EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 13, No. 3, 2020, 414-426
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Existence and uniqueness of solutions for the first order
non-linear differential equations with multi-point

boundary conditions

M.J. Mardanov1, Y.A.Sharifov1,2,∗, H.N. Aliyev3, R.A. Sardarova4

1 Institute of Mathematics and Mechanics, ANAS, Baku, Azerbaijan
2 Baku State University Baku, Azerbaijan
3 Baku Engineering University, Khirdalan city, Azerbaijan
4 Azerbaijan State University of Economics (UNEC), Baku,Azerbaijan

Abstract. This article discusses the existence and uniqueness of solutions for the system of non-
linear first order ordinary differential equations with multipoint boundary conditions. The Green
function is constructed, and the problem is reduced to the equivalent integral equation. Existence
and uniqueness of the solution to this problem is studied using the Banach contraction mapping
principle and Schaefer’s fixed point theorem.

2020 Mathematics Subject Classifications: 34A12, 34B10, 34B15

Key Words and Phrases: Multipoint boundary conditions, existence and uniqueness solutions,
fixed point theorems, first order differential equations, Schaefer’s fixed point theorem.

1. Introduction and Problem Statement

Multipoint boundary value problems for the ordinary differential equations (ODEs)
arise in modeling the broad class of natural processes. For example, if to consider the
dynamical system with n degrees of freedom, exactly n states observed at n different
instants of time, then the mathematical description of this system leads to the multi-
point boundary value problem. As another example we can note the vibrations of a
uniform cross-section string composed of N parts of different densities and also some
problems in the theory of elastic stability [29]. As another example we can note the
vibrations of a uniform cross-section string composed of N parts of different densities
and also some problems in the theory of elastic stability [29]. In some cases multipoint
boundary value problems also arise when discrediting the boundary value problems for
the partial differential equations. Due to these and many other strong relation with a

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v13i3.3698

Email addresses: misirmardanov@yahoo.com (M.J. Mardanov), sharifov22@rambler.ru (Y.A.Sharifov),
hualiyev@beu.edu.az (H.N. Aliyev), sardarova.rita.77@gmail.com (R.A. Sardarova )

https://www.ejpam.com 414 c© 2020 EJPAM All rights reserved.



Y.A.Sharifov et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 414-426 415

broad range of applications in different fields of physics and mathematics such problems
are under the intensive focus of many researchers [9, 10].

It should be noted that the multipoint boundary value problems have been well studied
for the second order differential equations (see [4, 11–13, 22, 24] and references therein).
These works were mainly initiated by Ilin and Moiseev [12]. Since then, nonlinear multi-
point boundary-value problems have been studied by several authors using the Leray-
Schauder Continuation Theorem, Leray-Schaudern nonlinear alternatives, coincidence de-
gree theory, and fixed point theorem in cones. However, for the first order differential
equations, such problems have been less studied. Examples of such works can be shown
[1, 3, 15–17, 19, 23, 25, 30, 31]

Similar problems for two-point and integral boundary value problems are considered
in [2, 5–8, 14, 18, 20, 21, 24, 26–28].

Note that the problem under consideration in this work was also studied by M. Urabe.
In [30] he gives the similar result. But those results were obtained under more strong
conditions. Thus he requires the existence of the approximate solution of the considered
problem with high enough accuracy that cannot be achieved in many cases. Moreover,
the fundamental matrix of some quasilinear system also should be known in [30] that is
difficult problem itself. The results in this work are obtained by only the initial data of
the problem and we do not need solving any auxiliary problem.

In this work for the first time Green function is constructed for the multi-point bound-
ary value problem. The considered problem is reduced to the equivalent integral equations.
Then the existence and uniqueness result are studied using the Banach contraction map-
ping principle. The existence of the solution is also proved by applying Schaefer’s fixed
point theorem.

Consider the existence and uniqueness of solutions of the nonlinear differential equa-
tions of the type

ẋ(t) = f(t, x), t ∈ [0, T ], (1)

with multi-point boundary conditions

m∑
i=0

lix(ti) = α, (2)

where li, i = 1, 2, ...,m are constant square matrices of order n such that detN 6= 0,

N =
m∑
i=0

li; f : [0, T ] × Rn → Rn is a given function; points ti, i = 1, 2, ...,m satisfy the

condition 0 = t0 < t1 < · · · < tm = T . We denote by C([0, T ];Rn) the Banach space of
all continuous functions from [0, T ] into Rn with the norm ‖x‖ = max {|x(t)| : t ∈ [0, T ]}
where |·| is the norm in the space Rn.

