Optimal System for non-linear Burger equation ut=uxx +uux.

Abstract: The paper discusses the optimal system for a nonlinear Burger equation whose coefﬁcients are dependent

on ﬁrst order spatial derivatives. The main purpose for the project is to determine the optimal system for the oper-

ators accepted by the equation. We construct the principal Lie algebra, calculate transformations for the generators

which provide one-parameter group of transformations for the operator using Lie equations. We construct optimal

systems for the equation where the method requires a simpliﬁcation of a vector to a general form for each of the

transformations of the generators. These are ﬁnally used to determine invariant solutions for some operators.

Key–Words: Principal Lie Algebra, Transformations, Invariant solution, One-dimensional optimal systems.

Received: March 15, 2024. Revised: July 8, 2024. Accepted: August 23, 2024. Published: September 25, 2024.

1 Introduction

The Burger’s equation

ut=uxx +uux(1)

is a basic non-linear partial differential equation that

is used to model propagation of shocks and solitons,

[1]. It also appears in many other applications includ-

ing plasma physics, non-linear harmonics and traf-

ﬁc ﬂow, [2]. Lie group method is one of most efﬁ-

cient computational methods to obtain exact analytic

as well as invariant solutions of nonlinear partial dif-

ferential equations. It was pioneered by Sophus Lie

in the 19th century (1849-1899), [2].An optimal rep-

resent the best or most favoured. In the context of

the project, the optimal system seek to determine the

minimal representation of the operators accepted by

the nonlinear Burger equation. An optimal system

of one-dimensional subalgebras is constructed using

Lie vectors. Optimal system of symmetry subalge-

bras is important in producing possible invariant solu-

tions through through Lie symmetry simpliﬁcation or

reduction, [3]. The results of other work on symme-

try method have been captured in several outstanding

literary, works, [4], [5], [6], [7].

The optimal systems for a general Burger’s equa-

tion

ut=f(x, u)u2

x+g(x, u)uxx

was determined by the method of symmetry group

classiﬁcation, [8]. The present work discusses the

optimal system of the nonlinear partial differential

equation (1). In this work we use the results of one-

dimensional optimal systems to calculate the invariant

solutions of some examples. The method followed in

the construction of the one-dimensional optimal sys-

tems is found in [3].

In this paper while constructing the principal Lie

algebra, we also show how to determine the Lie

point symmetries of (1). We proceed to construct

transformations for the generators which provide one-

parameter group of transformations for the operator

using Lie equations, [9], [10]. We also show the

method of determining invariant solutions,[6], [7].

The paper also illustrates the construction of one-

dimensional optimal systems of principal Lie algebras

L5. We conclude by calculating invariant solutions of

some one-dimensional subalgebras of each extended

algebra L5.

2 Symmetries of the Burgers equa-

tion

The Burgers equation is given by (1) in which the de-

pendent variable is uand independent variables tand

TSHIDISO MASEBEa , PETER MATHYEb

TUT

MSBE Department

2 Aubrey Matlala Rd, Soshanguve H

SOUTH AFRICA

aOrcid no. 0000-0002-3792-5213

bOrcid no. 0009-0007-4829-0827

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2.1 Prolongation formulas

Given that xand tare two independent variables, and

ua differential variable, then the total derivatives are

deﬁned by

Dx=∂

∂x +ux

∂

∂u +uxx

∂

∂ux

+uxt

∂

∂ut

+···.

Dt=∂

∂t +ut

∂

∂u +uxt

∂

∂ux

+utt

∂

∂ut

+···.

The inﬁnitesimal generator Xis given by

X=T(t, x, u)∂

∂t +ξ(t, x, u)∂

∂x + (fu +g)∂

∂u,(2)

where X[2] is the second prolongation of Xand is

given by

X(2) =X+ζ1

∂

∂ux

+ζ1

∂

∂ut

+ζ2

∂

∂uxx

(3)

+ζ2

∂

∂uxt

+ζ2

∂

∂utt

The coefﬁcients ζ1

x, ζ1

t, ζ2

tt, ζ2

tx and ζ2

xxare given by

ζ1

x=Dx(fu +g)−uxDx(ξ)−utDx(T)

=gx+ufx+ux(f−ξx)−utTx,

ζ1

t=Dt(fu +g)−uxDtξ−utDtT

=gt+uft+ut(f−Tt)−uxξt.

