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Monotone iterative method for first-order differential equations at resonance
Applied Mathematics and Computation 233 (2014) 20–28
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Monotone iterative method for first-order differential equations at resonance Tadeusz Jankowski ´ sk, Poland Gdansk University of Technology, Department of Differential Equations and Applied Mathematics, 11/12 G. Narutowicz Str., 80-233 Gdan
a r t i c l e
i n f o
a b s t r a c t
Keywords: Monotone iterative method First-order differential equations at resonance Stieltjes integrals Existence of solutions (extremal and quasisolutions)
This paper concerns the application of the monotone iterative technique for first-order differential equations involving Stieltjes integrals conditions. We discuss such problems at resonance when the measure in the Stieltjes integral is positive and also when this measure changes the sign. Sufficient conditions which guarantee the existence of extremal, unique and quasi-solutions are given. Three examples illustrate the results. Ó 2014 Elsevier Inc. All rights reserved.
1. Introduction In this paper, we study the existence of solutions of the following problem:
x0 ðtÞ ¼ f ðt; xðtÞÞ FxðtÞ;
t 2 J ¼ ½0; T;
xð0Þ ¼ k½x;
ð1Þ
where k denotes a linear functional on CðJÞ given by
k½x ¼
Z
T
xðsÞdAðsÞ
0
involving a Stieltjes integral with a suitable function A of bounded variation on J. We are interested in the case when problem (1) is at resonance, that is the homogeneous problem
x0 ðtÞ ¼ 0;
t 2 J;
xð0Þ ¼ k½x
ð2Þ
has nontrivial solutions. The resonance condition is
Z
T
dAðsÞ ¼ 1:
ð3Þ
0
Problems with integral (nonlocal) boundary conditions have some applications, for example, in termostat models [5], interaction problems [16], chemical engineering [2]. Positive (nonnegative) solutions for differential equations with nonlocal boundary conditions involving Stieltjes integrals have been discussed in many papers, for example, see papers [4,9–12,17–19] in which Stieltjes integrals appeared with the signed measure, see also [13]. Positive solutions for differential equations at resonance have been discussed, for example, in [1,3,6,15,20–24]. In this paper, we discuss problem (1) at E-mail address: [email protected] http://dx.doi.org/10.1016/j.amc.2014.01.123 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
21
resonance by using the monotone iterative method [14]. According to my knowledge, it is a first application of this method to such problems. 2. Case when the measure dA is positive (nonnegative) First, we study a linear problem. Lemma 1. Let A be a function of bounded variation. Suppose that K; M 2 CðJ; RÞ; p 2 C 1 ðJ; RÞ and
p0 ðtÞ ¼ KðtÞpðtÞ þ MðtÞ;
t 2 J;
ð4Þ
pð0Þ ¼ k½p: In addition, we assume that
Z
T
Z s exp KðsÞds dAðsÞ – 1:
0
ð5Þ
0
Then problem (4) has a unique solution. Proof. Indeed,
Z t Z pðtÞ ¼ exp KðsÞds pð0Þ þ 0
t
exp
Z
0
s
KðsÞds MðsÞds :
0
In view of the boundary condition from (4) and condition (5), we have the assertion. h In the next lemma, we shall discuss a differential inequality. Lemma 2. Let A be a function of bounded variation with a positive measure. Suppose that K 2 CðJ; RÞ; p 2 C 1 ðJ; RÞ and
p0 ðtÞ 6 KðtÞpðtÞ; pð0Þ 6 k½p:
t 2 J;
ð6Þ
In addition, we assume that
Z
T
Z s exp KðsÞds dAðsÞ < 1:
0
ð7Þ
0
Then pðtÞ 6 0 on J. Proof. Note that
Z t pðtÞ 6 pð0Þ exp KðsÞds : 0
Using the boundary condition from (6) and condition (7) we have the assertion. h Let us introduce the notion of lower and upper solutions of problem (1). We say that u 2 C 1 ðJ; RÞ is called a lower solution of (1) if
u0 ðtÞ 6 FuðtÞ;
t 2 J;
uð0Þ 6 k½u
and it is an upper solution of problem (1) if the above inequalities are reversed. We use the following assumption: H1 : f 2 CðJ R; RÞ; A is a function of bounded variation and condition (3) holds. The next theorem concerns the situation when problem (1) has extremal solutions. Theorem 1. Let Assumption H1 hold. Assume that the measure dA is positive. Let y0 ; z0 2 C 1 ðJ; RÞ be lower and upper solutions of (1), respectively and y0 ðtÞ 6 z0 ðtÞ; t 2 J. In addition, we assume that H2 : there exists a function K 2 CðJ; RÞ such that
f ðt; u1 Þ f ðt; v 1 Þ 6 KðtÞ½v 1 u1 ;
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T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
if y0 ðtÞ 6 u1 6 v 1 6 z0 ðtÞ, H3 : condition (7) holds. Then problem (1) has, in the sector ½y0 ; z0 , minimal and maximal solutions, where
½y0 ; z0 ¼ fw 2 C 1 ðJ; RÞ : y0 ðtÞ 6 wðtÞ 6 z0 ðtÞ; t 2 Jg:
Proof. For n ¼ 0; 1; . . ., let
y0nþ1 ðtÞ ¼ Fyn ðtÞ KðtÞ½ynþ1 ðtÞ yn ðtÞ;
t 2 J;
ynþ1 ð0Þ ¼ k½ynþ1 ;
