EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 3, 2021, 942-948
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Tensor Product and Certain Solutions of Fractional
Wave Type Equation

Ibtissem Benkemache1, Mohammed Al Horani1, Roshdi Khalil1,∗

1 Department of Mathematics, School of Science, The University of Jordan, Amman, Jordan

Abstract. In this paper we find certain solutions of some fractional partial differential equations.
Tensor product of Banach spaces is used to find some solutions where separation of variables does
not work. We solve the fractional wave type equation using fractional Fourier series

2020 Mathematics Subject Classifications: 26A33

Key Words and Phrases: Conformable derivative, fractional Fourier series, fractional wave type
equation.

1. Introduction

In [7], a definition of the so-called α−conformable fractional derivative was introduced:
Let α ∈ (0, 1), and f : E ⊆ (0,∞)→ R. For x ∈ E, let:

Dαf(x) = lim
ε→0

f(x+ εx1−α)− f(x)

ε
.

If the limit exists, then it is called the α−conformable fractional derivative of
f at x. If f is α−differentiable on (0, r) for some r > 0, and lim

x→0+
Dαf(x) exists then we

define
Dαf(0) = lim

x→0
Dαf(x).

For α ∈ (0, 1] and f, g are α−differentiable at a point t, one can easily see that the con-
formable derivative satisfies:

1. Dα(af + bg) = aDα(f) + bDα(g), for all a, b ∈ R.

2. Dα(λ) = 0, for all constant functions f(t) = λ.

3. Dα(fg) = fDα(g) + gDα(f).

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v14i3.4012

Email addresses: ibtissem19932017@gmail.com (I. Benkemache),
horani@ju.edu.jo (M. Al Horani), roshdi@ju.edu.jo (R. Khalil)

http://www.ejpam.com 942 © 2021 EJPAM All rights reserved.



I. Benkemache, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (3) (2021), 942-948 943

4. Dα(fg ) = gDα(f)−fDα(g)
g2

, g(t) 6= 0.

We list here the fractional derivatives of certain functions,

(i) Dα(tp) = p tp−α .

(ii) Dα(sin 1
α t
α) = cos 1

α t
α.

(iii) Dα(cos 1
α t
α) = − sin 1

α t
α.

(iv) Dα(e
1
α
tα) = e

1
α
tα .

On letting α = 1 in these derivatives, we get the corresponding classical rules for ordi-
nary derivatives. Further, one should notice that a function could be α−conformable differ-
entiable at a point but not differentiable, for example, take f(t) = 2

√
t, then D

1
2 (f)(0) = 1.

This is not the case for the known classical fractional derivatives, since D1(f)(0) does not
exist. For more on fractional calculus and its applications we refer to [7]-[6].

Many differential equations can be transformed to fractional form and can have many
applications in many branches of science. The main technique to solve partial differential
equations is using Fourier series. So, fractional Fourier series was introduced in [3]. Such
a concept proved to be very fruitful in solving fractional partial differential equations.

In this paper we will use fractional Fourier series to solve a fractional wave type equa-
tion. In Section 2 we introduce the atomic solution. The complete solution is given in
Section 3.

2. Atomic solution

Let X and Y be two Banach spaces and X∗ be the dual of X. Assume x ∈ X and
y ∈ Y. The operator T : X∗ → Y, defined by

T (x∗) = x∗(x)y

is a bounded one rank linear operator. We write x⊗y for T. Such operators are called
atoms. Atoms are among the main ingredient in the theory of tensor products. Atoms are
used in theory of best approximation in Banach spaces, see [2]. One of the known results,
see [4], that we need in our paper is that: If the sum of two atoms is an atom, then either
the first components are dependent or the second ones are dependent. For more on tensor
products of Banach spaces, we refer to [4].

Let us write Dα
xu to mean the partial α−derivative of u with respect to x. Further we

write D2α
x u to mean Dα

xD
α
xu. Similarly for derivatives with respect to y.

If f is a function of one variable, say x , we write fα, f2α to denote Dα
xf and D2α

x f
respectively.



I. Benkemache, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (3) (2021), 942-948 944

Our main object in this section is to find an atomic solution of the equation

D2α
x D2β

y u+Dα
xD

β
yu = 2u , (1)

where by an atomic solution we mean a solution of the form u(x, y) = P (x)Q(y).

