Example 7.1. For the 2-stage Runge-Kutta method of Example 6.11, we have Ψ(u, t, k) =f
u+1
2kf(u)
.
Iff is Lipschitz continuous with Lipschitz constantLthen Ψ has Lipschitz constant L0 =L+12kL2. The 1-step method (7.17) is consistentif
Ψ(u, t,0) =f(u, t) for allu, tand Ψ is continuous ink. The local truncation error is
τn =
u(tn+1)−u(tn) k
−Ψ(u(tn), tn, k).
We can show that any one-step methods satisfying these conditions is convergent. We have u(tn+1) =u(tn) +kΨ(u(tn), tn, k) +kτn
and subtracting this from (7.17) gives
En+1=En+k(Ψ(Un, tn, k)−Ψ(u(tn), tn, k))−kτn. Using the Lipschitz condition we obtain
|En+1| ≤ |En|+kL0|En|+k|τn|.
This has exactly the same form as (7.15) and the proof of convergence proceeds exactly as from there.
Table 7.1: SolutionUN to (7.20) with U0= 0, U1=kand various values ofk= 1/N.
N UN
5 4.2
10 258.4 20 1954408
with solution u(t)≡0. On this equation (7.18) takes the form
Un+2−3Un+1+ 2Un= 0. (7.20)
We need two starting valuesU0andU1. If we takeU0=U1= 0 then (7.20) generatesUn = 0 for alln and in this case we certainly converge to correct solution, and in fact we get the exact solution for any k.
But in general we will not have the exact value U1 available and will have to approximate this, introducing some error into the computation. Table 7.1 shows results obtained by applying this method with starting data U0 = 0, U1 = k. Since U1(k) → 0 as k → 0, this is valid starting data in the context of Definition 7.1.1 of convergence. If the method is convergent we should see that UN, the computed solution at time T = 1, converges to zero as k→0. Instead it blows up quite dramatically.
Similar results would be seen if we applied this method to an arbitrary equation u0 =f(u) and used any one-step method to compute U1from U0.
The homogeneous linear difference equation (7.20) can be solved explicitly for Un in terms of the starting values U0 andU1 We obtain
Un= 2U0−U1+ 2n(U1−U0). (7.21) It is easy to verify that this satisfies (7.20) and also the starting values. (We’ll see how to solve general linear difference equations in the next section.)
Sinceu(t) = 0, the error isEn =Un and we see that any initial errors in U1 or U0 are magnified by a factor 2n in the global error (except in the special case U1=U0). This exponential growth of the error is the instability that leads to nonconvergence. In order to rule out this sort of growth of errors, we need to be able to solve a general linear difference equation.
7.4.1 Solving linear difference equations
We briefly review the solution technique for linear difference equations. See Appendix A5 for more details. Consider the general homogeneous linear difference equation
r
X
j=0
αjUn+j = 0. (7.22)
Eventually we will look for a particular solution satisfying given initial conditionsU0, U1, . . . , Ur−1, but to begin with we will find the general solution of the difference equation in terms ofrfree parameters.
We will hypothesize that this equation has a solution of the form
Un=ζn (7.23)
for some value ofζ (hereζn is thenth power!). Plugging this into (7.22) gives
r
X
j=0
αjζn+j = 0
and dividing byζn yields
r
X
j=0
αjζj= 0. (7.24)
We see that (7.23) is a solution of the difference equation ifζsatisfies the equation (7.24), i.e., ifζ is a root of the polynomial
ρ(ζ) =
r
X
j=0
αjζj.
Note that this is just the first characteristic polynomial of the LMM introduced in (6.33). In general ρ(ζ) has rrootsζ1, ζ2, . . . , ζr and can be factored as
ρ(ζ) =αr(ζ−ζ1)(ζ−ζ2)· · ·(ζ−ζr).
Since the difference equation is linear, any linear combination of solutions is again a solution. If ζ1, ζ2, . . . , ζr are distinct (ζi 6=ζj fori6=j) then therdistinct solutionsζin are linearly independent and the general solution of (7.22) has the form
Un=c1ζ1n+c2ζ2n+· · ·+crζrn (7.25) wherec1, . . . , crare arbitrary constants. In this case, every solution of the difference equation (7.22) has this form. If initial conditionsU0, U1, . . . , Ur−1are specified, then the constantsc1, . . . , crcan be uniquely determined by solving ther×rlinear system
c1+c2+· · ·+cr = U0
c1ζ1+c2ζ2+· · ·+crζr = U1 (7.26) ... ...
c1ζ1r−1+c2ζ2r−1+· · ·+crζrr−1 = Ur−1
Example 7.3. The characteristic polynomial for the difference equation (7.20) is
ρ(ζ) = 2−3ζ+ζ2= (ζ−1)(ζ−2) (7.27) with rootsζ1= 1, ζ2= 2. The general solution has the form
Un=c1+c2·2n
and solving forc1andc2 fromU0andU1 gives the solution (7.21).
This example indicates that if ρ(ζ) has any roots that are greater than one in modulus, the method will not be convergent. It turns out that the converse is nearly true: If all of the roots have modulus no greater than one, then the method is convergent, with one proviso. There must be norepeatedroots with modulus equal to one. The next two examples illustrate this.