z0nþ1 ðtÞ ¼ Fzn ðtÞ KðtÞ½znþ1 ðtÞ zn ðtÞ;
t 2 J;
znþ1 ð0Þ ¼ k½znþ1 ;
with F defined as in (1). The sequences fyn ; zn g are well defined by Lemma 1. We first show that
y0 ðtÞ 6 y1 ðtÞ 6 z1 ðtÞ 6 z0 ðtÞ;
t 2 J:
ð8Þ
Put p ¼ y0 y1 ; q ¼ z1 z0 . This shows
pð0Þ 6 k½p;
qð0Þ 6 k½q;
p0 ðtÞ 6 Fy0 ðtÞ Fy0 ðtÞ þ KðtÞ½y1 ðtÞ y0 ðtÞ ¼ KðtÞpðtÞ; q0 ðtÞ 6 Fz0 ðtÞ KðtÞ½z1 ðtÞ z0 ðtÞ Fz0 ðtÞ ¼ KðtÞqðtÞ: By Lemma 2, y0 ðtÞ 6 y1 ðtÞ; z1 ðtÞ 6 z0 ðtÞ; t 2 J. Now, we put p ¼ y1 z1 . In view of Assumption H2 , we have
pð0Þ ¼ k½p; p0 ðtÞ ¼ Fy0 ðtÞ Fz0 ðtÞ KðtÞ½y1 ðtÞ y0 ðtÞ z1 ðtÞ þ z0 ðtÞ 6 KðtÞpðtÞ: Lemma 2 yields y1 ðtÞ 6 z1 ðtÞ on J. It proves (8). In the next step, we show that y1 ; z1 are lower and upper solutions of problem (1), respectively. Note that
y01 ðtÞ ¼ Fy0 ðtÞ KðtÞ½y1 ðtÞ y0 ðtÞ Fy1 ðtÞ þ Fy1 ðtÞ 6 Fy1 ðtÞ; z01 ðtÞ ¼ Fz0 ðtÞ KðtÞ½z1 ðtÞ z0 ðtÞ Fz1 ðtÞ þ Fz1 ðtÞ P Fz1 ðtÞ; by Assumption H2 . This proves that y1 ; z1 are lower and upper solutions of problem (1), respectively. Using the mathematical induction, we can show that
y0 ðtÞ 6 y1 ðtÞ 6 6 yn ðtÞ 6 ynþ1 ðtÞ 6 znþ1 ðtÞ 6 zn ðtÞ 6 6 z1 ðtÞ 6 z0 ðtÞ for t 2 J and n ¼ 0; 1; . . . . Now we shall prove that the sequences fyn ; zn g converge to their limit functions y; z, respectively. First, we need to show that the sequences are bounded and equicontinuous on J. Indeed,
A1 6 y0 ðtÞ 6 yn ðtÞ 6 zn ðtÞ 6 z0 ðtÞ 6 A2 ;
t 2 J; n ¼ 0; 1; . . . ;
so the sequences fyn ; zn g are uniformly bounded. Note that y0n and z0n are bounded on J by W > 0 because jf ðt; yn ðtÞÞj is bounded on J ½A1 ; A2 . Hence yn ; zn are equicontinuous because for > 0; t1 ; t2 2 J such that jt1 t 2 j < =W we have
jyn ðt 1 Þ yn ðt 2 Þj ¼ jy0n ðnÞjjt 1 t 2 j < jzn ðt 1 Þ zn ðt2 Þj < : The Arzela–Ascoli theorem guarantees the existence of subsequences fynk ; znk g of fyn ; zn g, respectively, and continuous functions y; z with ynk ; znk converging uniformly on J to y and z, respectively. Note that ynk ; znk satisfy the integral equations
8 R h i > < ynk þ1 ðtÞ ¼ exp 0t KðsÞds ynk þ1 ð0Þ þ Pynk ðtÞ ; t 2 J; R
> : znk þ1 ðtÞ ¼ exp 0t KðsÞds znk þ1 ð0Þ þ Pznk ðtÞ ; t 2 J
and
ynk þ1 ð0Þ ¼ k½ynk þ1 ; znk þ1 ð0Þ ¼ k½znk þ1 ;
with
Pynk ðtÞ ¼
Z 0
t
exp
Z 0
s
h i KðsÞds Fynk ðsÞ þ KðsÞynk ðsÞ ds:
If nk ! 1, then from the above relations, we have
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T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
8 R > < yðtÞ ¼ exp 0t KðsÞds ½yð0Þ þ PyðtÞ; t 2 J; yð0Þ ¼ k½y; R > : zðtÞ ¼ exp 0t KðsÞds ½zð0Þ þ PzðtÞ; t 2 J; zð0Þ ¼ k½z; because f is continuous. Thus y; z 2 C 1 ðJÞ and
y0 ðtÞ ¼ FyðtÞ;
z0 ðtÞ ¼ FzðtÞ; t 2 J:
It proves that y; z are solutions of problem (1) and
y0 ðtÞ 6 yðtÞ 6 zðtÞ 6 z0 ðtÞ;
t 2 J:
Now we need to show that y; z are extremal solutions of (1) in the sector ½y0 ; z0 . Let w 2 ½y0 ; z0 be any solution of (1). We assume that ym ðtÞ 6 wðtÞ 6 zm ðtÞ; t 2 J for some m. Let p ¼ ymþ1 w; q ¼ w zmþ1 . Then,
pð0Þ ¼ k½p;
qð0Þ ¼ k½q;
p0 ðtÞ ¼ Fym ðtÞ KðtÞ½ymþ1 ðtÞ ym ðtÞ FwðtÞ 6 KðtÞpðtÞ; q0 ðtÞ ¼ FwðtÞ Fzm ðtÞ þ KðtÞ½zmþ1 ðtÞ zm ðtÞ 6 KðtÞqðtÞ; by Assumption H2 . This and Lemma 2 give ymþ1 ðtÞ 6 wðtÞ 6 zmþ1 ðtÞ; t 2 J. By induction, yn ðtÞ 6 wðtÞ 6 zn ðtÞ; t 2 J; n ¼ 0; 1; . If n ! 1, then we have the assertion. This ends the proof. h Example 1. Consider the problem:
(
2 x0 ðtÞ ¼ exp t sin xðtÞ þ 2 cos xðtÞ FxðtÞ;
t 2 J ¼ ½0; ln 2;
ð9Þ
xð0Þ ¼ k½x; with
dAðtÞ ¼
p 2 ln 2
sin
tp dt: ln 2
Note that the measure dA is positive and
Z
ln 2
dAðtÞ ¼ 1;
0
so problem (9) satisfies the resonance condition (3). Indeed, KðtÞ ¼ 2ðln 2 þ 1Þ. Let y0 ðtÞ ¼ 0; z0 ðtÞ ¼ p; t 2 J. Note that
Fy0 ðtÞ ¼ 3 > 0 ¼ y00 ðtÞ;
t 2 J;
Fz0 ðtÞ ¼ 1 < 0 ¼ z00 ðtÞ; t 2 J; k½y0 ¼ 0 ¼ y0 ð0Þ; k½z0 ¼ p ¼ z0 ð0Þ: This proves that y0 ; z0 are lower and upper solutions of problem (9), respectively. Moreover,
Z 0
ln 2
Z s Z exp KðsÞds dAðsÞ < 0
ln 2
dAðsÞ ¼ 1; 0
so condition (7) holds. Hence, problem (9) has extremal solutions in the region ½y0 ; z0 , by Theorem 1. Our next theorem concerns the case when problem (1) has a unique solution at the resonance case. Theorem 2. Assume that all assumptions of Theorem 1 hold. In addition, we assume that H4 : there exists a function Q 2 CðJ; RÞ such that KðtÞ þ Q ðtÞ P 0; t 2 J, condition (7) holds with Q instead of K and
f ðt; uÞ f ðt; v Þ P Q ðtÞðv uÞ if y0 6 u 6 v 6 z0 : Then problem (1) has, in the sector ½y0 ; z0 , a unique solution. Proof. Theorem 1 guarantees that problem (1) has extremal solutions y; z and y0 ðtÞ 6 yðtÞ 6 zðtÞ 6 z0 ðtÞ; t 2 J. Put p ¼ z y, so pðtÞ P 0 on J and
p0 ðtÞ ¼ FzðtÞ FyðtÞ 6 Q ðtÞpðtÞ;
t 2 J; pð0Þ ¼ k½p:
This and Lemma 2 show that y ¼ z, so problem (1) has a unique solution. This ends the proof.