This paper is organized as follows. In Section 2 we introduce definitions and lemmas
which are the key tools for our main result. Section 3 focuses the theorems on the exis-
tence and uniqueness of the solution of problem (1)-(2) established under some sufficient
conditions on the nonlinear terms.



Y.A.Sharifov et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 414-426 416

2. Preliminaries

We define the solution of problem (1)-(2) as follows:

Definition 1. The function x ∈ C([0, T ], Rn) is called a solution of problem (1)-(2) if
ẋ(t) = f(t, x(t)) for each t ∈ [0, T ], and boundary conditions (2) are satisfied.

For the sake of simplicity, we can consider the following problem:

ẋ = y(t), t ∈ [0, T ], (3)

m∑
i=0

lix(ti) = α. (4)

Lemma 1. Let y ∈ C([0, T ], Rn). Then the unique solution x(t) ∈ C([0, T ], Rn) of the
boundary value problem for differential equation (3) with boundary conditions (4) is given
by

x(t) = N−1α+

T∫
0

G(t, τ)y(τ)dτ, (5)

where

G (t, τ) =


G1 (t, τ) , t ∈ [0, t1] ,
G2 (t, τ) , t ∈ (t1, t2] ,
................................

Gm (t, τ) , t ∈ (tm−1, T ] ,

with

Gi (t, τ) =



N−1l0, t0 ≤ τ ≤ t1,

N−1

(
1∑

k=0

lk

)
, t1 < τ ≤ t2,

..............................................

N−1

(
i−1∑
k=0

lk

)
, ti−1 < τ ≤ ti,

N−1

(
i∑

k=0

lk

)
, ti < τ ≤ t,

−N−1

(
m∑

k=i+1

li

)
, t < τ ≤ ti+1,

−N−1

(
m∑

k=i+2

li

)
, ti+1 < τ ≤ ti+2,

...................................................
−N−1lm, tm−1 < τ ≤ T,

i = 1, 2, ...,m.



Y.A.Sharifov et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 414-426 417

Proof. If the function x = x(·) is a solution of equation (3), then for t ∈ (0, T )

x(t) = x(0) +

t∫
0

y(τ)dτ, (6)

where x0 is an arbitrary constant vector. Now we define x0 so that, the function in equality
(6) satisfies condition (4). Then we have

m∑
i=0

li[x0 +

ti∫
0

y (s) ds] = α.

This obviously gives

x0 = N−1α−N−1

 m∑
i=1

li

ti∫
0

y (s) ds

 . (7)

Considering the value x0 determined from the equality (7) in (6) we get

x (t) = N−1α−N−1

 m∑
i=1

li

ti∫
0

y (s) ds

+

t∫
0

y (s) ds. (8)

Suppose that t ∈ [0, t1] Then equality (8) may be written as follows:

x(t) = N−1α−N−1

l1 t∫
0

y(τ)dτ + l1

t1∫
t

y(τ)dτ

−N−1

l2 t∫
0

y(τ)dτ + l2

t1∫
t

y(τ)dτ



−N−1l2

t2∫
t1

y (τ)dτ −N−1

l3 t∫
0

y (τ) dτ + l3

t1∫
t

y (τ) dτ

−N−1l3

 2∑
i=1

ti+1∫
ti

y (τ)dτ



−...−N−1

lm t∫
0

y (τ)dτ + lm

t1∫
t

y (τ) dτ

−N−1lm

 m∑
i=1

ti+1∫
ti

y (τ) dτ

+

t∫
0

y (τ) dτ.

