Int. J. Anal. Appl. (2022), 20:27

Applications of Standard Methods for Solving the Electric Train Mathematical Model

With Proportional Delay

S. M. Khaled∗

Department of Mathematics, Faculty of Science, Helwan University, P.O. Box 11795, Cairo, Egypt

∗Corresponding author: ksmahmoud@ut.edu.sa

Abstract. In electric trains, the current is collected via a certain device, called the Pantograph. The

governing mathematical model of such physical problem is well-known as the Pantograph delay dif-

ferential equation (PDDE): y ′(t) = ay(t)+ by (ct), where c is a proportional delay parameter. In

the literature, a special case of the PDDE was analyzed when c =−1. The objective of this paper is
to determine the general solution of the PDDE for arbitrary c. In addition, it will be shown that the

obtained general solution reduces to exact one at c = −1. Such exact solution is expressed in terms
of several types of functions, e.g., hyberbolic, Mittag-Leffler, and trigonometric functions. Moreover,

it is declared that the exact trigonometric solution is periodic with periodicity 2π√
b2−a2

which agrees

with the corresponding results in the literature. Furthermore, the solution of PDDE is provided at

almost all possible cases of the involved parameters a, b, and c. Finally, the solution Ambartsumian

delay equation (ADE), which has an application in the theory of surface brightness in the Milky Way,

will be recovered as a special case of our results.

1. Introduction

The field of Delay differential equations (DDEs) is a growing area of research due to its wide

applications in applied sciences. A specific and famous application of the DDEs is well-known as

the Pantograph by which the current can be collected in electric trains. The mathematical model

describing the Pantograph technique is called the PDDE [1-5]. Also, other DDEs-models arise in

astronomical applications, as an example, the Ambartsumian delay equation (ADE) [6-14] which is

Received: Apr. 21, 2022.

2010 Mathematics Subject Classification. 93A30.

Key words and phrases. pantograph; electric trains; mathematical model; series method; Mittag-Leffler.

https://doi.org/10.28924/2291-8639-20-2022-27
ISSN: 2291-8639

© 2022 the author(s).

https://doi.org/10.28924/2291-8639-20-2022-27


2 Int. J. Anal. Appl. (2022), 20:27

essential in studying the brightness in the Milky Way. To author’s knowledge, there are no standard

methods to solve DDEs in contrast to the Ordinary differential equations (ODEs). For examples,

the simple ODE y ′(t) = ay(t) can be easily solved via applying the separation technique, while the

simple DDE y ′(t) = ay (ωt) (ω 6= 1) or y ′(t) = ay (t −σ) (σ ∈ R) can’t be solved neither by the

separation technique nor by the other standard methods that dealing with ODEs. On the other hand,

the DDEs can be viewed as a generalization of the ODEs because of the existing of delay parameters.

For illustration, the DDEs y ′(t) = ay (ωt) and y ′(t) = ay (t −σ) reduces to the ordinary version

y ′(t) = ay (t) when ω = 1 and σ = 0, respectively. In this paper, an extended version of the above

DDEs in the form [15-19]:

y ′(t) = ay(t) + by (ct) , y(0) = λ, (1)

is to be analyzed for reals a,b,λ, and c. The model (1) is well-known in the literature [15-19] and

called the PDDE.

This model reduces to the ADE [6-14]]; y ′(t) = −y(t) + 1
q
y
(
t
q

)
, y(0) = λ at the particular

values a = −1 and b = c = 1
q
(q > 1). This last model and its several generalizations have been

addressed by numerous researchers [6-14, 20, 21]. In Ref. [22], the special case of the PDDE, i.e.,

y ′(t) = ay(t) + by (−t) (c = −1) has been solved. The main purpose of this work is to obtain the

solution of the PDDE at arbitrary c. Our analysis is based on the standard series method (SSM) and

the Maclaurin series expansion (MSE). Our results will also be used to derive previous results in the

literature as special cases of ours. The convergence of the obtained series will also be addressed.

