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Solvability of some singular nonlinear boundary value problems and existence of positive radial solutions of some nonlinear elliptic problems
Nonlinear Analysrs, Theory, Methods Printed in Great Britain.
& Apphcations,
Vol. 16, No. 9, pp. 781-790, 1991.
0
0362-546X/91 $3.00+ .M) 1991 Pergamon Press plc
.
SOLVABILITY OF SOME SINGULAR NONLINEAR BOUNDARY VALUE PROBLEMS AND EXISTENCE OF POSITIVE RADIAL SOLUTIONS OF SOME NONLINEAR ELLIPTIC PROBLEMS ZONGMING Department
of Mathematics,
(Received 21 February
University
Guo*
of Glasgow,
1990; received in revised form
University
Gardens,
Glasgow
G12 SQW, U.K.
1 August 1990; received for publication 20 September
Key words and phrases: Singular boundary value nonlinear elliptic problems, Leray-Schauder degree.
problems,
nonlinear,
positive
radial
1990)
solutions,
1. INTRODUCTION
IN THIS paper we study the existence and uniqueness of positive solutions of singular secondorder differential equations of the form
l E (ti >&I,
Y” + Ht)g(JJ) = 0,
(1.1)
Y(h) = Y(b) = 0,
where 0 I t, < t2 < + co; 4(t) E C’(t,, tz), 4(t) > 0 on (ti, tz), C#Iis allowed to have suitable singularities at t = tl and t = f2; g E C’(O, oo), g > 0 on (0, w) and g is also allowed to be singular at 0. Equations of this type arise in diffusion and osmotic flow theory, one particular example being the generalized Emden-Fowler equation (see [8, 141). They also arise from the study of positive radial solutions y(r), r = 1x1,x E RRN,N 2 3, in R, < r < R, and on r = R2
AY + g(y) = 0, i y=o
r=
(1 .a
R,.
Putting t = [(N - 2)rN-2]-’ k = (2N - 2)/(N - 2)
4(t) = ](N - 2)Kk,
Ri = [(N - 2)ti]-1’(N-2),
(1.3)
i = 1,2,
(where we put t, = 0 when R, = 03) gives equations of type (1.1). Problems of the type like (1.2) have been studied by numerous authors (see [ 1, 3, 5-7, 9-l 11) when g(y) is nondecreasing in (0, m). In this paper, we shall give some results for positive radial solutions of (1.2) when g(y) is decreasing. By a solution of (1.1) we mean a function y E C’[tr , f2] n C2(t, , t2) that satisfies (1.1). The existence of solutions for problem (1.1) has been considered by Callegari and Nachman [4], who effectively consider (1.1) with 4(t) = [min(t, 1 - t)], g(y) = y-l and t, = 0, t2 = 1; by Luning and Perry [8], who establish constructive results for g(y) = y-” when 0 < 01_= 1 and t, = 0, t, = 1; and by Taliaferro [13], who proves, in particular, that a necessary and sufficient *On leave from the Department
of Mathematics,
Henan
Normal 781
University,
Xinxiang,
People’s
Republic
of China.
782
ZONGMING Guo
condition for existence of solutions with first derivative continuous on [0, l] is that !
tP(l
- t)-“6(f)
dt < + co.
(1.4)
i0
In Section 2, we use the Leray-Schauder degree to seek positive solutions of (1.1). We will extend some of the results of [2, 121 by allowing a more general g(y) and at the same time generalize the sufficient condition of [13] to problems with a more general g(y). In [15], the author discusses problem (1.1) with nonlinear term g(t, y, y’). In Section 3, we apply the results obtained in Section 2 to (1.2) and obtain the e&istence of a positive radial solution of (1.2). 2. EXISTENCE
OF A POSITIVE
SOLUTION
OF (1.1)
1. Suppose that 6(t) E C’[t, , t,], 4(t) > 0 in (tl, f2), where 0 < t, < tz < co; g satisfies (A) g is continuous and nonincreasing on (0, co), (B) g(u) > 0 on (0, 00)~ forO< k-c 1, (C) s:: Ns)g(k(s - tl)(tz - s)) dJ < + =J, (D) lim g(t) 1; [g(s)]-’ ds = 00.
THEOREM
Then dro%em (1.1) has a positive solution. To prove this theorem, we consider two cases: (a) g has singularity at 0; and (b) g has no singularity at 0. We only prove theorem 1 in case (a), the proof of case (b) is similar to case (a). In order to avoid the possible singularity of g at 0, we consider, for each n E N = i 1,2, -. .) and 1 E [0, 11, the family of problems:
Y” + (1 - I)&,
+ Mt)g(Y) = 0,
v(tJ = Y(b) = l/n,
(2.lh,n
where 0 < 6 < 1 is a positive real number which is determined below, @I E C"[tl , t2] and +1 > 0 on (to, t,). We shall use Leray-Schauder degree to show that the existence of a solution for A = 0 implies the existence of a solution for 1 = 1; passage to the limit as n --t 00 will yield the existence for (1.1). We let (y( I l/n AYO? (2.2) gfl(y) = g(l/n), IYI < l/n and consider Y" + (1 - 4&,(t)
+ aw&(Y)
= 0,
y(tJ = y(t*) =
l/n.
(2.3),,,
From g, > 0 and & > 0, it follows that y 2’ l/n for any solution y of (2.3),,, , and hence any solution of (2.3),,, is a solution of (2.1&. We set u(t) = y(t) - l/n
(2.4)
Singular
nonlinear
boundary
783
value problems
to get that 21”+ (1 - k)&,(t)
+ Ag,(u + l/n)4(t)
u(t,) = V&) = 0
= 0,
(2.5),,,
therefore, any solution of (2.5),,, satisfies u 1 0. Now we establish the a priori bounds necessary for application of Leray-Schauder degree. LEMMA 1. There
exists a constant
M,, independently
of d E [0, 11, 0 < 6 < 1 and n E R\i
such that (2.6)
Ir(t)l 5 MCl for any solution Y of (2.5),,,.
