EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 4, 2022, 1444-1454
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Exact Solution of Burger’s Equation Using Tensor
Product Technique

Ameerah Al-Jarrah1,∗, Sharifa Alsharif1, Hasan Almefleh1

1 Department of Mathematics, Faculty of Science, Yarmouk University, Irbid, Jordan

Abstract. In this paper a new technique using tensor product is presented which yields an exact
solution to Burger’s equation

ut + αuux = υuxx

which is one of the very few nonlinear partial differential equations that can be solved analytically.
More over we give an atomic solution for linear partial differential equations with and without
variable coefficient terms.

2020 Mathematics Subject Classifications: 46M05, 35-xx, 46Bxx

Key Words and Phrases: Burger’s equation, tensor product, atom solution

1. Introduction

One of the well known partial differential equations which governs a wide variety
of mathematical models is the Burger’s equation which provides the simplest nonlinear
model of turbulence, and else occurring in various areas of applied mathematics such as
fluid mechanics, gas dynamics, and traffic flow. This equation was first introduced by
Harry Bateman in 1915, [1] and later studied by Johannes Martinus Burgers, [4] in 1948.

In this paper, we present a new way of solving the nonlinear (Burger equation) partial
differential equation

ut + αuux = υuxx

using tensor product technique.
General Burger’s Equation:
Consider the one-dimensional quasi-linear Burger’s equation with the following initial

and boundary conditions:

ut + αuux = υuxx

u(x, 0) = f(x), 0 ≤ x ≤ l

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v15i4.4468

Email addresses: amera.gassan@gmail.com (A. Al-Jarrah),
sharifa@yu.edu.jo (Sh. Alsharif), almefleh@yu.edu.jo (H. Almefleh)

https://www.ejpam.com 1444 © 2022 EJPAM All rights reserved.



A. Al-Jarrah, Sh. Alsharif, H. Almefleh / Eur. J. Pure Appl. Math, 15 (4) (2022), 1444-1454 1445

u(0, t) = f1(t)

ux(0, t) = f2(t), t > 0,

where u = u(x, t) is unknown function in some domain and the nonlinear term coefficient
α is an arbitrary constant and υ is the coefficient of the kinematics viscosity of fluids
which is equal to 1

R . Further, R is the Reynolds number, and when it is large the equation
describes shock wave behavior, where uux is the nonlinear term.

This equation has been solved in different methods, such as Homotopy Perturbation
Method [2], Linearized solution, and numerically like the least-squares quadratic B-spline
finite element method [5, 6, 12–14], explicit and exact-explicit finite difference methods,
variational iteration method (VIM) [3].

As well, tensor product used to solve one of the classical differential equations in Banach
spaces is called Abstract Cauchy Problem by Ziqan, Al-Horani, and Khalil [15], and also
Abdullah and Khalil [7] in a different conditions. Also, many of the non-homogeneous
second order partial differential equations has been solved by finding an atomic solution
u = u1 ⊗ x [10, 11].

2. Tensor Product

Let X, Y be two Banach spaces and X∗, Y ∗ denote their respective duals. For x ∈ X
and y ∈ Y , define the linear operator

x⊗ y : X∗ −→ Y

x⊗ y(x∗) = ⟨x, x∗⟩y,

where ⟨x, x∗⟩ is the value of x∗ at x, x⊗ y is called an atom. It is easy to see that x⊗ y
is a bounded linear operator with norm ∥x⊗ y∥ = ∥x∥ ∥y∥.

The tensor product X ⊗ Y := span{x⊗ y : x ∈ X, y ∈ Y }, where X ⊗ Y ⊆ L(X∗, Y );
L(X∗, Y ) is the space of bounded linear operators from X∗ into Y , X ⊗ Y is a linear
subspace of finite rank operators in L(X∗, Y ) [8].

Lemma 1. [9] For any x, z ∈ X, y, t ∈ Y and scaler β, the following are valid:

1- β(x⊗ y) = βx⊗ y = x⊗ βy.
2- (x+ z)⊗ y = x⊗ y + z ⊗ y.
3- x⊗ (y + t) = x⊗ y + x⊗ t.
4- x⊗ 0 = 0⊗ y = 0⊗ 0.
5- ||x⊗ y|| = ||x|| ||y||.
6- Any T ∈ X ⊗ Y can be written as

∑n
i=1 λi(xi ⊗ yi) with ||xi|| = ||yi|| = 1.