Remark 1. One should remark that not every linear partial differential equation (frac-
tional or not) can be solved using separation of variables. In such a case, the concept of
atomic solution is inevitable. In equation (1), the method of separation of variables is not
possible though the equation is linear. Hence we try to find an atomic solution of this
equation. In other words, we look for a solution of the form

u(x, y) = P (x)Q(y).

Procedure

Let u(x, y) = P (x)Q(y). Substitute in equation (1) to get:

P 2α(x)Q2β(y) + Pα(x)Qβ(y) = 2P (x)Q(y).

This can written in tensor product form as:

P 2α ⊗Q2β + Pα ⊗Qβ = P ⊗ 2Q . (2)

Let us consider the following conditions : P (0) = 0, Pα(0) = 1.
In equation (2), we have the situation: the sum of two atoms is an atom.
Hence we have two cases:

Case(i): P 2α = Pα. Using the result in [5], we get

P (x) = e
xα

α . (3)

Now, we substitute in (2) to get

ex ⊗ [Q2β +Qβ] = ex ⊗Q .

Hence, Q2β +Qβ = 2Q. Again, using the result in [5],

Q(y) = c1e
−2 y

β

β + c2e
yβ

β

Using the conditions Q(0) = Qβ(0) = 1, we get

Q(y) = −1

3
e
−2 y

β

β +
1

3
e
yβ

β . (4)

From (3) and (4), we obtain the atomic solution of (1) as:

u(x, y) = e
xα

α (−1

3
e
−2 y

β

β +
1

3
e
yβ

β ) . (5)



I. Benkemache, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (3) (2021), 942-948 945

One can easily check that the atom u in (5) satisfies (1).

Case (ii): Q2β = Qβ. Following the same steps as in case (i), we find the atomic solution
in the form

u(x, y) = (−1

3
e−2

xα

α +
1

3
e
xα

α )e
yβ

β

3. Complete Solution

Consider the fractional partial differential equation

D2β
x u− c2D2α

y u = Dα
y u (6)

with conditions

u(x, 0) = f(x) , u(x, 1) = 0, u(L, y) = 0, u(0, y) = 0 , 0 < α, β < 1.

Here c is a given constant. This is called fractional wave type equation. We will use
fractional Fourier series and separation of variables to solve equation (6).

Remark 2. One may attempt to use change of variables to transform it to an ordinary
partial differential equation. This is possible if in equation (1) and (6), the function u is
u = u(x

α

α ,
yα

α ). But the function u in equations in (1) and (6) is u = u(x, y). So any change
of variables will not simplify the problem. Further, the partial derivatives of u: ux and uy
need not to be exist even if Dα

xu and Dα
y u exist, see [7].

Let
u(x, y) = P (x)Q(y) .

Substitute in the equation (6) to get

P 2β(x)Q(y)− c2P (x)Q2α(y) = P (x)Qα(y) .

Simplifying to get

P 2β(x)Q(y)− c2P (x)Q2α(y)− P (x)Qα(y) = 0 ,

P 2β(x)Q(y)− P (x)
(
c2Q2α(y) +Qα(y)

)
= 0 ,

P 2β(x)Q(y) = P (x)
(
c2Q2α(y) +Qα(y)

)
.

From which we obtain
P 2β(x)

P (x)
=
c2Q2α(y) +Qα(y)

Q(y)
= λ .

Since x and y are independent variables, then we get

P 2β(x)

P (x)
= λ



I. Benkemache, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (3) (2021), 942-948 946

and
c2Q2α(y) +Qα(y)

Q(y)
= λ .

Simplifying to get
P 2β(x)− λP (x) = 0 (7)

and
c2Q2α(y) +Qα(y)− λQ(y) = 0 . (8)

Let us first deal with equation (7). There are three possibilities for λ:

Case 1: λ = 0

Then equation (7) becomes P 2β(x) = 0. Using the result in [1], we see that P (x) =

c1
xβ

β + c2. By using the condition u(0, y) = 0, we get c2 = 0. Another use of condition
u(L, y) = 0 we get c1 = 0. So, P (x) = 0. Thus λ = 0 gives the trivial solution.

Case 2: λ = µ2 > 0

Then equation (7) becomes
P 2β(x) = µ2P (x) .

Using the result in [5], we see that

P (x) = c1e
µx

β

β + c2e
−µx

β

β .

