On a k-dimensional nonlocal boundary value
problem at resonance
Katarzyna Szyma´
nska-D¸ebowska
Abstract. In this paper we study the existence of at least one solution
to the following system of nonlocal resonant boundary value problem
Z 1
x00 = f (t, x), x0 (0) = 0, x0 (1) =
x0 (s)dg(s),
0
k
k
k
where f : [0, 1] × R → R , g : [0, 1] → R .
Mathematics Subject Classification (2010). Primary 34B10; Secondary
34B15.
Keywords. nonlocal boundary value problem; boundary value problem
at resonance; the perturbation method; Neumann BVP.
1. Introduction
In the paper the following system of ordinary differential equations
Z 1
x00 = f (t, x), x0 (0) = 0, x0 (1) =
x0 (s)dg(s),
(1.1)
0
where f = (f1 , . . . , fk ) : [0, 1] × Rk → Rk is continuous, g = (g1 , . . . , gk ) :
[0, 1] → Rk has bounded variation, is studied.
Speaking precisely, (1.1) is the system of k BVPs
 00
 xi (t) = fi (t, x(t)),
x0 (0) = 0,
R1
 i0
xi (1) = 0 x0i (s)dgi (s),
R1
where t ∈ [0, 1], i = 1, . . . , k and the integrals 0 x0i (s)dgi (s) are meant in the
sense of Riemann-Stieltjes.
Our main goal is to show that the problem (1.1) has at least one solution.
We impose on the function f a sign condition, which we called: the asymptotic
2
Katarzyna Szyma´
nska-D¸ebowska
integral sign condition. The idea comes from [21], where the author shows
that the following problem
x0 = f (t, x),
x(0) = 0,
x(T ) = 0,
has periodic solutions. The method can be successfully applied to other BVPs
(not necessarily only for differential equations of the first or second order
but, for instance, involving p-Laplacians), for which the function f does not
depend on x0 .
As far as we are aware, the BVP (1.1) has not been studied in this
generality so far. Note that a special case of (1.1) is the Neumann BVP
x00 = f (t, x),
x0 (0) = 0,
x0 (1) = 0.
Until now, under suitable monotonicity conditions or nonresonance conditions, some existence or uniqueness theorems or methods for Neumann BVPs
have been presented (see, for instance, [1, 4, 18, 23, 22, 24, 26, 27, 25] and
the references therein).
For example, in [10], the authors study the Neumann boundary value
problem x00 + µ(t)x+ − ν(t)x− = p(t, x), x0 (0) = 0 = x0 (π), where µ, ν lie
in L1 (0, π), p(t, x) is a Carathodory function, p ≥ 0, x+ (t) = max(x(t), 0),
and x− (t) = max(−x(t), 0). They obtain several necessary and sufficient
conditions on p so that the Neumann problem has a positive solution or a
solution with a simple zero in (0, π).
In [11], the author uses phase plane and asymptotic techniques to discuss the number of solutions of the problems −x00 = f (t, x), x0 (0) = σ1 ,
x0 (π) = σ1 . It is assumed that f : [0, π] × R → R is a continuous jumping
nonlinearity with nonnegative asymptotic limits: x−1 f (t, x) → α as x → −∞
and x−1 f (t, x) → β as x → ∞. The limit problem where f (t, x) = αx− +βx+
plays a key role in his methods. The authors describe how the number of solutions of the problem depends on the four parameters: α, β, σ1 , σ2 . The results
differ from those obtained by various authors who were mainly concerned with
forcing the equation with large positive functions and keeping the boundary
conditions homogeneous.
BVP
x00 = f (t, x, x0 ), x0 (0) = 0, x0 (1) = 0,
is considered in [8]. The authors obtain some results of existence of solutions
assuming that there is a constant M > 0 such that yf (t, x, y) > 0 for |y| > M
and the function f satisfies the Bernstein growth condition (or the BernsteinNagumo growth condition).
The generalization of the Neumann problem (1.1) is a nonlocal problem.
BVPs with Riemann-Stieltjes integral boundary conditions include as special
cases multi-point and integral BVPs.
The multi-point and integral BCs are widely studied objects. The study
of multi-point BCs was initiated in 1908 by Picone [20]. Reviews on differential equations with BCs involving Stieltjes measures has been written in 1942
by Whyburn [29] and in 1967 by Conti.