Definition 1. [9] Let T =
∑n

i=1 xi ⊗ yi ∈ X ⊗ Y, define the injective norm on X ⊗ Y
as

||T ||∨ = sup

{
n∑

i=1

|⟨x, x∗⟩⟨y, y∗⟩|, x∗ ⊗ y∗ ∈ X∗ ⊗ Y ∗, ||x∗|| = ||y∗|| = 1

}
.



A. Al-Jarrah, Sh. Alsharif, H. Almefleh / Eur. J. Pure Appl. Math, 15 (4) (2022), 1444-1454 1446

The space (X ⊗ Y, ||.||∨) need not be complete. We let X
∨
⊗ Y denote the completion

of X ⊗ Y in L (X∗, Y ) with respect to the injective norm.

Theorem 1. [9]For any compact Hausdorff space I and a Banach space X, we have

C(I,X) is isometrically isomorphic to C(I)
∨
⊗X.

For more on tensor product we refer the reader to [9].

3. Atomic Solution of linear partial differential equations

In this section, we solve two kinds of partial differential equations by using tensor
product technique. We start by the following lemma:

Lemma 2. Let x1 ⊗ y1 and x2 ⊗ y2 be two non zero atoms in X
∨
⊗ Y . Then the following

are equivalent:
(i) x1 ⊗ y1+ x2 ⊗ y2 = x3 ⊗ y3 a non zero atom.
(ii) x1, x2 or y1, y2 are linearly dependent.

Proof. (i) → (ii) If x3 ⊗ y3 = 0, we are done. Assume x3 ⊗ y3 ̸= 0. Then there exists
t0 ∈ I and y∗ ∈ Y ∗ such that x3 (t0) ̸= 0 and y∗ (y3) ̸= 0. using (i) we have

x3 =
y∗ (y1)

y∗ (y3)
x1 +

y∗ (y2)

y∗ (y3)
x2 = c1x1 + c2x2,

y3 =
x1 (t0)

x3 (t0)
y1 +

x2 (t0)

x3 (t0)
y2 = b1y1 + b2y2.

Consequently

x3 ⊗ y3 = c1b1x1 ⊗ y1 + c1b2x1 ⊗ y2 + c2b1x2 ⊗ y1 + b2c2x2 ⊗ y2

= x1 ⊗ y1 + x2 ⊗ y2.

Hence

x1 ⊗ y1 (1− c1b1) + x2 ⊗ y2 (1− c2b2) + c1b2x1 ⊗ y2 + c2b1x2 ⊗ y1 = 0.

If x1, x2 and y1, y2 are linearly independent, it follows that

1− b1c1 = 1− b2c2 = b2c1 = b1c2 = 0,

which turns out to a contradiction 1 = b1c1, 1− b2c2, b2c1 = 0, b1c2 = 0. Hence the result.
(ii) → (i) If x1, x2 are linearly dependent, then x1 = λx2. Using (ii)

x3 ⊗ y3 = λx2 ⊗ y1 + x2 ⊗ y2 = x2 ⊗ (λy1 + y2)

which completes the proof.



A. Al-Jarrah, Sh. Alsharif, H. Almefleh / Eur. J. Pure Appl. Math, 15 (4) (2022), 1444-1454 1447

Theorem 2. Let u(x, t) ∈ C(I × J), where I, J = [0, 1] or [0,∞). If u has continuous
second partial derivatives, then the linear differential equation

ut + ux = uxx (1)

has an atomic solution.

Proof. Let u(x, t) = Φ ⊗ Ψ, where Φ is a function of x and Ψ is a function of t with
Φ(0) = 1, Φ′(0) = 1 and Ψ(0) = 1. Then ux = Φ′ ⊗ Ψ, ut = Φ ⊗ Ψ′ and uxx = Φ′′ ⊗ Ψ.
This implies that

Φ⊗Ψ′ +Φ′ ⊗Ψ = Φ′′ ⊗Ψ. (2)

By using Lemma (2) either Φ′ = λΦ or Ψ́ = µΨ. Without loss of generality we can assume
λ = µ = 1.
Case (1) If Φ′ = Φ = Φ′′, then

Φ′

Φ
= 1.

Integrating both sides,we get ∫
dΦ

Φ
=

∫
dx

ln |Φ| = x+ c

Φ = cex

since Φ(0) = 1 =⇒
Φ = ex.

Now, since the first and the second derivatives of Φ are equal, then equation (2) becomes

Φ⊗Ψ′ +Φ⊗Ψ = Φ⊗Ψ.

(Ψ′ +Ψ−Ψ)⊗ Φ = 0.