Proof. Similar to the proof of lemma 1 of [2]. We just give the idea. Let Y,,, be the maximum of Y(t) on [tr , f2J and suppose it occurs at t3. Then Y’(&) = 0. Integrate the inequality
Y” = -(l
- ~)&,(0
- Mt)g,(Y(t)
+ l/n) 2 -(1 - G%(t)
- M(t)g(Y(t))
from t3 to t > t, to obtain t
f
2
2 - (1 - A)6
y’(t)
s t1
41(t) dt - A
s t3
HMY(@) ds
5 L
-(l
- A)6
6(t) dt - MY(t)) j
’ 46) C-IS s
since 4 > 0, y 1 0 and g(y) is nonincreasing. Divide by g(y(t)) and integrate from t, to t2 to get f2
s ‘2
+ (t3 - t,)-’
(t - f&
- tM(t) dt.
t3
A similar argument on [tl , t3] yields
s r3
+
(t2
-
tp
(t
-
t&z
-
tM(t)
dt.
fl
From these two inequalities it follows that Y max
[g(u)]-’ du 5 0
d(t,
-
tl)
(j;;
h(t)
dr)([&w.~l-l) (t - ti)(tz - tM(t) dt,
(t - tr)(fz - t)+(t) (2.7)
784
ZONGMINC Guo
Here
From
(D) it follows
io 12 (t - t,)(tz - tM(t) dt, (t - tr)(t, ii .a I fl
C, = max
P = (t1 + M/2,
- tM(t)
1
that there exists a M,,, such that (2.8)
Ymax 5 MO. LEMMA 2. There exist constants
K > 0, K, > 0, such that
y(t) z (tz - tr)-$1 for any solution
We deduce
- A)6K, + AK](t
- t,)(tz - t)
(2.9)
of (2.5),,,.
Proof. From lemma y” = -(l
.
dt
1, we know that y I IV&,. Therefore, - Aqqt)g(y + l/n)
- n)&,(t)
by integration
- A)&,(t)
I -(l
- Ac#J(f)g(M, + 1).
that
y(t) 2 (tz - tI)-‘Kl
- n)se,(t)
+ MM,
(2.10)
+ l)Qt)ll
where t2 e(t) = (t - t1)
(tz - M(d Ii t ‘Pa f e,(t) = (t - tl) z (t2 - W(d /t From
te [tl,fJ.
f (s - ta#@) b, i fl ”t (s - t,)dh(d h ds + (f2 - t) ?t1
ds + (4 - t)
t
E
[tl, a.
[2], we know that there exists C, > 0, C, > 0 such that
Let K = g(M,
+ l)C,,
@(I) 2 C,(f - tl)(f2 - t),
(2.11)
e,(t) 2 c3(t - tl)(tz - t).
(2.12)
K, = C,, then we get (2.9).
LEMMA 3. There exist positive
constants
MI, M2 such that, when
0 < 6 < min(K/K, , (t2 - t,)/K, for any solutron
, 11,
y(t) of (2.5),,, (2.13)
lu’(t)l 5 Ml 9
(2.14)
k(t)u”(t)l 5 M2, where MI, M, are independent
of A and n; l(t)
= [g(GK,(t, - tl)-‘(t
Proof. We have from lemma 2 that, for any solution Iv”(t)1 5 (1 - ~w,(t)
+ Mt)g(dKl(t,
- tl)(t2 - t))]-‘.
y(t) of (2.5)x,,, - tl)-‘(t - tl)(t2 - t)).
Singular
nonlinear
boundary
785
value problems
By conditions g E C’(O, co), g > 0 and g has singularity at 0, we know that t(t) E C"[t, , t2] and \r(t)u”(t)I 5 (1 - A)s<(t)+,(t) + A+(t) I M2, M, is independent of 13.and n. Let y(t,) = max y(t); then tt1*tz1 f2
41(t) dt t1 tr)-‘((1 - AM,
+ AK)(t - fr)(fz - t)) dt
$(t)g((tz - t,)-‘G&(t
- tr)(tz - t)) dt.
Let +(t)gW&
- t,)-‘(t - tA(tz - t)) dt,
by (C), we get (2.13). The proof of theorem C1[tl, t2]? define
1, We introduce
as in [2]. For u E C2(tl , t2) fl
some notation
ll4lo= ls”IJ] lN)l? lM1 = ~axtl1409 II~‘lloh Ilull = ma4410,IIu’llo,suplWW)O. ct,,t21
Here r(t) = [g(X,(t,
- tJ’(t
IK = (1.4E
- tl)(t2 - t))]-‘. Set
C’(t,, tz) fl C’[t,, t,]: u(tl) = u(t,) = 0 and llu]lz < 00)
with norm II * [I2 and C = (U E c(tl,
t,): sup lu(t)l (fl.fZ)
< Q)]
with the obvious norm L: IK -+ C,
Lu(t) = r(t)u”(t);
j: IK + C’[t,, t2], Fh,n: C’Pl, f21 + c,
F,,,(W)
ju = u;
= MtM(tk,W)
+ l/n) + (1 - W3OW).
From [2] we know that IK and G are Banach spaces, j is completely continuous, tinuous and surely F,,, is continuous. Problem (2.5),,, is equivalent to (I + L-‘F,,.j)y for A E [0, 11, L-‘F,,,j:
IK + IK
= 0,
is a compact homotopy.
L-’ is con-
786
ZONOMINGGuo
Let A4 = max(M,,
+ 2 and Sz = (u E IK 1 I/u112I Ml,
Ml, M2, (6 max 0,(t))]
where M,,
MI,
tt1, t21 AI2
are as in (2.6), (2.13), (2.14). From
above we know that
(Z + Z-‘&&Y Therefore,
properties
of Leray-Schauder
G(Y)(t) = <(t)$(t)g,(Y(t) E SJ, we have
asz.
degree give
deg(Z + L-‘(&!(t)+,(t)), Here M,(t)
YE
# 0,
+ l/n).
Q, 0) = deg(Z + L-‘Gj, Using
deg(Z + Z-‘(R(0&(0),
the
fact
that
Q 0).