If the roots are not distinct, say ζ1=ζ2for simplicity, thenζ1n andζ2n are not linearly independent and theUn given by (7.25), while still a solution, is not the most general solution. The system (7.26) would be singular in this case. In addition toζ1nthere is also a solution of the formnζ1n and the general solution has the form
Un=c1ζ1n+c2nζ1n+c3ζ3n+· · ·+crζrn.
If in additionζ3=ζ1, then the third term would be replaced byc3n2ζ1n. Similar modifications are made for any other repeated roots. Note how similar this theory is to the standard solution technique for an r’th order linear ordinary differential equation. See Appendix A5 for more details.
Example 7.4. Applying the consistent LMM Un+2−2Un+1+Un= 1
2k(f(Un+2)−f(Un)) (7.28)
to the differential equationu0(t) = 0 gives the difference equation Un+2−2Un+1+Un = 0.
The characteristic polynomial is
ρ(ζ) =ζ2−2ζ+ 1 = (ζ−1)2 (7.29) so ζ1=ζ2= 1. The general solution is
Un =c1+c2n.
For particular starting valuesU0 andU1 the solution is Un=U0+ (U1−U0)n.
Again we see that the solution grows withn, though not as dramatically as in Example 7.2 (the growth is linear rather than exponential). But this growth is still enough to destroy convergence. If we take the same starting values as before,U0= 0 andU1=k, thenUn=knand so
lim
k→0
N k=T
UN =kN =T.
The method converges to the function v(t) =t rather than tou(t) = 0, and hence the LMM (7.28) is not convergent.
This example shows that if ρ(ζ) has a repeated root of modulus 1, then the method cannot be convergent.
Example 7.5. Now consider the consistent LMM Un+3−2Un+2+5
4Un+1−1
4Un= 1
4hf(Un). (7.30)
Applying this to (7.19) gives
Un+3−2Un+2+5
4Un+1−1 4Un= 0 and the characteristic polynomial is
ρ(ζ) =ζ3−2ζ2+5 4ζ−1
4 = (ζ−1)(ζ−0.5)2. (7.31)
Soζ1= 1, ζ2=ζ3= 1/2 and the general solution is Un=c1+c2
1 2
n
+c3n 1
2 n
.
Here there is a repeated root but with modulus less than 1. The linear growth of n should then be overwhelmed by the decay of (1/2)n.
For this 3-step method we need three starting valuesU0, U1,U2and we can findc1, c2, c3in terms of them by solving a linear system similar to (7.26). Eachciwill be a linear combination ofU0, U1, U2 and so ifUν(k)→0 ask→0 thenci(k)→0 ask→0 also. The valueUN computed at timeT with step sizek (wherekN =T) has the form
UN =c1(k) +c2(k) 1
2 N
+c3(k)N 1
2 N
. (7.32)
Now we see that
lim
k→0
N k=T
UN = 0
and so the method (7.30) converges onu0= 0 with arbitrary starting valuesUν(k) satisfyingUν(k)→0 ask→0. (In fact this LMM is convergent in general.)
More generally, if ρ(ζ) has a root ζj that is repeated r times, then UN will involve terms of the form NsζjN fors= 1, 2, , . . . , r. This converges to zero asN → ∞provided |ζj|<1. The algebraic growth of Ns is overwhelmed by the exponential decay ofζjN. This shows that repeated roots are not a problem as long as they have magnitude strictly less than 1.
With the above examples as motivation, we are ready to state the definition of zero-stability.
Definition 7.4.1 An r-step Linear Multistep Method is said to be zero-stable if the roots of the characteristic polynomial ρ(ζ)defined by (6.33) satisfy the following conditions:
|ζj| ≤1 for j= 1, 2, , . . . , r
Ifζj is a repeated root, then |ζj|<1. (7.33) If the conditions (7.33) are satisfied for all roots ofρ, then the polynomial is said to satisfy theroot condition.
Example 7.6. The Adams methods have the form Un+r=Un+r−1+k
r
X
j=1
βjf(Un+j) and hence
ρ(ζ) =ζr−ζr−1= (ζ−1)ζr−1.
The roots are ζ1 = 1 and ζ2 = · · · = ζr = 0. The root condition is clearly satisfied and all of the Adams-Bashforth and Adams-Moulton methods are zero stable.
The examples given above certainly do not prove that zero-stability as defined above is a sufficient condition for convergence. We only looked at the simplest possible ODEu0(t) = 0 and saw that things could go wrong if the root condition isnotsatisfied. It turns out, however, that the root condition is all that is needed to prove convergence on the general initial value problem (in the sense of Definition 7.1.1).
For the initial value problem we have the general result that
consistency + zero-stabilty ⇐⇒ convergence. (7.34) This is the analog of the statement (2.21) for the boundary value problem. A proof of this result can be found in [Hen62]
Note: A consistent linear multistep method always has one root equal to 1, say ζ1 = 1, called theprincipal root. This follows from (6.34). Hence a consistent 1-step LMM (such as Euler, backward Euler, trapezoidal) is certainly zero-stable. More generally we have proved in Section 7.3.4 that any consistent 1-step method (that is Lipschitz continuous) is convergent. Such methods are automatically
“zero-stable” and behave well ask→0. We can think of zero-stability as meaning “stable in the limit ask→0.
Although a consistent zero-stable method is convergent, it may have other stability problems that show up the time stepkis chosen too large in an actual computation. Additional stability considerations are the subject of the next chapter.