class_17

class_17 - CS205 Class 17 Covered in class: All Reading:...

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CS205 – Class 17 Covered in class : All Reading : 9.3.9 1. Backward (Implicit) Euler 1 1 1 ( , ) k k k k y y f t y h a. 1 st order accurate b. Backward Euler applied to the model equation y y   is 1 1 k k k y y h y i. So 1 1 (1 ) k k y h y and (1 ) k k o y h y ii. The error shrinks when 1 |(1 ) | 1 h iii. Thus, 0 h   is needed for stability iv. i.e. stable for all h or unconditionally stable c. Generally need to solve a nonlinear equation to find 1 k y i. Can use Newton iteration, i.e. linearize, solve, linearize, solve, etc. ii. For some applications, only one linearize and solve cycle is used d. One can take very large time steps since it is stable i. However it is not very accurate ii. As h   , 1 1 1 0 0 k k k k y y h y h y   or 1 0 k y iii. This is the long run solution for ' y y with 0 , but we get there too fast iv. Everything damps out too quick, i.e. not accurate 2. Trapezoidal rule 1 1 1 ( , ) ( , ) 2 k k k k k k y y f t y f t y h a. 2 nd order accurate b. Trapezoidal rule applied to the model equation is   1 1 2 k k k k h y y y y i. So 1 (1 /2)/(1 /2) k k y h h y and (1 /2) /(1 /2) k k k o y h h y ii. The error shrinks when |(1 /2)/(1 /2) | 1 h h iii. Thus, 0 h   is needed for stability iv. i.e. unconditionally stable c. Generally need to solve a nonlinear equation to find 1 k y i. One can take very large time steps since it is stable ii. However it is not very accurate iii. As h   ,     1 1 1 0 0 2 2 k k k k k k h h y y y y y y   or 1 k k y y   iv. This is NOT the long time solution for y y   v. Bad oscillatory behavior 3. 1 st order Runge-Kutta is Euler’s method 1 ( , ) k k k k y y f t y h
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4. 2 nd order Runge-Kutta 1 1 2 2 k k y y k k h a. 1 ( , ) k k k f t y and 2 1 ( , ) k k k f t h y hk 5. 4 th order Runge-Kutta 1 1 2 3 4 2 2 6 k k y y k k k k h a. 1
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class_17 - CS205 Class 17 Covered in class: All Reading:...

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