2. Solution by SSM

Based on the SSM, the solution of Eq. (1) is assumed as

y(t) =

∞∑
n=0

dnt
n. (2)

From Eq. (3) and Eq. (1), it then follows

∞∑
n=1

ndnt
n−1 = a

∞∑
n=0

dnt
n + b

∞∑
n=0

dnc
ntn, (3)



Int. J. Anal. Appl. (2022), 20:27 3

i.e.,
∞∑
n=0

(n + 1)dn+1t
n =

∞∑
n=0

(a + bcn) dnt
n, (4)

which yields
∞∑
n=0

[(n + 1)dn+1 − (a + bcn) dn] tn = 0. (5)

Therefore

dn+1 =

(
a + bcn

n + 1

)
dn, n ≥ 0. (6)

Accordingly,

d1 =
1

1!
(a + b) d0,

d2 =
1

2
(a + bc) d1 =

1

2
(a + b) (a + bc) d0 =

1

2!

1∏
k=0

(
a + bck

)
d0,

d3 =
1

3

(
a + bc2

)
d2 =

1

6
(a + b) (a + bc)

(
a + bc2

)
d0 =

1

3!

2∏
k=0

(
a + bck

)
d0,

d4 =
1

4

(
a + bc3

)
d3 =

1

24
(a + b) (a + bc)

(
a + bc2

)(
a + bc3

)
d0 =

1

4!

3∏
k=0

(
a + bck

)
d0,

...

dn =
1

n!
(a + b) (a + bc) . . .

(
a + bcn−2

)(
a + bcn−1

)
d0 =

1

n!

n−1∏
k=0

(
a + bck

)
d0. (7)

Hence

y(t) = d0 +

∞∑
n=1

dn t
n,

= d0 +

∞∑
n=1

1

n!

n−1∏
k=0

(
a + bck

)
d0 t

n,

= d0

[
1 +

∞∑
n=1

n−1∏
k=0

(
a + bck

) tn
n!

]
. (8)

Applying the initial condition y(0) = λ, yields

y(t) = λ

[
1 +

∞∑
n=1

tn

n!

n−1∏
k=0

(
a + bck

)]
. (9)



4 Int. J. Anal. Appl. (2022), 20:27

3. Solution by the MSE

According to the MSE, we have

y(t) =

∞∑
n=0

y(n)(0)

n!
tn. (10)

At t = 0, we get from Eq. (1) that

y(1)(0) = ay(0) + by (0) = λ (a + b) . (11)

Differentiating Eq. (11) w.r.t. to t, then

y(2)(t) = ay(1)(t) + bcy(1) (ct) , (12)

and hence,

y(2)(0) = (a + bc) y(1)(0) = λ (a + b) (a + bc) . (13)

Similarly, we have from Eq. (13) that

y(3)(t) = ay(2)(t) + bc2y(2) (ct) . (14)

Accordingly,

y(3)(0) =
(
a + bc2

)
y(2) (0) = λ (a + b) (a + bc)

(
a + bc2

)
. (15)

Repeating this procedure n-times, we get

y(n)(0) = λ (a + b) (a + bc)
(
a + bc2

)
. . . . . .

(
a + bcn−2

)(
a + bcn−1

)
,

= λ

n−1∏
k=0

(
a + bck

)
, n ≥ 1. (16)

Therefore

y(t) = y(0) +

∞∑
n=1

y(n)(0)
tn

n!
,

= λ + λ

∞∑
n=1

n−1∏
k=0

(
a + bck

) tn
n!
,

= λ

[
1 +

∞∑
n=1

tn

n!

n−1∏
k=0

(
a + bck

)]
, (17)

which is also the same solution obtained in the previous section.



Int. J. Anal. Appl. (2022), 20:27 5

4. Convergence analysis

Theorem 1. For a,b ∈R, the closed-form series solution:

y(t) = λ

[
1 +

∞∑
n=1

tn

n!

n−1∏
k=0

(
a + bck

)]
, (18)

has infinite radius of convergence ∀ c ∈ [−1, 1] and hence the series is uniformly convergent on any

compact interval on R.

Proof: Eq. (18) can be expressed as

y(t) = λ

[
1 +

∞∑
n=1

Cn(t)

]
, (19)

where Cn is defined by

Cn(t) =
tn

n!

n−1∏
k=0

(
a + bck

)
, n ≥ 1. (20)

Let ρ is the radius of convergence, by ratio test we can write

1

ρ
= lim
n→∞

∣∣∣∣Cn+1Cn
∣∣∣∣ = limn→∞

∣∣∣∣∣∣
tn+1

(n+1)!

∏n
k=0

(
a + bck

)
tn

n!