ζ2

xx =Dx(ζ1

x)−uxxDx(ξ)−uxtDx(T)

=gxx +ufxx +ux(2fx−ξxx)−utTxx

+uxx(f−2ξx)−2utxTx,

ζ2

tx =Dt(ζ1

x)−uxxDt(ξ)−uxtDt(T)

=gxt +ufxt +ux(ft−ξxt)−ut(fx−Txt)

+ξtuxx +utx(f−ξx−Tt)−uttTx,

ζ2

tt =Dt(ζ1

t)−uxtDt(ξ)−uttDt(T)

=gtt +uftt +ux(ξtt)−ut(2ft−Ttt)

+utt(f−2Tt)−2utxξt.

2.2 Determination of symmetries Burgers

equation

We solve the determining equations for symmetries of

the Burgers equation (1). The determining equation is

determined from the invariance condition

T(t, x, u)∂

∂t +ξ(t, x, u)∂

∂x + (fu +g)∂

∂u

+ζ1

∂

∂ux

+ζ1

∂

∂ut

+ζ2

∂

∂utt

+ζ2

∂

∂utx

+ζ2

∂

∂xx ut−uxx −uux



ut=uxx +uux

= 0,

or equivalently

(ζ1

t−ζ2

xx −uζ1

x−ux(fu +g))



ut=uxx +uux

= 0,(4)

After substituting for ζ1

t,ζ1

x,ζ2

xx and ut=uxx +uux

in equation (4), we obtain

gt+uft+ (uxx +uux)(f−Tt)(5)

−uxξt−gxx −ufxx

−ux(2fx−ξxx)+(uxx +uux)Txx

−uxx(f−2ξx)+2utxTx

−ugx−u2fx−uux(f−ξx)

+(uxx +uux)Tx−uuxf−uxg= 0.

Separation of coefﬁcients in equation (5) yields

C:gt= 0,(6)

u:ft−fxx −gx= 0,(7)

u2:fx= 0,(8)

ux:−uTt+ξxx +uξx−uf −g−ξt= 0,(9)

uxx :Tx+ 2ξx−Tt+Txx = 0.(10)

utx :Tx= 0.(11)

Integrating (11) with respect to xresults into

T=a(t)(12)

Substituting for Tin (10) and integrating with respect

to xresults in that

ξ=1

2atx+b(t),(13)

Differentiating (13) with respect to twe have that

ξt=1

2attx+bt.(14)

The determining equation (9) splits into

ξxx −g−ξt= 0 (15)

−Tt+ξx−f= 0 (16)

After differentiating equation (14) with respect to t,

and applying equation (6) we obtain that

ξtt =1

2atttx+btt = 0,(17)

whence attt = 0 and btt = 0,and thus we have that

a(t) = C1t2+ 2C2t+C3, b(t) = C4t+C5.(18)

It follows from equation (15) that

g=−ξt=−1

2attx−bt=−C1x−C4(19)

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We also have equation (16) which determines that the

function fis given by

f=Tt−ξx=−1

2at(t) = −C1t−C2(20)

The inﬁnitesimals are

T=C1t2+ 2C2t+C3,(21)

ξ=C1tx +C2x+C4t+C5,(22)

η=−(C1t+C2)u−C1x−C4(23)

2.2.1 Symmetries

For the symbol of inﬁnitesimal transformation or the

generator,

X= (T)∂

∂t + (ξ)∂

∂x + (fu +g)∂

∂u

the corresponding symmetries are given by

X1=t2∂

∂t +tx ∂

∂x −(x+tu)∂

∂u,(24)

X2= 2t∂

∂t +x∂

∂x −u∂

∂u,(25)

X3=∂

∂t,(26)

X4=t∂

∂x −∂

∂u,(27)

X5=∂

∂x.(28)