Ψ′ ⊗ Φ = 0.

Here Φ = 0 or Ψ′ = 0. If Φ = 0, then we have a contradiction since Φ ̸= 0. So Ψ′ = 0, this
implies Ψ = K, where K is a constant.

To verify equation (2), set u = Φ⊗Ψ, ux = Φ′ ⊗Ψ = Φ⊗Ψ, uxx = Φ′′ ⊗Ψ = Φ⊗Ψ,
ut = Φ⊗Ψ′ = Φ⊗ 0 = 0.

Then

ut + ux = Φ⊗ 0 + Φ⊗Ψ

= 0 + Φ⊗Ψ

= Φ⊗Ψ

= uxx.



A. Al-Jarrah, Sh. Alsharif, H. Almefleh / Eur. J. Pure Appl. Math, 15 (4) (2022), 1444-1454 1448

This implies u = Φ⊗Ψ, where Φ = ex and Ψ = K.
Case (2) If Ψ′ = Ψ, then

Ψ′

Ψ
= 1.

Integrating both sides, we get ∫
dΨ

Ψ
=

∫
dt

ln |Ψ| = t+ a

Ψ = aet.

since Ψ(0) = 1 =⇒
Ψ = et.

Now, Ψ′ = et = Ψ, then equation (2) becomes

Φ⊗Ψ+Φ′ ⊗Ψ = Φ′′ ⊗Ψ

[Φ + Φ′ − Φ′′]⊗Ψ = 0.

Using Lemma (1) Φ′′ −Φ′ −Φ = 0 or Ψ = 0. If Ψ = 0, then we have a contradiction since
Ψ ̸= 0. So

Φ′′ − Φ′ − Φ = 0. (3)

The characteristic equation of equation (3) is

λ2 − λ− 1 = 0,

with roots λ1 =
1+

√
5

2 and λ2 =
1−

√
5

2 . Hence

Φ = Geλ1x + Feλ2x,

since Φ(0) = 1 and Φ′(0) = 1, then F = λ1−1
λ1−λ2

and G = λ2−1
λ2−λ1

.

To verify equation (2), set u = Φ⊗Ψ, ux = Φ′⊗Ψ, uxx = Φ′′⊗Ψ, ut = Φ⊗Ψ′, where Φ =
λ2−1
λ2−λ1

eλ1x+ λ1−1
λ1−λ2

eλ2x, Φ′ = λ1
λ2−1
λ2−λ1

eλ1x+λ2
λ1−1
λ1−λ2

eλ2x, Φ′′ = λ2
1

λ2−1
λ2−λ1

eλ1x+λ2
2

λ1−1
λ1−λ2

eλ2x,

Ψ = et, and Ψ′ = et.
Then

ut + ux = Φ⊗Ψ′ +Φ′ ⊗Ψ

= Φ⊗Ψ+Φ′ ⊗Ψ

= [Φ + Φ′]⊗Ψ,

and so

Φ + Φ′ = (
λ2 − 1

λ2 − λ1
eλ1x +

λ1 − 1

λ1 − λ2
eλ2x + λ1

λ2 − 1

λ2 − λ1
eλ1x + λ2

λ1 − 1

λ1 − λ2
eλ2x)



A. Al-Jarrah, Sh. Alsharif, H. Almefleh / Eur. J. Pure Appl. Math, 15 (4) (2022), 1444-1454 1449

= (1 + λ1)(
λ2 − 1

λ2 − λ1
)eλ1x + (1 + λ2)(

λ1 − 1

λ1 − λ2
)eλ2x.

Now, 1 + λ1 = 1 + 1+
√
5

2 = 3+
√
5

2 and 1 + λ2 = 1 + 1−
√
5

2 = 3−
√
5

2 , but λ2
1 = (1+

√
5

2 )2 =
3+

√
5

2 and λ2
2 = (1−

√
5

2 )2 = 3−
√
5

2 so 1 + λ1 = λ2
1 and 1 + λ2 = λ2

2, then

Φ + Φ′ = (1 + λ1)e
λ1x + (1 + λ2)e

λ2x

= λ2
1e

λ1x + λ2
2e

λ2x

= Φ′′.

ut + ux = [Φ + Φ′]⊗Ψ

= Φ′′ ⊗Ψ

= uxx.

This implies that u = Φ ⊗ Ψ is a solution of equation (1), where Φ = ( λ2−1
λ2−λ1

)eλ1x +

( λ1−1
λ1−λ2

)eλ2x and Ψ = et.

