Solution of a nonlinear two-point boundary value problem with Neumann-type boundary data

JOURNAL

OF MATHEMATICAL

ANALYSIS

AND

APPLICATIONS

135, 691-701 (1988)

Solution of a Nonlinear Two-Point Boundary Value Problem with Neumann-Type Boundary Data JUKKA SARANEN AND SEPPO SEIKKALA Section of Mathematics, Faculty of Technology, University of Oulu, Linnanmaa, SF-90570 Oulu, Finland Submitted by V. Lakshmikantham Received April 13, 1987

In this article we propose a new method for solution of nonlinear two-point boundary value problems of Neumann type. Our approach is based on a generalized integral equation representation of the solution together with an additional control variable. The method is constructive using Picard iterations. Some numerical examples are given, 6 1988 Academic Press, Inc.

1. INTRODUCTION The Neumann condition is one of the basic boundary conditions appearing in mathematical physics. For example, in equilibrium problems concerning beams, columns, or strings the homogeneous Neumann condition describes freely supported boundaries. In fluid flow problems the Neumann condition for the velocity potential is used to describe impermeable boundaries. In the case of the heat transfer the homogeneous Neumann condition in turn describes isolated boundaries. In many simplified models of these physical problems we finally end up with a two-point boundary value problem with given Neumann condition at both ends of the interval. When nonlinearities are present the theoretical and numerical solution of the corresponding boundary value problem is well discussed for Dirichlet or also for more general mixed-type boundary conditions (cf. [ l]-[4]). Bailey, Shampine, and Waltman [l] use integral equation methods. An equivalent integral equation is derived by using Green’s functions for the linear boundary value problem. Ciarlet, Schultz, and Varga [24], on the other hand, employ a variational approach. They consider also some nonlinear boundary value problems of Neumann type [3]. Mimura, Tabata, and Hosono [S] discuss a two-component interaction-diffusion system subject to zero flux boundary conditions. In their article the singular perturbation method is combined with a variational approach. 691 0022-247X/88 $3.00 CopyrIght 0 IYXX by Academic Press, lx. All rights nl reproduction m any form reserved.

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AND

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Here we propose a new method to solve nonlinear two-point boundary value problems of Neumann type. Our approach is based on a generalized integral equation representation combined with an additional control variable. In order to present the basic ideas we do not aim at the most generality and therefore we consider the boundary value problem u”(t) =f(t, &&f(t)) u’(u) = A,

u’(b) = B.

Here A, BE R, f: [a, b] x R -+ R is square integrable in the first argument and Lipschitz continuous in the second argument and the operator d: &(a, b) + &(a, b) is Lipschitz continuous. By a solution of problem (1.1) we mean a function u E H*(a, b) satisfying the differential equation in the distributional sense. Recall that #“(a, b) is the Sobolev space of functions UE L,(a, b) such that U’ and U” also are square integrable, hence U’ E C[a, b]. The Neumann-type boundary value problem (1.1) differs from a corresponding initial value problem or Dirichlet-type boundary value problem in that a Lipschitz condition does not guarantee the existence nor the uniqueness of a solution. Even the simple linear problem u’(u) = A,

u”(t) =f(f),

u’(b) = B

is solvable only if

s

;j-(r)dt=

B-A

and the solution is unique up to an additive constant. Thus, in case of problem (l.l), it is more diflicult to find general conditions for existence and uniqueness. Furthermore, Green’s functions cannot be utilized in reformulating (1.1) as an integral equation. Our approach avoids these difficulties as follows. Instead of (1.1) we study the problem u”(t) + 6 =f(t, du(t)) u’(u) = A,

u’(b) = B

Ju=A,

where A is a real parameter, J is the mean value operator, Ju=-

1 b u(t) & s b-u a

(1.2)