On a k-dimensional nonlocal BVP at resonance
3
Since then, the existence of solutions for nonlocal nonlinear BVPs has
been studied by many authors by using, for instance, the Leray-Schauder
degree theory, the coincidence degree theory of Mawhin, the fixed point theorems for cones. For such problems and comments on their importance, we
refer the reader to [3, 5, 6, 7, 12, 13, 14, 15, 16, 28, 30, 31] and the references
therein.
2. The perturbed problem
First, we shall introduce notation and terminology. Throughout the paper
| · | will denote the Euclidean norm on Rk , while the scalar product in Rk
corresponding to the Euclidean norm will be denoted by ( · | · ). Denote by
C 1 ([0, 1], Rk ) the Banach space of all continuous functions x : [0, 1] → Rk
which have continuous first derivatives x0 with the norm
(
)
kxk = max
sup |x(t)| , sup |x0 (t)| .
t∈[0,1]
(2.1)
t∈[0,1]
The Lemma below, which is a straightforward consequence of the classical
Arzel`
a−Ascoli theorem, gives a compactness criterion in C 1 [0, 1], Rk .
Lemma 2.1. For a set Z ⊂ C 1 [0, 1], Rk to be relatively compact, it is necessary and sufficient that:
(1) there exists M > 0 such that for any x ∈ Z and t ∈ [0, 1] we have
|x (t)| ≤ M and |x0 (t)| ≤ M ;
(2) for every t0 ∈ [0, 1] the families Z := {x | x ∈ Z} and Z 0 := {x0 | x ∈ Z}
are equicontinuous at t0 .
Now, let us consider problem (1.1) and observe that the homogeneous
linear problem, i.e.,
Z 1
x00 = 0, x0 (0) = 0, x0 (1) =
x0 (s)dg(s),
0
has always nontrivial solutions, hence we deal with a resonant situation.
The following assumptions will be needed throughout the paper:
(i) f = (f1 , . . . , fk ) : [0, 1] × Rk → Rk is a continuous function.
(ii) g = (g1 , . . . , gk ) : [0, 1] → Rk has bounded variation on the interval
[0, 1].
(iii) There exists a uniform limit
h(t, ξ) := lim f (t, λ ξ)
λ→∞
k
with respect to ξ ∈ R , |ξ| = 1.
4
Katarzyna Szyma´
nska-D¸ebowska
(iv) Set
Z
1
1
Z
s
Z
h(u, ξ)du −
h0 (ξ) :=
h(u, ξ)dudg(s).
0
0
0
For every ξ ∈ Rk , |ξ| = 1, we have (ξ | h0 (ξ)) < 0.
The problem (1.1) is resonant. Hence, there is no equivalent integral
equation. The existence of a solution will be obtained by considering the
following perturbed boundary value problem:
x00 = f (t, x),
t ∈ [0, 1],
(2.2)
x0 (0) = 0,
Z
0
x (1) =
(2.3)
1
x0 (s)dg(s) + αn x(0),
αn ∈ (0, 1),
αn → 0.
(2.4)
0
Notice that the problem (2.2), (2.3), (2.4) is always nonresonant.
Now, let us consider the equation (2.2) and integrate it from 0 to t. By
(2.3), we get
Z t
x0 (t) =
f (u, x(u))du.
(2.5)
0
By (2.4) and (2.5), we obtain
Z 1
Z
f (u, x(u))du =
0
1
0
s
Z
f (u, x(u))dudg(s) + αn x(0),
0
so
x(0) =
1
αn
Z
1
Z
1
Z
f (u, x(u))du −
0
0
s
f (u, x(u))dudg(s) ,
0
Moreover, by (2.5), we have
Z tZ
x(t) = x(0) +
s
f (u, x(u))duds.
0
0
Now, it is easily seen that the following Lemma is true:
Lemma 2.2. A function x ∈ C 1 ([0, 1], Rk ) is a solution of the problem (2.2),
(2.3), (2.4) if and only if x satisfies the following integral equation
Z tZ s
x (t) =
f (u, x(u))duds +
0
0
Z 1
Z 1Z s
1
+
f (u, x(u))du −
f (u, x(u))dudg(s) .
αn 0
0
0
To search for solutions to the problem (2.2), (2.3), (2.4), we first reformulate the problem as an operator equation.
On a k-dimensional nonlocal BVP at resonance
5
Given x ∈ C 1 ([0, 1], Rk ) and fixed n ∈ N let
Z tZ s
f (u, x(u))duds +
(An x) (t) =
0
0
Z 1
Z 1Z s
1
+
f (u, x(u))dudg(s) .
f (u, x(u))du −
αn 0
0
0
Then
t
Z
0
(An x) (t) =
f (u, x(u))du.
