Shampine, Allen, Pruess - Fundamentals of Numerical Computing (523185), страница 50
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Dept., Southern Methodist University,Dallas, Tex., 1992.2. P. Brown, G. Byrne, and A. Hindmarsh, “VODE: a variable-coefficient ODE solver,” SIAM J.Sci. Stat. Comput., 10 (1989), pp. 1038-1051.3. E. Coddington, An Introduction to Ordinary Differential Equations, Prentice Hall, EnglewoodCliffs, N.J., 1961.4. W. Cowell, ed., Sources and Development of Mathematical Software, Prentice Hall, EnglewoodCliffs, N.J., 1984.5.
R. England, “Error estimates for Runge-Kutta type solutions to systems of ordinary differentialequations,” Computer Journal, 12 (1969), pp. 166-170.6. C. W. Gear, Numerical Initial Value Problems in Ordinary Differential Equations, Prentice Hall,Englewood Cliffs, N.J., 1971.REFERENCES 2497. Handbook of Mathematical Functions, M.
Abramowitz and I. Stegun, eds., Dover, Mineola,N.Y., 1964.8. L. Shampine, Numerical Solution of Ordinary Differential Equations, Chapman & Hall, NewYork, 1994.9. L. Shampine and C. W. Gear, “A user’s view of solving stiff ordinary differential equations,”SIAM Review, 21 (1979), pp. 1-17.10. L. Shampine and M. Gordon, Computer Solution of Ordinary Differential Equations: The InitialValue Problem, Freeman, San Francisco, 1975.MISCELLANEOUS EXERCISES FOR CHAPTER 6Unless otherwise indicated, use 10-5 for tolerancevalues and 10-7 for the threshold values for the computing exercises.k123456789106.25 Consider the problemy´ = 2|x|y, y(-1) = 1/eon the interval [-1,1].
Verify that the existence anduniqueness theorem (Theorem 6.1) applies to thisproblem. Verify that the solution to this problem iserrorh = 2/2 k1.0E - 11.0E - 28.2E - 45.6E - 53.6E - 62.3E - 71.4E - 89.0E - 105.6E - 113.5E - 12errorh = 2/3k2.4E - 12.3E - 22.5E - 32.8E - 43.1E - 53.4E - 63.8E - 74.2E - 84.7E - 95.2E - 106.26 In modeling circuits containing devices with electrical properties that depend on the current, differentialand further that y(x) has one continuous derivativeequations of the formon [-1,1] but does not have two.
Is Euler’s methodconvergent and O(h) for this problem? What about“higher order” Runge-Kutta methods? What are theanswers to these questions for the two problemsoccur. For the case wherey´ = 2|x|y, y(-1) = 1/eon [-1,0] andy´ = 2|x|y, y(0) = 1on [0,1]? Show that solving the original problem on[-1,1] with a mesh point at x = 0 is the same as solving these two problems.Explain the following numerical results: A fourthorder Runge-Kutta code was used to integrateandy´ = 2|x|y, y(-1) = 1/efrom x = -1 to x = 1 using a fixed step size h andthe true error at x = 1 computed from the analyticalsolution. Two computations were done.
One usedh = 2/2k and the other h = 2/3 k. The results in thefollowing table were obtained:6.27compute x(t),t = 0.4, 0.8, 1.2, . . . , 16. Plot x1(t),x2(t), and x3(t), 0 < t < 16, on separate graphs. (Hint:Use Factor/Solve in conjunction with Rke.)Approximate the solution to the nonlinear two-point250CHAPTER 6ORDINARY DIFFERENTIAL EQUATIONSboundary value problemy´´ = ey - 1, y(0) = 0, y(l) = 3at x = 0.1, 0.2,. . .
, 0.9. First use Rke and Zero to findthe missing initial condition y´(0). Then use Rke tosolve the resulting initial value problem at the desiredx values. (Hint: Denoting the solution ofthe velocity of the wind as a function of altitude. Assumptions are that the wind direction is constant, theground is level, and the balloonist is only allowed tocoast.
Lengths are measured in feet and time in seconds. The following data have been collected for v(a):y´´ = ey - l, y(0) = 0, y´(0) = sby y(t;s), the problem is to find s so that y(1;s ) = 3.Use Zero to find the root of G(s) = y(1;s) - 3 = 0.) Ifyou have worked Exercise 5.12, compare results.6.28 The following set of differential equations arises insemiconductor theory:Typical side conditions are n(0) = p(0), p(1) = 0, andn (1) = 1. For ε = 1 find E(1) such that n(0) = p(0).Print out your final value for E(1). What happens ifε = 10-2?6.29 The motion of a wind-blown balloon can be approximated by the solution of the initial value problemThe problem is to find the initial altitude, a(0) = A, sothat the balloon lands at x = 35,000 ft from its pointof origin, and the time of flight tf . At tf we know thata(tf ) = 0 and x(tf ) = 35,000 so the system can be integrated backwards in time until t = 0.
The value tfis to be chosen so that x(0) = 0; the initial altitude Awill be the value of a(0) for this tf .To solve the problem, perform the following stepsin the order indicated.(a) Assuming that v(a) = 20 (constant wind velocity)and tf = 2000, use Rke to calculate x(0) and a(0).Check your results with an analytical solution of thedifferential equations.(b) Fit the v(a) data using the complete cubic interpolating spline.