∏n−1
k=0 (a + bc

k)

∣∣∣∣∣∣ ,
= lim
n→∞

∣∣∣∣∣(a + bc
n)
∏n−1
k=0

(
a + bck

)
(n + 1)

∏n−1
k=0 (a + bc

k)

∣∣∣∣∣ |t| ,
= lim
n→∞

∣∣∣∣a + bcnn + 1
∣∣∣∣ |t| . (21)

For c ∈ (−1, 1), we have limn→∞cn = 0, then

1

ρ
= lim
n→∞

∣∣∣∣a + bcnn + 1
∣∣∣∣ |t| = 0, ∀ c ∈ (−1, 1), t ∈R. (22)

At c = 1 we have cn = 1 (∀n ∈N), then the limit in Eq. (22) reduces to

1

ρ
= lim
n→∞

∣∣∣∣a + bn + 1
∣∣∣∣ |t| = 0, where c = 1, t ∈R. (23)

At c = −1 we have cn = −1 (if n is odd) and cn = 1 (if n is even), Accordingly, the limit (22)

becomes

1

ρ
= lim
n→∞

∣∣∣∣a±bn + 1
∣∣∣∣ |t| = 0, where c = −1, t ∈R, (24)

which completes the proof. �



6 Int. J. Anal. Appl. (2022), 20:27

5. Special cases & Comparisons

It was shown in the previous section that the solution of the PDDE is given by the closed-form:

y(t) = λ

[
1 +

∞∑
n=1

tn

n!

n−1∏
k=0

(
a + bck

)]
. (25)

In this section, the solutions of several special cases are determined and some of them are compared

with those in the relevant literature.

5.1. λ = 1. At λ = 1, the PDDE (1) becomes

y ′(t) = ay(t) + by (ct) , y(0) = 1, (26)

which has been solved by Fox et. al [19]. The solution of Eqs. (26) can be directly obtained from

(18) by substituting λ = 1, and this gives

y(t) = 1 +

∞∑
n=1

tn

n!

n−1∏
k=0

(
a + bck

)
, (27)

which is the corresponding solution in [19].

5.2. b = 0. At b = 0, the present PDDE reduces to

y ′(t) = ay(t), y(0) = λ, (28)

which has the exact solution:

y(t) = λeat. (29)

Such exact solution can also be derived as a special case of our solution (18). To do that, we substitute

b = 0 into (18) to obtain

y(t) = λ

[
1 +

∞∑
n=1

tn

n!

n−1∏
k=0

a

]
, (30)

i.e.,

y(t) = λ

[
1 +

∞∑
n=1

antn

n!

]
, (31)

where
n−1∏
k=0

a = an, ∀ n ≥ 1. (32)

Eq. (32) can be written as

y(t) = λ

∞∑
n=0

(at)n

n!
= λeat, (33)



Int. J. Anal. Appl. (2022), 20:27 7

which agrees with the exact solution of given in Eq. (29).

5.3. a = 0. At a = 0, the PDDE yields

y ′(t) = by (ct) , y(0) = λ. (34)

The solution of such case can be deduced from (18) as follows. Inserting a = 0 into (18) leads to

y(t) = λ

[
1 +

∞∑
n=1

tn

n!

n−1∏
k=0

(
bck

)]
= λ

[
1 +

∞∑
n=1

c
1
2
n(n−1) (bt)

n

n!

]
, (35)

where
n−1∏
k=0

(
bck

)
= bnc

1
2
n(n−1), ∀ n ≥ 1. (36)

The solution (36) is equivalent to the form:

y(t) = λ

∞∑
n=0

c
1
2
n(n−1) (bt)

n

n!
. (37)

5.4. c = 1. At c = 1, we have PDDE becomes the ODE:

y ′(t) = (a + b)y(t), y(0) = λ, (38)

and its solution is known as

y(t) = λe(a+b)t. (39)

This solution can also be obtained by substituting c = 1 into (18) which gives

y(t) = λ

[
1 +

∞∑
n=1

tn

n!

n−1∏
k=0

(a + b)

]
, (40)

i.e.,

y(t) = λ

[
1 +

∞∑
n=1

(a + b)ntn

n!

]
, (41)

where
n−1∏
k=0

(a + b) = (a + b)n, ∀ n ≥ 1. (42)

Eq. (42) can be written as

y(t) = λ

∞∑
n=0

((a + b)t)n

n!
= λe(a+b)t, (43)

which is the same exact solution of the Eqs. (39).