In the following Theorem we use Tensor product technique to find an exact solution
of a general form of equation (1).

Theorem 3. Let u(x, t) ∈ C(I × J), where I, J = [0, 1] or [0,∞). If u has continuous
second partial derivatives and f any continuous function of t, then the differential equation

ut + fux = uxx (4)

can be solved by tensor product.

Proof. Put u = Φ⊗Ψ, with Φ(0) = 1 and Ψ(0) = 1

Φ⊗Ψ
′
+Φ′ ⊗ fΨ = Φ′′ ⊗Ψ. (5)

Since the sum of two atoms is an atom using Lemma (2), we have Φ′ = Φ or Ψ′ = fΨ.
Case (1) if Ψ′ = fΨ, then

Ψ′

Ψ
= f.

Integrating both sides, we get ∫ t

0

dΨ

Ψ
=

∫ t

0
fdu

lnΨ

∣∣∣∣t
0

=

∫ t

0
fdu

lnΨ− lnΨ(0) =

∫ t

0
fdu



A. Al-Jarrah, Sh. Alsharif, H. Almefleh / Eur. J. Pure Appl. Math, 15 (4) (2022), 1444-1454 1450

lnΨ− ln 1 =

∫ t

0
fdu

lnΨ =

∫ t

0
fdu

Ψ = e
∫ t
0 fdu.

Since Ψ′ = fΨ, then equation (5) becomes

Φ⊗ fΨ+Φ′ ⊗ fΨ = Φ′′ ⊗Ψ

[Φ + Φ′]⊗ fΨ = Φ′′ ⊗Ψ

[Φ + Φ′]⊗ [fΨ]− Φ′′ ⊗Ψ = 0

([Φ + Φ′]f − Φ′′)⊗Ψ = 0.

Using Lemma (1), we have [Φ + Φ′]f − Φ′′ = 0 or Ψ = 0. If Ψ = 0, then we have a
contradiction since Ψ ̸= 0. So

[Φ + Φ′]f − Φ′′ = 0

Φ′′

Φ′ +Φ
= f.

Contradiction, since Φ is a function depends only on x, which means that this case does
not hold.
Case(2) if Φ′ = Φ = Φ′′, then

Φ′

Φ
= 1.

Integrating both sides, we get: ∫
dΦ

Φ
=

∫
dx

ln |Φ| = x+ w

Φ = w1e
x,

since Φ(0) = 1 =⇒
Φ = ex.

Now, since the first and the second derivative of Φ are equal then equation (5) become

Φ⊗Ψ
′
+Φ⊗ fΨ = Φ⊗Ψ

Φ⊗ [Ψ
′
+ fΨ−Ψ] = 0

Φ⊗ [Ψ
′
+ (f − 1)Ψ] = 0.

Thus using Lemma (1) either Φ = 0 or Ψ
′
+ (f − 1)Ψ = 0. If Φ = 0, then we have a

contradiction since Φ ̸= 0. So

Ψ
′
+ (f − 1)Ψ = 0



A. Al-Jarrah, Sh. Alsharif, H. Almefleh / Eur. J. Pure Appl. Math, 15 (4) (2022), 1444-1454 1451

Ψ
′

Ψ
= 1− f∫

dΨ

Ψ
=

∫
(1− f)dt

ln |Ψ| =

∫
(1− f)dt

Ψ = e
∫
(1−f)dt.

To verify equation (5), set Φ = ex = Φ′ = Φ′′ and Ψ = e
∫
(1−f)dt

and Ψ′ = (1−f)e
∫
(1−f)dt

=
(1 − f)Ψ, then u = Φ ⊗ Ψ, ux = Φ′ ⊗ Ψ = Φ ⊗ Ψ, uxx = Φ′′ ⊗ Ψ = Φ ⊗ Ψ, and
ut = Φ⊗Ψ′ = Φ⊗ (1− f)Ψ.

Then

ut + fux = Φ⊗ (1− f)Ψ + f(Φ⊗Ψ)

= (1− f)(Φ⊗Ψ) + f(Φ⊗Ψ)

= (Φ⊗Ψ)− f(Φ⊗Ψ) + f(Φ⊗Ψ)

= Φ⊗Ψ

= uxx,

this implies u = Φ⊗Ψ is a solution of equation (4), where Φ = ex and Ψ = e
∫
(1−f)dt

.
Now, if f(t) = t2, then equation (4) becomes

ut + t2ux = uxx,

which has the solution u = Φ⊗Ψ, where Φ = ex and Ψ = e
∫
(1−t2)dt

= et−
t3

3 .