2.3 Commutator Table

Considering the operators

Xa=Ta

∂

∂t +ξa

∂

∂x +ηa

∂

∂u

Xb=Tb

∂

∂t +ξb

∂

∂x +ηb

∂

∂u

The commutator [Xa, Xb]of operators (24) to (28) is

a linear operator deﬁned by the formula

[Xa, Xb] = XaXb−XbXa

Furthermore we deﬁne

[Xa, Xb]=(Xa(Tb)−Xb(Ta)) ∂

∂t

+ (Xa(ξb)−Xb(ξa)) ∂

∂x

+ (Xa(ηb)−Xb(ηa)) ∂

∂u

As an illustration we determine the commutator

[X2, X3]=(X2(1) −X3(2t)) ∂

∂t (29)

+ (X2(0) −X3(x)) ∂

∂x

+ (X2(0) −X3(0)) ∂

∂u

=−2∂

∂t

=−2X3

The complete drawn out Τable of commutators is given

Table 1.

Table 1: Commutator Table of operators

[,]X1X2X3X4X5

X10−2X1−X20−X4

X22X10−2X3X4−X5

X3X22X30X50

X40−X4−X50 0

X5X4X50 0 0

The Lie Algebra L5spanned by the symmetries

(24 - 28) provide a possibility of ﬁnding invariant so-

lutions of equation (1) based on any one-dimensional

subalgebra of the algebra L5,i.e. on any operator

X∈L5.However, there are an inﬁnite number of

one-dimensional subalgebra of the algebra L5,since

an arbitrary operator from L5is expressed as

X=l1X1+l2X2+... +l5X5(30)

which depends upon arbitrary constants l1, l2, ..., l5.

2.4 Construction of an optimal system of

one-dimensional subalgebras.

The construction of optimal system of one-

dimensional subalgebras of Lie Algebra L5follow

from the method in [3]. The transformations of the

symmetry group with Lie algebra L5provide the 5 -

parameter group of transformations of the operators

X∈L5or, equivalently, linear transformations of the

vector

l= (l1, l2, ..., l5)(31)

To determine linear transformations we use their gen-

erators

Eα=ci

αβXii= 1,2,3,4,5.

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and the structure constant of the Lie Algebra L5is

given by ci

αβ which is deﬁned by

[Xα, Xβ] = ci

αβXi

Case 1: For α= 1,we have

E1=ci

1βlβ∂

∂li

where ci

αβ the structure constant of the Lie Algebra

L5is given by

[X1, Xβ] = ci

1βXi

Setting β= 2,we have i= 1.That is row (1) column

(2) from the commutator table, we have that

[X1, X2] = c1

12X1

The non-vanishing structure constant is

12 =−2

Setting β= 3,we have i= 2.That is row (1) column

(3) from the commutator table, we have that

[X1, X3] = c2

13X2

The non-vanishing structure constant is

13 =−1

Setting β= 5,we have i= 4.That is row (1) column

(5) from the commutator table, we have that

[X1, X5] = c4

15X4

The non-vanishing structure constant is

15 =−1

From equation (2.4) we have that

E1=c1

12l2X1+c2

13l3X2+c5

15l5X4

Thus we have that

E1=−2l2∂

∂l1−l3∂

∂l2−l5∂

∂l4(32)