ANONLINEARTWO-POINTBOIJNDARYVALUEPROBLEM

693

and (1.3) The role of condition JU = 1 is twofold. First, for fixed 2 it guarantees (with Lipschitz conditions on f and J&‘) the uniqueness of a solution to (1.2) and second, by letting 1.vary, we find a solution to the original problem (l.l), if there exists a I. such that 6 = s(n) = 0. On the other hand, if 6(A) # 0 for all AE R we get a nonexistence result. We will give some examples where the existence of a zero of 8(n) is deduced either numerically or theoretically. Our method of proof is constructive. It uses contraction mapping principle and Picard iterations. Simultaneously with a zero of the function h(1) we find a numerical solution U. We do not discuss the numerical error caused by the iterative method and by the use of numerical quadratures, on the other hand. Although there is no Green’s function for problem U” =O, u’(a) = A, u’(b) = B, problem (1.2) as well as problem (1.1) can be transformed into an equivalent integral equation. This is based on a representation result for u E H2(a, b), u(t)q

It-slU..(S)ds+;

[u(u)+u(b)]

a

-;[(a-z)u’(u)+(b-t)u’(b)].

(1.4)

2. EXISTENCE OF SOLUTIONS We prove first the integral representation (1.4) of a function u E H2(a, b) and then give an equivalent integral equation form of problem (1.2). LEMMA

2.1. Zf u E H*(u, b), then we haoe u(?)=f/”

It--S1

dQ)ds+f

[u(u)+u(b)]

a

-;

[(a - t) u’(u) + (b - t) u’(b)],

u
Proof. For a function u E C2[a, b] one can check (1.4) in an elementary way. Then (1.4) follows for a u E H*(u, b) since C*[u, b] is dense in H*(u, 6) and since the embedding H*(u, 6) c C’[u, b] is continuous. 1

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AND

SEIKKALA

We define the operator G on L,(a, 6) by Gu(t)=;Jb

adt
It-s1 u(s)ds,

u

(2.1)

and G, by G,u=Gu-JGu.

(2.2)

One easily verities that G, has the following properties: (i) (ii) (iii) (iv)

G,: Lz(a, b) -+ H2(a, b) is continuous, (G,u)” = U, (G,u)‘(t) = $( J: U(S)ds- Jf U(S)ds}, and JG, = 0.

From (iii) it follows that (v)

for any u E L,(a, 6) the function v = G,u- Ju. G,l

satisfies

v’(a) = v’(b) = 0.

Let c: [a, b] + R denote the solution of B-A df = b-a’

c’(b) = B,

c’(a) = A,

Jc = 0,

(2.3)

i.e., c is the quadratic polynomial c(t) = c0 + c, t + c2t* with cl=A--u,

1 B-A

B-A b-a

c*=Yj=

and a+b co= - -c,

2

-

a*+ab+b* c2.

3

2.2. A function ME L,(a, b) is a solution of problem (1.2) if and only if u is a solution of the operator equation LEMMA

u(t)= CGoS(.,~M.))l(t)-

Jft.7 d4.1).

CG,ll(~)+c(~)+A

a
(2.4) Proof: Suppose first that UE L,(a, 6) satisfies (2.4). Then, by (i), u~H~(a, 6) and by (2.3) and (ii) we have u”(t)=f(c

@‘4t))-Jf(;

W.))+b-a,

B-A

a
A NONLINEAR

TWO-POINT

BOUNDARY

VALUE

PROBLEM

695

i.e., u”(t) + 6 =f(t,

a
du(t)),

where

Moreover, by (v) and (2.3), u’(a)= c’(a) = A, u’(b) = c’(b) = B, and finally, by (iv) and (2.3), Ju = A. This proves that u is a solution of (1.2). On the other hand, if u is a solution of problem (1.2) then u E H*(a, b) and f( ., xZu(. )) E &(a, b). Applying J on both sides of the differential equation of (1.1) we get

Using this and Lemma 2.1 we obtain

+;

[u(a)+u(b)]

-;

[(a-

t)A + (b- t)B],

adt
or u(t)=

[W(., @W~))l(t)-Jf(~,

du(.))[Gl](t)+c(t),

(2.4)’

where F(t)= -AICl](t)+~[u(a)+u(b)]-;[(a-t)A+(b-t)B], b-a for t E [a, b]. Now applying J on (2.4)’ we get A = JGf( ., du( . )) - Jf( ., du( .)) JGl + JF. Subtracting this from the right-hand side of (2.4)’ and adding we get (2.4),

u(t)= [Got., ~4.))l(t)--Jf(~,

~~(.))CGoll(t)+c(t)+E.,

where c(t) = Z(t) - JC obviously satisfies (2.3). 1 Using (2.1) and (2.2) the operator equation (2.4) can be written as the integral equation

u(t) = j” k(t, S)f(S, du(s)) ds+ c(t)+ 2, u

(2.5)