(2.15)
L-‘(G<(t)&(t))
a, 0) = deg(Z, Q a&(t))
= -M,(t)
and (2.16)
= 1,
therefore, deg(Z + L-‘Gj, Sz, 0) = 1.
(2.17)
From this, we see that Y” + ti(f)&(Y
+ l/n)
= 0,
has at least one solution in Q, call it u, . The remainder proof of theorem 1 of [2].
Remark 1. For the proof of theorem
1 it is sufficient 6 below we will need a general &(t).
of theorem
If g has no singularity THEOREM
of theorem
at 0, we deduce
(2.18)
v(tr) = Y(f2) = 0
of the proof is exactly the same as the
to take &(t)
the following
= 1. However,
theorem.
2. Let 6 be as in theorem 1, 0 < t, < tz < 00, g satisfy the conditions 1 and g(0) # 03. Then problem (1.1) has a positive solution.
Proof. Similar to the proof of theorem
1. In this case, we can directly
y” + (1 - A)6 + Q(Z)g(y) and condition
(C) of theorem
1 automatically
in the proof
= 0,
consider
(A), (B), (D)
problem
Y(f1) = Y(&) = 0
holds.
Now we consider a special case g(y) = y-“. From [13] we know that when 0 < (Y 5 1, problem y” + +(t)y-” has a solution theorem. THEOREM 3.
= 0,
Y(0) = Y(1) = 0
in C’[O, l] if and only if (1.4) ho1d.s. When
Let 4 > 0 is continuous
(2.19)
01 > 1, we have the following
on (0, 1) and
‘1
(1 - t)-*t-“+(t)dt
< ~0.
(2.20)
10 Then problem
(2.19) has a positive
solution.
Proof. By (2.20), we know that $(t) has no singularities of theorem
1 holds.
Following
the steps of the proof
at 6 = 0 and I = 1 and condition (C) of theorem 1 we prove this theorem.
Singular nonlinear boundary value problems
787
Remark 2. When 4(t) has no singularities at t = 0 and t = 1, then:
(i) if g(y) has singularity at y = 0, then theorem 1 holds for t, = 0, t2 = 1; (ii) if g(y) has no singularity at y = 0, then theorem 2 holds for t, = 0, f2 = 1.
(tr , tz) and 00; g satisfies all the conditions of theorem 1, g is singular at 0. S::(t - fr)(fz - tM(t) dt < Then, the result of theorem 1 is still true. At this time, we only let 4r(t) = 1 and c(t) = [4(t)g(2-‘6(t - tl)(tz - t))]-‘. Now K, 6 as in the proof of theorem 1 are
Remark
3. If 4(t) E C"(f, , t2), I$ has singularities at I = t,, t = t2, 4(t) > 0 on
K = (tz - tl)-‘g(M,
+ l)C,
from [2] we know there exists C, > 0 and 0 < 6 = min[K, 1). If 4(t) has singularity at t = 0 and g has no singularity at y = 0. Then we have the following theorem. THEOREM4. Let 4(t) E C”(O, 11, 4(t) > 0 in (0, l] and 4(t) satisfy j: t+(t) dt < 00; g satisfy: (a) g is continuous and nonincreasing on [0, co); (b) g(y) > 0 on 10, 00); (c) him_g(t) s; k(w’ d.!?= UJ. Then problem Y” + 4(%(Y)
= 0,
Y(0) = Y(l) = 0
(2.21)
has a positive solution. Proof. Similar to the proof of remark 3, at this time, r(t) = [4(t)]-‘. If 4(t) has singularities at t = 0, t = 1 and g has singularity at y = 0, then we have the
following theorem. THEOREM5. Let 4(t) E C’(O, l), 4(t) > 0 in (0, 1) and 4(t) satisfy 1
t(1 - t)4(t) dt < co,
(2.22)
i0 satisfy the conditions of theorem 1 with to = 0, t, = 1. Then problem (1.1) has a positive solution.
g
Proof.
Similar to the proof of remark 3. Now, we give the following theorem 6, which shows that some technical restrictions on g in theorem 2 of [2] are unnecessary. THEOREM6. Suppose that g(t,y) is continuous and positive on (0, 1) x (0, m); g(t, y) is strictly decreasing in y, for y > 0 and t E (0, 1); there exists a nonnegative a(t) E C2[0, l] satisfying: (i) a”(t) + g(t, a(t)) > 0, on (0, l), 01(O)= cr(1) = 0, (ii) 1; g(& a(t)) dt < *, (iii) g(t, y)/g(t, a(t)) is continuous on [0, l] x (0, co).
788
ZONCMING Guo
Then the problem y” + g(t, Y) = 0, has a positive
(2.23)
y(0) = Y(1) = 0
solution.
Proof. Define &([,Y) then g, is strictly decreasing
=
IyI z l/n
g(t, IYI), i g(t, l/n)
+ l/n
IYI < l/n,
- Iyl,
for y 2 0. As in the preceding
y” + (1 - A)g(t, Cl(t) + l/n)
+ Ag,(t, y + l/n)
argument,
we consider
= 0,
the problem
Y(O) = ~(1) = 0
(2.24),,,
and prove the following lemmas which replace lemmas 1 and 3 in the proof of theorem 1. The remainder of the proof goes through with only minor alterations of theorem 1, and so will be omitted. LEMMA 4. Let the hypotheses Then
of theorem
6 hold, and yx = y,,,
denote
a solution
of (2.24),,,
for A E [0, 11, I E [0, 11.
yx(t) 2 a(t)
.
(2.25)
Proof. For n large enough, we claim that a”(f) + g(t, a(t) + l/n) 2 0, for t E (0, 1). Since a”(t) is bounded on [0, I], g(t, a(t)) is unbounded at I = 0 and d = 1, then there exist /? > 0, N, such that a”(t) + g(t, a(t) + l/n) 2 0 for n > N, t E (0, p) U (1 - p, 1). On [/3, 1 - p], a”(t) + g(t, a(t)) is bounded below by a positive constant and hence for n large enough, a”(t) + g(t, a(t) + l/n) I 0. To see that yx 2 a(1), on [0, 11, assume the contrary and let a(t) - yx(t) achieve its maximum positive value at I,. Then a”&)
- y$(t,) I 0.