One can easily rewrite this equality in the equivalent form:

x(t) = N−1α+

t∫
0

(
E −N−1

m∑
i=1

li

)
y(τ)dτ −N−1

t1∫
t

(
m∑
i=1

li

)
y(τ)dτ

−N−1

(
m∑
i=2

li

) t2∫
t1

y (τ) dτ −N−1

(
m∑
i=3

li

) t3∫
t2

y (τ)dτ − ...−N−1lm

T∫
tm−1

y (τ)dτ, (9)



Y.A.Sharifov et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 414-426 418

where E is an identity matrix. Since equality(
E −N−1

m∑
i=1

li

)
= N−1l0

is valid following function may be introduced

G1 (t, τ) =



N−1l0, t0 ≤ τ ≤ t,

−N−1

(
m∑
i=1

li

)
, t < τ ≤ t1,

−N−1

(
m∑
i=2

li

)
, t1 < τ ≤ t2,

−N−1

(
m∑
i=3

li

)
, t2 < τ ≤ t3,

.............................................
−N−1lm, tm−1 < τ ≤ T.

Considering the last one we can transfer equality (9) to the following an integral equation

x(t) = N−1α+

T∫
0

G1(t, τ)y(τ)dτ, t ∈ [0, t1] .

Assuming t ∈ (t1, t2] we can write equality (8) in the following form

x(t) = N−1α−N−1

(
m∑
i=1

li

) t1∫
0

y(t)dt−N−1

(
m∑
i=2

li

) t∫
t1

y(τ)dτ +

t2∫
t

y(τ)dτ



−N−1

(
m∑
i=3

li

) t3∫
t2

y (τ) d−N−1

(
m∑
i=4

li

) t4∫
t3

y (τ) dτ − ...−N−1lm

T∫
tm−1

y (τ) dτ

+

t1∫
0

y(t)dt+

t∫
t1

y(τ)dτ.

From this it is easy to derive

x(t) = N−1α+N−1l0

t1∫
0

y(t)dt+N−1

(
1∑
i=0

li

) t∫
t1

y(τ)dτ

−N−1

(
m∑
i=2

l

) t2∫
t

y (τ) dτ

−N−1

(
m∑
i=3

li

) t3∫
t2

y (τ) d−N−1

(
m∑
i=4

li

) t4∫
t3

y (τ) dτ − ...−N−1lm

T∫
tm−1

y (τ) dτ.



Y.A.Sharifov et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 414-426 419

In this step we again introduce a new function

G2 (t, τ) =



N−1l0, t0 ≤ τ ≤ t1,

N−1

(
1∑
i=0

li

)
, t1 < τ ≤ t,

−N−1

(
m∑
i=2

li

)
, t < τ ≤ t2,

−N−1

(
m∑
i=3

li

)
, t2 < τ ≤ t3,

..........................................
−N−1lm, tm−1 < τ ≤ T.

Therefore we conclude that if t ∈ (t1, t2] then the solution of the considered boundary
value problem can be presented in the form

x(t) = N−1α+

T∫
0

G2(t, τ)y(τ)dτ.

Continuing this process in a similar way, for the segment t ∈ (ti, ti+1] we get

Gi (t, τ) =



N−1l0, t0 ≤ τ ≤ t1,

N−1

(
1∑
i=0

li

)
, t1 < τ ≤ t2,

................................................

N−1

(
i−1∑
k=0

lk

)
, ti−1 < τ ≤ ti,

N−1

(
i∑

k=0

lk

)
, ti < τ ≤ t,

−N−1

(
m∑

k=i+1

li

)
, t < τ ≤ ti+1,

−N−1

(
m∑

k=i+2

li

)
, ti+1 < τ ≤ ti+2,

................................................
−N−1lm, tm−1 < τ ≤ T.

Finally we see that the solution of boundary value problem (1)-(2) may be presented in
the form

x(t) = N−1α+

T∫
0

G(t, τ)y(τ)dτ.

Proof is completed.

Lemma 2. Let f ∈ C([0, T ]×Rn;Rn). Then the function x(t) is a solution of boundary
value problem (1)-(2) if and only if x(t) is a solution of the integral equation



Y.A.Sharifov et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 414-426 420

x(t) = N−1α+

T∫
0

G(t, τ)f(τ, x(τ))dτ. (10)

Proof. Let x(t) be a solution of boundary value problem (1)-(2). This lemma can be
proved analogously to Lemma 1. By direct checking it is easy to justify that the solution
of integral equation (10) satisfies also boundary value problem (1)-(2). Lemma 2 is proved.

