Remarks on the shooting method for nonlinear two-point boundary-value problems

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VNU JOURNAL OF SCIENCE, Mathematics - Physics T.XIX, No3 - 2003

REMARKS ON THE SHOOTING METHOD FOR NONLINEAR

TWO-POINT BOUNDARY-VALUE PROBLEMS

Nguyen Trung Hieu Department of Mathematics, College of Science, VNU

Abstract In this note, we prove a convergent theorem for the shooting method combin- ing the explicit Euler’s scheme with the Newton method for solving nonlinear two-point boundary problems (TPBVPs) Some illustrative numerical examples are also considered

A convergent result obtained before by T Jankowski is a particular case of our result,

when the boundary condition (BC) becomes linear

1 Introduction

The shooting method for TPBVPs has been studied throughly in many works (c.f [1-10]) However, the convergence of the method did not receive adequate attention of researchers In 1995, T Jankowski gave an adequate proof for a convergent theorem of a shooting method

In this paper, we will generalize this result for nonlinear ordinary differential equa-

tions (ODEs) with nonlinear boundary conditions

Consider the problem

where f : JxR’ — R” is contimuons in t and continously differentiable in y, @: R” xR” > R’ continuously differentiable in both variables

If we denote y = y(t; 3) a solution of (1a) subject to initial condition

then the problem is reduced to that of finding s = 5 which solves the equation

(y(a; 5), U(b; s)) = Ó(s, y(b; s)) = 0 (2) This equation can be solved approximately by the Newton’s iteration

8341 = 8) — Ó(3;.9(b;s;)) '6(s;,w(b:3;)), 7 20, (3) where ¢'(s) = 61(s, y(b; s)) + $2(s, y(b; s)).y4(b; s) and ¢1,¢2 are partial derivatives of ó with respect to the first and the second variables, respectively In addition , Y(t; s) =

y,(t;s) can be found as a solution of the IVP

{ Y3) = f,(t-y(ts))-Y (G8)

Typeset by ÁA46-TEX

The remaining problem is to solve approximately the two IVPs (1 a,c) and (4) There are many methods for solving IVPs, but in this paper, we only use explicit Euler’s method to solve them ‘Thus we obtain a method using Euler’s scheme combined with Newton’s iteration In the following, this method will be discussed in detail Let the integration interval be subdivided by the points

(5)

First, applying the explicit Muler’s scheme mentioned above, we get

= Sh,j3 | Yn (bi+1s hj) = Yates Sng) +L (i Ya (tas 8n,9))5 (6a)

52 Ya(tiats Sng) = (E+ bSy (ti, yn (tis 8ng))]¥a (tis Saya)» (6b)

i=0,1, ,N-1,

p nts Sn0 = 80,80 € Ris an initial vector

yn (tos $n,

Yh(foiSn) ¢

Then we use the Newton method to improve the shooting vector

Shj~L = Sáu — [61(Sn,9,¥n(bs $n,3))

+¢02(8n,9+ Ya (0: 8n,3))¥n(b, 8h.9)] 7 'O($n9+ Ym (D5 Sn.) (6c)

j=0,1,

2 Convergence

In this note, we use the terminologies and notations of [4] However, the definition

of isolated solution should be stated as follows

Definition 1 A solution y(t) of (1) is said to be isolated if the following linear TPBVP

{ 2 = fy(t.u(t))z

$x(u(a),y(t))2(a) + o2(y(a),4(6))2(6) = s

has only trivial solution z(t) = 0

Lemma 1 The isolated solution is locally unique

Proof Consider the space C! (J R”) equipped with the norm ||y|| = max{||yl]max; {ly'Ilmax}:

the space C(J,R”) xR” equipped with the norm {|(Z, v2)|| = |\Ellmax +|+a|, where |2||max =

max|z()|, |.| is the Enelidian norm and the mapping ted

ES v— [dod y(b)) J? ted

Clearly,

F')h= xtc ),y(B))h(a) h- SE(ty)h + óa(0(a), v())h(ð)

and if y* is an isolated solution of (1), then KerF’(y*) = {0} Moreover, the shooting matrix 1 (y*(a), y*(b))U (a) + d2(y* (a), y*(b))U(0) is nonsingular, where U/(t) is a funda-

mental solution

Ư() =5‡(L*)00)

U(a) =I

It implies that ImF '(y*) = CJ, R’) xR” According to the Banach inverse mapping

theorem, [F’(y*)]~! exists and is continuous The inverse function theorem ensures that

Theorem 1 Let the BVP (1) have an isolated solution ọ Assưme thai

i) both f: Ix S-> R” and f; :J x SR” are continuous;

ii}the function f: Jx S74 R” has a bounded by a constant L > 0 derivative with respect to the second variable, and there exists a function 2: R, — Rx such that