8 Int. J. Anal. Appl. (2022), 20:27

5.5. c = −1, a 6= ±b. At c = −1, the PDDE becomes

y ′(t) = ay(t) + by(−t), y(0) = λ. (44)

Indeed, the model (44) is interesting and its exact solution is obtained here in terms of hyperbolic and

trigonometric functions. In addition, such exact solution will be expressed in an equivalent form via

Mittag-Leffler functions. Substituting c = −1 into (18), we obtain

y(t) = λ

[
1 +

∞∑
n=1

tn

n!
ψn

]
, (45)

where ψn is defined by

ψn =

n−1∏
k=0

(
a + b(−1)k

)
, n ≥ 1. (46)

Using the products rules, it can be shown that

ψn =

n−1∏
k=0

(
a + b(−1)k

)
=




(a2 −b2)
n
2 , if n even,

(a + b)
n+1
2 (a−b)

n−1
2 , if n odd.

. (47)

The series solution (45) can be written as

y(t) = λ

[
1 +

∞∑
n=1

t2n−1

(2n− 1)!
ψ2n−1 +

∞∑
n=1

t2n

(2n)!
ψ2n

]
, (48)

where ψ2n−1 and ψ2n are obtained from (47) by

ψ2n−1 = (a + b)
n(a−b)n−1, ψ2n = (a2 −b2)n. (49)

Accordingly, Eq. (48) gives

y(t) = λ

[
1 +

∞∑
n=1

t2n−1

(2n− 1)!
(a + b)n(a−b)n−1 +

∞∑
n=1

t2n

(2n)!
(a2 −b2)n

]
, (50)

i.e.,

y(t) = λ

[
1 +

∞∑
n=0

t2n+1

(2n + 1)!
(a + b)n+1(a−b)n +

∞∑
n=0

t2n+2

(2n + 2)!
(a2 −b2)n+1

]
. (51)



Int. J. Anal. Appl. (2022), 20:27 9

5.5.1. Hyperbolic functions (a > b). Eq. (51) gives

y(t) = λ

[
1 + (a + b)

∞∑
n=0

(√
a2 −b2

)2n
t2n+1

(2n + 1)!
+

∞∑
n=0

(√
a2 −b2

)2n+2
t2n+2

(2n + 2)!

]
,

= λ

[
1 +

a + b
√
a2 −b2

∞∑
n=0

(√
a2 −b2

)2n+1
t2n+1

(2n + 1)!
+

∞∑
n=0

(√
a2 −b2 t

)2n+2
(2n + 2)!

]
,

= λ

[
1 +

√
a + b

a−b

∞∑
n=0

(√
a2 −b2 t

)2n+1
(2n + 1)!

+

∞∑
n=0

(√
a2 −b2 t

)2n+2
(2n + 2)!

]
,

= λ

[
1 +

√
a + b

a−b
sinh

(√
a2 −b2 t

)
+

∞∑
n=1

(√
a2 −b2 t

)2n
(2n)!

]
,

= λ

[√
a + b

a−b
sinh

(√
a2 −b2 t

)
+ cosh

(√
a2 −b2 t

)]
, a > b. (52)

The curves of the hyperbolic solution in Eq. (52) are depicted in Fig. 1 at three different sets for the

values of a and b.

0.0 0.5 1.0 1.5 2.0
t

10

20

30

40

50

60

yHtL

a=4, b=3

a=3, b=2

a=2, b=1

Figure 1. Plots of the hyperbolic solution in Eq. (52) at a = 2,b = 1, a = 3,b = 2,

and a = 4,b = 3.

5.5.2. Mittag-Leffler functions. Here, it is noted that Eq. (51) can be written using the Gamma

function in the form:

y(t) = λ

[
1 +

∞∑
n=0

t2n+1

Γ(2n + 2)
(a + b)n+1(a−b)n +

∞∑
n=0

t2n+2

Γ(2n + 3)
(a2 −b2)n+1

]
, (53)



10 Int. J. Anal. Appl. (2022), 20:27

or

y(t) = λ

[
1 + (a + b)t

∞∑
n=0

(
(a2 −b2)t2

)n
Γ(2n + 2)

+ (a2 −b2)t2
∞∑
n=0

(
(a2 −b2)t2

)n
Γ(2n + 3)

]
. (54)

Using the definition of the two-parameter Mittag-Leffler function:

Eα,β(z) =

∞∑
n=0

zn

Γ(αn + β)
, (55)

then Eq. (54) takes the following final form:

y(t) = λ
[
1 + (a + b)tE2,2

(
(a2 −b2)t2

)
+ (a2 −b2)t2E2,3

(
(a2 −b2)t2

)]
. (56)

This last form can also be used to establish the solution in terms of trigonometric functions via some

properties of the Mittag-Leffler functions as shown below.

