A singular nonlinear boundary value problem in the nonlinear circular membrane under normal pressure

A SING.ULAR NONLINEAR BOUNDARY VALUE PROBLEM IN THE NONLINEAR CIRCULAR

MEMBRANE UNDER NORMAL PRESSURE

JUN YONG SHIN

1. Introduction

The nonlinear boundary value problem

0< x < ^1,

(1.1 )

" / ( ') 3, g(x) y = x,y,y = --y - - - ,

X y2

y'(O) =0, and either (H) : y(l) = .\ > 0 or (S) :y'(l) +(1 - v)y(l) = 0,1 - v > 0,

9 E G[O,l],k ~ g(x) ~ K on [0, 1J for some k,K >

°

arises in the nonlinear circular membrane under normal pressure [2,3J.

By a positive solution of (1.1) we mean a positive function y(x) E

GI[O,1J nC²(0, 1J that satisfies (1.1).

Previous existence and uniqueness theorems for (1.1) under the con- dition (H) or (S), using the iterative methods and the shooting method, have been given in [1-3J. Our emphasis in this paper is on treating (1.1) directly as a boundary value problem and on obtaining the existence of a positive solution via psitive solutions of perturbations of (1.1). The idea which is used here can be easily adapted to handle more general singular nonlinear problems.

We call a function (3 EG2[0,1J a positive lower solution of (1.1) if {J" ~ I(x,(J, (J') on (0,1),(3 >0 on [0,1],(J'(O) ~ 0,

(H) : (J(1) ::; A((S) :(3'(1)+(1 - v)(3(l) ::; 0).

Received October 6, 1994.

1991 AMS Subject Classification: 34B15.

Key words: Singular boundary value problem, upper and lower solutions.

The definition of a positive upper solution of (1.1) is given in a similar way. Similar definitions hold for a perturbation of (1.1) which will be given in section 2 or section3.

Insection 2, we consider the problem (1.1) under the condition(H).

And the problem (1.1) under the condition (S) is considered in section 3.

2. Existence and uniqueness theorems under the condition (H)

For each positive number rn, we consider the nonlinear boundary value problem

0< x < 1,

!

(2.1)m

Y" = ^{3 y' g(x)}

x +^{-!. -} -y-^{2- '}

m

y'(O) = 0, and (H) : y(l) =^oX >0,

9EC[O,1},k ~g(x) ~K on [0,1} for somek,K >0, which may be viewedas a perturbation of (1.1).

To prove the existence of a positive solution of (1.1), we establish the existence of a positive solution of (2.1)m.

LEMMA 2.1. Yl= oX is a positive lower solution of(2.1)m.

Proof. It is clear that

Thus Yl is a positive lower solution of (2.1)m, which completes the proof.

LEMMA 2.2. Yum = -b{(x+ ~? - (1+ ~)2} +oX is a positive upper solution of(2.1)m.

Proof. It is clear that

Yum(x) >0on (0,1),Yum(l) = oX, andY~m(O) ~ o.

Since ,X ~ Yum(x) and k ~ g(x) ~ K on [0,1], g(x) < K Y;m - ,X2

and so

11 _ _2K < 6K _ g(x) _ _ 3 , _ g(x ) Yum - 8,X2 - 8,X2 2 - +..LYum 2 '

Yum X m Yum

(2.1)

(2.2)

Thus Yum is a positive upper solution of (2.)m, which completes the proof.

The following result is clear from an application of Schauder's Fixed Point Theorem.

LEMMA 2.3. There exists a solution Ym of (2.1)m, Ym E C²[0,1], such that

Yl(X) ~ Ym(x) ~ Yum(x) on [0,1],

where Yl and Yum are given in Lemma 2.1 and Lemma 2.2.

