From a Differential Equation to a Stability Question
For the scalar test equation \(y'=\lambda y\), the exact solution starting from \(y(0)=y_0\) is \(y(t)=e^{\lambda t}y_0\). If \(\lambda\) is real and negative, this solution decays as time increases. More generally, for complex \(\lambda\), its magnitude is \(|y(t)|=e^{(\operatorname{Re}\lambda)t}|y_0|\), so it decays when \(\operatorname{Re}\lambda<0\).
Forward Euler does not follow the exact exponential curve. With a fixed positive step size \(h\), its update is \(y_{n+1}=y_n+h\lambda y_n\). The question is whether repeated numerical steps shrink the values, as a decaying exact solution should. The answer depends on the dimensionless complex quantity \(z=h\lambda\).
Why the Amplification Factor Controls Decay
Substituting the right-hand side \(f(t,y)=\lambda y\) into the forward Euler formula gives
Thus every step multiplies the current value by the same factor \(1+z\). Repeated substitution gives \(y_n=(1+z)^n y_0\). Taking magnitudes yields \(|y_n|=|1+z|^n|y_0|\). If \(|1+z|<1\), the powers tend to zero, so every nonzero initial value decays to zero under the numerical recurrence. If \(|1+z|>1\), the magnitudes grow geometrically when \(y_0\ne0\). If \(|1+z|=1\), the magnitudes remain constant; the values do not decay to zero.
This is a statement about repeated Euler steps for the test equation. It does not say that every differential equation is accurately approximated whenever its parameter lies in the region. Stability and accuracy are different questions: stability describes the behavior of repeated numerical amplification, while accuracy concerns how close the computed values are to the exact solution.
Deriving the Stability Disk
The strict decay condition is \(|1+z|<1\). In the complex plane, this says that the distance from \(z\) to \(-1\) is less than \(1\), because \(|1+z|=|z-(-1)|\). Therefore the stability region is the open disk centered at \(-1\) with radius \(1\).
To express the disk in coordinates, write \(z=x+iy\), where \(x,y\in\mathbb R\). Then \(1+z=(1+x)+iy\), and the modulus condition becomes
The boundary circle is \((x+1)^2+y^2=1\). It is not part of the strict decay region: on the boundary, \(|1+z|=1\), so each step preserves the magnitude. Outside the closed disk, \(|1+z|>1\), and nonzero values grow in magnitude under repeated steps.
Form \(z=h\lambda\), using the same step size and equation parameter as in the Euler update.
Calculate \(R(z)=1+z\).
If \(|R(z)|<1\), the numerical values decay; if it equals one, they do not decay; if it exceeds one, they grow for nonzero initial data.
Worked Examples
Worked Example: A Real Parameter Inside the Region
Apply forward Euler with \(h=0.25\) to \(y'=-4y\), starting from \(y_0=3\). The parameter is
Therefore \(R(z)=1+z=1-1=0\), and \(|R(z)|=0<1\). The first Euler step is
Every later value is also zero because \(y_{n+1}=0\cdot y_n=0\). This verifies decay for the numerical recurrence. As a check, substituting \(z=-1\) into the disk condition gives \((x+1)^2+y^2=(0)^2+0^2=0<1\), so the parameter is indeed inside the region. The exact solution is \(y(t)=3e^{-4t}\); its decay is consistent with the direction of the numerical behavior, although this Euler step reaches zero immediately rather than following the exact curve.
Worked Example: A Complex Parameter Inside the Region
Let \(\lambda=-2+2i\), \(h=0.25\), and \(y_0=1\). Then
The squared modulus is \(0.5^2+0.5^2=0.5\), so \(|R(z)|=\sqrt{0.5}=1/\sqrt{2}<1\). The first two steps are
As a check, \(|y_1|=\sqrt{0.5}\) and \(|y_2|=0.5\), matching \(|y_n|=(1/\sqrt{2})^n|y_0|\). The disk test gives \((x+1)^2+y^2=(0.5)^2+(0.5)^2=0.5<1\), confirming that this complex parameter is inside the stability region.