Plot v as a function of a.where for each time t, a(t) is the altitude of the balloon above ground, x(t) is the horizontal distance ofthe balloon from a fixed starting point, and v(a) is(c) Using v(a) from part (b), again calculate x(0) anda (0), this time for tf = 1000, 1500, 2000, 2500. Tabulate tf , x(0), and a(0) for each case.(d) Modify your program in (c) to use Zero to findthe time of flight tf so that x(0) = 0. What is A ?Previous Home NextAPPENDIX ANOTATION AND SOME THEOREMSFROM THE CALCULUSWe assume that the reader is familiar with the topics normally covered in the undergraduate analytical geometry and calculus sequence. For purpose of reference, wepresent some standard notation used in this book and a few theorems from the calculus.A.1 NOTATION[a,b ], the interval consisting of the real numbers x such that a < x < b.(a,b), the interval consisting of the real numbers x such that a < x < b.lies in the interval [a, b].belongs to the class of functions having a first derivative continuous onthe interval [a,b].belongs to the class of functions having an nth derivative continuous on[a,b]The norm ||f|| of a function f(x) continuous on the interval [ a,b] is the maximumvalue of |f(x)| over the interval.and so on, denotepartial differentiation.e0.d1 d2 .
. . dn(e) stands for 0.d1 d2 . . . dn × 10 ; for example, 0.123(4) means 0.123 ×410 . More often, we use 0.123E4 for this.251252APPENDIX ANOTATION AND SOME THEOREMS FROM THE CALCULUSis approximately equal to.f(x ) = limit of f(x + η) aswith η > 0, the limit from the right.f(x ) = limit of f( x - η ) aswith η > 0, the limit from the left.+-<<, much less.v = (v1v2··· vn ), notation for a vector.The norm ||v|| of a vector v is the maximum magnitude component of the vector.q(h) = 0(hk) asmeans that there are (unknown) constants h0, K such that|q(h)| < Khk for all 0 < h < h0.A.2 THEOREMSTheorem.
Intermediate Value Theorem. Let f(x) be a continuous function onthe interval [a,b]. If for some number a and for some x1 ,x2 [a, b] we have f(xl) <α < f(x2), then there is a point c[a,b] such that a = f(c).Theorem. Rolle’s Theorem. Let f(x) be continuous on the finite interval [a,b]and differentiable on (a, b).
If f(a) = f(b) = 0, there is a point c (a,b) such thatf´(c) = 0.Theorem. Mean Value Theorem for Integrals. Let g(x) be a nonnegative function integrable on the interval [a,b]. If f(x) is continuous on [a,b], then there is a pointc [a,b] such thatTheorem. Mean Value Theorem for Derivatives.Let f(x) be continuous onthe finite interval [a,b] and differentiable on (a,b). Then there is a point c (a,b)such thatTheorem. Taylor’s Theorem (for f(x)). Let f(x) have n + 1 continuous derivatives on (a,b) for some n > 0 and let x,x0 (a, b).
Thenf(x) = Pn(x) + Rn+1(x),whereandA.2 THEOREMS253for some z between x0 and x.Theorem. Taylor’s Theorem (for f(x,y)).Let (x0,y0) and (x0 + ξy0 + η) begiven points and assume that f(x,y) is n + 1 times continuously differentiable in someneighborhood of the line L(x0, y0;x0 + ξ,y0 + η) connecting (x0,y0) and (x0 + ξ,y0 +η). Thenfor some 0 < θ < 1. [The point (x0 + θξ,y0 + θη) is an unknown point on the line L.]For n = 2, the formula without remainder becomesLet f(x) be a continuous function on the finite interval [a,b]. ThenTheorem.f(x) assumes its maximum and minimum values on [a, b]; that is, there are pointsx1 ,x2 [a,b] such thatf(x1) < f(x) < f(x2)for all x[a,b].Theorem.
Integration by Parts.Let f(x) and g(x) be real-valued functionswith derivatives continuous on [a,b]. ThenTheorem. Fundamental Theorem of the Integral Calculus.tinuous on the interval [a,b], and letLet f(x) be con-Then F(x) is differentiable on (a, b) andF´(x) = f(x).Theorem. Common Maclaurin Series.For any real number x,254APPENDIX ANOTATION AND SOME THEOREMS FROM THE CALCULUSFor |x| < 1,For discussions and proofs of these theorems, see A.
Taylor, Advanced Calculus, Wiley,New York, 1983.Previous HomeANSWERS TO SELECTED EXERCISESExercise Set 1.1 (Page 11)1.2 (a) In six-digit decimal rounded arithmeticU = 0.00001 = 10-5(b) Same as those in part (a).1.3 (a) With n = 3 in six-digit decimal choppedarithmetic x - y = 0.000456.1.5The last quantity becomes infinite as x approaches1.Exercise Set 1.2 (Page 23)1.7 With s =factor |c/s| = |cot( θ )| < 1.
In this range, the valueof s is as accurate as the value of c, and it will be ofcomparable accuracy for θ not greatly smaller thanπ/4. However, for θ << 1, the factor cotis very large and s computed in this manner ismuch less accurate than c. The relative error1.11 (a) In IEEE single precision the final k = 46 withS = 0.00008138; its relative error is -0.793 or79%.(c) The series is a better algorithm for positive xbecause there is no cancellation error.1.13 Some information is lost in the subtraction offrom 1 becauseshowing that there issevere cancellation when this quantity is subtractedfrom 1.
The information lost is small in magnitude,but it becomes important due to cancellation,especially after it is multiplied by the large value ofN to form the approximation NL2N/2 to π. Therearranged form avoids the cancellation in theoriginal form of the recurrence.Miscellaneous Exercises for Chapter 1 (Page29)1.15 (a) Let x = 8.01, y = 1.25, z = 80.8, then(c) Let x = 200., y = -60.0, z = 6.03, then-10700.
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