h
24
T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
Example 2. Consider the following problem:
(
K < 0; t 2 J ¼ ½0; p;
x0 ðtÞ ¼ KxðtÞ þ 3 FxðtÞ; Rp xð0Þ ¼ 12 0 xðsÞ sin tdt:
ð10Þ
We see that now
k½x ¼
1 2
Z p
xðtÞ sin tdt:
0
Note that
1 2
Z p
sin sds ¼ 1;
0
so condition (3) holds and problem (10) is at resonance. It is easy to verify that xðtÞ ¼ K3 ; t 2 J is the unique solution of problem (10). Indeed, KðtÞ ¼ Q ðtÞ ¼ jKj; t 2 J. Put y0 ðtÞ ¼ K4 ; z0 ðtÞ ¼ K4 ; t 2 J. Then
Fy0 ðtÞ ¼ 7 > 0 ¼ y00 ðtÞ;
t 2 J;
Fz0 ðtÞ ¼ 1 < 0 ¼ z00 ðtÞ; k½y0 ¼ y0 ð0Þ;
t 2 J;
k½z0 ¼ z0 ð0Þ;
so y0 ; z0 are lower and upper solutions of problem (10), respectively. Moreover, for KðtÞ ¼ Q ðtÞ ¼ jKj, we have
Z p 0
Z s Z 1 p exp KðsÞds dAðsÞ < sin sds ¼ 1; 2 0 0
so condition (7) holds. Hence, problem (10) has a unique solution at resonance in the region ½y0 ; z0 , by Theorem 2. 3. Case when dA is a signed measure Assume that
AðtÞ 6 0; t 2 ½0; h;
and AðtÞ P 0; t 2 ½h; T:
ð11Þ
1
We say that u; v 2 C ðJ; RÞ are called coupled lower and upper solutions of problem (1) if
(
u0 ðtÞ 6 FuðtÞ; t 2 J;
v 0 ðtÞ P F v ðtÞ;
t 2 J;
uð0Þ 6
Rh
v ðsÞdAðsÞ þ
0
v ð0Þ P
Rh 0
RT
uðsÞdAðsÞ; RT v ðsÞdAðsÞ: h
h
uðsÞdAðsÞ þ
We say that X; Y 2 C 1 ðJ; RÞ are called quasi-solutions of (1) if
(
X 0 ðtÞ ¼ FXðtÞ; t 2 J; 0
Y ðtÞ ¼ FYðtÞ; t 2 J;
Xð0Þ ¼ Yð0Þ ¼
Rh 0 Rh 0
YðsÞdAðsÞ þ XðsÞdAðsÞ þ
RT h
XðsÞdAðsÞ;
h
YðsÞdAðsÞ:
RT
We first formulate conditions under which a corresponding linear system has a unique solution. Lemma 3. Let A be a function of bounded variation and let condition (11) hold. Suppose that K; M 1 ; M 2 2 CðJ; RÞ; p; q 2 C 1 ðJ; RÞ and
( (
p0 ðtÞ ¼ KðtÞpðtÞ þ M 1 ðtÞ; t 2 J; Rh RT pð0Þ ¼ 0 qðsÞdAðsÞ þ h pðsÞdAðsÞ;
ð12Þ
q0 ðtÞ ¼ KðtÞqðtÞ þ M2 ðtÞ; t 2 J; Rh RT qð0Þ ¼ 0 pðsÞdAðsÞ þ h qðsÞdAðsÞ:
In addition, we assume that
Z 1
T
2 CðsÞdAðsÞ –
Z
h
0
with
Z t CðtÞ ¼ exp KðsÞds : 0
h
!2 CðsÞdAðsÞ
;
ð13Þ
25
T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
Then system (12) has a unique solution ðp; qÞ. Proof. Put
Di ðtÞ ¼
Z
t
exp
Z
0
s
KðsÞds M i ðsÞds;
i ¼ 1; 2:
0
Then,
pðtÞ ¼ CðtÞ½pð0Þ þ D1 ðtÞ;
qðtÞ ¼ CðtÞ½qð0Þ þ D2 ðtÞ; t 2 J
and
(
pð0Þ ¼ qð0Þ ¼
Rh 0
CðsÞ½qð0Þ þ D2 ðsÞdAðsÞ þ
0
CðsÞ½pð0Þ þ D1 ðsÞdAðsÞ þ
Rh
RT h
CðsÞ½pð0Þ þ D1 ðsÞdAðsÞ;
h
CðsÞ½qð0Þ þ D2 ðsÞdAðsÞ:
RT
ð14Þ
System (14) has a unique solution with respect to pð0Þ; qð0Þ, in view of condition (13). It proves that the assertion holds. This ends the proof. h Lemma 4. Let A be a function of bounded variation and let condition (11) hold. Suppose that K 2 CðJ; RÞ; p; q 2 C 1 ðJ; RÞ and
8 0 p ðtÞ 6 KðtÞpðtÞ; q0 ðtÞ 6 KðtÞqðtÞ; t 2 J; > > < Rh RT pð0Þ 6 0 qðsÞdAðsÞ þ h pðsÞdAðsÞ; > > : qð0Þ 6 R h pðsÞdAðsÞ þ R T qðsÞdAðsÞ: 0 h
ð15Þ
In addition, we assume that
Z
h
CðsÞdAðsÞ þ
0
Z
T
CðsÞdAðsÞ < 1;
ð16Þ
h
with C defined as in Lemma 3. Then pðtÞ 6 0; qðtÞ 6 0 on J. Proof. Note that
pðtÞ 6 CðtÞpð0Þ; and
pð0Þ 6 qð0Þ qð0Þ 6
Rh
0 Rh pð0Þ 0
qðtÞ 6 CðtÞqð0Þ; t 2 J
CðsÞdAðsÞ þ pð0Þ CðsÞdAðsÞ þ qð0Þ
RT
ð17Þ
h
CðsÞdAðsÞ;
h
CðsÞdAðsÞ:
RT
This and condition (16) show that pð0Þ 0; qð0Þ 6 0. Combining this with (17) we have the assertion of Lemma 4. This ends the proof. h Similarly, we can prove the following. Lemma 5. Let A be a function of bounded variation and let condition (11) hold. Suppose that K 2 CðJ; RÞ; p 2 C 1 ðJ; RÞ and
p0 ðtÞ 6 KðtÞpðtÞ; t 2 J;
pð0Þ 6
Z
h
pðsÞdAðsÞ þ 0
Z
T
pðsÞdAðsÞ:
h
In addition, we assume that condition (16) holds. Then pðtÞ 6 0 on J. Theorem 3. Let assumptions H1 ; H2 hold. Let y0 ; z0 2 C 1 ðJ; RÞ be coupled lower and upper solutions of (1) and y0 ðtÞ 6 z0 ðtÞ; t 2 J. Assume that conditions (11) and (16) are satisfied. Then problem(1) has, in the sector ½y0 ; z0 , coupled quasi-solutions. Proof. For n ¼ 0; 1; . . ., let
y0nþ1 ðtÞ ¼ Fyn ðtÞ KðtÞ½ynþ1 ðtÞ yn ðtÞ;
t 2 J;
z0nþ1 ðtÞ ¼ Fzn ðtÞ KðtÞ½znþ1 ðtÞ zn ðtÞ; t 2 J; Rh RT ynþ1 ð0Þ ¼ 0 znþ1 ðsÞdAðsÞ þ h ynþ1 ðsÞdAðsÞ; Rh RT znþ1 ð0Þ ¼ 0 ynþ1 ðsÞdAðsÞ þ h znþ1 ðsÞdAðsÞ:
26
T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
Observe that functions y1 ; z1 are well defined by Lemma 3. We first show that
y0 ðtÞ 6 y1 ðtÞ 6 z1 ðtÞ 6 z0 ðtÞ;
t 2 J:
ð18Þ
Put p ¼ y0 y1 ; q ¼ z1 z0 . This shows
p0 ðtÞ 6 KðtÞpðtÞ; q0 ðtÞ 6 KðtÞqðtÞ; Z h Z T Z pð0Þ 6 z0 ðsÞdAðsÞ þ y0 ðsÞdAðsÞ 0
qð0Þ 6
Z
h
pðsÞdAðsÞ þ
0
z1 ðsÞdAðsÞ
Z
0
Z
h
h
T
y1 ðsÞdAðsÞ ¼
Z
h
h
qðsÞdAðsÞ þ
Z
0
T
pðsÞdAðsÞ;
h
T
qðsÞdAðsÞ:
h
By Lemma 4, y0 ðtÞ 6 y1 ðtÞ; z1 ðtÞ 6 z0 ðtÞ; t 2 J. Now, we put p ¼ y1 z1 , so
pð0Þ ¼
Z
h
z1 ðsÞdAðsÞ þ
Z
0
T
h
y1 ðsÞdAðsÞ
Z
h
0
y1 ðsÞdAðsÞ
Z
T
z1 ðsÞdAðsÞ ¼
h
Z
h
0
pðsÞdAðsÞ þ
Z
T
pðsÞdAðsÞ:
h
Moreover,
p0 ðtÞ ¼ Fy0 ðtÞ Fz0 ðtÞ KðtÞ½y1 ðtÞ y0 ðtÞ z1 ðtÞ þ z0 ðtÞ 6 KðtÞpðtÞ; by Assumption H2 . Lemma 5 yields y1 ðtÞ 6 z1 ðtÞ on J. It proves (18). It is easy to show that y1 ; z1 are coupled lower and upper solutions of problem (1). Moreover, using the mathematical induction, we can show that
y0 ðtÞ 6 y1 ðtÞ 6 6 yn ðtÞ 6 ynþ1 ðtÞ 6 znþ1 ðtÞ 6 zn ðtÞ 6 6 z1 ðtÞ 6 z0 ðtÞ for t 2 J and n ¼ 0; 1; . The sequences fyn ; zn g converge to their limit functions y; z, respectively. Indeed, y; z are quasi-solutions of problem (1) and y0 6 y 6 z 6 z0 . It ends the proof. h Example 3. Let
(
Indeed,
x0 ðtÞ ¼ expðxðtÞÞ þ 10 cos x þ 12 t ; t 2 J ¼ ½0; 1; R1 xð0Þ ¼ 0 xðtÞdAðtÞ with dAðtÞ ¼ 2ð3t 1Þdt:
R1 0
dAðtÞ ¼ 1 and 2ð3t 1Þ 6 0; t 2 0; 13 ;
2ð3t 1Þ P 0; t 2
ð19Þ 1 3
;1 .
