.4)^n\), whose magnitude tends to zero. Check the update using the original Euler formula: at \(y_0=10\), the slope is \(-3(10)=-30\), and \(y_1=10+0.2(-30)=4\), as obtained from the multiplier.
Worked Example: The Zero-Multiplier Endpoint
Take \(y'=-2y\), \(y_0=7\), and \(h=0.5\). Then \(kh=2(0.5)=1\), and
The Euler values are \(y_1=0(7)=0\), \(y_2=0(0)=0\), and \(y_3=0(0)=0\). The magnitude falls from \(7\) to \(0\), then remains \(0\); it is nonincreasing and tends to zero, but it is not strictly decreasing after the first step. Check directly: the first slope is \(-2(7)=-14\), so \(y_1=7+0.5(-14)=0\). At \(y_1=0\), the slope is zero, confirming that subsequent Euler values remain zero.
Worked Example: Alternating Decay
Let \(y'=-4y\), \(y_0=8\), and \(h=0.375\). First compute \(kh=4(0.375)=1.5\). Therefore
The recurrence gives \(y_1=(-0.5)(8)=-4\), \(y_2=(-0.5)(-4)=2\), and \(y_3=(-0.5)(2)=-1\). The signs alternate, while the magnitudes are \(8,4,2,1\), shrinking by a factor of \(0.5\) each step. The general expression \(y_n=8(-0.5)^n\) confirms that the magnitudes tend to zero. Check the first update directly: the initial slope is \(-4(8)=-32\), and \(8+0.375(-32)=8-12=-4\), agreeing with the recurrence.
Why This Is a Stability Question
The exact solution of \(y'=-ky\) with initial value \(y(t_0)=y_0\) is \(y(t)=y_0e^{-k(t-t_0)}\). For \(k>0\), its magnitude decreases toward zero as time increases, and its sign does not change when \(y_0\ne0\). The Euler multiplier provides a direct test of whether the discrete values reproduce those qualitative features.
If \(0<kh<1\), Euler preserves the sign and decreases the magnitude at every step. If \(kh=1\), it reaches zero in one step. If \(1<kh<2\), the values still decay in magnitude, but they reverse sign each step, unlike the exact solution. Thus shrinking magnitude alone does not guarantee that the numerical trajectory resembles the qualitative behavior of the exact solution.
This analysis concerns the recurrence's stability behavior, not a complete estimate of its numerical error. A step size can yield bounded, decaying values and still produce noticeable error over a chosen time interval. Stability and accuracy answer different questions: stability asks whether perturbations or values grow under repeated steps, while accuracy asks how closely the numerical values approximate the exact solution.
A Common Endpoint Mistake
A common error is to say that \(0\le q<1\) always gives strictly decreasing magnitudes. For example, take \(y_0=5\) and \(q=0\). Then the magnitudes are \(5,0,0,0,\ldots\). They decrease on the first step but not on later steps. The correction is to use \(0<q<1\) for strictly decreasing magnitudes, and \(0\le q<1\) for nonincreasing magnitudes that tend to zero. In step-size terms, these are respectively \(0<kh<1\) and \(0<kh\le1\).
A related endpoint mistake is to include \(kh=2\) among step sizes that give decaying magnitudes. At that value \(q=-1\), so \(y_n=(-1)^ny_0\) and \(|y_n|=|y_0|\) for every \(n\). The values alternate, but their magnitudes do not decrease. Strict decay in magnitude, whether or not signs alternate, requires \(|q|<1\), equivalently \(0<kh<2\).
Summary of the Step-Size Test
For \(k>0\) and \(h>0\), calculate the dimensionless product \(kh\), then use \(q=1-kh\). To obtain decay without sign alternation, choose \(0<kh\le1\); use \(0<kh<1\) if the magnitude must strictly decrease at each step. To obtain decay in magnitude while allowing alternation, the wider condition is \(0<kh<2\). The boundary \(kh=2\) is neutral, not decaying, and larger products produce growing alternating magnitudes.
Check Your Understanding
For each question, identify the multiplier first, then use its sign and magnitude to describe the Euler values.
- For \(y'=-5y\) and \(h=0.1\), do the values decay without sign alternation? Is the magnitude strictly decreasing?
- For \(y'=-3y\) and \(h=\frac{2}{3}\), what happens after the first Euler step, and why is this not strict decay at every step?
- For \(y'=-2y\) and \(h=0.75\), do the magnitudes tend to zero? Do the signs alternate?
- What behavior occurs when \(kh=2\)? Explain using \(q=1-kh\).
- State the condition on \(kh\) for magnitudes to strictly decrease to zero when sign alternation is allowed.