Using the condition u(0, y) = 0, we get c1 = −c2. So, P (x) = 2c1 sinh(µx
β

β ). Another use

of condition u(L, y) = 0 we get 2c1 sinh(µL
β

β ) = 0. Hence, c1 6= 0 and so µ = 0. Thus,
P (x) = 0. Therefore λ > 0 gives the trivial solution.

Case 3: λ = −µ2 < 0

Then equation (7) becomes
P 2β(x) + µ2P (x) = 0 .

Using results in [5], we get

P (x) = c1 cos(µ
xβ

β
) + c2 sin(µ

xβ

β
).

Applying the condition u(0, y) = 0 we get c1 = 0. So, P (x) = c2 sin(µx
β

β ). Another use of

condition u(L, y) = 0 gives c2 sin(µL
β

β ) = 0. Then c2 6= 0 and so sin(µL
β

β ) = 0. Hence,

µ = nπ
β

Lβ
. (9)

So,

P (x) = cn sin(nπ
xβ

Lβ
) , n = 1, 2, ... . (10)



I. Benkemache, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (3) (2021), 942-948 947

Now, we go back to equation (8). Substituting the value of µ that we got in (9), equation
(8) becomes

c2Q2α(y) +Qα(y) + µ2Q(y) = 0 .

Another use of the result in [5], we get two cases under consideration:

Case i: 1− 4µ2c2 > 0.

µ2 < 1, | µ |< 1, 1− 4µ2c2 = (
√

1− 4µ2c2)2. Then we get,

r =
−1±

√
1− 4µ2c2

2c2
.

So

Q(y) = c1e
−1+
√

1−4µ2c2

2c2
yα

α + c2e
−1−
√

1−4µ2c2

2c2
yα

α .

Using condition u(x, 0) = 0, we get c1 = −c2. So,

Q(y) = 2c1 sinh

(
−1 +

√
1− 4µ2c2

2c2
yα

α

)
. (11)

Thus, combining (10) and (11) we get:

u(x, y) =

∞∑
n=1

bn sin

(
nπ

xβ

Lβ

)
sinh

−1 +
√

1− 4(nβπ
Lβ

)2c2

2c2
yα

α

 .

By using Dβ
y (x, 0) = f(x) we get

f(x) =
∞∑
n=1

bn

−1 +
√

1− 4(nβπ
Lβ

)2c2

2c2

 sin

(
nπ

xβ

Lβ

)
.

So

bn =
2β

P β

(
−1+

√
1−4(nβπ

Lβ
)2c2

2c2

) ∫ P

0
f(x) sin

(
nπ

xβ

Lβ

)
dx

x1−β
.

Case ii: 1− 4µ2c2 < 0.

µ2 > 1, | µ |> 1, 1− 4µ2c2 = −(4µ2c2 − 1) = (
√

4µ2c2 − 1
2
) . Then we get,

r =
−1± i

√
4µ2c2 − 1

2c2
.

So

Q(y) = c1 cos

(
−1 +

√
4µ2c2 − 1

2c2
yα

α

)
+ c2 sin

(
−1 +

√
4µ2c2 − 1

2c2
yα

α

)



REFERENCES 948

By using condition u(x, 0) = 0 we get c1 = 0. Then,

Q(y) = c2 sin

(
−1 +

√
4µ2c2 − 1

2c2
yα

α

)
. (12)

Thus, (10) and (12) gives

u(x, y) =
∞∑
n=1

bn sin

(
nπ

xβ

Lβ

)
sin

(
−1 +

√
4µ2c2 − 1

2c2
yα

α

)
. (13)

By using condition uy(x, 0) = f(x) we deduce that

f(x) =

∞∑
n=1

bn

−1 +
√

4(nβπ
Lβ

)2c2 − 1

2c2

 sin

(
nπ

xβ

Lβ

)
.

Hence, using [3], we find

bn =
2β

P β

(
−1+

√
4(nβπ

Lβ
)2c2−1

2c2

) ∫ P

0
f(x)sin(nπ

xβ

Lβ
)
dx

x1−β

where P is a period of the function f which equals to (2βπ)
1
β . So we got the complete

solution of the differential equations (6).

References

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[4] R. Khalil. Isometries of lp ∗
∧
⊗lp. Tam. J. Math., 16:77–85, 1985.

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