(2.26)
But, a”(M
- Y&)
2 -g(t,,
&I)
+ l/n)
464 + l/n) + k&
+ (1 - ~)&I,
7Yx(hJ + l/n)
= a!&l , Yx (43)+ l/n) - g(4J9dh) + I/n)]. By a(&,) > yx(t,J, we know a”(&) - y{(to) > 0. This contradicts LEMMA 5. There exist M3 > 0, AI4 > 0 independent
IWYX~)l here t(t)
= [g(a(t))]-‘,
(2.26).
of ,I, n, such that
5 M3 9
and
IlY;llc~5 M4.
(2.27)
Proof. From lemma 4 we know yx(t) 2 a(t), hence, 1y{(t)1 5 2g(t, a(t)) and Mj = 2. Since each yx has a zero derivative somewhere in (0, l), we have 1
IYiWl 5 (1 - 2)
i
l g(t, a(l)
here we use condition
i
g,(& Y, + l/n)
a(t))
dt
1
g,(t, a(t) + l/n)di
+
dt
0
‘1 g(t,
3 0
dt + A
0
‘1 I
+ l/n)
1 0
(ii) and g is strictly
I 2
g(t, a(t)) dt 5 M I 0
decreasing
in y.
789
Singular nonlinear boundary value problems
From
(2.27), we also have
Remark 4. From theorem
3 of [2] we know that if g(y) is strictly decreasing on (0, oo), then the solution of (1.1) is unique. Therefore, the solution of theorem 3 and theorem 6 is unique. Also uniqueness holds in theorems 1, 2, 4 and 5, if g(y) is strictly decreasing. 3. APPLICATIONS
TO EXISTENCE
OF RADIAL
SOLUTIONS
THEOREM 7.
Let g satisfy (a) g is continuous and strictly decreasing (b) g(y) > 0 on (0, ~0) and g(0) # m; .
Cc) pysct,
s:,MW
on (0, 00);
ds = 00.
Then problem (1.2) has a unique 0 < R, < R, < ~0.
positive
Proof. This can be easily obtained
radial solution
from theorem
y(r) E Cz(R2, R,) n C’[R2, RI], where
2 and remark
4.
THEOREM 8.
Let 0 < (115 1, g(y) = y-“. Then problem (1.2) has a unique tion y(r) E C’(R,, R,) n C’[R,, R,], where 0 < R2 < R, < a.
Proof. This can be obtained THEOREM 9.
from theorem
1 and remark
Let CY> 0; R, > 0, r(r) E C”[R3, 00); r(r) >
positive
radial solu-
4.
0 on tR3,a) and
1f3 (t3 - t)-*t-*&(t)[(N
- 2)t]-(ZN-2)‘(N-2)
dt < ~0.
!0 Then,
problem Ay + [(r)ywu = 0,
r>R,,Nr3 on r = R,,
( Y = 0, has a unique Where
positive
t = [(N - 2)rN-‘]-‘;
radial
solution
r = 1x1;
Proof. This can be obtained THEOREM
t, = [(N - 2)Ry-‘1-l;
10. Let R3, r, r(r), t3 be as in theorem C,(t)[(N i
Let g (a) (b) (c)
y(r) E C2(R3, co) c7 C’[R,, a~) that tends to 0 as r -+ a.
from [13, theorem
t3 0
(3.1)
- Wl-
satisfy: g is continuous and strictly decreasing g(y) > 0 on (0, 00); g(0) f 00; ,“y g(t) s:, kw -’ ds = co.
31 and remark 9; rl(t)
t,(t) = ((((N - 2)t)-1”N-2’). 4.
satisfy
W’-W(N-2) dt < o3_
on (0, co);
(3.2)
ZONGMING Guo
790
Then,
problem AY + &-My)
= 0,
y=o has a unique
positive
radial
solution
Proof. This can be obtained
r>R,,Nl3 on r = R,
(3.3)
in C'(R, , CD)fl C'[R, , co) that tends to 0 as r + 00.
from theorem
4 and remark
4.
THEOREM 11. Let R,, t, rl(t), t3 be as in theorem 9; T(r) E C’(R,, co) and r(r) > 0 on (0, co). Letf(t) = c,(t)[(N - 2)t]-(2N-2)‘(N-2) andf(t) satisfy the conditions of qS(t) of theorem 5; g(y) satisfy the conditions of theorem 5 and be strictly decreasing. Then, problem (3.3) has a unique positive radial solution in C2(R3, co) II C’[R,, m) that tends to 0 as r + 00.
Proof. This can be obtained Acknowledgements-The sions. We are especially
author grateful
from theorem
5 and remark
4.
would like to thank Professor J. R. L. Webb for encouragement to the referee for valuable suggestions and corrections.