Case 2: For α= 2,we have

E2=ci

2βlβ∂

∂li

where ci

αβ the structure constant of the Lie Algebra

L5is given by

[X2, Xβ] = ci

2βXi

Setting β= 2,we have i= 1.That is row (2) column

(1) from the commutator table, we have

[X2, X1] = c1

21X1

The non-vanishing structure constant is

21 = 2

Setting β= 3,we have i= 3.That is row (2) column

(3) from the commutator table, we have that

[X2, X3] = c3

23X3

The non-vanishing structure constants are

23 =−2

Setting β= 4,we have i= 4.That is row (2) column

(4) from the commutator table, we have that

[X2, X4] = c4

24X4

The non-vanishing structure constant is

24 = 1

Setting β= 5,we have i= 5.That is row (2) column

(5) from the commutator table, we have that

[X2, X5] = c5

25X5

The non-vanishing structure constant is

25 =−1

From equation (2.4) we have that

E2=c1

21l1X1+c3

23l3X3+c4

24l4X4+c5

25l5X5

Thus we have that

E2= 2l1∂

∂l1−2l3∂

∂l3+l4∂

∂l4−l5∂

∂l5

Case 3: For α= 3,we have

E3=ci

3βlβ∂

∂li

where ci

αβ the structure constant of the Lie Algebra

L5is given by

[X3, Xβ] = ci

3βXi

Setting β= 1,we have i= 2 That is row (3) column

(1) from the commutator table, we have that

[X3, X1] = c2

31X2

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The non-vanishing structure constant is

31 = 1

Setting β= 2,we have i= 3.That is row (3) column

(2) from the commutator table, we have that

[X3, X2] = c3

32X3

The non-vanishing structure constant is

32 = 2

Setting β= 4,we have i= 5.That is row (3) column

(4) from the commutator table, we have that

[X3, X4] = c5

34X5

The non-vanishing structure constant is

34 = 1

From equation (2.4) we have that

E3=c2

31l1X2+c3

32l2X3+c5

34l4X5

Thus we have that

E3=l1∂

∂l2+ 2l2∂

∂l3+l4∂

∂l5

Case 4: For α= 4,we have

E4=ci

4βlβ∂

∂li

where ci

αβ the structure constant of the Lie Algebra

L5is given by

[X4, Xβ] = ci

4βXi

Setting β= 2,we have i= 4.That is row (4) column

(2) from the commutator table, we have that

[X4, X2] = c4

42X4

The non-vanishing structure constants are

42 =−1

Setting β= 3,we have i= 5.That is row (4) column

(3) from the commutator table, we have that

[X4, X3] = c5

43X5

The non-vanishing structure constant is

43 =−1

From equation (2.4) we have that

E4=c4

42l2X4+c5

43l3X5

Thus we have that

E4=−l2∂

∂l4−l3∂

∂l5

Case 5: For α= 5,we have

E5=ci

5βlβ∂

∂li

where ci

αβ the structure constant of the Lie Algebra

L5is given by

[X5, Xβ] = ci

5βXi

Setting β= 1,we have i= 4.That is row (5) column

(1) from the commutator table, we have that

[X5, X1] = c4

51X4

The non-vanishing structure constants are

51 = 1

Setting β= 2,we have i= 5.That is row (5) column

(2) from the commutator table, we have that

[X5, X2] = c5

52X5

] The non-vanishing structure constant is

52 = 1

From equation (2.4) we have that

E5=c4

51l1X4+c5

52l2X5

Thus we have that

E5=l1∂

∂l4+l2∂

∂l5

In summary we have the following linear transforma-

tions

E1=−2l2∂

∂l1−l3∂

∂l2−l5∂

∂l4,(33)

E2= 2l1∂

∂l1−2l3∂

∂l3+l4∂

∂l4−l5∂

∂l5,(34)

E3=l1∂

∂l2+ 2l2∂

∂l3+l4∂

∂l5,(35)

E4=−l2∂

∂l4−l3∂

∂l5,(36)

E5=l1∂

∂l4+l2∂

∂l5(37)

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2.5 Construction of Lie Equations

We determine the transformations provided by the

generators (33) to (37). For the generator E1,the Lie

equations for the parameter a1are written

d¯

da1

=−2¯

l2d¯

da1

=−¯

l3(38)

d¯

da1

=−¯

l5d¯

da1

= 0 d¯

da1

= 0

We integrate all ﬁve equations of (38) using the initial

condition

l|a1=0 =l(39)

We proceed as follows. For E1we have that

d¯

da1= 0

⇒¯

l3=l3;d¯

da1= 0

⇒¯

l5=l5;

d¯

da1=−¯

l3⇒¯

l2=−l3a1+l2;

d¯

da1=−2[−l3a1+l2]

⇒Rd¯

l1=R{2l3a1−2l2}da1

⇒¯

l1=l3a2

1−2a1l2+l1;

⇒¯

l4=−l5a1+l4;¯

l5=l5

Thus for E1we have the following transformations

l1=l3a2

1−2a1l2+l1;¯

l2=−l3a1+l2;(40)

l3=l3¯

l4=−l5a1+l4;¯

l5=l5

For the generator E2we have the transformations

l1=l1a2

2;¯

l2=l2;¯

l3=l3a−2

2(41)

l4=l4a2;¯

l5=−l5a2.