LEMMA 2.4. IfYl and Y2 are two positive solutions of(2.1)m, then Yl == Y2·

Proof. Let Yl andY2 be positive solutions of (2.1)m. Then we obtain ( (_X+ -1 )3 ,(Yl - Y2)' ) '= (x + ~)32 2^g(x)(Yl -² Y2) on (0,1).²

m YIY2

If we multiply both sides of (2.1) by (Yl - Y2), then we obtain ((x+ ~r(y~ -Y~))'(YI-Y2)

(X+~)3g(x) ^{2 2}

= 2 2 (Yl +Y2)(Yl - Y2) on (0,1).

YIY2

Therefore, if we integrate both sides of (2.2) from

°

Thus we obtainyi-y~ =

°

Note IfYm is a positive solution of (2.1)m for each m > 0, then Ym(X) _~ Aon [0,1] and hence all positive solutions of(2.1)m, m >0, are bounded below by Aon [0,1].

LEMMA 2.5. IfYm is a positive solutionof(2.1)m, then y:n <

°

(0,1] andVex) > y(l) > 0 on [0,1).

Proof. Since

(x+7k)3_z ^g(x) <

°

Ym

we obtain that (x+ ~)3y:nis strictly decreasing on [0,1], which implies that

y:n <

°

This completes the proof.

LEMMA 2.6. IfmI ~ mz and Yml andYm2^arepositive solutions of (2.1)ml and(2.1)m27 respectively, then Yml(X) _~ Ym2(X) ^on [0,1).

Proof. It is clear from the fact that Y_m2 is an upper solution of (2.1)ml·

THEOREM 2.7(EXISTENCE). IfYm is the positive solution of(2.1)m foreach m = 1,2,3,·· . ,then the sequence{Ym} convergesto apositive solution Yof (1.1).

Proof. To prove this theorem, we prove the following steps:

Step 1 : Ym -+ Y as m -+ 00.

Step 2 : yE GI[O,1] n G²(0, 1].

Step 3: Y is a solution of(2.1).

Step 1 : From Lemma2.3 and Lemma2.6, we know that the sequene {Ym} is monotone decreasing in m and bounded below y A. Therefore,

Ym -+Yas m -+ 00 and Vex) ~ Aon [0,1].

Step 2 : Ifwe integrate ((x+ ~)3y:n)' from

°

l

:rl:;(e)de

and (2.4)

Ifwe integrate both sides of (2.4) from 1 to x, then we obtain

(2.5)

and

(2.6)

Ifwe let m -+ 00 in both sides of (2.5) and (2.6), then by Lebesque's Dominated Convergence Theorem, we obtain

and

(2.8)

(2.9)

and so (2.10)

which implies y E G2(O,1]. Since the second term of the right side of (2.8) converges to 0 as x approaches 0,yis continuous at 0. From (2.7) and (2.8), we obtain

lim y(x) - y(o)

X-+O+ x

= ^lim (_1

r

x-+O+ 2x³ 10 ^{y2(e) 2x}10 ^y2(e)

which implies y'(O) =0. Ifwe take the first derivative of both sides of (2.8), we have .

, 1

r:

y (x) = - x3 10 ^{y2(e) de on (0,1]}

lim y'()x = li^{m - 3}1

LX

°

%-+0+ x-+Q+ x 0 Y ^e"

which implies y E GI[O, 1]nG²(0,1].

Step 3 : It is clear from (2.8) and (2.9) that y(l) = A and y'(O) = 0.

IT we take the derivative of both sides of (2.10), then we get

3 g(x) 3 3 2 r^x ~g<e)d&

y"(x)= ^x

m

x 6 y2(x) x

which implies that y is a solution of (1.1). This completes the proof.

THEOREM 2.8 (UNIQUENESS). Assume that YI andY2 are positive solutions of(1.1). Then YI ==Y2.

Proof. The proof of this theorem is similar to that of Lemma 2.4.