and helpful
discus-
REFERENCES 1. BANDLE C., COFFMAN C. V. & MARCUS M., Nonlinear elliptic problems in annular domains, J. di.J Eqns 69, 322-345 (1987). of some nonlinear boundary value problems, 2. BOBISUD L. E., O’REAGAN D. & ROYALTY W. D., Solvability Nonlinear Analysis 12, 855-869 (1988). 3. BREZIS H. & NIRENBERG L., Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, Communs pure appl. Math. 36, 431-477 (1983). equations arising in boundary layer theory, 4. CALLEGARI A. & NACHMAN A., Some singular nonlinear differential J. math. Analysis Applic. 64, 96-105 (1978). equa5. COFFMAN C. V., On the positive solutions of boundary value problems for a class of nonlinear differential tions, J. di’J Eqns 3, 92-111 (1967). 6. GIDAS B., NI W. M. & NIRENBERG L., Symmetry and related properties in almost spherically symmetric domains, Archs ration Mech. Analysis 96, 167-197 (1986). to singular nonlinear boundary value problems, J. dij.. I. KURTZ J. C., Weighted Sobolev spaces with applications Eqns 49, 105-123 (1983). boundary value 8. LUNING C. D. & PERRY W. L., Positive solutions of negative exponent generalized Emden-Fowler problems, SIAM J. math. Analysis 12, 874-879 (1981). eauations. Trans. Am. math. Sot. 95, 101-123 (1960). 9. NEHARI Z.. On a class of second order differential of solutions of nonlinear Dirihlet problems, J. d$f. Eqns 50, 289-304 (1983). 10. NI W. M.,‘Uniqueness and non-uniqueness for positive radial solutions of Au + f(u, r) = 0, 11. NI W. M. & NUSSBAUM R., Uniqueness Communspure app/. math. 38, 67-108 (1985). Second-order boundary value problems, J. math. 12. O’REGAN D., Positive solutions to singular and non-singular Analysis Applic. 142, 40-52 (1989). value problem, Nonlinear Analysis 3, 897-904 (1979). 13. TALIAFERRO S. D., A nonlinear singular boundary equations, SIAM Review 17, 339-360 (1975). 14. WONG J. S. W., On the generalized Emden-Fowler 15. ZONGMING Guo, Existence of positive radial solutions for a class of nonlinear singular elliptic problems in annular domains (preprint).
& Apphcations,
Vol. 16, No. 9, pp. 781-790, 1991.
0
0362-546X/91 $3.00+ .M) 1991 Pergamon Press plc
.
SOLVABILITY OF SOME SINGULAR NONLINEAR BOUNDARY VALUE PROBLEMS AND EXISTENCE OF POSITIVE RADIAL SOLUTIONS OF SOME NONLINEAR ELLIPTIC PROBLEMS ZONGMING Department
of Mathematics,
(Received 21 February
University
Guo*
of Glasgow,
1990; received in revised form
University
Gardens,
Glasgow
G12 SQW, U.K.
1 August 1990; received for publication 20 September
Key words and phrases: Singular boundary value nonlinear elliptic problems, Leray-Schauder degree.
problems,
nonlinear,
positive
radial
1990)
solutions,
1. INTRODUCTION
IN THIS paper we study the existence and uniqueness of positive solutions of singular secondorder differential equations of the form
l E (ti >&I,
Y” + Ht)g(JJ) = 0,
(1.1)
Y(h) = Y(b) = 0,
where 0 I t, < t2 < + co; 4(t) E C’(t,, tz), 4(t) > 0 on (ti, tz), C#Iis allowed to have suitable singularities at t = tl and t = f2; g E C’(O, oo), g > 0 on (0, w) and g is also allowed to be singular at 0. Equations of this type arise in diffusion and osmotic flow theory, one particular example being the generalized Emden-Fowler equation (see [8, 141). They also arise from the study of positive radial solutions y(r), r = 1x1,x E RRN,N 2 3, in R, < r < R, and on r = R2
AY + g(y) = 0, i y=o
r=
(1 .a
R,.
Putting t = [(N - 2)rN-2]-’ k = (2N - 2)/(N - 2)
4(t) = ](N - 2)Kk,
Ri = [(N - 2)ti]-1’(N-2),
(1.3)
i = 1,2,
(where we put t, = 0 when R, = 03) gives equations of type (1.1). Problems of the type like (1.2) have been studied by numerous authors (see [ 1, 3, 5-7, 9-l 11) when g(y) is nondecreasing in (0, m). In this paper, we shall give some results for positive radial solutions of (1.2) when g(y) is decreasing. By a solution of (1.1) we mean a function y E C’[tr , f2] n C2(t, , t2) that satisfies (1.1). The existence of solutions for problem (1.1) has been considered by Callegari and Nachman [4], who effectively consider (1.1) with 4(t) = [min(t, 1 - t)], g(y) = y-l and t, = 0, t2 = 1; by Luning and Perry [8], who establish constructive results for g(y) = y-” when 0 < 01_= 1 and t, = 0, t, = 1; and by Taliaferro [13], who proves, in particular, that a necessary and sufficient *On leave from the Department
of Mathematics,
Henan
Normal 781
University,
Xinxiang,
People’s
Republic
of China.
782
ZONGMING Guo
condition for existence of solutions with first derivative continuous on [0, l] is that !
tP(l
- t)-“6(f)
dt < + co.
(1.4)
i0
In Section 2, we use the Leray-Schauder degree to seek positive solutions of (1.1). We will extend some of the results of [2, 121 by allowing a more general g(y) and at the same time generalize the sufficient condition of [13] to problems with a more general g(y). In [15], the author discusses problem (1.1) with nonlinear term g(t, y, y’). In Section 3, we apply the results obtained in Section 2 to (1.2) and obtain the e&istence of a positive radial solution of (1.2). 2. EXISTENCE
OF A POSITIVE
SOLUTION
OF (1.1)
1. Suppose that 6(t) E C’[t, , t,], 4(t) > 0 in (tl, f2), where 0 < t, < tz < co; g satisfies (A) g is continuous and nonincreasing on (0, co), (B) g(u) > 0 on (0, 00)~ forO< k-c 1, (C) s:: Ns)g(k(s - tl)(tz - s)) dJ < + =J, (D) lim g(t) 1; [g(s)]-’ ds = 00.
THEOREM
Then dro%em (1.1) has a positive solution. To prove this theorem, we consider two cases: (a) g has singularity at 0; and (b) g has no singularity at 0. We only prove theorem 1 in case (a), the proof of case (b) is similar to case (a). In order to avoid the possible singularity of g at 0, we consider, for each n E N = i 1,2, -. .) and 1 E [0, 11, the family of problems:
Y” + (1 - I)&,
+ Mt)g(Y) = 0,
v(tJ = Y(b) = l/n,
(2.lh,n
where 0 < 6 < 1 is a positive real number which is determined below, @I E C"[tl , t2] and +1 > 0 on (to, t,). We shall use Leray-Schauder degree to show that the existence of a solution for A = 0 implies the existence of a solution for 1 = 1; passage to the limit as n --t 00 will yield the existence for (1.1). We let (y( I l/n AYO? (2.2) gfl(y) = g(l/n), IYI < l/n and consider Y" + (1 - 4&,(t)
+ aw&(Y)
= 0,
y(tJ = y(t*) =
l/n.