IIZz(,#) — ;(t,®)|| < 9(z — 3|),

where the matria norm is consistent with the vector norm;

iti) Q is continuous, 2(0) = 0 and Q is non-decreasing;

iv) The Euler scheme is consistent with (1a), i.e there ewists a function «: H >

R*,H = |0,h*] for some h* > 0,c(h) — 0 as h 4 0 such that

IF Ct, e() + #Œ) — v(t + h)|| < he(h), for t € [a,b — A];

v) There exists a function 6: H + Rt, 6(h) + 0 ash > 0, such that the following

condition holds

lữ +f¿(,£(9)|Y (,e(e)) — Y( + hịg(a))||< h: ố{h) for t € |a,b — hị;

vi) The function ¢ in the boundary condition has Lipschitz continuous partial derivatives

llới(, ì) — 1 (23, yo) ll S Ki (| — 51] + Ilya — voll);

W4o(%, yx) — #i(#;, we)|| < K2(|# — a5l] + lyn — yoll)-

Moreover

|la(s, w)|| < K›

Then for a sufficiently small h, the shooting method (6) is convergent

Let

Uh = yn(tni ng) — P(tn); T= Ya(tn, $n9 [80,5 — 9(@)] — wh;

Ta := hf (tas @(ta)) + (ta) — (tai); AR = T+ hfi (tn, ualtns sny))s

đi = Ya(tns 81,3) — ¥ (tn, 9(a))3 Cr = MO“); Cp = (Cy — 1)/L- Introducing the norm ||¿|| = max: et) lu(t)||, : u € C(J,R”) and using the Gronwall-

Bellman inequality we can show that the problem (4) has an unique solution Y is bounded

Lemma 2 Under the conditions in the theorem 1, we have

a) ell < Cull + Gach)

ii) TRI S Ca: v(t, Heals

it) [lah] < C2: [Cys Q(Cy [fed] + Cse(h)) + 6(A)],

where v(h, €) = A(C\E + Cze(h))(Ci€é + Cse(h)) + c(h)

Proof The proof is carried out analogously to the proof of the convergence theorem in

Lermma 3 Lef

Quíu) = 6i (u, (bị )) + óa(u, (bị ))Yn (b, 4);

Q(u) = br (u, y(b; u)) + óa(u, y(b:4)) Y (b, u)

Then

Qn(sna)es! = beTh, — l@(Œ) — O25) — ở (8) = xj)},

tphere #j = (sa,j Wh(b; sh,j));# = (@(4),€(b)) and te hauc an estimate

l2) ~ ø(+,) = ở (,)(Œ ~ z;)|I < Called? + ah) Gl + a),

tphere cị: H — l.,: c¡(h) — 0 as h — 0,:i= 1,2 and C3 is a nonnegative constant Proof The above-mentioned relation is easily obtained Furthermore,

lø(Œ) - #(&;) ~ #'(/)Œ — #j)|l = II | [6 (xy + (@ — 25) — # (x5)|(@ — 25)

IA ew 4

;ilg- ll?

S Fons ~ eCa)ll + ly(bs 90.3) = e(8)|)?

Now, we go on proving the theorem 1 It is known that, assuming that ¢ is an isolated

sohition of (1) then the matrix Q(y(a)) is nonsingular(see, for example, [5]) Let ||Q7'(e(a))|| < 6) From Lemma 3 and condition (vi) in theorem 1, it follows

that

HQn(sn.3) — Q((2))|| = llới (sà,¿, (Bi sẽ,;)) — ớì ((8), #())+

+ 92(5h.;.Un(bì 8h,j))Yn(b, sh.j) — b2(8n,5 9m (5 $n,3))¥ (b, (a) + + b2(Sn5s¥)n(d; $n,3))¥ (0, (a) — đa((a), @@))Y (b, e(2))I]

< Ki(|sa,¿ — ¢(@)|] + [lyn (bs 8,5) — e@)II)+

+ Ky [l¥a(b, sn5) — ¥(6, #(4))|| +

4h

+ K¿(|ss.¿ — #(4)|| + |lua(b; sa¿) — g()|ÙDC:

= Millsjl|+ Mae(h) + Kola

l-Therefore

Q7! (Pa) )Qn(sn.5) ~ Q(¢(a))]I S Bi{ Milled] + Mae(h) + Koll ll}

< A{M, |ludl| + Moe(h) + KyC2[C,Q(Cy [val] + Coe(h)) + 6(A)]} =: P2) — (8) Since Q is continuous and 2(0) = 0, one can choose p = ||v$|| = ||so — #(4)|| << 1, such

that

Po(h) Sa < 1 and Me+7 "qa0@) +o9(Œ\ø) <Ø <1 for h << 1, (9)

where 9 = 3) KyC2C,/(1— a), M = ‘maxi M,, My = 8,C3/(1 — a)}

Using Lemma 4.4.14 from [1], one conclides that the matrix

14+ Q7!(¢(a))[Qu(sn,o) — Q(¢(a))]

is nonsingular Furthermore, Q),(so) is nonsingular and {Qj '(so)|| < G1/(1-a) Applying Lemma 3, one has

