Abstract: In this paper and in the first part of it, homotopy perturbation method is applied to solve second order

differential equation with non-constant coefficients. The method yields solutions in convergent series forms with

easily computable terms (the convergence of this series is demonstrated in this paper). The result shows that

this method is very convenient and can be applied to large class of problems. As for the second part, we found a

solution of Telegraph equation using the Laplace transform and Stehfest algorithm method. Next, we used method

of Homotopy perturbation. Finally, we give some examples for illustration.

Key-Words: Ordinary/Partial differential equations, Laplace transform method, Stehfest algorithm, Homotopy

Perturbation Method.

1 Introduction

Homotopy perturbation method (HPM) is a semi-

analytical technique for solving linear as well as non-

linear ordinary/partial differential equations. The

method may also be used to solve a system of coupled

linear and nonlinear differential equations. The HPM

was proposed by J. He in 1999 [10]. This method was

developed by making use of artificial parameters [11].

Interested readers may go through Refs. [[8], [5]] for

further details.

Almost all traditional perturbation methods are

based on small parameter assumption. Liu [11] pro-

posed artificial parameter method and Liao [[13],

[12]] contributed homotopy analysis method to elim-

inate small parameter assumption. Further, He [10]

developed an effective technique viz. The HPM

method in which no small parameter assumptions are

required.

In this paper and in the first part of it, He’s ho-

motopy perturbation method is applied to solve lin-

ear and nonlinear second order differential equations

with non-constant coefficients. The method yields so-

lutions in convergent series forms with easily com-

putable terms, for the information that the proof of the

convergence of this series has not been demonstrated

in the articles which have been published before, we

in this article have demonstrated this convergence in

a simple way.

In the second part of this paper is to establish the

solution of Telegraph equations with Dirichlet bound-

ary conditions. The proofs are based on a Homo-

topy perturbation method and Laplace transforma-

tion technique with Stehfest algorithm. Furthermore,

some examples are given to compare between the ap-

proximate and the exact solutions and to show the ef-

ficiency of this method.

2 Homotopy-perturbation method

The Homotopy-perturbation method was proposed by

Ji-Huan He in 1998 [5] and was developed and im-

proved by himself [9, 7, 4]. To illustrate the basic

ideas of this method, we consider the following non-

linear functional equation

A(u)−f(r) = 0; r∈Ω,(1)

with the boundary condition of

B(u;∂u

∂η ) = 0; r∈Γ,(2)

where Arepresents a general differential operator,

Bis a boundary operator, Γis the boundary of the do-

main Ω, and f(r)is known analytic function. The op-

erator Acan be decomposed into two parts viz. linear

Land nonlinear N. Therefore, Eq. (1) may be written

in the following form

L(u) + N(u)−f(r) = 0.(3)

Using homotopy technique, proposed by He [10]

and Liao [[13], [12]], we construct a homotopy

v(r;p) : Ω ×[0; 1] →R. to Eq. (3) which satisfies

H(v;p)=Θ=0,(4)

Received: June 28, 2022. Revised: February 21, 2023. Accepted: March 8, 2023. Published: March 23, 2023.

Application of the Homotopy Perturbation Method for Differential

Equations

NECIB ABDELHALIM, REZZOUG IMAD, BENBRAHIM ABDELOUAHAB

Department of Mathematics and Computer Science

University of Oum El Bouaghi, Algeria.

Laboratory of Dynamics systems and control

ALGERIA.

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Θ = L(v)−L(u0) + p[L(u0) + N(v)−f(r)] ,

and

H(v;p) = Θ = 0; p∈[0; 1]; r∈Ω,(5)

Θ = (1 −p)[L(v)−L(u0)] + p[A(v)−f(r)],

where p∈[0; 1] is the embedding parameter (also

called as an artificial parameter), u0is an initial ap-

proximation of Eq. (1) which satisfies the given con-

ditions.

Obviously, considering Eqs. (5) and (4) we have

H(v; 0) = L(v)−L(u0) = 0,(6)

H(v; 1) = A(v)−f(r) = 0.(7)

As pchanges from zero to unity, v(r;p)changes

from u0(r)to u(r). In topology, this is called defor-

mation and L(v)−L(u0)and A(v)−f(r)are homo-

topic to each other. Due to the fact that p∈[0; 1] is a

small parameter, we consider the solution of Eq. (5)

as a power series in p as below

v=v0+pv1+p2v2+· · · (8)

and the best approximation is

u=p→1limv=v0+v1+v2+· · · (9)

The combination of the perturbation method and

the Homotopy method is called the Homotopy-

perturbation method (HPM), which has eliminated the

limitations of traditional perturbation techniques. The

series (9) is convergent for several cases. Certain cri-

teria are suggested for the convergence of the series

(9), in [4].