(2.3),,,
From g, > 0 and & > 0, it follows that y 2’ l/n for any solution y of (2.3),,, , and hence any solution of (2.3),,, is a solution of (2.1&. We set u(t) = y(t) - l/n
(2.4)
Singular
nonlinear
boundary
783
value problems
to get that 21”+ (1 - k)&,(t)
+ Ag,(u + l/n)4(t)
u(t,) = V&) = 0
= 0,
(2.5),,,
therefore, any solution of (2.5),,, satisfies u 1 0. Now we establish the a priori bounds necessary for application of Leray-Schauder degree. LEMMA 1. There
exists a constant
M,, independently
of d E [0, 11, 0 < 6 < 1 and n E R\i
such that (2.6)
Ir(t)l 5 MCl for any solution Y of (2.5),,,.
Proof. Similar to the proof of lemma 1 of [2]. We just give the idea. Let Y,,, be the maximum of Y(t) on [tr , f2J and suppose it occurs at t3. Then Y’(&) = 0. Integrate the inequality
Y” = -(l
- ~)&,(0
- Mt)g,(Y(t)
+ l/n) 2 -(1 - G%(t)
- M(t)g(Y(t))
from t3 to t > t, to obtain t
f
2
2 - (1 - A)6
y’(t)
s t1
41(t) dt - A
s t3
HMY(@) ds
5 L
-(l
- A)6
6(t) dt - MY(t)) j
’ 46) C-IS s
since 4 > 0, y 1 0 and g(y) is nonincreasing. Divide by g(y(t)) and integrate from t, to t2 to get f2
s ‘2
+ (t3 - t,)-’
(t - f&
- tM(t) dt.
t3
A similar argument on [tl , t3] yields
s r3
+
(t2
-
tp
(t
-
t&z
-
tM(t)
dt.
fl
From these two inequalities it follows that Y max
[g(u)]-’ du 5 0
d(t,
-
tl)
(j;;
h(t)
dr)([&w.~l-l) (t - ti)(tz - tM(t) dt,
(t - tr)(fz - t)+(t) (2.7)
784
ZONGMINC Guo
Here
From
(D) it follows
io 12 (t - t,)(tz - tM(t) dt, (t - tr)(t, ii .a I fl
C, = max
P = (t1 + M/2,
- tM(t)
1
that there exists a M,,, such that (2.8)
Ymax 5 MO. LEMMA 2. There exist constants
K > 0, K, > 0, such that
y(t) z (tz - tr)-$1 for any solution
We deduce
- A)6K, + AK](t
- t,)(tz - t)
(2.9)
of (2.5),,,.
Proof. From lemma y” = -(l
.
dt
1, we know that y I IV&,. Therefore, - Aqqt)g(y + l/n)
- n)&,(t)
by integration
- A)&,(t)
I -(l
- Ac#J(f)g(M, + 1).
that
y(t) 2 (tz - tI)-‘Kl
- n)se,(t)
+ MM,
(2.10)
+ l)Qt)ll
where t2 e(t) = (t - t1)
(tz - M(d Ii t ‘Pa f e,(t) = (t - tl) z (t2 - W(d /t From
te [tl,fJ.
f (s - ta#@) b, i fl ”t (s - t,)dh(d h ds + (f2 - t) ?t1
ds + (4 - t)
t
E
[tl, a.
[2], we know that there exists C, > 0, C, > 0 such that
Let K = g(M,
+ l)C,,
@(I) 2 C,(f - tl)(f2 - t),
(2.11)
e,(t) 2 c3(t - tl)(tz - t).
(2.12)
K, = C,, then we get (2.9).
LEMMA 3. There exist positive
constants
MI, M2 such that, when
0 < 6 < min(K/K, , (t2 - t,)/K, for any solutron
, 11,
y(t) of (2.5),,, (2.13)
lu’(t)l 5 Ml 9
(2.14)
k(t)u”(t)l 5 M2, where MI, M, are independent
of A and n; l(t)
= [g(GK,(t, - tl)-‘(t
Proof. We have from lemma 2 that, for any solution Iv”(t)1 5 (1 - ~w,(t)
+ Mt)g(dKl(t,
- tl)(t2 - t))]-‘.
y(t) of (2.5)x,,, - tl)-‘(t - tl)(t2 - t)).
Singular
nonlinear
boundary
785
value problems
By conditions g E C’(O, co), g > 0 and g has singularity at 0, we know that t(t) E C"[t, , t2] and \r(t)u”(t)I 5 (1 - A)s<(t)+,(t) + A+(t) I M2, M, is independent of 13.and n. Let y(t,) = max y(t); then tt1*tz1 f2
41(t) dt t1 tr)-‘((1 - AM,
+ AK)(t - fr)(fz - t)) dt
$(t)g((tz - t,)-‘G&(t
- tr)(tz - t)) dt.
Let +(t)gW&
- t,)-‘(t - tA(tz - t)) dt,
by (C), we get (2.13). The proof of theorem C1[tl, t2]? define
1, We introduce
as in [2]. For u E C2(tl , t2) fl
some notation
ll4lo= ls”IJ] lN)l? lM1 = ~axtl1409 II~‘lloh Ilull = ma4410,IIu’llo,suplWW)O. ct,,t21
Here r(t) = [g(X,(t,
- tJ’(t
IK = (1.4E
- tl)(t2 - t))]-‘. Set
C’(t,, tz) fl C’[t,, t,]: u(tl) = u(t,) = 0 and llu]lz < 00)
with norm II * [I2 and C = (U E c(tl,
t,): sup lu(t)l (fl.fZ)
< Q)]
with the obvious norm L: IK -+ C,
Lu(t) = r(t)u”(t);
j: IK + C’[t,, t2], Fh,n: C’Pl, f21 + c,
F,,,(W)
ju = u;
= MtM(tk,W)
+ l/n) + (1 - W3OW).
