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lecture39

# lecture39 - CMPSC/MATH 451 Numerical Computations Lecture...

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Unformatted text preview: CMPSC/MATH 451 Numerical Computations Lecture 39 Dec 2, 2011 Prof. Kamesh Madduri Class Overview • Euler’s method • Slides from textbook follow. 2 Ordinary Differential Equations Numerical Solution of ODEs Additional Numerical Methods Euler’s Method Accuracy and Stability Implicit Methods and Stiffness Euler’s Method For general system of ODEs y = f ( t, y ) , consider Taylor series y ( t + h ) = y ( t ) + h y ( t ) + h 2 2 y 00 ( t ) + ··· = y ( t ) + h f ( t, y ( t )) + h 2 2 y 00 ( t ) + ··· Euler’s method results from dropping terms of second and higher order to obtain approximate solution value y k +1 = y k + h k f ( t k , y k ) Euler’s method advances solution by extrapolating along straight line whose slope is given by f ( t k , y k ) Euler’s method is single-step method because it depends on information at only one point in time to advance to next point Michael T. Heath Scientific Computing 22 / 84 Ordinary Differential Equations Numerical Solution of ODEs Additional Numerical Methods Euler’s Method Accuracy and Stability Implicit Methods and Stiffness Example: Euler’s Method Applying Euler’s method to ODE y = y with step size h , we advance solution from time t = 0 to time t 1 = t + h y 1 = y + hy = y + hy = (1 + h ) y Value for solution we obtain at t 1 is not exact, y 1 6 = y ( t 1 ) For example, if t = 0 , y = 1 , and h = 0 . 5 , then y 1 = 1 . 5 , whereas exact solution for this initial value is y (0 . 5) = exp(0 . 5) ≈ 1 . 649 Thus, y 1 lies on different solution from one we started on Michael T. Heath Scientific Computing 23 / 84 Ordinary Differential Equations Numerical Solution of ODEs Additional Numerical Methods Euler’s Method Accuracy and Stability Implicit Methods and Stiffness Example, continued To continue numerical solution process, we take another step from t 1 to t 2 = t 1 + h = 1 . , obtaining y 2 = y 1 + hy 1 = 1 . 5 + (0 . 5)(1 . 5) = 2 . 25 Now y 2 differs not only from true solution of original problem at t = 1 , y (1) = exp(1) ≈ 2 . 718 , but it also differs from solution through previous point ( t 1 , y 1 ) , which has approximate value 2 . 473 at t = 1 Thus, we have moved to still another solution for this ODE Michael T. Heath Scientific Computing 24 / 84 Ordinary Differential Equations...
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lecture39 - CMPSC/MATH 451 Numerical Computations Lecture...

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