Question

Part A: What is the (forward) Euler method to solve the IVP y(t) = f(t, y(t)) te [0.tfinal] y(0) = 1 Part B: Derive the (forward) Euler method using an integration rule or by a Taylor series argument. Part C: Based on that derivation, state the local error (order of accuracy) for this Euler method. Part D: Assume that you apply this Euler method n times over an interval [a,b]. What is the global error here? Show your work.

Answer 1

Part A Consider the differential equation : 4'htt- dy = f (t, y(47) te[o, afinal, y 107-1 le y = I at tuo Let too and yo=1 as shown in dotted in figure We are interested in finding successively yeg a Tes where you is corresponding to t= twee tot ruh, his small. Using the property that in a small interval, a curve is nearly a straight line, we approximate the core sy the tangent at the point (to yo) Eqt of tangent at (to, yo) is! y yo = no (t-to) - @ (to, to) where m=slope of tangent at point = Code). = f(to, jo) ett sto, yo) - Eq" ☺ becomes 7-do = f (to, yo) (t-to) y = yot (t-to) of (to, jo) 9. = yot (to-to) of (to, yo) y = yet hf (to, yo ) fitt to t to the

Er gives approximate value of Il further approximate the curre in the next internal [ty, tz] by a line through (t, y) with slope f (th) In general, we have Yonen = Yurt hf (tre, Jul 6 where tre = 0+ (m-1) h Y uy = y mongth of (tuere duen) and initially taking yo=1 Derivation of Euler's Method: Using Taylor expansion : Taylor expansion of the function y around to: yltoth) = y1to) + hy'to) + shty"to) + O(1) The differential equation is y flty), y(tol=&_6 If this is substituted in the Taylor expansion and the quadratic and higher under termes are ignored then we are left with: y to th) = y to) + hy' (to)

as y Hol=yo and t = toth where h is the length difference between initial pt and upcoming next port. and het ylti) = (toth) = 21 YI= Yo th f (to yo) Continuing the same way, Inet - Yvet hof (tre, Tue) which is Fuler's method. (6) Local truncation error : error made in a single step. It is the difference between the numerical solution after one step, y, and the exact solution at time ti= toth. The numerical solution is given by y = Yoth f (to, yo) for the exact solution, we use Taylor expansion mentioned in Part (6)

y (toth) = yltos + hy' tos & he geto ) + 0(4): The LTE (Local Truncation error) in the difference between these two equations: LTE = y to the y red the y"(ta) + 0(13) A slightly different formulation for the LTE can be tron) obtained by using Lagrange force of remainder in Taylor's theoren. If y has a continuary second derivative, then there exists a [to, toth] 84.1 on 14/40) LTE = y (toth) -21 = I hey" (?) Proceeding like in Put cole after a steps: (tu, y(tullt h (he) (10) y (tri) = y talt n'f (tasy (tallt heute Substituting flta, & (tul) = f (taiful & fylte (Itul gen) = f (tu, gal - to (A, B ) Eu.. Put si Gº@:

. Condition, hence, Substitute these bounds in Gr 13. after (1) steps, we get : .. | Em I < (1+kw) lea) + mai C We begin Euleis method with the initial there is no error at oep o. from the enor bound, 1 ?, Continuing, we have 16V (14 kh) memang memp and ers 161 < (+kaya maua y cerras met meget porta ceramajor there 5 Carnot The quantity Sch) is a finite Geometric series, and can be calculated by a trick. We hare: kn schl= (1+kh) shisch) - = (1+kh) ~ = (1+kh) tulen The result simplifier, global coror bound to El & Mh [ (it kuth_1] We can apply dei (It kng lh _ ekty th ak

and to show that for Khal. This for Cirkh) tlh is a decreasing function of h a small we have the result : lend is more ces tro). which is required Global Error Result Truncation

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