From [2] we know that IK and G are Banach spaces, j is completely continuous, tinuous and surely F,,, is continuous. Problem (2.5),,, is equivalent to (I + L-‘F,,.j)y for A E [0, 11, L-‘F,,,j:
IK + IK
= 0,
is a compact homotopy.
L-’ is con-
786
ZONOMINGGuo
Let A4 = max(M,,
+ 2 and Sz = (u E IK 1 I/u112I Ml,
Ml, M2, (6 max 0,(t))]
where M,,
MI,
tt1, t21 AI2
are as in (2.6), (2.13), (2.14). From
above we know that
(Z + Z-‘&&Y Therefore,
properties
of Leray-Schauder
G(Y)(t) = <(t)$(t)g,(Y(t) E SJ, we have
asz.
degree give
deg(Z + L-‘(&!(t)+,(t)), Here M,(t)
YE
# 0,
+ l/n).
Q, 0) = deg(Z + L-‘Gj, Using
deg(Z + Z-‘(R(0&(0),
the
fact
that
Q 0).
(2.15)
L-‘(G<(t)&(t))
a, 0) = deg(Z, Q a&(t))
= -M,(t)
and (2.16)
= 1,
therefore, deg(Z + L-‘Gj, Sz, 0) = 1.
(2.17)
From this, we see that Y” + ti(f)&(Y
+ l/n)
= 0,
has at least one solution in Q, call it u, . The remainder proof of theorem 1 of [2].
Remark 1. For the proof of theorem
1 it is sufficient 6 below we will need a general &(t).
of theorem
If g has no singularity THEOREM
of theorem
at 0, we deduce
(2.18)
v(tr) = Y(f2) = 0
of the proof is exactly the same as the
to take &(t)
the following
= 1. However,
theorem.
2. Let 6 be as in theorem 1, 0 < t, < tz < 00, g satisfy the conditions 1 and g(0) # 03. Then problem (1.1) has a positive solution.
Proof. Similar to the proof of theorem
1. In this case, we can directly
y” + (1 - A)6 + Q(Z)g(y) and condition
(C) of theorem
1 automatically
in the proof
= 0,
consider
(A), (B), (D)
problem
Y(f1) = Y(&) = 0
holds.
Now we consider a special case g(y) = y-“. From [13] we know that when 0 < (Y 5 1, problem y” + +(t)y-” has a solution theorem. THEOREM 3.
= 0,
Y(0) = Y(1) = 0
in C’[O, l] if and only if (1.4) ho1d.s. When
Let 4 > 0 is continuous
(2.19)
01 > 1, we have the following
on (0, 1) and
‘1
(1 - t)-*t-“+(t)dt
< ~0.
(2.20)
10 Then problem
(2.19) has a positive
solution.
Proof. By (2.20), we know that $(t) has no singularities of theorem
1 holds.
Following
the steps of the proof
at 6 = 0 and I = 1 and condition (C) of theorem 1 we prove this theorem.
Singular nonlinear boundary value problems
787
Remark 2. When 4(t) has no singularities at t = 0 and t = 1, then:
(i) if g(y) has singularity at y = 0, then theorem 1 holds for t, = 0, t2 = 1; (ii) if g(y) has no singularity at y = 0, then theorem 2 holds for t, = 0, f2 = 1.
(tr , tz) and 00; g satisfies all the conditions of theorem 1, g is singular at 0. S::(t - fr)(fz - tM(t) dt < Then, the result of theorem 1 is still true. At this time, we only let 4r(t) = 1 and c(t) = [4(t)g(2-‘6(t - tl)(tz - t))]-‘. Now K, 6 as in the proof of theorem 1 are
Remark
3. If 4(t) E C"(f, , t2), I$ has singularities at I = t,, t = t2, 4(t) > 0 on
K = (tz - tl)-‘g(M,
+ l)C,
from [2] we know there exists C, > 0 and 0 < 6 = min[K, 1). If 4(t) has singularity at t = 0 and g has no singularity at y = 0. Then we have the following theorem. THEOREM4. Let 4(t) E C”(O, 11, 4(t) > 0 in (0, l] and 4(t) satisfy j: t+(t) dt < 00; g satisfy: (a) g is continuous and nonincreasing on [0, co); (b) g(y) > 0 on 10, 00); (c) him_g(t) s; k(w’ d.!?= UJ. Then problem Y” + 4(%(Y)
= 0,
Y(0) = Y(l) = 0
(2.21)
has a positive solution. Proof. Similar to the proof of remark 3, at this time, r(t) = [4(t)]-‘. If 4(t) has singularities at t = 0, t = 1 and g has singularity at y = 0, then we have the
following theorem. THEOREM5. Let 4(t) E C’(O, l), 4(t) > 0 in (0, 1) and 4(t) satisfy 1
t(1 - t)4(t) dt < co,
(2.22)
i0 satisfy the conditions of theorem 1 with to = 0, t, = 1. Then problem (1.1) has a positive solution.
g
Proof.
Similar to the proof of remark 3. Now, we give the following theorem 6, which shows that some technical restrictions on g in theorem 2 of [2] are unnecessary. THEOREM6. Suppose that g(t,y) is continuous and positive on (0, 1) x (0, m); g(t, y) is strictly decreasing in y, for y > 0 and t E (0, 1); there exists a nonnegative a(t) E C2[0, l] satisfying: (i) a”(t) + g(t, a(t)) > 0, on (0, l), 01(O)= cr(1) = 0, (ii) 1; g(& a(t)) dt < *, (iii) g(t, y)/g(t, a(t)) is continuous on [0, l] x (0, co).
788
ZONCMING Guo
Then the problem y” + g(t, Y) = 0, has a positive
(2.23)
y(0) = Y(1) = 0
solution.