3 Homotopy perturbation method

for a second-order differential

equation with non-constant

coefficients

3.1 Statement of the problem

We consider the following problem

u00 =A(t)du

dt +B(t)u+C(t),(10)

u(α) = a, (11)

du (α)

dt =b, (12)

where A(t), B (t)and C(t)are continuous func-

tions on Ian interval contains α.

3.1.1 Homotopy-perturbation method

We can construct the following homotopy

v(t, p) : I×[0,1] →R;d2v

dt2= Θ,(13)

Θ = pA(t)dv

dt +B(t)v+C(t),

suppose that the solution of (13) is written as the

following series

h(t) =

∞

i=0

pivi(t),(14)

we put (14) in (13), we get

dt2 ∞

i=0

pivi!=p





A(t)d

dt ∞

i=0

pivi!

+B(t) ∞

i=0

pivi!+C(t)







(15)

By conformity, we find

p0:d2v0

dt2= 0

v0(α) = a

dv0(α)

dt =b.

Then

v0=a−bα +bt

p1:d2v1

dt2=A(t)d

dt v0+B(t)v0+C(t)

v1(α) = 0

dv1(α)

dt = 0.

Then

v1=Rt

αRξ

αA(t)d

dt v0+B(t)v0+C(t)dtdξ

p2:d2v2

dt2=A(t)d

dt v1+B(t)v1

v2(α) = 0

dv2(α)

dt = 0.

Then

v2=Rt

αRξ

αA(t)d

dt v1+B(t)v1dtdξ

· · ·

pn:d2vn

dt2=A(t)d

dt (vn−1) + B(t)vn−1

vn−1(α) = 0

dvn−1(α)

dt = 0.

Then

vn=Rt

αRξ

αA(t)d

dt vn−1+B(t)vn−1dtdξ.

Where

v0=a−bα +bt (16)

v1=Zt

αZξ

αA(t)d

dtv0+B(t)v0+C(t)dtdξ

(17)

vn=Zt

αZξ

αA(t)d

dtvn−1+B(t)vn−1dtdξ, ∀n≥2.

(18)

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When p→1, (14) gives the approximate solution

of the problem (10)-(12), i.e.

h(t) =

∞

i=0

vi(t).(19)

3.1.2 Study of the convergence of the previous

series

Theorem 3.1 If t∈Isup |A(t)|<∞,

t∈Isup |B(t)|<∞and t∈Isup |C(t)|<∞,

then the series defined by (16)-(19) is converging

towards h(t). Such that h(t)is solution of the

problem (10)-(12).

Proof. We pose

I0v=v(t)

I1v=Rt

αv(ξ)dξ

In+1v=I1(Inv)

and

k=Max {t∈ISup |A(t)|, t ∈ISup |B(t)|}

M=t∈Isup n

dt v1(t)+|v1(t)|o

with



dt v1(t)+|v1(t)|=

Rt

αA(ξ)b+B(ξ)

(a−bα +bξ) + C(ξ)dξ

+Rt

αRξ

αA(t)b+B(t)

(a−bα +bt) + C(t)dtdξ

We have

v2(t) = I2A(t)d

dt v1(t) + B(t)v1(t).

Where

v2(t)≤kI21I0

dt v1(t)+|v1(t)|

v3(t) = I2A(t)d

dt v2(t) + v2(t)

=I2





A(t)d

dt I2A(t)d

dt v1(t)

+B(t)v1(t)

+B(t)I2A(t)d

dt v1(t)

+B(t)v1(t)





=I2





A(t)I1A(t)d

dt v1(t)

+B(t)v1(t)

+B(t)I2A(t)d

dt v1(t)

+B(t)v1(t)





Where

|v3(t)| ≤ k2I31I0+ 1I1

dt v1(t)+|v1(t)|

v4(t) = I2A(t)d

dt v3(t) + B(t)v3(t)

=I2

Ad

dt 

I2



AI1Ad

dt v1+Bv1

+BI2Ad

dt v1+Bv1





+I2B

I2



AI1Ad

dt v1+Bv1

+BI2Ad

dt v1+Bv1





=I2

AI1



AI1Ad

dt v1+Bv1

+BI2Ad

dt v1+Bv1





+I2B

I2



AI1Ad

dt v1+Bv1

+BI2Ad

dt v1+Bv1



.