(2.6)
0
It is clear that An x, (An x)0 : [0, 1] → Rk are continuous. It follows that the
operator
An : C 1 [0, 1], Rk → C 1 [0, 1], Rk
is well-defined.
By assumption (iii), function f is bounded and we put
M :=
sup
|f (t, x)|.
t∈[0,1],x∈Rk
By (2.6), we have
sup |(An x)0 (t)| ≤ M.
(2.7)
t∈[0,1]
Moreover, we get
sup |(An x)(t)| ≤ M +
t∈[0,1]
1
(M + M Var(g)) ,
αn
(2.8)
where Var(g) means the variation of g on the interval [0, 1].
From (ii), L := Var(g) < ∞. Put Mn := M + α1n (M + M L), then
kAn xk ≤ Mn for every n ∈ N. Moreover, (An x)00 (t) and (An x)0 (t), t ∈ [0, 1],
are bounded, hence the families (An x)0 and (An x) are equicontinuous. Now,
by Lemma 2.1, the following Lemma holds:
Lemma 2.3. The operator An is completely continuous.
Let Bn := x ∈ C 1 [0, 1], Rk kxk ≤ Mn . Now, considering operator
An : Bn → Bn ,
by Schauder’s fixed point Theorem, we get that the operator An has a fixed
point in Bn for every n.
We have proved the following
Lemma 2.4. Under assumptions (i) − (iii) problem (2.2), (2.3), (2.4) has at
least one solution for every n ∈ N.
6
Katarzyna Szyma´
nska-D¸ebowska
3. The main result
Let ϕn be a solution of the problem (2.2), (2.3), (2.4), where n is fixed.
Lemma 3.1. The sequence (ϕn ) is bounded in C 1 [0, 1], Rk .
Proof. Assume that the sequence (ϕn ) is unbounded. Then, passing to a
subsequence if necessary, we have kϕn k → ∞. We can proceed analogously
as in (2.7) to show that
sup |(ϕn )0 (t)| ≤ M,
t∈[0,1]
for every n. Hence, supt∈[0,1] |ϕn (t)| → ∞, when n → ∞.
ϕn
) ⊂ C 1 [0, 1], Rk and noLet us consider the following sequence ( kϕ
nk
tice that the norm of the sequence equals 1. Hence, the sequence is bounded.
ϕ0
ϕn
Moreover, the family ( kϕ
) (and simultaneously ( kϕnn k )) is equicontinuous,
nk
ϕ0 (t)
ϕ00 (t)
since kϕnn k (or kϕnn k ) is bounded. By Lemma 2.1, there exists a convergent
ϕn
subsequence of ( kϕ
). To simplify the notation, let us denote this subsenk
ϕn
quence as ( kϕn k ).
First, observe that
ϕ0n (t)
kϕn k
→ 0 ∈ Rk . Now, we shall show that
ϕn (t)
→ ξ,
kϕn k
(3.1)
where ξ = (ξ1 , . . . , ξk ) does not depend on t and |ξ| = 1.
Indeed, notice that
ϕn (t)
kϕn k
ϕn (t)
kϕn k
is given by
RtRs
=
f (u, ϕn (u))duds
+
kϕn k
R1Rs
R1
f (u, ϕn (u))du − 0 0 f (u, ϕn (u))dudg(s)
0
+
.
αn kϕn k
0
0
(3.2)
Since f is bounded, we obtain
RtRs
lim
n→∞
0
0
f (u, ϕn (u))duds
= 0 ∈ Rk .
kϕn k
(3.3)
Now, by (3.2) and (3.3), we can easily observe that the limit (3.1) does not
ϕn
depend on t. The norm of the sequence ( kϕ
) equals 1. Hence ϕkϕnn(t)k → ξ,
nk
where |ξ| = 1.
On a k-dimensional nonlocal BVP at resonance
7
On the other hand, we get
RtRs
f (u, ϕn (u))duds
ϕn (t)
ξ = lim
= 0 0
+
n→∞ kϕn k
kϕn k
R1
R1Rs
f (u, ϕn (u))du − 0 0 f (u, ϕn (u))dudg(s)
+ 0
=
αn kϕn k
R
1
n (u)
f (u, kϕn k ϕkϕ
)du
0
nk

−
= lim
n→∞
αn kϕn k

R1Rs
n (u)
)dudg(s)
f (u, kϕn k ϕkϕ
k
0 0
n
.