3. Main results

Let us set the following conditions:
(H1) The function f ∈ C([0, T ]×Rn;Rn) is continuous;
(H2) There exist a constant M ≥ 0 such that

|f (t, x)− f (t, y)| ≤M |x− y|

for t ∈ [0, T ] each and all x, y ∈ Rn;
(H3) There exists a constant K ≥ 0 such that |f(t, x)| ≤ K for each t ∈ [0, T ] and all
x ∈ Rn.

We give here the following uniqueness result.

Theorem 1. Assume that, assumptions(H1) and (H2) hold and

L = TSM < 1, (11)

where
S = max

[0,T ]×[0,T ]
‖G (t, τ)‖ .

Then boundary value problem (1)-(2) has a unique solution on [0, T ].

Proof. To prove the statement of the above theorem we transform the boundary value
problem (1)- (2) into a fixed point problem. Consider the operator

(Fx) (t) = N−1α+

T∫
0

G(t, τ)f(τ, x(τ))dτ. (12)

It is not difficult to see that

F : C ([0, T ] ;Rn)→ C ([0, T ] ;Rn)

Obviously, the fixed points of the operator F are solutions of boundary problem (1)-(2).
Setting

max
[0,T ]
|f(t, 0)| = Mf



Y.A.Sharifov et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 414-426 421

we take

r ≥
∥∥N−1d

∥∥+MfTS

1− L
We show that FBr ⊂ Br, where

Br = {x ∈ C([0, T ]Rn) : ‖x‖ ≤ r}

For x ∈ Br, using (H1), we get

‖(Fx)(t)‖ ≤
∥∥N−1α

∥∥+

T∫
0

|G(t, τ)| (|f(τ, x(τ))− f(τ, 0)|+ |f(τ, 0)|)dτ

≤
∥∥N−1d

∥∥+ S

T∫
0

(M |x|+Mf )dt ≤
∥∥N−1d

∥∥+ SMrT +MfTS ≤
∥∥N−1α

∥∥+MfTS

1− L
≤ r.

In order to show that the operator F is a contraction, for any x, y ∈ Br we have

|Fx− Fy| ≤
T∫
0

|G(t, τ) (f(τ, x(τ))− f(τ, y(τ))|dτ ≤
T∫
0

|G(t, τ)| |f(τ, x(τ))− f(τ, y(τ))| dτ

≤ SM
T∫
0

|x(t)− y(t)| dt ≤SMT max
[0,T ]
|x(t)− y(t)| ≤ SMT ‖x− y‖

or
‖Fx− Fy‖ ≤ L ‖x− y‖ .

As one can see F is contraction by condition (11). So, boundary value problem (1)- (2)
has a unique solution.

Now we give a theorem on the existence of solutions for the considered problem.

Theorem 2. Assume conditions(H1) and (H3) hold. Then boundary value problem (1)-
(2) has at least one solution on [0, T ].

Proof. Let F be the operator defined by (12). We use Schaefer’s fixed point theorem
to prove that F has a fixed point.

First we show that F is continuous. To do this suppose that be {xn} a sequence such
that xn → x in C ([0, T ];Rn). Then for each t ∈ [0, T ]

|(Fx) (t)− (Fxn) (t)| =

∣∣∣∣∣∣
T∫
0

G (t, τ) (f (τ, x (τ))− f (τ, xn (τ)))dτ

∣∣∣∣∣∣
≤ TSM |x (t)− xn (t)| ≤ L ‖x− xn‖ .



Y.A.Sharifov et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 414-426 422

It gives ‖(Fx) (t)− (Fxn) (t)‖ → 0 as n → ∞, which implies that the operator F is
continuous.

The next step is to show that F maps bounded sets from C ([0, T ];Rn) into bounded
sets in C ([0, T ];Rn). To do this it is enough to show that for any η > 0 there exists a
positive constant ω such that for each x ∈ Bη = {x ∈ C ([0, T ] ;Rn) : ‖x‖ ≤ η} we have
‖F (x)‖ ≤ ω. For each t ∈ [0, T ] we have

|(Fx) (t)| ≤
∣∣N−1α

∥∥+ TSK.

From this we obtain
‖(Fx) (t)‖ ≤

∥∥N−1α
∥∥+ TSK = ω.