Where

|v4(t)| ≤ k3I41I0+ 2I1+ 1I2



dξ v1(t)+|v1(t)|

v5(t) = I2A(t)d

dt v4(t) + B(t)v4(t).

Where

|v5(t)| ≤ k4I51I0+ 3I1+ 3I2+ 1I2



dt v1(t)+|v1(t)|.

· · ·

|vn(t)| ≤ kn−1In

n−2

p=0

n−2Ip 

dξ v1(t)

+|v1(t)|!

≤kn−1

n−2

p=0

n−2In+p 

dξ v1(t)

+|v1(t)|!

≤kn−1

n−2

p=0

n−21

(n+p−1)! Rt

αΘ (ξ)dξ

Θ (ξ) = (t−ξ)n+p−1

dξ v1(ξ)+|v1(ξ)|

≤Mkn−1

n−2

p=0

n−21

(n+p−1)! Rt

α(t−ξ)n+p−1dξ

≤Mkn−1

n−2

p=0

n−21

(n+p)! [−(t−ξ)p]ξ=t

ξ=α

≤Mkn−1

n−2

p=0

n−21

(n+p)! (t−α)n+i

≤Mkn−1

n−2

p=0

n−21

(n)! (t−α)n+p

≤Mkn−1(t−α)n

(n)!

n−2

p=0

n−2(t−α)p1n−2−p

≤Mkn−1(t−α)n

(n)! (t−α+ 1)n−2.

Thus

|vn(t)| ≤ Mkn−1(t−α)n

(n)! (t−α+ 1)n−2=

wn(t).

According to D’Alembert’s rule we have

lim wn+1(t)

wn(t)=lim k(t−α)(t−α+1)

n+1 = 0 <1.

Then the series

∞

i=0

wi(t)is convergent, therefore

the series

∞

i=0

vn(t)is also convergent.

We now come to prove that (h(t)) is a solution to

the problem (10)-(12).

Indeed

Step 1 : Take the limit of the two sides of (15)

when p−→ 1, we get d2h(t)

dt2=A(t)dh(t)

dt +

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B(t)h(t) + C(t),so h(t)is a solution of the equa-

tion (10).

Step 2 : We have ∀n≥1 : vn(α)=0and

dvn

dt (α) = 0,thus h(α) = v0(α) = aand dh

dt (α) =

dv0

dt (α) = b.

3.1.3 Special case : The coefficients Aand Bare

constant

By replacing the previous data in (16), (17) and (18)

we obtain

v0=a−bα +bt

v1=Rt

αRξ

αAb +B(a−bα)

+Bbt +C(t)dtdξ

=Rt

α(t−ξ)Ab +B(a−bα)

+Bbξ +C(ξ)dξ

· · ·

vn=Rt

αRξ

αAd

dt (vn−1(t)) dt

+Bvn−1(t)dtdξ

=Rt

α(t−ξ) Ad

dξ (vn−1(ξ)) dξ

+Bvn−1(ξ)!dξ.

We pose

f(t) = Ab +B(a−bα) + Bbt +C(t)

p0:v0=a−bα +bt

p1:v1=Rt

α(t−ξ)f(ξ)dξ

p2:v2=ARt

αv1(ξ)dξ +BRt

αRµ

αv1(ξ)dξdµ

=ARt

αRµ

α(t−ξ)f(ξ)dξdµ +

BRt

αRµ

αRx

α(t−ξ)f(ξ)dξdxdµ

2! ARt

α(t−ξ)2f(ξ)dξ +

3! BRt

α(t−ξ)3f(ξ)dξ

p3:v3=1

3! A2Rt

α(t−ξ)3f(ξ)dξ

+2 1

4! AB Rt

α(t−ξ)4f(ξ)dξ +

5! B2Rt

α(t−ξ)5f(ξ)dξ

p4:v4=1

4! A3Rt

α(t−ξ)4f(ξ)dξ +

5! A2BRt

α(t−ξ)5f(ξ)dξ

6! 3AB2Rt

α(t−ξ)6f(ξ)dξ +

7! B3Rt

α(t−ξ)7f(ξ)dξ

p5:v5=1

5! A4Rt

α(t−ξ)5f(ξ)dξ +

6! 4A3BRt

α(t−ξ)6f(ξ)dξ

7! 6A2B2Rt

α(t−ξ)7f(ξ)dξ +

8! 4AB3Rt

α(t−ξ)8f(ξ)dξ

9! B4Rt

α(t−ξ)9f(ξ)dξ

· · ·

pn:vn=

n−1

i=0

(n+i)! Ci

n−1An−i−1BiRt

αΘ (ξ)dξ

Θ (ξ) = (t−ξ)n+if(ξ).