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SARANENANDSEIKKALA

where $--[(t+)*+(s-?)‘I-?}

(2.6)

or equivalently (2.7) Here J, It-,ss( denotes the mean value of t + It --sI on [a, b]. THEOREM 2.1. Assume that f: [a, b] x R -+ R is square integrable in the first argument and in the second argument satisfies the Lipschitz condition

If(t, Y)-f(t,

a
Al GLIY-Yl?

y, PER,

(2.8)

and that A : L,(a, b) + L,(a, b) satisfies u, UE L,(a, 6).

lI~~-~~ll,d~~Il~-~ll,,

Then, if LA, < 3@/(b

(2.9)

- a)*, problem (1.2) has a unique solution u for each

AER.

Proof. From

the

assumption

on f

and

d

it

follows

that

f( ., &u(. )) E L,(a, b) for each u E L,(a, b). Thus the equation Ku(t)=~‘k(t,s)f(s, u

&‘u(s))ds+c(t)+1

(2.10)

defines an operator K on L,(a, 6). Obviously K maps L,(a, 6) into itself. Hence, by Lemma 2.2, by the equivalence of (2.4) and (2.5), and by the contraction mapping principle it is enough to show that K is a contraction on L,(a, b). Let U, GELJa, b). Then, by (2.8), (2.9) and Cauchy-Schwarz inequality, /Ku(t) - Ku(t)1
I&‘u(s)-&‘ti(s)i”ds]l-2

697

ANONLINEARTWO-POINTBOUNDARYVALUEPROBLEM

and so IKu-KG~2
dL41u-41,p,

jbk2(t,s)dsdt]“2 (1 u (b-a)’

(2.11)

a

which proves that K is a contraction when LA (h-0’.

i

O 345

When the assumptions of Theorem 2.1 are satisfied then Eq. (1.3),

b(A)=Jf(.,du;(.))-E, where u:, is the unique solution on (1.2) with mean value 1, defines a function 6: R -+ R and the existence of a solution to the original problem (1.11, u”(t) =f(t, du(t)) u’(a) = A,

u’(b) = B,

is equivalent to the existence of a zero of s(J). We have COROLLARY 2.1. If the assumptions of Theorem 2.1 are satisfied then 6 is Lipschitz continuous and hence problem (1.1) has a solution provided there exist 1, and I., such that &A,) 6(2,)<0. Moreover, A -+ uj, is Lipschitz continuous.

Proof:

For A, LER,

bf(S, &u>.(s)) ds-&jbf(S, <&

0

dui(s)) ds

j” ldu,(s) - dux(s)l ds a

and as (2.11) was obtained we get u>-ulll2+Jb-a

IL-Xl,

698

SARANEN ANDSEIKKALA

i.e.,

Thus,

IS(A)- &Gl G

LA, 1 - LA,( (b - a)‘/345

and so 6 is Lipschitz condtinuous.

(A-AI

1

An immediate consequence of Corollary 2.1 is COROLLARY 2.2 In addition to the assumptions of Theorem 2.1 suppose that ,I= Ju -+ CQ implies Jf( ., du( .)) + cc (resp. -GO) and Ju + -CO implies Jf( ., &u( .)) + -CC (resp. + CO). Then problem (1.1) has a solution for all A, BE R.

Also we get COROLLARY 2.3. Let du = u and let the assumptions of Theorem 2.1 be satisfied. If f(t, y) is p-periodic in y for all t E [a, b], then 6(l) is p-periodic and hence problem ( 1.1) has either infinitely many or no solutions.