3. Existence and uniqueness theorems under the condition (S)

For each positive number rn, we consider the nonlinear boundary value problem

y" = _ ^{3 Y' _ g(x ) 0} <^x <1

x+^{.!. y2 ' ,}

(3.1)m y'(O)=^{0, : d}(S) :y'(l)+^(1-v)y(l)= 0, 1-v> 0, 9 EG[O,1], k:5 g(x) :5K on [0,1] for some k, K >0, which may also be viewed as a perturbation of(1.1).

LEMMA 3.1. Yl = -(:g:::::~:)i. ^{(x2 -}

:=:)

Proof. It is clear that

Yl(X) >0 on (0,1),

y;(x) =-2x

(~g = _~~:)~ ,

yHO) =0,y;(1) +(1 - v)y/(l) =0, and

( 2)

y"(x) = ^-2 k(l- v)

I 8(3-v)2

Since Yl(X) ~

(:g:::::?:)i. (:=:)

)2)

- > -8

Yf - (kfl-V?;)i(3-V)2 - 8(3 - v)2

8 3-v I-v

So we have

( 2)!

Y;' =^-2

~g = _~~2

.1 .1

( k(1-V)2)S (k(1-V)2)S X)

> -2 -6 ( 1 - - -

- 8(3 - V)2 8(3 - V)2 X +~

3 , g(x)

> - - - Y / - - - ·

- X + ~ Yf

Thus Yl is a positive lower solution of (3.1)m, which completes the proof.

.1

LEMMA 32_{. .} Y_um = _(K(I-V)2)₃₂ ^{S •((x} +.1..)2 __m 3-V(1_I-v +.1..)2)_m is a positive upper solution of(3.1)m.

Proof. It is clear that

Yum(x) >⁰ on (0,1),

, (x)=-2( ~)((1-v)2)i

Yum x + ^{m 32 '}

Y~m(O) ~ 0,Y~m(1) +(1 - v)Yum(l) ~ 0,

and

<

11 ( )⁼ -2(K(l-V)2)t

Y^um X 32

Since Yum(x) 2:: _(K(~;v)2)t .(1+~^?(1- ~=:)^andK 2:: g(x) on [0,1], we have

1

g(x) K 8(K(l-V)2)^"8

- - < <

y² - (K(I-V)2)i . (1+.1.)4(_2_)2 - 32

um 32 ^{m I-v}

Therefore we obtain

2 1

11 ( )= -2(K(l - v) )

Y^{um X} 32

< (K(I-vy~)l _ (K(1-V)2)l

_6. ~ ⁸ ~

3 , g(x)

+^{.1.Yum - - y2 ..}

X m um

Thus Yum is a positive upper solution of(3.1)m. This completes the proof.

The following result is clear from an application of Schauder's Fixed Point Theorem.

LEMMA 3.3. Tbere exists a positive solution Ym of(3.1)m, Ym E

C²[0,I], such tbat.

YI(X) ::; Ym(x) ::; Yum(x) on [0,1],

wbere YI andYum are giveninLemma 3.1 and Lemma 3.2.

LEMMA 3.4. IfYI andY2 are two positive solutions of(3.1)m' tben YI ==Y2·

Proof. Let YI andY2be positive solutions of(3.1)m. Then we obtain

(3.1) « ^{1)3(' , ))' (x}+_~)3g(x)(2 2) (0 1)

x + - ^{YI -} Y2 = 2 2 YI - Y2 on , .

m Y2Y2

Hwe multiply both sides of(3.1) by (YI - Y2), then we obtain

(3.2)

Therefore, if we integrate both sides of (3.2) from 0 to 1, then we obtain

and so

which implies that

YI(I) - Y2(1) = 0 and yl' - Y~ == 0 on [0,1].

Thus YI == Y2 on [0,1], which completes the proof.

Note. IfYm is a positive solution of(3.1)m for each m> 0, then

1

(

k(1-V)2)3 2

Y^{m (}X) _{~ B(}3 _ v)2 . 1 _ Von [0, 1]

and so all positive solutions of (3.1)m, m > 0, are bounded below by some positive number on [0,1].