Hebl) < HQn(sn.0)~ "I {Ilbe(sn,0,4n( i sn o))I TRI + [o(z) — O25) — ở (z)Œ = z))l}

Pr ex(h) = uh, (B= A KeCa/( = a)),

<8: u(h,p) + MpỀ + I “na +

It can be proved that there exists a function €, : €(h) > 0 as h - 0 such that

8

+ —a(0) -a

Well < uh = Bu(h, p) + Mp? + ra palh) +7

< BopU(Cip) + Mp? # a Peat) + 6(h)

<ap+c(h) < p:= ub le h “<1

By an induction and argue as above, one sees that Q7'(sp,j) is nonsingular Moreover

a fog supe! <i,

x *(sas)Il S mm

where uit! = By(h, uj) + Ms)? + Tượng tế, cị (h) + Py a(h)

The sequence {u}}, is nonincreasing and nonnegative, so up, = limjsoo uj, exits h 6 i h and up satisfies the equation

un = Bo(h, up) + Muy)? + = un ex(h) + A eo(h) 1

Moreover, if u, + u as h > 0, then u is a non-negative solution of

Since 0 < uy < ue = p, it implies u < p The estimate JoQ(C,u) + Mu < & < | ensures

that u = 0 is an unique solution of (10) This and (i) in Lemma 2 yield the

which was to be proved

3 Numerical Experiments

In this section, we consider three examples In each example, step sizes and an initial vector are Lê ‘The convergence of the method in each example is expressed by

maxlyj() — y2~'(4,)|, where yp (t:) denotes yp (t) 8ny)-

Example 1 Consider the problem

y! = —4ty"/?, te (0,1),

sin(y(0)) + cos(y(1)) — sin avy — cos B5: =

This problem has been studied in [4] with the linear boundary condition (0) — 2(1) =

In both cases, it has an exact solution y(t) = (?+1+V0))~ ? and y(0) = 0.1715 Tae

Furthermore, Silty = —6ty'/? is not bounded However, it is bounded in a bounded

neighbourhood of the isolated solution We can put Q(v) = 6v'/? Thus Q and the

function in the boundary condition satisfy the conditions in the theorem 1

10 | 0.17173830227481 | O.11s 10-4

40 | 0.17161116830044 | 0.275 10-5

150| 0.17158289850619 | 0.885 TÚ”

Table 1: Numerical result of example 1 with initial vector so = 0.5 Starting from the initial vector so = 0.5 we get the improved shooting points given

in Table 1 The convergent speed of the method (N = 10) is given in Table 2

j max|yh (ts) — "(tI

6 ` 111x10"!19

Table 2: The convergence of the method in example 1 with N = 10 Example 2 Consider the problem

yl=are=y te 01),

7 991Y"() + cog y(1) — e7 2! — cos(e-!) = 0,

This problem has an exact solution y = e~', : z = —e~t and y(0) = 1,: z(0) = —1 It is

easy to show that the function in the boundary condition satisfies all the conditions in the theorem 1

2000 1 —0.99992170530796 10.93s 10-5

Table 3: Numerical result of example 2 with initial vector sọ = [1.5; —1.5]

j max|y} (ts) — gh (ti) | max|zj (ti) — 2471 (ti)|

9 7.22 x 10716 7.22 x 10716

Table 4: The convergence of the method in example 2 with N = 30 Choose the initial vector so = (1.5; —1.5] we have Table 3 The convergent speed of

the method (N = 30) is given in Table 4

Example 3 Consider the problem

3 y= 59", te (0,1), e70-00147(0) — 70.016 _ 0, e70-001y7(1) _ ¿7.0001 _ g Put z = y’, the equation becomes

HỆ = [say te [0,1]

The partial derivative As is not bounded in the whole space but it is bounded in a neigh-

borhood of an isolated solution The remaining conditions of the theorem 1 can be easily

checked

4000 4 —8.00029758256503 30.35 10-5

Table 5: Numerical result of example 3 with initial vector sọ = |4.5; —8.5]

This problem has an exact solution y = 4/(1 + t)? and y(0) = 4,: 2(0) = —8 Choose the initial vector sp = [4.5;—-8.5] we have Table 5 The convergent speed of the

method (N = 40) is shown in Table 6

j max|y, (ts) — ¿` (R)| max|z}(t,) — z¿—'(H)|

Table 6: The convergence of the method in example 3 with N = 40 References

uì

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(3

(

(6)

(7

{8

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