Let y0 ðtÞ ¼ t; z0 ðtÞ ¼ 3 t; t 2 J. Note that
3 3 > 1 ¼ y00 ðtÞ; t P expð1Þ þ 10 cos 2 2 1 5 < 1 ¼ z00 ðtÞ: Fz0 ðtÞ ¼ expðt 3Þ þ 10 cos 3 t 6 expð2Þ þ 10 cos 2 2 Z 1 Z 1 3 2 ð3 tÞð6t 2Þdt þ tð6t 2Þdt ¼ > 0 ¼ y0 ð0Þ; 1 27 0 3 Z 1 Z 1 3 79 tð6t 2Þdt þ ð3 tÞð6t 2Þdt ¼ < 3 ¼ z0 ð0Þ: 1 27 0 3
Fy0 ðtÞ ¼ expðtÞ þ 10 cos
Moreover KðtÞ ¼ 11. Condition (16) holds too. In view of Theorem 3, problem (19) has coupled quasi solutions. 4. Some comments 1. Let a; b P 0 and
8 t 2 ½0; hÞ; > < 0; AðtÞ ¼ a; t 2 ½h; TÞ; > : b a; t ¼ T: Then,
k½x ¼ axðhÞ þ bxðTÞ; so problem (1) takes the form
T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
x0 ðtÞ ¼ f ðt; xðtÞÞ FxðtÞ; t 2 J ¼ ½0; T; xð0Þ ¼ axðhÞ þ bxðTÞ; a; b > 0:
27
ð20Þ
Such problems without condition (3) have been discussed, for example in [7,8]. Similarly as before, we define the notions of coupled lower and upper solutions and quasi-solutions of problem (20); so we say that u; v 2 C 1 ðJ; RÞ are called coupled lower and upper solutions of problem (20) if
uð0Þ 6 av ðhÞ þ buðTÞ;
u0 ðtÞ 6 FuðtÞ; t 2 J;
v 0 ðtÞ P F v ðtÞ;
t 2 J;
v ð0Þ P auðhÞ þ bv ðTÞ:
Let y0 ; z0 be coupled lower and upper solutions of problem (20). Then, we can show that the sequences fyn ; zn g:
y0nþ1 ðtÞ ¼ Fyn ðtÞ KðtÞ½ynþ1 ðtÞ yn ðtÞ;
ynþ1 ð0Þ ¼ aznþ1 ðhÞ þ bynþ1 ðTÞ;
z0nþ1 ðtÞ ¼ Fzn ðtÞ KðtÞ½znþ1 ðtÞ zn ðtÞ;
znþ1 ð0Þ ¼ aynþ1 ðhÞ þ bznþ1 ðTÞ
converge to quasi-solutions of problem (20) provided that
aCðhÞ þ bCðTÞ < 1;
R t with CðtÞ ¼ exp 0 KðsÞds . Then, in view of Theorem 3, problem (20) has quasi-solutions in ½y0 ; z0 . 2. Let
k½x ¼
ð21Þ
r X
ai xðti Þ; ai 2 R; 0 < t1 < t2 < < tr1 < tr ¼ T:
i¼1
Then condition (21) is replaced by r X jai jCðti Þ < 1: i¼1
3. Let AðtÞ ¼ A1 ðtÞ þ A2 ðtÞ, where A2 ðtÞ 6 0; t 2 ½0; h; A2 ðtÞ P 0; t 2 ½h; T and
8 0; > > > > > a1 ; > > > ; > > > > > ar1 þ ar2 ; > > : ar þ ar1 ;
t 2 ½0; t 1 Þ; t 2 ½t1 ; t 2 Þ; t 2 ½t2 ; t 3 Þ; t 2 ½tr1 ; tr Þ; t ¼ tr ¼ T:
For ai 2 R; 0 < t 1 < t 2 < < tr1 < t r ¼ T we have
k½x ¼
r X
ai xðti Þ þ
Z
h
xðsÞdA2 ðsÞ þ
0
i¼1
Z
T
xðsÞdA2 ðsÞ:
h
Then condition (21) is replaced by
Z r X jai jCðti Þ i¼1
h
CðsÞdA2 ðsÞ þ 0
Z
T
CðsÞdA2 ðsÞ < 1:
h
We can also consider the case when A2 is negative in k subintervals of ½0; T; k > 1.
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
Ch. Bai, J. Fang, Existence of positive solutions for three-point boundary value problems at resonance, J. Math. Anal. Appl. 291 (2004) 538–549. Y.S. Choi, K.Y. Chan, A parabolic equation with nonlocal boundary conditions arising from electrochemistry, Nonlinear Anal. 18 (1992) 317–331. X. Han, Positive solutions for a three-point boundary value problem at resonance, J. Math. Anal. Appl. 336 (2007) 556–568. G. Infante, P. Pietramala, M. Zima, Positive solutions for a class of nonlocal impulsive BVPs via fixed point index, Topol. Methods Nonlinear Anal. 36 (2010) 263–284. G. Infante, J.R.L. Webb, Loss of positivity in a nonlinear scalar heat equation, NoDEA Nonlinear Differ. Equ. Appl. 13 (2006) 249–261. G. Infante, M. Zima, Positive solutions of multi-point boundary value problems at resonance, Nonlinear Anal. 69 (2008) 2458–2465. T. Jankowski, Existence of solutions of differential equations with nonlinear multipoint boundary conditions, Comput. Math. Appl. 47 (2004) 1095– 1103. T. Jankowski, Multipoint boundary value problems for ODEs, Part II, Czechoslovak Math. J. 54 (2004) 843–854. T. Jankowski, Positive solutions for second order impulsive differential equations involving Stieltjes integral conditions, Nonlinear Anal. 74 (2011) 3775–3785. T. Jankowski, Existence of positive solutions to third order differential equations with advanced arguments and nonlocal boundary conditions, Nonlinear Anal. 75 (2012) 913–923. T. Jankowski, Positive solutions to second-order differential equations with dependence on the first-order derivative and nonlocal boundary conditions, Boundary Value Prob. 2013 (2013) 8.