LEMMA 3.5. H Yisa positive solution of(3.1)m, then y'(x) <0 on (0,1] andVex) > y(l) >

°

Proof. Since

(( 1)3, )'

X + - ^{Ym = - 2} <0 on (0,1),

m Y_m

we obtain that(x+ ~)3y:nis strictly decreasing on[0,1], whic h implies that

y:n < 0 on (0,1] andVex) > y(l) on[0,1).

Since y'(l) <

°

°

on [0, 1), which completes the proof.

LEMMA 3.6. HmI .~m2 andYml "and Ym^{2 are}positiv.e solutionsof (3.1)ml and (3.1)m2' respectively, then Yml(X) _~ Ym2(X) on [0,1].

Proof. It is clear from the fact that Ym² is an upper solution of (3.1)ml·

THEOREM 3.7 (EXISTENCE). HYm is a positive solution of(3.1)m for eachm =1,2,3,··· ,then the sequence {Ym} convergestoapositive solution Y of (1.1).

Proof. To prove this theorem, we prove the following steps:

Step 1: Ym ^-+ Yas m ^{-+ 00.}

Step 2: yE GI[O,1]nG2(0, 1].

Step 3: Yis a solution of (1.1).

Step 1: From Lemma3.3 and Lemma 3.6, we know that the sequence bm}is monotone decreasing in m and bounded below by some positive number. Therefore,

Ym -+Y as m -+ 00 andVex) >

°

Step 2: IT we integrate((x+

;J

°

and

Ifwe integrate both sides of (3.4) from 1 to:t, thenwe obtain

(3.5)

and

(3.6)

Ifwe let m -+ 00 in both sides of (3.5) and (3.6), then by Lebesque's Dominated Convergence Theorem, we obtain

(3.7) (0) = (1) _! [,~^{W dt} ! ^{[I eg(e)dt'}

Y Y 210 '" +210 ^{y2(e) '"}

and

(3.8)

(3.9)

which implies y E G²(O,1]. Since the second term of the right side of (3.8) converges to 0asx approaches 0,yis continuous at o. From (3.7) and (3.8), we obtain

y'(O) = ^lim y(x) - y(O) .x-+O+ X

_ lim

(_1 r

=0.

Ifwe take the first derivative of both sides of (3.8), we obtain (3.10)

and so

lim Y'(x) = ^lim-a fxe~~i)^de=0,

X-+O+ x-+O+ x 10 ^Y

which implies y E Gl[O, 1]nG2(0, 1].

Step 3: It is clear from (3.9) and (3.10) that

~'(O)

-1

~~~e;

On the other hand, from(3.4), we obtain

y'(l) =

-1

>. (1+ ~)310 y~(e) .

=_m-+co^lim y~(l)=^{-(1-'v)y(l:):}

Ifwe take the derivative of both sides of (3.10), then we get x3 g(x) x3 _ 3x2 fX eag(e)dc ( ) , y"(x) = _ ^{y2(X) Jo y2(e) r"} =_.!!....!..- _! ^(x),

x⁶ y2(x) y

which implies that y is a solution of (1.1). This completes the proof.

THEOREM 3.8 (UNIQUENESS). Assume that Yl andY2 arepositive solutionsof(1.1). Then Yl ==Y2.

Proof. The proof of this theorem is similar to that of Lemma 3.4.

References

1. E. Bohl, On two boundary vale problems in nonlinear elasticity from a numer- ical viewpoint, In: Lecture Notes in Mathematics No. 679, Ed.: R. Ansorge,

W. Toring p. 1-14, Springer, Belin, 1974.

2. A. J. Callegari and E. L. Reiss, Non-linear boundary value problems for the circular membrane, Arch. Rat. Mech. Anal. 31 (1970), 390-400.

3. R. W. Dickey, The plane circular elastic surface under normal pressure, Arch.

Rat. Mech. AnI. 26 (1967), 219-236.

Department of Natural Sciences

Pusan National University of Technology Pusan 608-739, KOREA