Lecture9

Lecture9 - Initial Value Problems (IVP) EGN5455 Lecture 9 1...

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Initial Value Problems (IVP) EGN5455 Lecture 9 1
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Euler’s Method (Backward Integration) ± Initial value problem with initial condition )) ( , ( ) ( ) 1 ( t y t f t y = . ) ( 0 0 y t y = ± We approximate the function with the line e denote then the above equation ) )( , ( 0 1 0 0 0 1 t t y t f y y + = We denote then the above equation simplifies to 0 1 t t h = ) , ( ) ( 0 0 0 1 1 y t hf y t y y + = 2
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Example1 ± Given the IVP and initial value approximate y(1 5) and y(1) ) ( 1 ) ( ) 1 ( t y t y = = , approximate y(1.5) and y(1). Then evaluate the results by dividing h with 2. 5 . 2 ) 5 . 0 ( y 3
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Example1 . 5 . 0 , 5 . 2 , 5 . 0 1 ) 5 . 2 1 ( 1 5 . 2 ) 1 ( ) 5 . 1 ( . 1 , 5 . 2 , 5 . 0 0 0 1 0 0 = = = = + = + = = = = h y t y h y y h y t 75 . 1 ) 5 . 2 1 ( 5 . 0 5 . 2 ) 1 ( 1 0 0 = × + = y 4
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Example1 ± The answer to the differential equation is: 1 5 . 1 5 . 0 + = + t e y ± The real values of 1598 . 0 , 9098 . 1 ) 1 ( 5518 . 0 , 5518 . 1 1 5 . 1 ) 5 . 1 ( 1 1 1 = = = = + = error y error e y 5
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Example1 h Approx. Error 1 1 0.5518 0.5 1.75 0.1598 0.25 2.1250 0.0432 ± By dividing h by 2, error is divided by 4. 6
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Error Analysis ± Taylor series 1 e now from IVP at: 2 ) 2 ( ) 1 ( ) ( 2 ) ( ) ( ) ( h y h t y t y h t y τ + + = + ± We know from IVP that: ) ( ) ( ) ( ) 1 ( t hy t y h t y + = + ± Therefore the error is: 2 ) 2 ( 7 ) ( 2 / 1 h y
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Example 2 ± Given an IVP with an initial condition if the second derivative is bounded by , ) 0 ( y y = y , on how large an interval can we estimate y(t) if we want to ensure that the 0 8 ) ( 8 ) 2 ( < < t y error is less than 0.0001? 8
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Example 2 0 4 ) 2 ( 2 005 . 0 005 . 0 10 4 / 1 10 2 / 1 4 2 < < × < < h h y h 9
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Heun’s Method (Trapezoidal Integration) ± From Euler’s method, k 0 0 0 0 0 1 1 ) , ( ) ( y t f k hk y y t y = + = = ± Let’s now sample the slope at 0 1 t t h = ) , ( 1 1 y t = ± We can use the average of two slopes ) , ( 1 1 1 y t f k 10 ( ) h k k y y 1 0 0 1 2 1 + + =
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Heun’s Method ± Process ) , ( ) , ( 0 0 1 0 0 0 h y h t f k y t f k + + = = 2 1 2 0 1 k k h y y + + = 11
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Example1 ± Given the IVP and initial value , approximate y(1.5) and y(1) with ) ( 1 ) ( ) 1 ( t y t y = 5 . 2 ) 5 . 0 ( = y ,p p y ( ) y ( ) Heun’s method. Then evaluate the results by ividing h with 2 (compare it with the results of dividing h with 2 (compare it with the results of Euler’s method). 12
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Example1 1 1 1 1 1 2 1 5 . 1 5 . 2 1 ) 5 . 2 , 5 . 0 ( . 1 , 5 . 2 , 5 . 0 0 0 0 = = = = = = = f k h y t 75 . 1 75 . 0 5 . 2 75 . 0 ) 5 . 1 ( 75 . 0 ) ( 2 / 1 0 ) , 5 . ( ) 5 . 5 . , 5 . ( 0 1 1 0 1 = = × + = = + h y y k k f f k 5 . 1 ) 5 . 2 , 5 . 0 ( 5 . 0 , 5 . 2 , 5 . 0 0 0 0 = = = = = f k h y t 375 250 250 1250 . 1 ) 5 . 1 75 . 0 ( 2 / 1 ) ( 2 / 1 75 . 0 ) 75 . 1 , 1 ( ) 5 . 1 5 . 0 5 . 2 , 1 ( 1 0 1 = = + = = × = k k f f k 13 9375 . 1 1250 . 1 5 . 0 5 . 2 1250 . 1 ) 1 ( 0 1 = × = × + = h y y
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Example1 h Approx. Error 1 1.75 0.1982 0.5 1.9375 0.0277 0.25 2.1719 0.0037 y dividing h by 2, error is divided by 8. 14 ± By dividing h by 2, error is divided by 8.
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Error Analysis ± Taylor series approximation (1) (2) 2 (3) 3 11 ) () t h yt y t h y y h = + + + 00 0 0 (1) (3) 0 ) ( 26 ( ) 1 2 h y h h τ += +− =− 10 0 0 0 (1) 0 0 ( , ) (, ) k y tf t y h y hk h f t h y h k == +≈+ +≈ + + 100 0 0 ( , ) 1 2 kf t h y hk kk y h h =++ ≈− 3 1 1 k 15 h y 2 0 3 01 24 6 1 ( ) 21 2 kh h y h y h h y h y h + + +
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4 th -order Runge-Kutta Method ± It is an improvement on Heun’s method similar to Simpson’s rule to trapezoid method.
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This note was uploaded on 07/23/2011 for the course EGN 5455 taught by Professor Miralles during the Spring '11 term at FIU.

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Lecture9 - Initial Value Problems (IVP) EGN5455 Lecture 9 1...

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