−
αn kϕn k
Now, observe, that there exist a uniform limits of
Z 1
ϕn (u)
)du
f (u, kϕn k
kϕn k
0
and
1
Z
Z
s
f (u, kϕn k
0
0
ϕn (u)
)dudg(s)
kϕn k
Moreover, by (iv), the sum of the limits is different from zero. Hence, since
(3.1) holds, there exists γ ∈ (0, ∞) such that γ := limn→∞ 1/(αn kϕn k).
Now, by assumption (iii), we obtain
ξ
ϕn (t)
lim
=
kϕn k
Z 1
Z
= γ
h(u, ξ)du −
=
n→∞
0
0
1
Z
s
h(u, ξ)dudg(s) .
Finally, by (3.4) and (iv), we get
Z 1
Z
1 = (ξ | ξ) = γ ξ h(u, ξ)du −
0
=
(3.4)
0
0
1
Z
s
h(u, ξ)dudg(s)
0
γ(ξ | h0 (ξ)) < 0
a contradiction. Hence, the sequence (ϕn ) is bounded.
Now, it is easy to see that the following lemma is true:
Lemma 3.2. The set Z = {ϕn | n ∈ N} is relatively compact in C 1 [0, 1], Rk .
By the above Lemmas, we get the proof of the following
Theorem 3.3. Under assumptions (i) − (iv) problem (1.1) has at least one
solution.
8
Katarzyna Szyma´
nska-D¸ebowska
Proof. Lemma 3.2 implies that (ϕn ) has a convergent subsequence (ϕnl ),
ϕnl → ϕ. We know that ϕnl (ϕ0nl ) converges uniformly to ϕ (ϕ0 ) on [0, 1].
Since (ϕnl ) is equibounded and f is uniformly continuous on compact sets,
one can see that f (t, ϕnl ) is uniformly convergent to f (t, ϕ). Since
ϕ00nl (t) = f (t, ϕnl (t)),
the sequence ϕ00nl (t) is also uniformly convergent. Moreover, ϕ00nl (t) converges
uniformly to ϕ00 (t).
Note that we have actually proved that function ϕ ∈ C 1 [0, 1], Rk is
a solution of the equation of problem (1.1). By (2.3) and (2.4), it is easy
to see that ϕ satisfies boundary conditions of problem (1.1). This ends the
proof.
4. Applications
To illustrate our result we shall present some examples.
Example 1. Let us consider the Neumann BVP
x00 = f (t, x),
x0 (0) = 0,
x0 (1) = 0.
In this case gi (t) = constant, i = 1, . . . , k, t ∈ [0, 1] and condition (ii) always
holds. Moreover, we have
Z 1
h0 (ξ) =
h(s, ξ)ds.
0
Hence for any f which satisfies conditions (i), (iii) and (iv) the Neumann
BVP has at least one solution.
Example 2. Let k = 1, g(t) = t and f (t, x) = t−|x|x
x2 +1 . We have
−1,
ξ=1
h(t, ξ) = lim f (t, λ ξ) =
.
1,
ξ = −1
λ→∞
Then h0 (1) = − 12 and h0 (−1) = 12 and we get (ξ|h0 (ξ)) < 0. Hence, problem
(1.1) has at least one nontrivial solution.
Example 3. Let k = 3, g(t) = (t, t, t) and
f1 (t, x1 , x2 , x3 )
=
f2 (t, x1 , x2 , x3 )
=
f3 (t, x1 , x2 , x3 )
=
−x1
p
x21 + x22 + x23 + sin2 t + 1
−x2 − t
p
,
2
x1 + x22 + x23 + 1
−x3 + arctan(x2 − t)
p
.
x21 + x22 + x23 + 1
For every ξ = (ξ1 , ξ2 , ξ3 ), |ξ| = 1, we get
ξ1
ξ2
ξ3
h(t, ξ) = lim f (t, λ ξ) = − , − , −
λ→∞
|ξ| |ξ| |ξ|
,
On a k-dimensional nonlocal BVP at resonance
and
9
ξ2
ξ3
ξ1
,−
,−
.
h0 (ξ) = −
2|ξ| 2|ξ| 2|ξ|
Then
ξ2
ξ2
ξ12
+ 2 + 3 = − 12 |ξ| < 0.
|ξ| |ξ| |ξ|
Hence, problem (1.1) has at least one nontrivial solution.
(ξ|h0 (ξ)) = − 21
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On a k-dimensional nonlocal BVP at resonance
Katarzyna Szyma´
nska-D¸ebowska
Institute of Mathematics
Technical University of L´
od´z
90-924 L´
od´z, ul. W´
olcza´
nska 215, Poland
e-mail: [email protected]
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