Zero stability

Zero-stability, also known as D-stability in honor of Germund Dahlquist,{{cite journal |last1=Dahlquist |first1=Germund |title=Convergence and stability in the numerical integration of ordinary differential equations |journal=Mathematica Scandinavica |date=1956 |volume=4 |issue=4 |pages=33–53 |doi=10.7146/math.scand.a-10454 |jstor=24490010 |doi-access=free }} refers to the stability of a numerical scheme applied to the simple initial value problem $y'(x) = 0$ .

A linear multistep method is zero-stable if all roots of the characteristic equation that arises on applying the method to $y'(x) = 0$ have magnitude less than or equal to unity, and that all roots with unit magnitude are simple.{{cite book |last1=Hairer |first1=Ernst |last2=Nørsett |first2=Syvert |last3=Wanner |first3=Gerhard |title=Solving Ordinary Differential Equations I |date=1987 |publisher=Springer-Verlag |location=Berlin |pages=326–328}} This is called the root condition{{cite book |last1=Butcher |first1=John C |title=The Numerical Analysis of Ordinary Differential Equations |date=1987 |publisher=Wiley |page=11}} and means that the parasitic solutions of the recurrence relation will not grow exponentially.

Example

The following third-order method has the highest order possible for any explicit two-step method for solving $y'(x) = f(x)$ :

$y_{n+2} + 4 y_{n+1} - 5y_n = h(4f_{n+1} + 2 f_n).$

If $f(x)=0$ identically, this gives a linear recurrence relation with characteristic equation

$r^2 + 4r - 5=(r-1)(r+5) = 0.$

The roots of this equation are $r=1$ and $r=-5$ and so the general solution to the recurrence relation is $y_n = c_1\cdot 1^n + c_2 (-5)^n$ . Rounding errors in the computation of $y_1$ would mean a nonzero (though small) value of $c_2$ so that eventually the parasitic solution $(-5)^n$ would dominate. Therefore, this method is not zero-stable.

References

Category:Numerical differential equations