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[12] T. Jankowski, Positive solutions to Sturm–Liouville problems with nonlocal boundary conditions, in: Proceedings of the Royal Society of Edinburgh, Section A, vol. 144, 2014. [13] T. Jankowski, R. Jankowski, Monotone iterative method to second order differential equations with deviating arguments involving Stieltjes integral boundary conditions, Dyn. Syst. Appl. 21 (2012) 17–31. [14] G.S. Ladde, V. Lakshmikantham, A.S. Vatsala, Monotone Iterative Techniques for Nonlinear Differential Equations, Pitman, Boston, 1985. [15] S. Liang, L. Mu, Multiplicity of positive solutions for singular three-point boundary value problems at resonance, Nonlinear Anal. 71 (2009) 2497–2505. [16] F.A. Mehmeti, S. Nicaise, Nonlinear interaction problems, Nonlinear Anal. 20 (1993) 27–61. [17] J.R.L. Webb, G. Infante, Positive solutions of nonlocal boundary value problems: a unified approach, J. London Math. Soc. 74 (2006) 673–693. [18] J.R.L. Webb, G. Infante, Positive solutions of nonlocal boundary value problems involving integral conditions, NoDEA Nonlinear Differ. Equ. Appl. 15 (2008) 45–67. [19] J.R.L. Webb, G. Infante, Non-local boundary value problems of arbitrary order, J. London Math. Soc. 79 (2009) 238–259. [20] J.R.L. Webb, Remarks on nonlocal boundary value problems at resonance, Appl. Math. Comput. 216 (2010) 497–500. [21] J.R.L. Webb, M. Zima, Multiple positive solutions of resonant and non-resonant non-local fourth-order boundary value problems, Glasgow Math. J. 54 (2012) 225–240. [22] L. Yang, Ch. Shen, On the existence of positive solution for a kind of multi-point boundary value problem at resonance, Nonlinear Anal. 72 (2010) 4211– 4220. [23] X. Zhang, M. Feng, W. Ge, Existence result of second-order differential equations with integral boundary conditions at resonance, J. Math. Anal. Appl. 353 (2009) 311–319. [24] M. Zima, Positive solutions for first-order boundary value problems at resonance, Commun. Appl. Anal. 13 (2009) 671–680.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Monotone iterative method for first-order differential equations at resonance Tadeusz Jankowski ´ sk, Poland Gdansk University of Technology, Department of Differential Equations and Applied Mathematics, 11/12 G. Narutowicz Str., 80-233 Gdan
a r t i c l e
i n f o
a b s t r a c t
Keywords: Monotone iterative method First-order differential equations at resonance Stieltjes integrals Existence of solutions (extremal and quasisolutions)
This paper concerns the application of the monotone iterative technique for first-order differential equations involving Stieltjes integrals conditions. We discuss such problems at resonance when the measure in the Stieltjes integral is positive and also when this measure changes the sign. Sufficient conditions which guarantee the existence of extremal, unique and quasi-solutions are given. Three examples illustrate the results. Ó 2014 Elsevier Inc. All rights reserved.
1. Introduction In this paper, we study the existence of solutions of the following problem:
x0 ðtÞ ¼ f ðt; xðtÞÞ FxðtÞ;
t 2 J ¼ ½0; T;
xð0Þ ¼ k½x;
ð1Þ
where k denotes a linear functional on CðJÞ given by
k½x ¼
Z
T
xðsÞdAðsÞ
0
involving a Stieltjes integral with a suitable function A of bounded variation on J. We are interested in the case when problem (1) is at resonance, that is the homogeneous problem
x0 ðtÞ ¼ 0;
t 2 J;
xð0Þ ¼ k½x
ð2Þ
has nontrivial solutions. The resonance condition is
Z
T
dAðsÞ ¼ 1:
ð3Þ
0
Problems with integral (nonlocal) boundary conditions have some applications, for example, in termostat models [5], interaction problems [16], chemical engineering [2]. Positive (nonnegative) solutions for differential equations with nonlocal boundary conditions involving Stieltjes integrals have been discussed in many papers, for example, see papers [4,9–12,17–19] in which Stieltjes integrals appeared with the signed measure, see also [13]. Positive solutions for differential equations at resonance have been discussed, for example, in [1,3,6,15,20–24]. In this paper, we discuss problem (1) at E-mail address: [email protected] http://dx.doi.org/10.1016/j.amc.2014.01.123 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
21
resonance by using the monotone iterative method [14]. According to my knowledge, it is a first application of this method to such problems. 2. Case when the measure dA is positive (nonnegative) First, we study a linear problem. Lemma 1. Let A be a function of bounded variation. Suppose that K; M 2 CðJ; RÞ; p 2 C 1 ðJ; RÞ and
p0 ðtÞ ¼ KðtÞpðtÞ þ MðtÞ;
t 2 J;
ð4Þ
pð0Þ ¼ k½p: In addition, we assume that
Z
T
Z s exp KðsÞds dAðsÞ – 1:
0
ð5Þ
0
Then problem (4) has a unique solution. Proof. Indeed,
Z t Z pðtÞ ¼ exp KðsÞds pð0Þ þ 0
t
exp
Z
0
s
KðsÞds MðsÞds :
0
In view of the boundary condition from (4) and condition (5), we have the assertion. h In the next lemma, we shall discuss a differential inequality. Lemma 2. Let A be a function of bounded variation with a positive measure. Suppose that K 2 CðJ; RÞ; p 2 C 1 ðJ; RÞ and
p0 ðtÞ 6 KðtÞpðtÞ; pð0Þ 6 k½p:
t 2 J;
ð6Þ
In addition, we assume that
Z
T
Z s exp KðsÞds dAðsÞ < 1:
0
ð7Þ
0
Then pðtÞ 6 0 on J. Proof. Note that
Z t pðtÞ 6 pð0Þ exp KðsÞds : 0
Using the boundary condition from (6) and condition (7) we have the assertion. h Let us introduce the notion of lower and upper solutions of problem (1). We say that u 2 C 1 ðJ; RÞ is called a lower solution of (1) if
u0 ðtÞ 6 FuðtÞ;
t 2 J;
uð0Þ 6 k½u