Where

v0=a+bt

vn=

n−1

i=0

n−1An−i−1Bi

(n+i)! Rt

αΘ (ξ)dξ

Θ (ξ) = (t−ξ)n+i(Ab +Ba +Bbξ +C(ξ)) .

Thus hn(t) = v0+

j=1

vj.

We obtain

hn(t) = Θ (20)

Θ =

j=1







j−1

i=0

j−1Aj−i−1Bi

(j+i)!

α(t−ξ)j+iAb +B(a−bα)

+Bbξ +C(ξ)dξ







+a−bα +bt.

If Cis constant, then using integration by parts,

we get

hn(t) = Θ (21)

Θ =

j=1







j−1

i=0

j−1Aj−i−1Bi

Ab+Ba+C

(j+i+1)! (t−α)j+i+1

+Bb

(j+i+2)! (t−α)j+i+2 !







+a−bα +bt.

3.2 Numerical example

Example 3.2 A(t) = −et

α= 0

B(t) = −sin t

C(t) = et(t+ 2) + e2t(t+ 1) + tetsin t.

So we have the following problem :

d2v

dt2=−etdv

dt −(sin t)v+et(t+ 2)+e2t(t+ 1)+

tetsin t

v(0) = 0

dt (0) = 1.

The exact solution to this problem is given by :

vexa =tet.

We are now looking for the solution by the method

(HPM), by replacing the previous data in (16)-(18),

we get

v0=t. (22)

v1= Θ (23)

Θ = 2 cos t−3

4t−et+1

4te2t+tet+tsin t

2(cos t)et+1

2etsin t−1

2t(cos t)et−3

vn=Zt

0Zξ

0etd

dtvn−1+ (sin t)vn−1dtdξ, ∀n≥2

(24)

and

hn(t) =

i=0

vi(t).(25)

For n= 3, n = 6 and n= 9 we compute hn(t)

withe the relations (22) −(25) and compare the Tay-

lor development of hn(t)and vexa (t)in the neigh-

borhood of 0 (atorder10) .

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DL(h3(t)) = t+t2+1

2t3+1

6t4+7

120 t5

80 t6+5

144 t7+89

3360 t8+5791

362 880 t9+9349

1209 600 t10 +

◦t10.

DL(h6(t)) = t+t2+1

2t3+1

6t4+1

24 t5

120 t6+1

720 t7+1

6720 t8−1

7560 t9−29

103 680 t10 +

◦t10.

DL(h9(t)) = t+t2+1

2t3+1

6t4+1

24 t5

120 t6+1

720 t7+1

5040 t8+1

40 320 t9+1

362 880 t10 +

◦t10

DL(vexa) = t+t2+1

2t3+1

6t4+1

24 t5

120 t6+1

720 t7+1

5040 t8+1

40 320 t9+1

362 880 t10 +

◦t10

Example 3.3 A(t) = −t

B(t) = t2

C(t) = et−t2+t+ 1, α = 1

a=e

b=e.

So we have the following problem :

d2v

dt2=−etdv

dt +t2vet−t2+t+ 1, v (1) = e

dt (1)=e.

The exact solution to this problem is given by :

vexa =et.

v0=e×t

v1=43

15 e−7et−1

6t3e+1

20 t5e−t2et+1

4te + 5tet

vn=Rt

1Rξ

1−td

dt vn−1+t2vn−1dtdξ

hn(t) =

∞

i=0

vi(t).

For n= 3, n = 6 and n= 9 we compute hn(t)

and compare the Taylor development of hn(t)and

vexa (t)in the neighborhood of 1 (atorder10) .

