(1.1) for n > 2 (resonance at an eigenvalue greater than the first one). Considering the linear ... (a) There exist 7, C G R, 0 < 7 < (n + l)2 - n2, such that. \g(u)\ < -)\u\ +C,. uGR. .... exists a subset of (0, tt) with positive measure in which p(x) sin+ nx > n2 sin+ nx or i(x)sin~ nx ..... growth for a class of second order elliptic boundary ...
PROCEEDINGS OF THE AMERICAN MATHEMATICAL Volume
97, Number
SOCIETY
1, May 1986
EXISTENCE RESULTS ON THE ONE-DIMENSIONAL DIRICHLET PROBLEM SUGGESTED BY THE PIECEWISE LINEAR CASE M. ARIAS1 ABSTRACT. We study the existence of solutions of a two-point boundary value problem at resonance in which the nonlinearity grows at most linearly. Sharp results for the linear growth of the nonlinearity in each direction are obtained.
1. Introduction. We study the existence of solutions in the sense of Carathéodory of the nonlinear boundary value problem at resonance
(1.1)
u" +n2u + g(u) = f(x),
u(0) = u(m) = 0,
where g: R —>R is a continuous function which may grow linearly, / G L2(0,7r) and n is an integer number. One of the first results related to this problem, but in the nonresonant case, was
due to Loud [8] (see also [7]). When n = 1 (resonance at the first eigenvalue), (1.1) has been studied by many authors (see, for instance, [1, 3 and 9]), but few (see [4]) have studied problem (1.1) for n > 2 (resonance at an eigenvalue greater than the first one). Considering the linear problem, i.e., g(u) = 71t in (1.1), it is known that (1.1) is solvable for each / € L2(0, n) provided that 0 < 7 < (n + l)2 - n2. Then, one can hope that if the nonlinearity g behaves like qu with 0 < ^ < (n + 1)2 —n2, the result holds in the nonlinear case. In this line we obtain, as a consequence of our main theorem, the following result:
THEOREM 1. Assume that (a) There exist 7, C G R, 0 < 7 < (n + l)2 - n2, such that
\g(u)\ < -)\u\ +C,
uGR.
(b) For n > 1 there exist s > 0 and L G R such that
g(u) < g(v) + L,
if v —u > s.
(c) g(—oo) I
Jo
sin+nxdx
—n, ag(u)>-(v2-n2)\u\+k,
u 1, (n + 2)/2p + n/2u > 1 if n is even, (ii) (n + l)/2p + (n + l)¡2v > 1 if n is odd. Then (1.1) has at least one solution.
Note that if p = v = 7, condition (a') is condition (a) of Theorem 1. Condition (a') is also sharp, as we will show in the final remarks and, in contrast to condition (a), allows g to have a very large linear growth in the positive direction provided that it has sufficiently small growth in the negative one, and vice versa. Our proof is based on establishing a priori bounds for possible solutions of (1.1) and takes some ideas from [1], where Ahmad, by comparison with the linear problem, proves Theorem 1 in the case n = 1. The author wishes to thank Professor R. Kannan for suggesting the problem to her and Professors R. Ortega and P. Martínez Amores for their valuable comments and suggestions about this work. 2. The piecewise
(2.1)
linear
case.
Consider the piecewise linear problem
u" + p2u+ - v2u~ = f,
u(0) = u(n) = 0.
This problem has been extensively studied by Dancer [4]. There he proves that (2.1) has a nontrivial solution for / = 0 if and only if OO
(p, u) G A0 =
r /
U \^u
k=i L *•
p
(p,v)^R2:—
V
p
+ - = l v
lit*,")
U -i (u, u) G R¿ : - + p
V
Moreover, he proves that if (p, v) G Ax, where Ax is the shaded set in Figure 1, (2.1) has a solution for an arbitrary / G L2(0, it), whereas if (p, u) G R2 - Ax there exists / G C°°[0, it] such that (2.1) has no solution.
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EXISTENCE
RESULTS ON THE DIRICHLET
PROBLEM
123
Figure i These results suggest the following
LEMMA 1. Let p,v G L2(0, tt). Suppose there exist nonnegative constants p,v satisfying (i) and (ii) of hypothesis (a') of Theorem 2 and such that n2 < p(x) < p2, a.e. (0,tt), n2 < v(x) < u2, a.e. (0,tt). Then the problem u" + p(x)u+ - ù(x)vr
= 0
a.e. (0,m),
u(0) = u(tt) = 0, has a nontrivial ,
.
solution if and only if (asinnx)+p(x) (trsinnx)~¿'(a;)
= (crsinnx)+n2 = (fjsinna;)_n2
a.e. (0,7r), a.e. (0,7r),
where o = ±1. In this case if is a nontrivial solution of (2.2), 0 or 0. The other case is analogous. By the Sturm comparison theorem (S.C.T), there exists ti G [m/p, ir/n] such that "+ p(x)4> = 0,
a.e. (0,ii), '(h) < 0. Again by S.T.C. there exists t2 G
[n/p + tt/v, 2-rr/n]such that 0" + v(x)(ti)—fifa) —0. In this way we found that there exists tn G {mr/2p + mr/2i', it] such that = 0, a.e. (ín,ín+i),
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(tn) = 4>(tn+x) = 0,
M. ARIAS
124
but this is impossible because, by S.CT., -.1
,
«n+1 > -
p
«. (n + 2)7T
+ ¿n >
0
2p
nm + TT > 7r-
2v
On the other hand, if (2.3) is verified, 0, is a nontrivial solution of (2.2). By a similar argument we can prove that these are the only
solutions of (2.2) when (2.3) holds. 3. Proof of Theorem
2. We need
LEMMA 2. Let g: R —»R be a continuous function satisfying hypothesis (b) of Theorem 1 (for n—1, suppose there exist a,ß G R such that g(u) > a, u > 0 and g(u) < ß, u < 0), and let {u¿}¿ejv be a sequence in C1[0,7r] such that ||u¿||—>¿_,ooOO and Ui/\\ui\\ —►crsin(n-), o — ±1, in Cx[0,7r] (i.e., uí(x)/\\uí\\ —»osmnx and u'¿(x)/||u¿|| —»nacosnx uniformly in [0, tt]). TTien
lim inf o /
g(+oo)
/
7o
(trsinn)+xdx
- g(—oo) /
Jo
(asinn)
xdx.
(Here, \\u\\ = maxie[0i7r] \u(x)\.)
PROOF. For simplicity we only consider the cases n = 1 and n = 2. Suppose (7=1. If 0 there exists ie G N such that
