ENGI 9420 Lecture Notes 3 - Numerical Methods Page 3.01

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1 ENGI 940 Lecture Notes 3 - Numerical Methods Page Numerical Methods The majority of equatios of iterest i actual practice do ot admit ay aalytic solutio. Eve equatios as simple as = e ad I = e d have o eact solutio. Such cases require umerical methods. Oly a very brief survey is preseted here. Sectios i this Chapter: 3.0 Bisectio 3.0 Newto s Method 3.03 Euler s Method for First Order ODEs 3.04 Fourth Order Ruge-Kutta Procedure (RK4)

2 ENGI Bisectio Page Bisectio Eample 3.0. Fid the solutio of e =, correct to 4 decimal places. From a sketch of the two curves y = ad y = e, it is obvious that the oly solutio is somewhere i the iterval (0, ). Let f ( ) = e. Clearly f (0) = < 0 ad f () = /e > 0 f () is cotiuous ad chages sig oly oce iside (0, ). Halve the iterval repeatedly ad retai the half with a sig chage: f ( ) = < 0 root is i ( ,.00000) f ( ) = > 0 root is i ( , ) f (0.6500) = > 0 root is i ( , ) f (0.5650) = < 0 root is i (0.5650, )

3 ENGI Bisectio Page 3.03 Eample 3.0. (cotiued) This method is slow ad requires eightee steps before the chage i is small eough to leave the fourth decimal place udisturbed with certaity: f (0.5674) = < 0 root is i (0.5674, ) This method is equivalet to zoomig i graphically by repeated factors of util the desired accuracy is obtaied. The result of a faster graphical zoom, sufficiet to determie the solutio to five decimal places, is displayed here: Correct to four decimal places, the solutio to e = is = A calculator quickly cofirms that e A spreadsheet to demostrate the bisectio method for this eample is available from the course web site, at "

4 ENGI Newto s Method Page Newto s Method dy From the defiitio of the derivative, d dy we obtai y or, equivaletly, d y. f ( ) = y lim, 0 The taget lie to the curve y = f () at the poit P(, y ) has slope = f ' ( ). Follow the taget lie dow to its ais itercept. That itercept is the et approimatio +. y = y+ y = 0 y = f ( ) ad = + f ( ) + = f ( ) If is the th approimatio to the equatio f () = 0, the a better approimatio may be f ( ) + = f which is Newto s method. ( )

5 ENGI Newto s Method Page 3.05 Eample 3.0. Fid the solutio of e =, correct to 4 decimal places. From a sketch of the two curves y = ad y = e, it is obvious that the oly solutio is somewhere i the iterval (0, ). A reasoable first guess is = 0. f ( ) = e ( ) f ( ) e = f ( ) + e Table of cosecutive values: f = + e ( ) f e. f = + e = ( ) f f ( ) ( ) Correct to four decimal places, the solutio to = e is = I fact, we have the root correct to si decimal places, = A spreadsheet to demostrate Newto s method for this eample is available from the course web site, at " This method coverges much more rapidly tha bisectio, but requires more computatioal effort. Note that Newto s method ca fail if f ' () = 0 i the eighbourhood of the root. A shallow taget lie could result i a sequece of approimatios that fails to coverge to the correct value.

6 ENGI Euler s Method Page Euler s Method for First Order ODEs Oe of the simplest methods for obtaiig the umerical values of solutios of iitial value problems of the form is Euler s method. (, ), ( ) y = f y y = y 0 0 dy y dy From the defiitio of the derivative, = lim, we obtai y. d 0 d If we seek values of the solutio y() at successive evely spaced values of, the we have y = y+ y y+ = y + y y + f (, y). With (by covetio) h=, we have the iterative scheme ( ) y y hf y + = +, However, errors propagate rapidly uless the step size h is very small, which requires a proportioate icrease i the umber of computatios. Several modificatios to Euler s y = f, y by a weighted method have bee proposed, that replace the derivative ( ) average of values of f at poits aroud (, y ). Oe of the most popular modificatios is the fourth order Ruge-Kutta method (RK4).

7 ENGI RK4 Method Page Fourth Order Ruge-Kutta Procedure (RK4) Values (, ) y [with = 0 + h] of the solutio y() to the iitial value problem are give by the iterative scheme (, ), ( ) y = f y y = y (, ) k = f y 0 0 (, ) (, ) 3 (, ) 4 3 k = f + h y + hk k = f + h y + hk k = f + h y + hk h y+ = y + k + k + k + k 6 ( ) 3 4 Eample Use the RK4 procedure with step size h = 0. to obtai a approimatio to y(.5) for the solutio of the iitial value problem y' = y, y() =. 0 =, h = 0. ad we wat y(.5)..5 = , so we eed to fid y 5. ( 0, y 0 ) = (, ) ad f (, y) = y. For = 0: k = f, y = y = = k k 3 ( 0 0) 0 0 ( ( ))( ( ) ) ( ( )) ( ) ( )( ( ) ) = =.3 ( ) = =.3455 k4 = =.7536 h 0. y = y0 + k + k + k + k = Therefore y(.) y = ( 3 4) ( (.3) (.3455).7536) We ca proceed with a similar chai of calculatios to fid y, y3, y 4 ad fially y 5.

8 ENGI RK4 Method Page 3.08 Eample (cotiued) y k k k 3 k Therefore y(.5) This iitial value problem happes to have a eact solutio, y = e. We ca therefore test the accuracy of the RK4 procedure i this case. The eact value of y(.5) is , a absolute error of less tha ad a relative error of less tha 0.0%. Euler s method, i cotrast, has a error eceedig 6%! END OF CHAPTER 3