Heath - Scientific Computing (523150), страница 76
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Thus, Euler’s method advances thesolution by extrapolating along a straight line whose slope is given by f (tk , yk ). Euler’smethod is called a single-step method because it depends on information at only one pointin time to advance to the next point.Example 9.7 Euler’s Method. We previously considered the equation y 0 = y, whichis easily solved analytically, but for illustration let us apply Euler’s method to solve itnumerically. For some stepsize h, we advance the solution from time t0 = 0 to time t1 =t0 + h:y1 = y0 + y00 h = y0 + y0 h = y0 (1 + h).Note that the value for the solution we obtain at t1 is not exact (i.e., y1 6= y(t1 )). Forexample, if t0 = 0, y0 = 1, and h = 0.5, then we get y1 = 1.5, whereas the exact solutionfor this initial value is y(0.5) = exp(0.5) ≈ 1.649.From the Taylor series used to derive Euler’s method, we know that the error is proportional to h2 , so we can reduce the error for this step by a factor of 41 by reducing the stepsizeby a factor of 21 , provided rounding error is negligible.
For any nonzero error, however, thevalue y1 lies on a different member of the family of solution curves from the one on whichwe started.y................................... ............... ........... .......... ... ....... ...... ................. ... ...
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... ........ ................................................................................................................................................0 ...........................................................................................................................................................................................................................................................................................................................................................................................................................•y =y••y •t0•t1t2t3t4tFigure 9.4: Euler’s method for the ODE y 0 = y.To continue the numerical solution process, we take another step from t1 to t2 = t1 +h =1.0, obtaining y2 = y1 + y1 h = 1.5 + (1.5)(0.5) = 2.25. Note that y2 differs not only fromthe true solution of the original problem at t = 1, namely, y(1) = exp(1) ≈ 2.718 but it282CHAPTER 9.
INITIAL VALUE PROBLEMS FOR ODESalso differs from the solution curve passing through the previous point (t1 , y1 ), which hasthe approximate value 2.473 at t = 1. Thus, we have moved to still another member of thefamily of solution curves for this ODE. We can continue to take additional steps, generatinga table of discrete values of the approximate solution over whatever interval we desire. Aswe do so, we will hop from one member of the solution family to another at each step.For this unstable equation, the errors we make in the numerical method are amplifiedwith time as a result of the divergence of the solution curves, as shown in Fig.
9.4. For astable equation such as y 0 = −y, on the other hand, the errors in the numerical solutionmay diminish with time, as shown in Fig. 9.5.y..................0 ............... ...... ........ .............................. ..... .................... ....... ..... ............0.... .... ................ ..... ............ ... ......... ............. ... ......... .............. .... ..................... ........ ............ ................. .........................
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............................. .......... .......... ...... .......... ...................... ......... ................ .......... ..................... ........... ................ .......... ......................... ............ .......... ............ ............................
............. ............ ............. .............................. ............... ................. ............... .................................. ............... ........................ ................ ................................... ................................................. .................. .......................................... ...................................................................................................................................................................................................................................................................................................................................................................................................................................................................y •y = −y•••t0t1t2t3•t4tFigure 9.5: Euler’s method for the ODE y 0 = −y.9.39.3.1Accuracy and StabilityOrder of AccuracyLike other methods that replace derivatives with finite differences, a numerical procedurefor solving an ODE suffers from two distinct sources of error:• Rounding error , which is due to the finite precision of floating-point arithmetic• Truncation error (or discretization error), which is due to the method used, and whichwould remain even if all arithmetic could be performed exactlyAlthough they arise from different sources, these two types of errors are not independentof each other.
For example, the truncation error can usually be reduced by using a smallerstepsize h, but doing so may incur greater rounding error (see Example 1.11). In mostpractical situations, however, truncation error is the dominant factor in determining theaccuracy of numerical solutions of ODEs, and we shall henceforth ignore rounding error.The truncation error at step k of a numerical solution of an ODE can be further brokendown into:• Local truncation error , denoted by Lk , which is the error made in one step of the numerical9.3.
ACCURACY AND STABILITY283method. More precisely,Lk = yk − uk−1 (tk ),where yk is the computed solution at tk , and uk−1 is the member of the family of truesolutions to the ODE that passes through the previous point (tk−1 , yk−1 ).• Global truncation error , denoted by Ek , which is the difference between the computedsolution and the true solution determined by the initial data at t0 . More precisely,Ek = yk − u0 (tk ) = yk − y(tk ).The global error is not necessarily the same as the sum of the local errors.
The global errorwill generally be greater than the sum of the local errors if the equation is unstable butmay be less than that sum if the equation is stable, as shown in Figs. 9.6 and 9.7, wherethe local errors are indicated by small vertical bars between solution curves and the globalerror is indicated by a bar at the end. Having a small global error is obviously what wewant, but we can control only the local error directly.y.
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............................................................................................................................................................................... .... .. ...... ..................................................................................................................................0 ...................................................................................................................................................................................................................................................................................................................................................................................................................global errory =yyt0t1t2t3t4tFigure 9.6: Local and global errors in Euler’s method for the ODE y 0 = y.The accuracy of a numerical method is said to be of order p ifLk = O(hp+1k ).The motivation for this definition, with the order one less than the exponent of the stepsizein the local error, is that if the local error is of order p + 1, then the sum of the local errorsfrom t0 to tk will betk − t0O(hp+1 ) = O(hp ),hwhere h is the average stepsize, and this gives a rough approximation of the global errorEk .Example 9.8 Accuracy of Euler’s Method.
Consider the Taylor seriesy(t + h) = y(t) + y 0 (t)h + O(h2 ) = y(t) + f (t, y(t))h + O(h2 ).284CHAPTER 9. INITIAL VALUE PROBLEMS FOR ODESIf we take t = tk and h = hk , we gety(tk+1 ) = y(tk ) + f (tk , y(tk ))hk + O(h2k ).If we now subtract this from Euler’s method we getyk+1 − y(tk+1 ) = [yk − y(tk )] + [f (tk , yk ) − f (tk , y(tk ))]hk − O(h2k ).The difference on the left side preceding is the global error Ek+1 . If there were no priorerrors, then we would have yk = y(tk ), and the first two differences on the right side wouldbe zero, leaving only the O(h2k ) term, which is the local truncation error.
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