and it is an upper solution of problem (1) if the above inequalities are reversed. We use the following assumption: H1 : f 2 CðJ R; RÞ; A is a function of bounded variation and condition (3) holds. The next theorem concerns the situation when problem (1) has extremal solutions. Theorem 1. Let Assumption H1 hold. Assume that the measure dA is positive. Let y0 ; z0 2 C 1 ðJ; RÞ be lower and upper solutions of (1), respectively and y0 ðtÞ 6 z0 ðtÞ; t 2 J. In addition, we assume that H2 : there exists a function K 2 CðJ; RÞ such that
f ðt; u1 Þ f ðt; v 1 Þ 6 KðtÞ½v 1 u1 ;
22
T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
if y0 ðtÞ 6 u1 6 v 1 6 z0 ðtÞ, H3 : condition (7) holds. Then problem (1) has, in the sector ½y0 ; z0 , minimal and maximal solutions, where
½y0 ; z0 ¼ fw 2 C 1 ðJ; RÞ : y0 ðtÞ 6 wðtÞ 6 z0 ðtÞ; t 2 Jg:
Proof. For n ¼ 0; 1; . . ., let
y0nþ1 ðtÞ ¼ Fyn ðtÞ KðtÞ½ynþ1 ðtÞ yn ðtÞ;
t 2 J;
ynþ1 ð0Þ ¼ k½ynþ1 ;
z0nþ1 ðtÞ ¼ Fzn ðtÞ KðtÞ½znþ1 ðtÞ zn ðtÞ;
t 2 J;
znþ1 ð0Þ ¼ k½znþ1 ;
with F defined as in (1). The sequences fyn ; zn g are well defined by Lemma 1. We first show that
y0 ðtÞ 6 y1 ðtÞ 6 z1 ðtÞ 6 z0 ðtÞ;
t 2 J:
ð8Þ
Put p ¼ y0 y1 ; q ¼ z1 z0 . This shows
pð0Þ 6 k½p;
qð0Þ 6 k½q;
p0 ðtÞ 6 Fy0 ðtÞ Fy0 ðtÞ þ KðtÞ½y1 ðtÞ y0 ðtÞ ¼ KðtÞpðtÞ; q0 ðtÞ 6 Fz0 ðtÞ KðtÞ½z1 ðtÞ z0 ðtÞ Fz0 ðtÞ ¼ KðtÞqðtÞ: By Lemma 2, y0 ðtÞ 6 y1 ðtÞ; z1 ðtÞ 6 z0 ðtÞ; t 2 J. Now, we put p ¼ y1 z1 . In view of Assumption H2 , we have
pð0Þ ¼ k½p; p0 ðtÞ ¼ Fy0 ðtÞ Fz0 ðtÞ KðtÞ½y1 ðtÞ y0 ðtÞ z1 ðtÞ þ z0 ðtÞ 6 KðtÞpðtÞ: Lemma 2 yields y1 ðtÞ 6 z1 ðtÞ on J. It proves (8). In the next step, we show that y1 ; z1 are lower and upper solutions of problem (1), respectively. Note that
y01 ðtÞ ¼ Fy0 ðtÞ KðtÞ½y1 ðtÞ y0 ðtÞ Fy1 ðtÞ þ Fy1 ðtÞ 6 Fy1 ðtÞ; z01 ðtÞ ¼ Fz0 ðtÞ KðtÞ½z1 ðtÞ z0 ðtÞ Fz1 ðtÞ þ Fz1 ðtÞ P Fz1 ðtÞ; by Assumption H2 . This proves that y1 ; z1 are lower and upper solutions of problem (1), respectively. Using the mathematical induction, we can show that
y0 ðtÞ 6 y1 ðtÞ 6 6 yn ðtÞ 6 ynþ1 ðtÞ 6 znþ1 ðtÞ 6 zn ðtÞ 6 6 z1 ðtÞ 6 z0 ðtÞ for t 2 J and n ¼ 0; 1; . . . . Now we shall prove that the sequences fyn ; zn g converge to their limit functions y; z, respectively. First, we need to show that the sequences are bounded and equicontinuous on J. Indeed,
A1 6 y0 ðtÞ 6 yn ðtÞ 6 zn ðtÞ 6 z0 ðtÞ 6 A2 ;
t 2 J; n ¼ 0; 1; . . . ;
so the sequences fyn ; zn g are uniformly bounded. Note that y0n and z0n are bounded on J by W > 0 because jf ðt; yn ðtÞÞj is bounded on J ½A1 ; A2 . Hence yn ; zn are equicontinuous because for > 0; t1 ; t2 2 J such that jt1 t 2 j < =W we have
jyn ðt 1 Þ yn ðt 2 Þj ¼ jy0n ðnÞjjt 1 t 2 j < jzn ðt 1 Þ zn ðt2 Þj < : The Arzela–Ascoli theorem guarantees the existence of subsequences fynk ; znk g of fyn ; zn g, respectively, and continuous functions y; z with ynk ; znk converging uniformly on J to y and z, respectively. Note that ynk ; znk satisfy the integral equations
8 R h i > < ynk þ1 ðtÞ ¼ exp 0t KðsÞds ynk þ1 ð0Þ þ Pynk ðtÞ ; t 2 J; R
> : znk þ1 ðtÞ ¼ exp 0t KðsÞds znk þ1 ð0Þ þ Pznk ðtÞ ; t 2 J
and
ynk þ1 ð0Þ ¼ k½ynk þ1 ; znk þ1 ð0Þ ¼ k½znk þ1 ;
with
Pynk ðtÞ ¼
Z 0
t
exp
Z 0
s
h i KðsÞds Fynk ðsÞ þ KðsÞynk ðsÞ ds:
If nk ! 1, then from the above relations, we have
23
T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
8 R > < yðtÞ ¼ exp 0t KðsÞds ½yð0Þ þ PyðtÞ; t 2 J; yð0Þ ¼ k½y; R > : zðtÞ ¼ exp 0t KðsÞds ½zð0Þ þ PzðtÞ; t 2 J; zð0Þ ¼ k½z; because f is continuous. Thus y; z 2 C 1 ðJÞ and
y0 ðtÞ ¼ FyðtÞ;
z0 ðtÞ ¼ FzðtÞ; t 2 J:
It proves that y; z are solutions of problem (1) and
y0 ðtÞ 6 yðtÞ 6 zðtÞ 6 z0 ðtÞ;
t 2 J:
Now we need to show that y; z are extremal solutions of (1) in the sector ½y0 ; z0 . Let w 2 ½y0 ; z0 be any solution of (1). We assume that ym ðtÞ 6 wðtÞ 6 zm ðtÞ; t 2 J for some m. Let p ¼ ymþ1 w; q ¼ w zmþ1 . Then,
pð0Þ ¼ k½p;
qð0Þ ¼ k½q;
p0 ðtÞ ¼ Fym ðtÞ KðtÞ½ymþ1 ðtÞ ym ðtÞ FwðtÞ 6 KðtÞpðtÞ; q0 ðtÞ ¼ FwðtÞ Fzm ðtÞ þ KðtÞ½zmþ1 ðtÞ zm ðtÞ 6 KðtÞqðtÞ; by Assumption H2 . This and Lemma 2 give ymþ1 ðtÞ 6 wðtÞ 6 zmþ1 ðtÞ; t 2 J. By induction, yn ðtÞ 6 wðtÞ 6 zn ðtÞ; t 2 J; n ¼ 0; 1; . If n ! 1, then we have the assertion. This ends the proof. h Example 1. Consider the problem:
(
2 x0 ðtÞ ¼ exp t sin xðtÞ þ 2 cos xðtÞ FxðtÞ;
t 2 J ¼ ½0; ln 2;
ð9Þ
xð0Þ ¼ k½x; with
dAðtÞ ¼
p 2 ln 2
sin
tp dt: ln 2
Note that the measure dA is positive and
Z
ln 2
dAðtÞ ¼ 1;
0
so problem (9) satisfies the resonance condition (3). Indeed, KðtÞ ¼ 2ðln 2 þ 1Þ. Let y0 ðtÞ ¼ 0; z0 ðtÞ ¼ p; t 2 J. Note that
Fy0 ðtÞ ¼ 3 > 0 ¼ y00 ðtÞ;
t 2 J;
Fz0 ðtÞ ¼ 1 < 0 ¼ z00 ðtÞ; t 2 J; k½y0 ¼ 0 ¼ y0 ð0Þ; k½z0 ¼ p ¼ z0 ð0Þ: This proves that y0 ; z0 are lower and upper solutions of problem (9), respectively. Moreover,
Z 0
ln 2
Z s Z exp KðsÞds dAðsÞ < 0
ln 2
dAðsÞ ¼ 1; 0
so condition (7) holds. Hence, problem (9) has extremal solutions in the region ½y0 ; z0 , by Theorem 1. Our next theorem concerns the case when problem (1) has a unique solution at the resonance case. Theorem 2. Assume that all assumptions of Theorem 1 hold. In addition, we assume that H4 : there exists a function Q 2 CðJ; RÞ such that KðtÞ þ Q ðtÞ P 0; t 2 J, condition (7) holds with Q instead of K and
f ðt; uÞ f ðt; v Þ P Q ðtÞðv uÞ if y0 6 u 6 v 6 z0 : Then problem (1) has, in the sector ½y0 ; z0 , a unique solution. Proof. Theorem 1 guarantees that problem (1) has extremal solutions y; z and y0 ðtÞ 6 yðtÞ 6 zðtÞ 6 z0 ðtÞ; t 2 J. Put p ¼ z y, so pðtÞ P 0 on J and
p0 ðtÞ ¼ FzðtÞ FyðtÞ 6 Q ðtÞpðtÞ;
t 2 J; pð0Þ ¼ k½p:
This and Lemma 2 show that y ¼ z, so problem (1) has a unique solution. This ends the proof.