Worked Example: A Complex Parameter Outside the Region
Take \(h=0.25\), \(\lambda=-4+4.8i\), and \(y_0=1\). Then
Here \(|R(z)|=|1.2i|=1.2>1\), so the numerical values grow in magnitude. In fact,
The magnitudes are \(1.2\), \(1.44\), and \(1.728\), respectively, and each is \(1.2\) times the preceding magnitude. The coordinate test confirms the same result: \((x+1)^2+y^2=(0)^2+(1.2)^2=1.44>1\). Thus \(z\) lies outside the disk.
Worked Example: A Parameter on the Boundary
Let \(h=0.25\), \(\lambda=-4+4i\), and \(y_0=1\). Then
The sequence begins \(y_0=1\), \(y_1=i\), \(y_2=-1\), and \(y_3=-i\). Each value has magnitude \(1\), so the sequence does not decay to zero. The disk equation gives \((x+1)^2+y^2=0^2+1^2=1\), confirming that \(z\) lies on the boundary. This check also illustrates why the stability region is open rather than closed.
How the Real Case Fits the Disk
If \(\lambda\) and \(h\) are real, then \(z=h\lambda\) lies on the real axis, so \(y=0\). The disk condition reduces to \((x+1)^2<1\), which is equivalent to \(-2<x<0\). Thus the real-axis portion of the strict stability region consists of negative parameters between \(-2\) and \(0\). This agrees with the scalar exponential-decay case: for a negative real \(\lambda\), the step must be small enough that \(h\lambda\) remains in that interval.
For complex \(\lambda\), checking only the real part is not enough. The imaginary component contributes to the distance from \(-1\), and therefore to the modulus of the Euler amplification factor. A parameter can have a negative real part while still lying outside the disk, as the previous example demonstrates. Continuous decay requires \(\operatorname{Re}\lambda<0\); Euler decay requires the stronger step-dependent test \(|1+h\lambda|<1\).
Common Pitfall: Comparing Real Parts Instead of Moduli
A tempting but incorrect test is to decide stability by checking whether the real part of \(1+z\) is less than \(1\). For example, with \(z=-1+2i\), the amplification factor is \(R(z)=2i\). Its real part is \(0<1\), so that incorrect test would label it stable. But the relevant quantity is the modulus:
The Euler values therefore grow in magnitude for any nonzero starting value. The correct comparison is between \(|1+z|\) and \(1\), not between the real part of \(1+z\) and \(1\). Modulus measures the size of a complex multiplier and captures the effect of both its real and imaginary components.
Why the Stability Region Matters
The stability region converts repeated numerical behavior into a geometric test. Once \(h\lambda\) is located relative to the disk, the scalar test equation's long-term Euler behavior follows directly. This is useful because a differential equation may contain decaying modes that the numerical method can nevertheless amplify when the chosen step is too large. The stability disk makes that step-size restriction visible.
The test equation isolates one scalar mode and gives a precise criterion for its numerical decay. Applying this criterion to a more complicated problem requires identifying the relevant modes and parameters; the scalar disk alone does not certify accuracy or stability for every problem. Its value here is the transparent derivation: Euler multiplies each value by \(1+h\lambda\), and decay occurs exactly when that multiplier has magnitude less than one.
Check Your Understanding
For each question, use the amplification factor and its modulus rather than relying on the sign of a real part alone.
- For \(z=-0.4\), calculate \(R(z)\) and decide whether the Euler values decay.
- For \(z=-1+0.6i\), compute \(|1+z|\) and determine whether \(z\) is inside, on, or outside the stability region.
- Why does a point on the boundary circle fail the strict decay condition?
- Give the interval of real \(z\)-values in the forward Euler stability region.
- For \(z=-0.5+1.5i\), use the coordinate disk condition to classify the parameter.