For the generator E3we have the transformations

l1=l1;¯

l2=l1a3+l2;¯

l3=l1a2

3+ 2l2a3+l3;(42)

l4=l4;¯

l5=l4a3+l5.

For the generator E4we have the transformations

l1=l1;¯

l2=l2;¯

l3=l3;(43)

l4=−l2a4+l4;¯

l5=−l3a4+l5.

For the generator E5we have the transformations

l1=l1;¯

l2=l2;¯

l3=l3;(44)

l4=l1a5+l4;¯

l5=l2a5+l5.

2.6 One functionally independent invariant

The assertion that the 5×5matrix ||cλ

µν lν|| of coef-

ﬁcients of operators (33) to (37) has rank four, means

that the transformations (40) to (44) have precisely

one functionally independent invariant. The integra-

tion of the equations (33) to (37)

Eµ(J)=0 µ= 1,2,3.., 5(45)

will help determine the invariant. From equation (33)

we have

E1(J) = −2l2∂J

∂l1−l3∂J

∂l2−l5∂J

∂l4= 0 (46)

The characteristic equation of equation (46) is given

dl1

−2l2=dl2

−l3=dl4

−l5(47)

Solving the linear equation

dl1

−2l2=dl2

−l3(48)

yields that

C= (l2)2−l1l3

Similarly from (35) we have

E3(J) = l1∂J

∂l2+ 2l2∂J

∂l3+l4∂J

∂l5= 0 (49)

The characteristic equation of equation (49) is given

dl2

l1=dl3

2l2=dl5

l4(50)

Solving the linear equation

dl2

l1=dl3

2l2(51)

yields that

C= (l2)2−l1l3

A similar integration with equations (34) and (36) will

yield that l1l3=l4l5=C, l4l3−l1l5=Cetc.

The logical conclusion is that the transformations (40)

to (44) have one functionally independent invariant

given by

J= (l2)2−l1l3(52)

This invariant helps to simplify the vector used to de-

termine the optimal system. We can exclude the oper-

ator X1from the operators providing for the optimal

system where possible. This is done by eliminating ¯

from the transformations of the optimal system from

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transformation (40). To accomplished this we solve

the quadratic equation

l3a2

1−2l2a1+l1= 0

for a1from the transformation (40). The results is that

a1=l2±√J

l3(53)

where Jis given by equation (52). We apply equation

(53) only if J≥0.

The method to determine the optimal system re-

quires the simpliﬁcation of the vector (31) by means

of transformations (40) to (44).

2.7 Cases

The construction of optimal system is divided into

several cases.

(1) The case l3= 0.We subdivide this into cases

namely (a) l3= 0 , l26= 0 and (b) l3= 0 , l2=

(a) We discuss the case when l3= 0, l26= 0.

The vector (30) takes the form

(l1, l2,0, l4, l5),where l26= 0.

We use l2to reduce the given vector (31). From

equation (40) if we set a1=l1

2l2,then l1= 0.