h
24
T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
Example 2. Consider the following problem:
(
K < 0; t 2 J ¼ ½0; p;
x0 ðtÞ ¼ KxðtÞ þ 3 FxðtÞ; Rp xð0Þ ¼ 12 0 xðsÞ sin tdt:
ð10Þ
We see that now
k½x ¼
1 2
Z p
xðtÞ sin tdt:
0
Note that
1 2
Z p
sin sds ¼ 1;
0
so condition (3) holds and problem (10) is at resonance. It is easy to verify that xðtÞ ¼ K3 ; t 2 J is the unique solution of problem (10). Indeed, KðtÞ ¼ Q ðtÞ ¼ jKj; t 2 J. Put y0 ðtÞ ¼ K4 ; z0 ðtÞ ¼ K4 ; t 2 J. Then
Fy0 ðtÞ ¼ 7 > 0 ¼ y00 ðtÞ;
t 2 J;
Fz0 ðtÞ ¼ 1 < 0 ¼ z00 ðtÞ; k½y0 ¼ y0 ð0Þ;
t 2 J;
k½z0 ¼ z0 ð0Þ;
so y0 ; z0 are lower and upper solutions of problem (10), respectively. Moreover, for KðtÞ ¼ Q ðtÞ ¼ jKj, we have
Z p 0
Z s Z 1 p exp KðsÞds dAðsÞ < sin sds ¼ 1; 2 0 0
so condition (7) holds. Hence, problem (10) has a unique solution at resonance in the region ½y0 ; z0 , by Theorem 2. 3. Case when dA is a signed measure Assume that
AðtÞ 6 0; t 2 ½0; h;
and AðtÞ P 0; t 2 ½h; T:
ð11Þ
1
We say that u; v 2 C ðJ; RÞ are called coupled lower and upper solutions of problem (1) if
(
u0 ðtÞ 6 FuðtÞ; t 2 J;
v 0 ðtÞ P F v ðtÞ;
t 2 J;
uð0Þ 6
Rh
v ðsÞdAðsÞ þ
0
v ð0Þ P
Rh 0
RT
uðsÞdAðsÞ; RT v ðsÞdAðsÞ: h
h
uðsÞdAðsÞ þ
We say that X; Y 2 C 1 ðJ; RÞ are called quasi-solutions of (1) if
(
X 0 ðtÞ ¼ FXðtÞ; t 2 J; 0
Y ðtÞ ¼ FYðtÞ; t 2 J;
Xð0Þ ¼ Yð0Þ ¼
Rh 0 Rh 0
YðsÞdAðsÞ þ XðsÞdAðsÞ þ
RT h
XðsÞdAðsÞ;
h
YðsÞdAðsÞ:
RT
We first formulate conditions under which a corresponding linear system has a unique solution. Lemma 3. Let A be a function of bounded variation and let condition (11) hold. Suppose that K; M 1 ; M 2 2 CðJ; RÞ; p; q 2 C 1 ðJ; RÞ and
( (
p0 ðtÞ ¼ KðtÞpðtÞ þ M 1 ðtÞ; t 2 J; Rh RT pð0Þ ¼ 0 qðsÞdAðsÞ þ h pðsÞdAðsÞ;
ð12Þ
q0 ðtÞ ¼ KðtÞqðtÞ þ M2 ðtÞ; t 2 J; Rh RT qð0Þ ¼ 0 pðsÞdAðsÞ þ h qðsÞdAðsÞ:
In addition, we assume that
Z 1
T
2 CðsÞdAðsÞ –
Z
h
0
with
Z t CðtÞ ¼ exp KðsÞds : 0
h
!2 CðsÞdAðsÞ
;
ð13Þ
25
T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
Then system (12) has a unique solution ðp; qÞ. Proof. Put
Di ðtÞ ¼
Z
t
exp
Z
0
s
KðsÞds M i ðsÞds;
i ¼ 1; 2:
0
Then,
pðtÞ ¼ CðtÞ½pð0Þ þ D1 ðtÞ;
qðtÞ ¼ CðtÞ½qð0Þ þ D2 ðtÞ; t 2 J
and
(
pð0Þ ¼ qð0Þ ¼
Rh 0
CðsÞ½qð0Þ þ D2 ðsÞdAðsÞ þ
0
CðsÞ½pð0Þ þ D1 ðsÞdAðsÞ þ
Rh
RT h
CðsÞ½pð0Þ þ D1 ðsÞdAðsÞ;
h
CðsÞ½qð0Þ þ D2 ðsÞdAðsÞ:
RT
ð14Þ
System (14) has a unique solution with respect to pð0Þ; qð0Þ, in view of condition (13). It proves that the assertion holds. This ends the proof. h Lemma 4. Let A be a function of bounded variation and let condition (11) hold. Suppose that K 2 CðJ; RÞ; p; q 2 C 1 ðJ; RÞ and
8 0 p ðtÞ 6 KðtÞpðtÞ; q0 ðtÞ 6 KðtÞqðtÞ; t 2 J; > > < Rh RT pð0Þ 6 0 qðsÞdAðsÞ þ h pðsÞdAðsÞ; > > : qð0Þ 6 R h pðsÞdAðsÞ þ R T qðsÞdAðsÞ: 0 h
ð15Þ
In addition, we assume that
Z
h
CðsÞdAðsÞ þ
0
Z
T
CðsÞdAðsÞ < 1;
ð16Þ
h
with C defined as in Lemma 3. Then pðtÞ 6 0; qðtÞ 6 0 on J. Proof. Note that
pðtÞ 6 CðtÞpð0Þ; and
pð0Þ 6 qð0Þ qð0Þ 6
Rh
0 Rh pð0Þ 0
qðtÞ 6 CðtÞqð0Þ; t 2 J
CðsÞdAðsÞ þ pð0Þ CðsÞdAðsÞ þ qð0Þ
RT
ð17Þ
h
CðsÞdAðsÞ;
h
CðsÞdAðsÞ:
RT
This and condition (16) show that pð0Þ 0; qð0Þ 6 0. Combining this with (17) we have the assertion of Lemma 4. This ends the proof. h Similarly, we can prove the following. Lemma 5. Let A be a function of bounded variation and let condition (11) hold. Suppose that K 2 CðJ; RÞ; p 2 C 1 ðJ; RÞ and
p0 ðtÞ 6 KðtÞpðtÞ; t 2 J;
pð0Þ 6
Z
h
pðsÞdAðsÞ þ 0
Z
T
pðsÞdAðsÞ:
h
In addition, we assume that condition (16) holds. Then pðtÞ 6 0 on J. Theorem 3. Let assumptions H1 ; H2 hold. Let y0 ; z0 2 C 1 ðJ; RÞ be coupled lower and upper solutions of (1) and y0 ðtÞ 6 z0 ðtÞ; t 2 J. Assume that conditions (11) and (16) are satisfied. Then problem(1) has, in the sector ½y0 ; z0 , coupled quasi-solutions. Proof. For n ¼ 0; 1; . . ., let
y0nþ1 ðtÞ ¼ Fyn ðtÞ KðtÞ½ynþ1 ðtÞ yn ðtÞ;
t 2 J;