Proof. Define &([,Y) then g, is strictly decreasing
=
IyI z l/n
g(t, IYI), i g(t, l/n)
+ l/n
IYI < l/n,
- Iyl,
for y 2 0. As in the preceding
y” + (1 - A)g(t, Cl(t) + l/n)
+ Ag,(t, y + l/n)
argument,
we consider
= 0,
the problem
Y(O) = ~(1) = 0
(2.24),,,
and prove the following lemmas which replace lemmas 1 and 3 in the proof of theorem 1. The remainder of the proof goes through with only minor alterations of theorem 1, and so will be omitted. LEMMA 4. Let the hypotheses Then
of theorem
6 hold, and yx = y,,,
denote
a solution
of (2.24),,,
for A E [0, 11, I E [0, 11.
yx(t) 2 a(t)
.
(2.25)
Proof. For n large enough, we claim that a”(f) + g(t, a(t) + l/n) 2 0, for t E (0, 1). Since a”(t) is bounded on [0, I], g(t, a(t)) is unbounded at I = 0 and d = 1, then there exist /? > 0, N, such that a”(t) + g(t, a(t) + l/n) 2 0 for n > N, t E (0, p) U (1 - p, 1). On [/3, 1 - p], a”(t) + g(t, a(t)) is bounded below by a positive constant and hence for n large enough, a”(t) + g(t, a(t) + l/n) I 0. To see that yx 2 a(1), on [0, 11, assume the contrary and let a(t) - yx(t) achieve its maximum positive value at I,. Then a”&)
- y$(t,) I 0.
(2.26)
But, a”(M
- Y&)
2 -g(t,,
&I)
+ l/n)
464 + l/n) + k&
+ (1 - ~)&I,
7Yx(hJ + l/n)
= a!&l , Yx (43)+ l/n) - g(4J9dh) + I/n)]. By a(&,) > yx(t,J, we know a”(&) - y{(to) > 0. This contradicts LEMMA 5. There exist M3 > 0, AI4 > 0 independent
IWYX~)l here t(t)
= [g(a(t))]-‘,
(2.26).
of ,I, n, such that
5 M3 9
and
IlY;llc~5 M4.
(2.27)
Proof. From lemma 4 we know yx(t) 2 a(t), hence, 1y{(t)1 5 2g(t, a(t)) and Mj = 2. Since each yx has a zero derivative somewhere in (0, l), we have 1
IYiWl 5 (1 - 2)
i
l g(t, a(l)
here we use condition
i
g,(& Y, + l/n)
a(t))
dt
1
g,(t, a(t) + l/n)di
+
dt
0
‘1 g(t,
3 0
dt + A
0
‘1 I
+ l/n)
1 0
(ii) and g is strictly
I 2
g(t, a(t)) dt 5 M I 0
decreasing
in y.
789
Singular nonlinear boundary value problems
From
(2.27), we also have
Remark 4. From theorem
3 of [2] we know that if g(y) is strictly decreasing on (0, oo), then the solution of (1.1) is unique. Therefore, the solution of theorem 3 and theorem 6 is unique. Also uniqueness holds in theorems 1, 2, 4 and 5, if g(y) is strictly decreasing. 3. APPLICATIONS
TO EXISTENCE
OF RADIAL
SOLUTIONS
THEOREM 7.
Let g satisfy (a) g is continuous and strictly decreasing (b) g(y) > 0 on (0, ~0) and g(0) # m; .
Cc) pysct,
s:,MW
on (0, 00);
ds = 00.
Then problem (1.2) has a unique 0 < R, < R, < ~0.
positive
Proof. This can be easily obtained
radial solution
from theorem
y(r) E Cz(R2, R,) n C’[R2, RI], where
2 and remark
4.
THEOREM 8.
Let 0 < (115 1, g(y) = y-“. Then problem (1.2) has a unique tion y(r) E C’(R,, R,) n C’[R,, R,], where 0 < R2 < R, < a.
Proof. This can be obtained THEOREM 9.
from theorem
1 and remark
Let CY> 0; R, > 0, r(r) E C”[R3, 00); r(r) >
positive
radial solu-
4.
0 on tR3,a) and
1f3 (t3 - t)-*t-*&(t)[(N
- 2)t]-(ZN-2)‘(N-2)
dt < ~0.
!0 Then,
problem Ay + [(r)ywu = 0,
r>R,,Nr3 on r = R,,
( Y = 0, has a unique Where
positive
t = [(N - 2)rN-‘]-‘;
radial
solution
r = 1x1;
Proof. This can be obtained THEOREM
t, = [(N - 2)Ry-‘1-l;
10. Let R3, r, r(r), t3 be as in theorem C,(t)[(N i
Let g (a) (b) (c)
y(r) E C2(R3, co) c7 C’[R,, a~) that tends to 0 as r -+ a.
from [13, theorem
t3 0
(3.1)
- Wl-
satisfy: g is continuous and strictly decreasing g(y) > 0 on (0, 00); g(0) f 00; ,“y g(t) s:, kw -’ ds = co.
31 and remark 9; rl(t)
t,(t) = ((((N - 2)t)-1”N-2’). 4.
satisfy
W’-W(N-2) dt < o3_
on (0, co);
(3.2)
ZONGMING Guo
790
Then,
problem AY + &-My)
= 0,
y=o has a unique
positive
radial
solution
Proof. This can be obtained
r>R,,Nl3 on r = R,
(3.3)
in C'(R, , CD)fl C'[R, , co) that tends to 0 as r + 00.
from theorem
4 and remark
4.
THEOREM 11. Let R,, t, rl(t), t3 be as in theorem 9; T(r) E C’(R,, co) and r(r) > 0 on (0, co). Letf(t) = c,(t)[(N - 2)t]-(2N-2)‘(N-2) andf(t) satisfy the conditions of qS(t) of theorem 5; g(y) satisfy the conditions of theorem 5 and be strictly decreasing. Then, problem (3.3) has a unique positive radial solution in C2(R3, co) II C’[R,, m) that tends to 0 as r + 00.
Proof. This can be obtained Acknowledgements-The sions. We are especially
author grateful
from theorem
5 and remark
4.
would like to thank Professor J. R. L. Webb for encouragement to the referee for valuable suggestions and corrections.
and helpful
discus-
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