Deturck, Wilf - Lecture Notes on Numerical Analysis (523142), страница 4
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The initial conditions tell us thatc1 = 1 and c2 = 0, hence the solution of our problem is y(t) = e−t , and it represents a18Differential and Difference Equationsfunction that decays rapidly to zero with increasing t. In fact, when t = 10, the solutionhas the value 0.000045.Now let’s change the initial data (1.5.2) just a bit, by asking for a solution with y 0 (0) =−0.999. It’s easy to check that the solution is nowy(t) = (0.999666 .
. .)e−t + (0.000333 . . .)e2t(1.5.3)instead of just y(t) = e−t . If we want the value of the solution at t = 10, we would findthat it has changed from 0.000045 to about 7.34.At t = 20 the change is even more impressive, from 0.00000002 to 161, 720+, just fromchanging the initial value of y 0 from −1 to −0.999. Let’s hope that there are no phenomenain nature that behave in this way, or our lives hang by a slender thread indeed!Now exactly what is the reason for the observed instability of the equation (1.5.1)?The general solution of the equation contains a falling exponential term c1 e−t , and a risingexponential term c2 e2t .
By prescribing the initial data (1.5.2) we suppressed the growingterm, and picked out only the decreasing one. A small change in the initial data, however,results in the presence of both terms in the solution.Now it’s time for a formalDefinition: A differential equation is said to be stable if for every set of initial data (att = 0) the solution of the differential equation remains bounded as t approaches infinity.A differential equation is called strongly stable if, for every set of initial data (at t = 0)the solution not only remains bounded, but approaches zero as t approaches infinity.What makes the equation (1.5.1) unstable, then, is the presence of a rising exponentialin its general solution.
In other words, if we have a differential equation whose generalsolution contains a term eαt in which α is positive, that equation is unstable.Let’s restrict attention now to linear differential equations with constant coefficients.We know from section 1.2 that the general solution of such an equation is a sum of termsof the form(polynomial in t)eαt .(1.5.4)Under what circumstances does such a term remain bounded as t becomes large and positive?Certainly if α is negative then the term stays bounded. Likewise, if α is a complexnumber and its real part is negative, then the term remains bounded. If α has positive realpart the term is unbounded.This takes care of all of the possibilities except the case where α is zero, or more generally,the complex number α has zero real part (is purely imaginary).
In that case the questionof whether (polynomial in t)eαt remains bounded depend on whether the “polynomial in t”is of degree zero (a constant polynomial) or of higher degree. If the polynomial is constantthen the term does indeed remain bounded for large positive t, whereas otherwise the termwill grow as t gets large, for some values of the initial conditions, thereby violating thedefinition of stability.1.6 Stability theory of difference equations19Now recall that the “polynomial in t” is in fact a constant if the root α is a simpleroot of the characteristic equation of the differential equation, and otherwise it is of higherdegree. This observation completes the proof of the following:Theorem 1.5.1 A linear differential equation with constant coefficients is stable if andonly if all of the roots of its characteristic equation lie in the left half plane, and those thatlie on the imaginary axis, if any, are simple.
Such an equation is strongly stable if and onlyif all of the roots of its characteristic equation lie in the left half plane, and none lie on theimaginary axis.Exercises 1.51. Determine for each of the following differential equations whether it is strongly stable,stable, or unstable.(a) y 00 − 5y 0 + 6y = 0(b) y 00 + 5y 0 + 6y = 0(c) y 00 + 3y = 0(d) (D + 3)3 (D + 1)y = 0(e) (D + 1)2 (D 2 + 1)2 y = 0(f) (D4 + 1)y = 02. Make a list of some natural phenomena that you think are unstable.
Discuss.3. The differential equation y 00 −y = 0 is to be solved with the initial conditions y(0) = 1,y 0 (0) = −1, and then solved again with y(0) = 1, y 0 (0) = −0.99. Compare the twosolutions when x = 20.4. For exactly which real values of the parameter λ is each of the following differentialequations stable? . . . strongly stable?(a) y 00 + (2 + λ)y 0 + y = 0(b) y 00 + λy 0 + y = 0(c) y 0 + λy = 11.6Stability theory of difference equationsIn the previous section we discussed the stability of differential equations. The key ideaswere that such an equation is stable if every one of its solutions remains bounded as tapproaches infinity, and strongly stable if the solutions actually approach zero.Similar considerations apply to difference equations, and for similar reasons.
As anexample, take the equation5yn+1 = yn − yn−12(n ≥ 1)(1.6.1)20Differential and Difference Equationsalong with the initial equationsy0 = 1;y1 = 0.5 .(1.6.2)It’s easy to see that the solution is yn = 2−n , and of course, this is a function thatrapidly approaches zero with increasing n.Now let’s change the initial data (1.6.2), say toy0 = 1;y1 = 0.50000001(1.6.3)instead of (1.6.2).The solution of the difference equation with these new data isy = (0.0000000066 . .
