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Introduction to Applied Mathematics

by: Braeden Lind

Introduction to Applied Mathematics MA 325

Marketplace > North Carolina State University > Mathematics (M) > MA 325 > Introduction to Applied Mathematics
Braeden Lind
GPA 3.93

Robert White

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Robert White
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This 5 page Class Notes was uploaded by Braeden Lind on Thursday October 15, 2015. The Class Notes belongs to MA 325 at North Carolina State University taught by Robert White in Fall. Since its upload, it has received 9 views. For similar materials see /class/223679/ma-325-north-carolina-state-university in Mathematics (M) at North Carolina State University.

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Date Created: 10/15/15
Lecture 3 Analysis of Discretization Error Introduction Initial value problems have the form ut ftu and u0 given 1 The simplest cases can be solved by separation of variable but in general they do not have to have closed form solutions Therefore one is forced to consider various approximation methods In this lecture continue to consider the Euler nite difference method and we will give an analysis of the discretization error This will require the extended mean value theorem We shall see that the error made by this approximation is proportional to h Model The continuous model of Newton39s empirical law of cooling states that the rate of change of the temperature is proportional to the difference in the surrounding temperature and the temperature of the liquid ut cusur 7 u If c 0 then there is perfect insulation and the liquid temperature must remain at its initial value For large c the liquids temperature will rapidly approach the surrounding temperature Euler s method involves the approximation of ut by the finite difference uk1 ukh where h TK uk is an approximation of ukh and f is evaluated at khuk If T is not fmite then h will be xed and k may range over all of the positive integers The differential equation 1 can be replaced 111 1 ukh fkhuk 2 The scheme given by 2 is call Euler39s method and it is a discrete model of heat transfer In the previous lecture we observed that as the time step decreases the solution from the discrete model approached the solution from the continuous model This was depicted in both graphical and table form Here we will give a careful mathematical analysis which can be applied to a larger class of initial value problems The main mathematical tool is the mean value theorem and some of it variations Mean Value Theorems In order to give an explanation of the discretization error we must review the mean value theorem and some of its variations The mean value theorem like the intermediate value theorem will appear to be clearly true once one draws the picture associated with it However it is a very powerful tool which can help us control the size of numerical errors in approximating hard integrals and differential e quati ons fX f39C fb f21 0 a L I I V X Figure Function Mean Value Drawing the picture does make some assumptions For example consider the function given by fx l lxl Here there is a quotcomerquot in the graph at x 0 that is fx does not have a derivative at x 0 Moreover there is no mean value x c as depicted in the above figure Mean Value Theorem Let f ab gtlR be continuous on ab If f has a derivative at each point of ab then there is a c in ab such that f39C fb femb a 3 Proof It suffices to prove the case where fa 0 fb because we can replace fx by fX fb f21W b aXa t f21 This special case is known as Rolle39s theorem Furthermore in order to keep the details at a minimum we will assume that the derivative is continuous on ab Suppose f 39X is not zero for all X in ab There are three possible cases First f 39X gt 0 and as fa 0 fb must be positive which is a contradiction Second f 39X lt 0 also gives a contradiction Third f 39X must change sign Here we apply the intermediate value theorem to the derivative function f 39X So there must be a c in ab such that f 39c 0 Often the conclusion is written with b X and c some point between a and X as fX fa f cXa 4 Another variation on the mean value theorem is to apply it directly to integrals The following figure indicates that there is a particular rectangle with height fc which has the same area as under the curve of fX The choice of c will depend on f as well as a and b 190 area under fX fc b a I a c b Figure Integral Mean Value Integral Form of Mean Value Theorem Let f and w be continuous on ab and let w gt 0 on ab Then there is some c between a and b such that l7 l7 Ifmwwt fc wow 5 The second form of the mean value theorem s conclusion is fX fa f39c X a for some c between a and X provided f 39X is continuous on ab The eXtended mean value theorem will conclude that fX fa f39a X a f quotcX a2 2 for some c between a and X provided f quotX is continuous on ab The eXtended mean value conclusion follows in a natural way from integration by parts and the integral form of the mean value theorem fx fa from fa r 00 x j r39 00 wait fa rax ar39cjx 0dr fafax a f 0 Extended Mean Value Theorem If f ab gtR has two continuous derivatives on ab then there is a c between a and X such fX fa f39a X a f quotcX a2 2 6 Here X is in ab and c will depend on the choice of X Analysis of the Discretiztion Error Discretization error E uk ukh where uk is from Euler39s algorithm 2 with no roundoff error and ukh is from the eXact continuum solution 1 Now we will give the discretization error analysis The relevant terms for the error analysis are u1kh cusur 11141 7 uk1 uk hcusur uk and 8 Use the eXtended mean value theorem along with 7 and also use 8 Ef uk uk1h uk hcusur ukn ukh Cusur ukhh um cw h2 2 aE bk hZz where a 1 7 ch gt 0 and bk1 unck1 Let a 1 ch r and ibk1i lt M2 M Use a quottelescopingquot argument and the partial sums ofthe geometric series 1 r r2 rk rm1 1r 1 to get E lt r Ef MhzZ lt rrE 391 MhzZ MhzZ lt rk i E2 rk1 lr 1 MhzZ 9 Assume E 0 and r l 7 ch gt 0 with h TK to obtain Em lt o 1 7 chk1 1ch MhzZ lt 1ch MhzZ M2c h 10 Euler Error Theorem Consider Euler39s algorithm 8 for the initial value problem 7 which has a unique solution with two continuous derivatives If E 0 h TK maxiuni lt M for t in 0T then Efli lt M2c h Remark The theorem can be generalized to initial value problems of the form 1 where additional assumptions on ftu must be made In many cases the exact solution may not be known but the numerical method can be used to approximate the solution The error approximations imply that the numerical solution is close to the unknown exact solution Homework 1 Fill in the steps leading to 9 2 In order to prove 10 we assumed T was nite Suppose a l 7 ch gt 0 and for all t the second derivative of the solution is bounded by M Show for all k Ef1lt1ch MhzZ 3 Prove the integral form of the mean value theorem by using the mean value theorem with fx replaced by Fx j ftwtdt K j wtdt and K de ned by Fb 0


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