z0nþ1 ðtÞ ¼ Fzn ðtÞ KðtÞ½znþ1 ðtÞ zn ðtÞ; t 2 J; Rh RT ynþ1 ð0Þ ¼ 0 znþ1 ðsÞdAðsÞ þ h ynþ1 ðsÞdAðsÞ; Rh RT znþ1 ð0Þ ¼ 0 ynþ1 ðsÞdAðsÞ þ h znþ1 ðsÞdAðsÞ:
26
T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
Observe that functions y1 ; z1 are well defined by Lemma 3. We first show that
y0 ðtÞ 6 y1 ðtÞ 6 z1 ðtÞ 6 z0 ðtÞ;
t 2 J:
ð18Þ
Put p ¼ y0 y1 ; q ¼ z1 z0 . This shows
p0 ðtÞ 6 KðtÞpðtÞ; q0 ðtÞ 6 KðtÞqðtÞ; Z h Z T Z pð0Þ 6 z0 ðsÞdAðsÞ þ y0 ðsÞdAðsÞ 0
qð0Þ 6
Z
h
pðsÞdAðsÞ þ
0
z1 ðsÞdAðsÞ
Z
0
Z
h
h
T
y1 ðsÞdAðsÞ ¼
Z
h
h
qðsÞdAðsÞ þ
Z
0
T
pðsÞdAðsÞ;
h
T
qðsÞdAðsÞ:
h
By Lemma 4, y0 ðtÞ 6 y1 ðtÞ; z1 ðtÞ 6 z0 ðtÞ; t 2 J. Now, we put p ¼ y1 z1 , so
pð0Þ ¼
Z
h
z1 ðsÞdAðsÞ þ
Z
0
T
h
y1 ðsÞdAðsÞ
Z
h
0
y1 ðsÞdAðsÞ
Z
T
z1 ðsÞdAðsÞ ¼
h
Z
h
0
pðsÞdAðsÞ þ
Z
T
pðsÞdAðsÞ:
h
Moreover,
p0 ðtÞ ¼ Fy0 ðtÞ Fz0 ðtÞ KðtÞ½y1 ðtÞ y0 ðtÞ z1 ðtÞ þ z0 ðtÞ 6 KðtÞpðtÞ; by Assumption H2 . Lemma 5 yields y1 ðtÞ 6 z1 ðtÞ on J. It proves (18). It is easy to show that y1 ; z1 are coupled lower and upper solutions of problem (1). Moreover, using the mathematical induction, we can show that
y0 ðtÞ 6 y1 ðtÞ 6 6 yn ðtÞ 6 ynþ1 ðtÞ 6 znþ1 ðtÞ 6 zn ðtÞ 6 6 z1 ðtÞ 6 z0 ðtÞ for t 2 J and n ¼ 0; 1; . The sequences fyn ; zn g converge to their limit functions y; z, respectively. Indeed, y; z are quasi-solutions of problem (1) and y0 6 y 6 z 6 z0 . It ends the proof. h Example 3. Let
(
Indeed,
x0 ðtÞ ¼ expðxðtÞÞ þ 10 cos x þ 12 t ; t 2 J ¼ ½0; 1; R1 xð0Þ ¼ 0 xðtÞdAðtÞ with dAðtÞ ¼ 2ð3t 1Þdt:
R1 0
dAðtÞ ¼ 1 and 2ð3t 1Þ 6 0; t 2 0; 13 ;
2ð3t 1Þ P 0; t 2
ð19Þ 1 3
;1 .
Let y0 ðtÞ ¼ t; z0 ðtÞ ¼ 3 t; t 2 J. Note that
3 3 > 1 ¼ y00 ðtÞ; t P expð1Þ þ 10 cos 2 2 1 5 < 1 ¼ z00 ðtÞ: Fz0 ðtÞ ¼ expðt 3Þ þ 10 cos 3 t 6 expð2Þ þ 10 cos 2 2 Z 1 Z 1 3 2 ð3 tÞð6t 2Þdt þ tð6t 2Þdt ¼ > 0 ¼ y0 ð0Þ; 1 27 0 3 Z 1 Z 1 3 79 tð6t 2Þdt þ ð3 tÞð6t 2Þdt ¼ < 3 ¼ z0 ð0Þ: 1 27 0 3
Fy0 ðtÞ ¼ expðtÞ þ 10 cos
Moreover KðtÞ ¼ 11. Condition (16) holds too. In view of Theorem 3, problem (19) has coupled quasi solutions. 4. Some comments 1. Let a; b P 0 and
8 t 2 ½0; hÞ; > < 0; AðtÞ ¼ a; t 2 ½h; TÞ; > : b a; t ¼ T: Then,
k½x ¼ axðhÞ þ bxðTÞ; so problem (1) takes the form
T. Jankowski / Applied Mathematics and Computation 233 (2014) 20–28
x0 ðtÞ ¼ f ðt; xðtÞÞ FxðtÞ; t 2 J ¼ ½0; T; xð0Þ ¼ axðhÞ þ bxðTÞ; a; b > 0:
27
ð20Þ
Such problems without condition (3) have been discussed, for example in [7,8]. Similarly as before, we define the notions of coupled lower and upper solutions and quasi-solutions of problem (20); so we say that u; v 2 C 1 ðJ; RÞ are called coupled lower and upper solutions of problem (20) if
uð0Þ 6 av ðhÞ þ buðTÞ;
u0 ðtÞ 6 FuðtÞ; t 2 J;
v 0 ðtÞ P F v ðtÞ;
t 2 J;
v ð0Þ P auðhÞ þ bv ðTÞ:
Let y0 ; z0 be coupled lower and upper solutions of problem (20). Then, we can show that the sequences fyn ; zn g:
y0nþ1 ðtÞ ¼ Fyn ðtÞ KðtÞ½ynþ1 ðtÞ yn ðtÞ;
ynþ1 ð0Þ ¼ aznþ1 ðhÞ þ bynþ1 ðTÞ;
z0nþ1 ðtÞ ¼ Fzn ðtÞ KðtÞ½znþ1 ðtÞ zn ðtÞ;
znþ1 ð0Þ ¼ aynþ1 ðhÞ þ bznþ1 ðTÞ
converge to quasi-solutions of problem (20) provided that
aCðhÞ þ bCðTÞ < 1;
R t with CðtÞ ¼ exp 0 KðsÞds . Then, in view of Theorem 3, problem (20) has quasi-solutions in ½y0 ; z0 . 2. Let
k½x ¼
ð21Þ
r X
ai xðti Þ; ai 2 R; 0 < t1 < t2 < < tr1 < tr ¼ T:
i¼1
Then condition (21) is replaced by r X jai jCðti Þ < 1: i¼1
3. Let AðtÞ ¼ A1 ðtÞ þ A2 ðtÞ, where A2 ðtÞ 6 0; t 2 ½0; h; A2 ðtÞ P 0; t 2 ½h; T and
8 0; > > > > > a1 ; > > > ; > > > > > ar1 þ ar2 ; > > : ar þ ar1 ;
t 2 ½0; t 1 Þ; t 2 ½t1 ; t 2 Þ; t 2 ½t2 ; t 3 Þ; t 2 ½tr1 ; tr Þ; t ¼ tr ¼ T:
For ai 2 R; 0 < t 1 < t 2 < < tr1 < t r ¼ T we have
k½x ¼
r X
ai xðti Þ þ
Z
h
xðsÞdA2 ðsÞ þ
0
i¼1
Z
T
xðsÞdA2 ðsÞ:
h
Then condition (21) is replaced by
Z r X jai jCðti Þ i¼1
h
CðsÞdA2 ðsÞ þ 0
Z
T
CðsÞdA2 ðsÞ < 1:
h
We can also consider the case when A2 is negative in k subintervals of ½0; T; k > 1.
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