4. Atomic Solution of Burger equation

Theorem 4. Let u(x, t) ∈ C(I × J), where I, J = [0, 1] or [0,∞). If u has continuous
second partial derivatives, then the differential equation

ut + αuux = υuxx (6)

has an atomic solution.

Proof. Put u = Φ⊗Ψ where Φ(0) = 0 and Ψ(1) = 1, to get

Φ⊗Ψ′ + α[Φ⊗Ψ][Φ′ ⊗Ψ] = υΦ′′ ⊗Ψ. (7)

The product of two atoms is one atom (Φ⊗Ψ)(Φ′⊗Ψ) = (ΦΦ′⊗ΨΨ), so that equation
(6) becomes

Φ⊗Ψ′ + αΦΦ′ ⊗Ψ2 = υΦ′′ ⊗Ψ, (8)



A. Al-Jarrah, Sh. Alsharif, H. Almefleh / Eur. J. Pure Appl. Math, 15 (4) (2022), 1444-1454 1452

and the sum of two atoms is an atom using Lemma 1, we have αΦ′Φ = Φ or Ψ′ = Ψ2.
Case (1) If Ψ′ = Ψ2, then

Ψ′

Ψ2
= 1.

Integrating both sides, we get ∫
dΨ

Ψ2
=

∫
dt

1

Ψ
= −t+ g

Ψ = −1

t
+ g1,

since Ψ(1) = 1 =⇒
Ψ = −1

t
+ 2.

Now, Ψ2 = 1
t2
− 4(1t − 1) ̸= 1

t2
= Ψ′, which means this case dose not hold.

But, if we take g = 0 to hold this case we get

Ψ = −1

t
,

so Ψ′ = 1
t2

= Ψ2, then equation (8) becomes

Φ⊗Ψ2 + αΦΦ′ ⊗Ψ2 = υΦ′′ ⊗Ψ

[Φ + αΦΦ′]⊗Ψ2 = υΦ′′ ⊗Ψ.

Thus Φ + αΦΦ′ = υΦ′′ and Ψ2 = Ψ, when Ψ2 = Ψ implies Ψ = 1. Contradiction, since
Ψ = −1

t , which means that this case does not hold.
Case (2) If αΦΦ′ = Φ, then

Φ′ =
1

α∫
dΦ =

∫
1

α
dx

Φ =
x

α
+ q,

since Φ(0) = 0 =⇒
Φ =

x

α
.

Now, Φ′ = 1
α and Φ′′ = 0 , then equation (8) becomes

Φ⊗Ψ
′
+ α

1

α
Φ⊗Ψ2 = 0⊗ υΨ

[Ψ
′
+Ψ2]⊗ Φ = 0.



REFERENCES 1453

So, Φ = 0 or Ψ
′
+Ψ2 = 0. If Φ = 0, it is a contradiction since Φ ̸= 0. Hence

Ψ
′
+Ψ2 = 0∫
−dΨ

Ψ2
=

∫
dt

1

Ψ
= t+ c,

since Ψ(1) = 1 =⇒
Ψ =

1

t
.

To verify equation (6), set Φ = x
α , Φ

′ = 1
α , Φ

′′ = 0, and Ψ = 1
t , Ψ

′ = −1
t2

= − Ψ2, then
u = Φ⊗Ψ, ux = Φ′⊗Ψ, υuxx = υ(Φ′′⊗Ψ) = υ(0⊗Ψ) = υ0⊗Ψ = 0⊗Ψ = 0, ut = Φ⊗Ψ′.

ut + αuux = Φ⊗Ψ′ + α(Φ⊗Ψ)(Φ′ ⊗Ψ)

= Φ⊗−Ψ2 + α(Φ⊗Ψ)(Φ′ ⊗Ψ)

= Φ⊗−Ψ2 + α(ΦΦ′ ⊗Ψ2)

= −Φ⊗Ψ2 + α(ΦΦ′ ⊗Ψ2)

= (−Φ+ αΦΦ′)⊗Ψ2.

Consequently, −Φ+ αΦΦ′ = − x
α + α x

α
1
α = − x

α + x
α = 0. So

ut + αuux = 0⊗Ψ2

= 0

= υuxx.

This implies u = Φ⊗Ψ is a solution of equation (6), where Φ = x
α and Ψ = 1

t .

Conclusion

In this paper we find an exact solution for secand order partial diffrential equation of
linear type. Further, exact solution using tensor product technique of Burger equation is
presented.

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