DLh3(t) = e+e(t−1) + 1

2e(t−1)2+1

(t−1)3+1

24 e(t−1)4

60 e(t−1)5+1

90 e(t−1)6+1

2520 e(t−1)7−

4032 e(t−1)8

−317

181 440 e(t−1)9+1

5760 e(t−1)10 +

o(t−1)10

DLh6(t) = e+e(t−1) + 1

2e(t−1)2+1

(t−1)3

24 e(t−1)4+1

120 e(t−1)5+1

720 e(t−1)6+

5040 e(t−1)7

−1

17 280 e(t−1)9−19

403 200 e(t−1)10 +

o(t−1)10

DLh9(t) = e+e(t−1) + 1

2e(t−1)2+1

(t−1)3+1

24 e(t−1)4

120 e(t−1)5+1

720 e(t−1)6+1

5040 e(t−1)7

40 320 e(t−1)8+1

362 880 e(t−1)9+1

3628 800 +

o(t−1)10

DLvexa (t) = e+e(t−1) + 1

2e(t−1)2+1

(t−1)3+1

24 e(t−1)4

120 e(t−1)5+1

720 e(t−1)6+1

5040 e(t−1)7+

40 320 e(t−1)8

362 880 e(t−1)9+1

3628 800 e(t−1)10 +

o(t−1)10

Example 3.4 A(t) = t2

B(t) = cos t

C(t) = et

a∈R, b ∈R

α= 0.

So we have the following problem :

d2v

dt2=t2dv

dt +vcos t+et, v (0) = a

dt (0) = b.

In this problem we do not know in advance the ex-

act solution. We are now looking for the solution by

the method (HPM), by replacing the previous data in

(16)-(18), we get

v0=a+bt

v1=a−t+et+1

12 bt4−bt −acos t+ 2bsin t−

bt cos t−1

vn=Rt

0Rξ

0t2d

dt vn−1(t) + vn−1(t)cos tdtdξ,

and

hn(t) =

i=0

vi(t).

For n= 4, n = 5 and n= 6 we compute hn(t)

and compare the Taylor development of hn(t)and

vexa (t)in the neighborhood of 0 (atorder10) .

DL(h4(t)) = a+bt +1

2a+1

2t2+

1

6b+1

6t3+1

12 b+1

12 t4

+1

20 a−1

60 b+1

15 t5+7

360 b−1

144 a+1

80 t6

+1

840 a+11

1680 b+13

1680 t7+

3

640 a−19

10 080 b+47

8064 t8

+691

362 880 b−11

12 096 a+19

24 192 t9+

91

518 400 a+19

45 360 b+2279

3628 800 t10.

DL(h5(t)) = a+bt +1

2a+1

2t2+

1

6b+1

6t3+1

12 b+1

12 t4

+1

20 a−1

60 b+1

15 t5+7

360 b−1

144 a+1

80 t6

+1

840 a+11

1680 b+13

1680 t7+

3

640 a−19

10 080 b+47

8064 t8

+691

362 880 b−11

12 096 a+19

24 192 t9+

319

1814 400 a+19

45 360 b+19

30 240 t10.

DL(h6(t)) =DL(h5(t)) + ◦t10.

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3.2.1 Homotopy perturbation method for a

nonlinear second-order differential

equation with nonconstant coefficients

We consider the following problem







d2v

dt2

v(α)

dt (α)

A(t)dv

dt +B(t)vm+C(t), m ∈N∗/{1},

We can build the following homotopy

v(t, p) : I×[0,1] →R

dt2 ∞

i=0

pivi!=p





A(t)d

dt ∞

i=0

pivi!

+B(t) ∞

i=0

pivi!m

+C(t)







By conformity, we find

p0:d2v0

dt2= 0

v0(α) = a

dv0(α)

dt =b

v0=a−bα +bt

p1:d2v1

dt2=

(A(t)b+B(t) (a−bα +bt)m+C(t))

v1(α) = 0

dv1(α)

dt = 0

v1=Rt

αRξ

α(A(t)b+B(t) (a−bα +bt)m+C(t)) dtdξ

p2:d2v2

dt2=A(t)d

dt v1+B(t)C1

mv1vm−1

v2(α) = 0

dv2(α)

dt = 0

v2=Rt

αRξ

αA(t)d

dt v1+B(t)mv1vm−1

0dtdξ

p3:d2v3

dt2=A(t)d

dt v2+

B(t)C1

mv2vm−1

0+C2

mv2

1vm−2

0

v3(α) = 0

dv3(α)

dt = 0

v3=Rt

αRξ

α





A(t)d

dt v1

+B(t) mv2vm−1

+m(m−1)

2v2

1vm−2

0!