ProoJ: By Theorem 2.1 problem ( 1.2) has a unique solution (uj,, s(n)) for each 2 E R. By the periodicity of f(t, y) in y the pair (ui. + p, s(n)) is the solution of (1.2) with the mean value ;I + p. Hence we have Remark 2.1. The condition Ju = 1 above plays a role of a control to obtain S(1) = 0 for some 2, and hence to obtain a solution to problem (1.1). If some other control is used, then the kernel k(t, s) in the integral equation (2.5) differs from (2.6) and accordingly the upper bound for the Lipschitz constant LA, changes. For example, by the above method condition u(a) = i leads to the constraint LA, < ,/‘%/(b-a)’ and condition u((a + b)/2) = 2 respectively to LA, < J80/(b - a)‘, both of which are more stringent than the constraint LAO < 3fi/(b -a)’ obtained when using the control Ju = ;i.

3. EXAMPLES EXAMPLE

3.1. Consider the boundary value problem u”(t) = u(t) + sin u(t) + h(t), u’(a) = A,

u’(b) = B,

a
(3.1)

A NONLINEAR

TWO-POINT

BOUNDARY

VALUE

699

PROBLEM

1.2 /=&-3Y. FIG.

1.

Behavior

of s(1)

on the interval

[0, 11

where heL,(a, 6) and b-a
0.0 -0.271

EXAMPLE

0.1

0.2

-0.174

-0.079

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.016

0.115

0.219

0.334

0.460

0.603

0.766

1.0 0.952

3.2. Consider the boundary value problem u”(t) = sin[27cu(t)] + t-j (3.2)

u’(0) = 0, u’( 1) = 0.

By Corollary 2.3 the function s(A) is l-periodic. Figure 1 shows the calculated behavior of s(A) on the interval [0, 11. We find two zeros, il, ~0.554 and I, ~0.951, in this interval. The corresponding approximate solutions ur and uq are shows in Fig. 2 below.

1.0

FIG. 2.

Functions

u,(f)

and u>(t).

700

SARANEN AND SEIKKALA

FIG. 3. Function 6(i), -2 < 1 Q 3.

By Corollary 2.3 the problem (3.2) has infinitely many solutions u; = u1 + n and us = u2 + n corresponding to the values 2; = 1, + n and A; = I, + n, n E Z, of the control variable A. EXAMPLE

3.3. The boundary value problem -s-(l+cost)3-coSf, [u(t)]3-(1 +cos t+cos 8-(1+cost)3-co~t,

t,

if u(t)< -2 if lu(t)l <2 if u(t)>2

(3.3)

with u’( 0) = 0,

u’( 1) = -sin 1,

has the solution u,,(t) = 1 + cos t. Theorem 2.1 is applicable in this case. Figure 3 shows the behavior of the function s(A), -2 < A< 3, where it has a zero at the point 1, z 1.84147.This is the correct value of 1 + sin 1 = Ju, by five decimals. The maximum error E=max((u,(t,)-u,,(ti)l:Odi,
ACKNOWLEDGMENT The authors are indebted to Dr. J. Mannermaa

for carrying out the numerical calculations.

Note added in proof: A slight modification of the proof of Theorem 2.1 shows that the upper bound 3,,/%/(b - a)2 for LA, can be replaced by the optimal upper bound n2/(b - a)*.

ANONLINEARTWO-POINTBOUNDARYVALUEPROBLEM

701

REFERENCES I. P. B. BAILEY, L. F. SHAMPINE,AND P. E. WALTMAN, “Nonlinear Two Point Boundary Value Problems,” Academic Press, New York, 1968. 2. P. G. CIARLET, M. H. SCHULTZ,AND R. S. VARGA, Numerical methods of high-order accuracy for nonlinear boundary value problems. I. One dimensional problem, Numer. Math. 9 (1967), 394430. 3. P. G. CIARLET, M. H. SCHULTZ,AND R. S. VARGA, Numerical methods of high-order accuracy for nonlinear boundary value problems. II. Nonlinear boundary conditions, Numer. Math. 11 (1968). 331-345. 4. P. G. CIARLET, M. H. SCHULTZ,AND R. S. VARGA, Numerical methods of high-order accuracy for nonlinear boundary value problems. IV. Periodic boundary conditions, Numer. Math. 12 (1968), 266-279. 5. M. MIMURA, M. TABATA,AND Y. HOSONO,Multiple solutions of two-point boundary value problems of Neumann type with a small parameter, Siam J. Math. Anal. 11 (1980), 613-631.

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