5.5.3. Trigonometric functions (b > a). Suppose that b > a, then we can rewrite (56) as

y(t) = λ
[

1 + (a + b)tE2,2

(
−(
√
b2 −a2 t)2

)
− (b2 −a2)t2E2,3

(
−(
√
b2 −a2 t)2

)]
. (57)

Applying the following properties [21]:

E2,2(−z2) =
sin(z)

z
, E2,3(−z2) =

1 − cos(z)
z2

, (58)

for z =
√
b2 −a2 t we have

E2,2

(
−(
√
b2 −a2 t)2

)
=

sin
(√
b2 −a2 t

)
√
b2 −a2 t

, (59)

E2,3

(
−(
√
b2 −a2 t)2

)
=

1 − cos
(√
b2 −a2 t

)
(b2 −a2)t2

. (60)

Substituting (59) and (60) into (57) and simplifying, we obtain

y(t) = λ

[√
b + a

b−a
sin
(√

b2 −a2 t
)

+ cos
(√

b2 −a2 t
)]
, b > a. (61)

This also the same solution reported by Ebaid and Al-Jeaid [22]. It can be easily seen that the solution

is periodic with periodicity P = 2π√
b2−a2

. Fig. 2 shows three different periodic solutions with P = 2π,π,

and π/2.



Int. J. Anal. Appl. (2022), 20:27 11

2 4 6 8 10
t

-3

-2

-1

1

2

3

yHtL

P=
Π

2

P=Π

P=2Π

Figure 2. Plots of the periodic solution in Eq. (61) at a = 1,b =
√

2 (P = 2π),

a =
√

3,b =
√

7 (P = π), and a = 1,b =
√

17 (P = π
2
).

6. The ADE (a = −1,b = c = 1
q
,q > 1)

At a = −1 and b = c = 1
q
(q > 1), the PDDE (1) becomes the ADE [10-12]:

y ′(t) = −y(t) +
1

q
y

(
t

q

)
, y(0) = λ, (62)

Substituting the above values into (18) implies that

y(t) = λ

[
1 +

∞∑
n=1

tn

n!

n−1∏
k=0

(
q−(k+1) − 1

)]
, (63)

i.e.,

y(t) = λ

[
1 +

∞∑
n=1

tn

n!

n∏
k=1

(
q−k − 1

)]
. (64)

which is in full agreement with the corresponding result in [10].

Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the

publication of this paper.

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https://doi.org/10.1243/pime_proc_1964_179_050_02
https://doi.org/10.1093/comjnl/13.4.363


12 Int. J. Anal. Appl. (2022), 20:27

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https://doi.org/10.1049/piee.1966.0078
https://doi.org/10.1049/piee.1969.0226
https://doi.org/10.1098/rspa.1971.0078
https://doi.org/10.1007/s40009-017-0565-2
https://doi.org/10.17265/2159-5291/2018.07.002
https://doi.org/10.3390/math6120331
https://doi.org/10.17265/2159-5291/2019.02.003
https://doi.org/10.3390/mca24010007
https://doi.org/10.3390/mca24010007
https://doi.org/10.4236/jamp.2020.83033
https://doi.org/10.4236/jamp.2020.83033
https://doi.org/10.3934/math.2020016
https://doi.org/10.3390/math7050425
https://doi.org/10.1017/s0956792500000966
https://doi.org/10.1006/jmaa.1997.5483
https://doi.org/10.1515/psr-2016-9103
https://doi.org/10.1093/imamat/8.3.271
https://doi.org/10.1140/epjp/i2018-12081-3
https://doi.org/10.1186/s13662-021-03235-w
https://doi.org/10.17654/0974324322003

	1. Introduction
	2. Solution by SSM 
	3. Solution by the MSE
	4. Convergence analysis
	5. Special cases & Comparisons
	5.1. =1
	5.2. b=0
	5.3. a=0
	5.4. c=1
	5.5. c=-1, a=b

	6. The ADE (a=-1,b=c=1q,q>1)
	References