dtdξ

· · ·

pn:d2vn

dt2=A(t)d

dt (vn−1)

+B(t)

X

k0+k1+.......+km−1=n−1 m−1

i=0

vki!



vn−1(α) = 0

dvn−1(α)

dt = 0

vn=Rt

αRξ

α





A(t)d

dt (vn−1)

+B(t)

X

n−1 m−1

i=0

vki!









dtdξ

k0+k1+.......+km−1=n−1

Thus

v0=a−bα +bt (26)

v1=Zt

αZξ

(A(t)b+B(t) (a−bα +bt)m+C(t)) dtdξ

(27)

∀n≥2 : vn=Zt

αZξ

(Θ (t, ξ)) dtdξ (28)

Θ (t, ξ) = A(t)d

dt (vn−1)

+B(t)X

k0+···+km−1=n−1 m−1

i=0

vki!,

and

hn(t) =

i=0

vi(t).(29)

Example 3.5 m= 2, α = 0, A (t) = t, B (t) =

et, C (t) = 2 −2t2−t4et, a = 0 and b= 0.

So we have the following problem :

d2v

dt2=tdv

dt +etv2+ 2 −2t2−t4et, v (0) = 0 and

dt (0)= 0.

The exact solution to this problem is given by :

vexa =t2.

We are now looking for the solution by the method

(HPM), by replacing the previous data in (26)-(29),

we get

v0= 0

v1=Rt

0Rξ

02−2t2−t4etdtdξ

vn=Rt

0Rξ

0





dt (vn−1)

+etX

k0+k1=n−1 1

i=0

vki!





dtdξ,

and

hn(t) =

i=0

vi(t).

Where

v0= 0

v1=Rt

0Rξ

02−2t2−t4etdtdξ

vn=Rt

0Rξ

0

d

dt (vn−1) +

n−1

k0=0

vk0vn−k0−1

dtdξ.

Thus

v0(t) = 0

v1(t) = Rt

0Rξ

02−2t2−t4etdtdξ

v2(t) = Rt

0Rξ

0td

dt (v1)+2etv0v1dtdξ

v3(t) = Rt

0Rξ

0td

dt v2+et2v0v2+v2

1dtdξ

v4(t) = Rt

0Rξ

0td

dt v3+ 2et(v0v3+v1v2)dtdξ

v5(t) = Rt

0Rξ

0td

dt v4+et2v0v4+ 2v1v3+v2

2dtdξ

v6(t) = Rt

0Rξ

0td

dt v5+ 2et(v0v5+v1v4+v2v3)dtdξ.

For n= 4, n = 5 and n= 6 we compute hn(t)

and compare the Taylor development of hn(t)and

vexa (t)in the neighborhood of 0(atorder13).

DL(h4(t)) = t2−59

22 680 t10 −251

110 880 t11

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−239

187 110 t12 −2323

3931 200 t13 +ot13

DL(h5(t)) = t2−155

299 376 t12 −2693

4717 440 t13 +ot13

DL(h6(t)) = t2+ot13

DL(vexa) = t2+ot13.

Example 3.6 m= 3

A(t) = t2

B(t) = 1

C(t) = ett2+ 4t+ 2−t6e3t−t3et(t+ 2)

a= 0 and b= 0.

So we have the following problem :

d2v

dt2=t2dv

dt +v3+ett2+ 4t+ 2−t6e3t−

t3et(t+ 2)

v(0) = 0 and dv

dt (0) = 0.

The exact solution to this problem is given by :

vexa =t2et.

We are now looking for the solution by the method

(HPM), by replacing the previous data in (26)-(28),

we get

v0=a+bt

v1=Rt

0Rξ

0A(t)b+B(t) (a+bt)3+C(t)dtdξ

vn=Rt

0Rξ

0





A(t)d

dt (vn−1)

+B(t)X

k0+k1+k2=2 2

i=0

vki!





dtdξ.

Where

v0= 0

v1=Rt

0Rξ

0ett2+ 4t+ 2−t6e3t−t3et(t+ 2)dtdξ

vn=Rt

0Rξ

0





t2d

dt (vn−1)

+



n−1

k1=0

k0=0

vn−k1−1vk1−k0vk0









dtdξ,

and

hn(t) =

i=0

vi(t).