.)2n + (0.9999999933 . . .)2−n .(1.6.4)The point is that the coefficient of the growing term 2n is small, but 2n grows so fast thatafter a while the first term in (1.6.4) will be dominant. For example, when n = 30, thesolution is y30 = 7.16, compared to the value y30 = 0.0000000009 of the solution with theoriginal initial data (1.6.2). A change of one part in fifty million in the initial conditionproduced, thirty steps later, an answer one billion times as large.The fault lies with the difference equation, because it has both rising and falling components to its general solution. It should be clear that it is hopeless to do extended computation with an unstable difference equation, since a small roundoff error may alter thesolution beyond recognition several steps later.As in the case of differential equations, we’ll say that a difference equation is stableif every solution remains bounded as n grows large, and that it is strongly stable if everysolution approaches zero as n grows large.
Again, we emphasize that every solution mustbe well behaved, not just the solution that is picked out by a certain set of initial data. Inother words, the stability, or lack of it, is a property of the equation and not of the startingvalues.Now consider the case where the difference equation is linear with constant coefficients.The we know that the general solution is a sum of terms of the form(polynomial in n)αn .(1.6.5)Under what circumstances will such a term remain bounded or approach zero?Suppose |α| < 1.
Then the powers of α approach zero, and multiplication by a polynomial in n does not alter that conclusion. Suppose |α| > 1. Then the sequence of powersgrows unboundedly, and multiplication by a nonzero polynomial only speeds the partingguest.Finally suppose the complex number α has absolute value 1. Then the sequence of itspowers remains bounded (in fact they all have absolute value 1), but if we multiply by anonconstant polynomial, the resulting expression would grow without bound.To summarize then, the term (1.6.5), if the polynomial is not identically zero, approacheszero with increasing n if and only if |α| < 1.
It remains bounded as n increases if and only1.6 Stability theory of difference equations21if either (a) |α| < 1 or (b) |α| = 1 and the polynomial is of degree zero (a constant). Nowwe have proved:Theorem 1.6.1 A linear difference equation with constant coefficients is stable if and onlyif all of the roots of its characteristic equation have absolute value at most 1, and those ofabsolute value 1 are simple. The equation is strongly stable if and only if all of the rootshave absolute value strictly less than 1.Exercises 1.61.
Determine, for each of the following difference equations whether it is strongly stable,stable, or unstable.(a) yn+2 − 5yn+1 + 6yn = 0(b) 8yn+2 + 2yn+1 − 3yn = 0(c) 3yn+2 + yn = 0(d) 3yn+3 + 9yn+2 − yn+1 − 3yn = 0(e) 4yn+4 + 5yn+2 + yn = 02. The difference equation 2yn+2 + 3yn+1 − 2yn = 0 is to be solved with the initialconditions y0 = 2, y1 = 1, and then solved again with y0 = 2, y1 = 0.99.
Compare y20for the two solutions.3. For exactly which real values of the parameter λ is each of the following differenceequations stable? . . . strongly stable?(a) yn+2 + λyn+1 + yn = 0(b) yn+1 + λyn = 1(c) yn+2 + yn+1 + λyn = 04. (a) Consider the (constant-coefficient) difference equationa0 yn+p + a1 yn+p−1 + a2 yn+p−2 + · · · + ap yn = 0.(1.6.6)Show that this difference equation cannot be stable if |ap /a0 | > 1.(b) Give an example to show that the converse of the statement in part (a) is false.Namely, exhibit a difference equation for which |ap /a0 | < 1 but the equation is unstable anyway.22Differential and Difference EquationsChapter 2The Numerical Solution ofDifferential Equations2.1Euler’s methodOur study of numerical methods will begin with a very simple procedure, due to Euler.We will state it as a method for solving a single differential equation of first order.
One ofthe nice features of the subject of numerical integration of differential equations is that thetechniques that are developed for just one first order differential equation will apply, withvery little change, both to systems of simultaneous first order equations and to equations ofhigher order. Hence the consideration of a single equation of first order, seemingly a veryspecial case, turns out to be quite general.By an initial-value problem we mean a differential equation together with enough givenvalues of the unknown function and its derivatives at an initial point x0 to determine thesolution uniquely.Let’s suppose that we are given an initial-value problem of the formy 0 = f (x, y);y(x0 ) = y0 .(2.1.1)Our job is to find numerical approximate values of the unknown function y at points xto the right of (larger than) x0 .What we actually will find will be approximate values of the unknown function at adiscrete set of points x0 , x1 = x0 + h, x2 = x0 + 2h, x3 = x0 + 3h, etc.
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