Now we show that F maps bounded sets from C ([0, T ];Rn) into equicontinuous sets in
C ([0, T ];Rn). Take ξ1, ξ2 ∈ [0, T ] , ξ1 < ξ2, and assume that Bη is a bounded set in
C ([0, T ];Rn) and let x ∈ Bη.

Here two cases should be considered
Case 1. Let ξ1, ξ2 ∈ [ti, ti + 1] . Then

F (x (ξ2))− F (x (ξ1)) =

ξ2∫
ti

N−1

(
i∑

k=0

li

)
f (τ, x (τ))dτ

−
ti+1∫
ξ2

N−1

(
m∑

k=i+1

li

)
f (τ, x (τ)) dτ −

ξ1∫
ti

N−1

(
i∑

k=0

li

)
f (τ, x (τ))dτ

+

ti+1∫
ξ1

N−1

(
m∑

k=i+1

li

)
f (τ, x (τ)) dτ =

ξ2∫
ξ1

N−1

(
i∑

k=0

li

)
f (τ, x (τ)) dτ

+

ξ2∫
ξ1

N−1

(
m∑

k=i+1

li

)
f (τ, x (τ)) dτ =

ξ2∫
ξ1

f (τ, x (τ))dτ.

Case 2. In this case let ξ1 ∈ [ti−1, ti) , ξ2 ∈ [ti, ti+1] . Then

F (x (ξ2))− F (x (ξ1)) =

ti∫
ti−1

N−1

(
i−1∑
k=0

li

)
f (τ, x (τ)) dτ

+

ξ2∫
ti

N−1

(
i∑

k=0

li

)
f (τ, x (τ))dτ −

ti+1∫
ξ2

N−1

(
m∑

k=i+1

li

)
f (τ, x (τ)) dτ

−
ξ1∫

ti−1

N−1

(
i−1∑
k=0

li

)
f (τ, x (τ))dτ +

ti∫
ξ1

N−1

(
m∑
k=i

li

)
f (τ, x (τ)) dτ



Y.A.Sharifov et al. / Eur. J. Pure Appl. Math, 13 (3) (2020), 414-426 423

+

ti+1∫
ti

N−1

(
m∑

k=i+1

li

)
f (τ, x (τ)) dτ =

ti∫
ξ1

f (τ, x (τ))dτ

+

ξ2∫
ti

f (τ, x (τ)) dτ =

ξ2∫
ξ1

f (τ, x (τ)) dτ.

As t2 → t1, the right-hand side of both above equalities tends to zero. Considering the
above results and the Arzela-Ascoli theorem, we can conclude that F : C ([0, T ] ;Rn) →
C ([0, T ] ;Rn) is completely continuous.

Here we establish apriori bounds i.e. we show that the set ∆ = {x ∈ C ([0, T ] ;Rn) :
x = λF (x)} for some 0 < λ < 1 is bounded. Let x ∈ ∆. Then x = λF (x) for some
0 < λ < 1. Thus, for each t ∈ [0, T ] we have

x(t) = λN−1α+ λ

T∫
0

G(t, τ)f(τ, x(τ))dτ.

From here
‖x‖ ≤

∥∥N−1α
∥∥+ SKT.

Therefore, the set ∆ is bounded. The statement of the Schaefer’s fixed point theorem
may be applied and derived that the operator F has at least one fixed point. So, there
exists at least one solution for problems (1)- (2) on [0, T ].

4. Conclusion

It should be noted that the method considered in this paper are general enough and
can be transformed to the different forms to cover a wide class of problems. We established
here the results on the existence and uniqueness of the solutions for the first order nonlinear
differential equations with multi-point boundary conditions. Given in the paper method
can be used in similar multi-point problems for the ordinary differential equations

ẋ = f(t, x), t ∈ [0, T ],

with multi-point and integral boundary conditions of the form

m∑
i=0

lix (ti) +

T∫
0

n (t)x (t) dt = α.

Here 0 = t0 < t1 < ... < tm−1 < tm = T ; n (t) ∈ Rn×n is a given function; li ∈ Rn×n, i =
1, 2, ...,m are given matrices; α ∈ Rn is a given vector; and

detN 6= 0, N =
m∑
i=0

li +

T∫
0

n (t) dt.



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