Thus

p0:v0(t) = 0

p1:v1(t) =

0Rξ

0ett2+ 4t+ 2−t6e3t−t3et(t+ 2)dtdξ

p2:v2(t) = Rt

0Rξ

0t2d

dt (v1)+3v2

0v1dtdξ

p3:v3(t) =

0Rξ

0t2d

dt v2+ 3v2v2

0+ 3v0v2

1dtdξ

p4:v4(t) =

0Rξ

0t2d

dt v3+ 3v3v2

0+ 6v2v0v1+v3

1dtdξ

p5:v5(t) =

0Rξ

0t2d

dt v4+ 3v4v2

0+ 6v3v0v1

+3v0v2

2+ 3v2

1v2dtdξ.

For n= 4, n = 5 and n= 6 we compute hn(t)

and compare the Taylor development of hn(t)and

vexa (t)in the neighborhood of 0 (atorder13) .

DLh3(t) = t2+t3+1

2t4+1

6t5+1

24 t6

120 t7−83

5040 t8−209

5040 t9−403

8064 t10

−15 551

362 880 t11 −1156 669

39 916 800 t12 −8522 347

518 918 400 t13 +ot13

DLh4(t) = t2+t3+1

2t4+1

6t5+1

24 t6

120 t7+1

720 t8+1

5040 t9+1

40 320 t10

−11 471

2851 200 t11−385 549

39 916 800 t12−6130 067

518 918 400 t13+ot13

DLh5(t) = t2+t3+1

2t4+1

6t5+1

24 t6

120 t7+1

720 t8+1

5040 t9+1

40 320 t10

362 880 t11 +1

3628 800 t12 +1

39 916 800 t13 +ot13

DL(vexa) = t2+t3+1

2t4+1

6t5+1

24 t6

120 t7+1

720 t8+1

5040 t9+1

40 320 t10

362 880 t11 +1

3628 800 t12 +1

39 916 800 t13 +ot13.

4 Laplace transform and

Homotopy-perturbation method

for solving the Telegraph equation

In the following example we expose the Homotopy-

perturbation method (HPM) combined with the

Laplace transformation (LT) to solve a partial differ-

ential equation of order 2.

In the rectangular domain

Q= (0,1) ×(0, T )or T < ∞.

We consider the following Telegraph equation

∂2v

∂t2−α∂2v

∂x2+β∂v

∂t =f(x, t),(30)

with the initial conditions

v(x, 0) = ϕ(x),0< x < 1,(31)

vt(x, 0) = ψ(x),0< x < 1,(32)

and the boundary conditions

v(0, t) = n(t),0< t ≤T, (33)

vx(0, t) = m(t),0< t ≤T, (34)

where f, ϕ, ψ, n and mare known functions. α, β

and Tare positive constants, moreover the functions

ϕ(x)and ψ(x)satisfying the following compatibility

conditions

ϕ(0) = n(0), ϕ0

x(0) = m0

t(0),(35)

and ψ(0) = n(0), ψ0

x(0) = m0

t(0).

Let us suppose that the function v(x, t)is defined

and of exponential order for t≥0, that is to say

that there exists A, γ > 0and t0>0such that

|v(t)| ≤ Aexp (γt)for t≥t0. Suppose also that the

Laplace transformation V(x, s)is exists and given by

the following formula

V(x, s) = L {v(x, t) ; t−→ s}

=R∞

0v(x, t)exp (−st)dt,

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where sis a positive parameter. Taking the

Laplace transformations on both sides of (30), we get

d2V

dx2(x, s) = Θ,(36)

Θ = (s2+βs

α)V(x, s)−F(x,s)+ψ(x)+(s+β)ϕ(x)

α,

where F(x, s) = L {g(x, t) ; t−→ s}. Similarly,

we have

V(0, s) = a(s),(37)

Vx(0, s) = b(s),(38)

where

a(s) = L {n(t); t−→ s},(39)

b(s) = L {m(t); t−→ s}.(40)

By replacing the previous data in (20), i.e.

t→x

a→a(s)

b→b(s)

A→0

B→s2+βs

C(ξ)→ −F(ξ,s)+ψ(ξ)+(s+β)ϕ(ξ)

α.

We obtain

Hn(x, s) =

j=1

Θj(41)

Θj=1

(2j−1)! s2+βs

αj−1Rx

0Θ (ξ)dξ

+a(s) + b(s)x.

Θ (ξ) = (x−ξ)2j−1

s2+βs

α(a(s) + b(s)ξ)−F(x,s)+ψ(x)+(s+β)ϕ(x)

α.

The solution hn(x, t)can be recovered approxi-

mately from Hn(x, s)by the analytical method or ac-

cording to the Stehfest algorithm [6]. By taking the

inverse L−1on both sides of (41), we obtain the ap-

proximate solution of the problem (30)-(34).

4.1 Numerical example

By take

α= 1

β= 1,

f(x, t) = et2x2+ 2t+ 1

ϕ(x) = x2

ψ(x) = x2+ 1

n(t) = tetand m(t) = 0.

So we have the following problem











∂2v

∂t2−α∂2v

∂x2+β∂v

∂t

v(x, 0)

vt(x, 0)

v(0, t)

vx(0, t)

f(x, t),

x2,

x2+ 1,

tet,

Thus

f(x, t) = et2x2+ 2t+ 1→F(x, s) =

(s−1)2s+ 2sx2−2x2+ 1

n(x, t) = tet→a(s) = 1

(s−1)2

m(x, t) = 0 →b(s) = 0.

The exact solution to this problem is given by

vexa (x, t) = t+x2et.

We are now looking for the solution by the method

(LT-HPM), by replacing the previous data in (41), we

obtain

Hn(x, s) =

j=1 1

(2j−1)! s2+sj−1Rx

0Θ (ξ)dξ+

(s−1)2

Θ (ξ) = (x−ξ)2j−1−1

s−1s2ξ2+sξ2−2.

Thus

hn(x, t) = L−1



j=1 1

(2j−1)! s2+sj−1

0Θ (ξ)dξ !

+

tet

Θ (ξ) = (x−ξ)2j−1−1

s−1s2ξ2+sξ2−2

For t∈ {0.2,0.6,2,10}, x ∈

{0.2,0.4,0.6,0.8,1},and n∈ {3,5},we com-

pute hn(x, t)numerically and compare this result

with the exact solution in the tables Tab1, Tab2.

t/x x = 0.2x= 0.4

t= 0.2veax = 0.293 14

hn= 0.293 14

0.439 7

t= 0.6veax = 1.166 2

hn= 1.166 2

1.384 8

t= 2 veax = 15.074

hn= 15.074

15.96

t= 10 veax =2.211 5

10−5

hn=2.211 5

10−5

2.237 9

10−5

2.237 9

10−5

t/x x = 0.6x= 0.8x= 1

t= 0.20.683 99

0.683 98

1.026 0

1.025 9

1.465 7

1.465 2

t= 0.61.749 2

1.749 2

2.259 4

2.259 3

2.915 4

2.914 7

t= 2 17.438

17.438

19.507

22.167

22.164

t= 10

2.281 9

10−5

2.281 9

10−5

2.343 6

10−5

2.343 6

10−5

2.422 9

10−5

2.422 8

10−5

Tab1:n= 3.

t/x x = 0.2x= 0.4

t= 0.2veax = 0.293 14

hn= 0.293 14

0.439 7

t= 0.6veax = 1.166 2

hn= 1.166 2

1.384 8

t= 2 veax = 15.074

hn= 15.074

15.96

t= 10 veax =2.211 5

10−5

hn=2.211 5

10−5

2.237 9

10−5

2.237 9

10−5

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t/x x = 0.6x= 0.8x= 1

t= 0.20.683 99

0.683 99

1.026 0

1.465 7

t= 0.61.749 2

1.749 2

2.259 4

2.915 4

t= 2 17.438

17.438

19.507

22.167

t= 10

2.281 9

10−5

2.281 9

10−5

2.343 6

10−5

2.343 6

10−5

2.422 9

10−5

2.422 9

10−5

Tab2:n= 5.

4.2 Conclusion

In this paper, He’s homotopy perturbation method has

been successfully applied to find the solution of the

second order differential equation with non-constant

coefficients. The method is reliable and easy to use.

The main advantage of the method is the fact that it

provides its user with an analytical approximation, in

many cases an exact solution, in a rapidly conver-

gent sequence with elegantly computed term. Know-

ing that the convergence of this series is demonstrated

in this paper. Then, by using this method and by in-

troducing the Laplace transformation technique with

the Stehfest algorithm, we were able to solve the tele-

graph problem with the Dirichlet boundary condi-

tions.

Acknowledgements

The authors thank the referees for their careful

reading and their precious comments. Their help is

much appreciated.

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formulation of the problem to the final findings and

solution.

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Scientific Article or Scientific Article Itself

No funding was received for conducting this study.

Conflict of Interest

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