The vector reduces to

(0, l2,0, l4, l5)

Since l26= 0 then from equation (43) we have

l4=−l2a4+l4.Setting a4=l4

l2,we have that

l4= 0.If we let a5=l4

l2from equation (44) in

l5=l2a5+l5,we get l5= 0.The vector (31)

reduces to

(0, l2,0,0,0)

Since l26= 0 we can divide the vector (31) by

l2and obtain the following representation for the

optimal system

X2(54)

(b) The case l3= 0, l2= 0.results in the the vector

(31) taking the form

(l1,0,0, l4, l5)

If l16= 0 we use transformation (43) with a4=

−l4

l1,and have that l4= 0.The vector (31) re-

duces to

(l1,0,0,0, l5,0)

. If l56= 0,we can assume that l5= 1,use

transformation (41) and make l1±1.Taking into

account the possibility that l5= 0,we thus ob-

tain the following representation for the optimal

system

X5, X1+X5, X1−X5(55)

If l5= 0 and l16= 0,we set l1= 1.We apply

transformation (44) with a5=−l4and obtain

the vector

(1,0,0,0,0)

If l1= 1,we get the vector

(0,0,0,1,0)

The contribution to the optimal system is pro-

vided by the vectors

X1, X4(56)

(2) The case l36= 0, J > 0.

We now deﬁne a1in terms of equation (2.6) and

eliminate l1.We thus have the vector given by

(0, l2, l3, l4, l5),where l36= 0.

Jis an invariant for the transformations (40) to

(44), and the condition J > 0with l36= 0 im-

plies that l26= 0.We use the transformation (44)

with ¯

l5=l2a5+l5,and set a5=−l5

l2and have

l5= 0.We also use the transformation (43) with

l4=−l2a4+l4and set a4=l4

l2resulting in

l4= 0.We set a3=−l3

l2in the transformation

(42) where ¯

l3= 2l2a3+l3,and get l3= 0.The

vector (31) reduces to

(0, l2,0,0,0).

The representation for the optimal system is sim-

ilar to equation (54).

(3) The case l36= 0, J = 0.

The case reduces equation (52) to a1=l2

l3.When

we apply transformation (40) we have that l2=

0.Due to the invariance of J= (l2)2−l1l3with

l2= 0,it follows that l1= 0.The vector (31)

becomes

(0,0, l3, l4, l5)

We set a4=l5

l3,in the transformation (43) and

have that l5= 0.This simpliﬁes the vector to

(0,0, l3, l4,0,0)

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If l46= 0,we apply the transformation (41) and

approximate a2to make l4=±1and l3= 1 with

the possibility that l4= 0 to obtain the represen-

tation for the optimal system as

X3X3+X4X3−X4(57)

(3) The case l36= 0, J < 0.

The condition J= (l2)2−l1l3<0implies that

l16= 0 since l36= 0.From the transformation

(40) with a1=l2

l3we get l2= 0.In a similar

fashion from the transformation (43) with a4=

l3we get l5= 0.We also apply transformation

(44) with a5=−l4

l1and get l4= 0.The vector

(31) reduces to

(l1,0, l3,0,0).

The condition J < 0,with l2= 0 suggests that

l1and l3should have a common sign. We can

approximate a2in transformation (41) such that

when we divide the vector (31) by an appropriate

constant we get that l1=l3= 1.The represen-

tation for the optimal system as given by

X1+X3(58)

We ﬁnally collect all the operators (24) to (28) to-

gether with the operators (54), (55), (56), (57) and

(58) to form the optimal system

X1, X2, X3, X4, , X5,(59)

X1+X5, X1−X5,

X3−X4, X3+X4,

X1+X3

3 Invariant Solutions for equation

(59)

A useful feature of a symmetry is that it preserves

the solutions of a differential equation. This means

that if a differential equation has a symmetry then

the solutions of the differential equation remain un-

changed under symmetry transformations. The sym-

metry transformations merely permute the integral

curves among themselves. Such integral curves are

termed invariant solutions. To construct an optimal

system of invariant solutions we have to determine the

invariant solution for each of the operators of the op-

timal system (59).

3.1 Invariant Solution for the operator X1

The operator X1=t2∂

∂t +tx ∂

∂x −(x+tu)∂

∂u has the

characteristic equation given by

t2=dx

tx =−du

(x+tu)(60)

There are two linear equations that are formed from

the characteristic equation (60). The ﬁrst such linear

equation is

t2=dx

tx (61)

Integrating equation (61) yields that x

t=C1where

C1is the constant of integration. Hence one of the

invariants is x

t=λ1(62)

Checking the invariant λ1we have that

X1(λ1) = t2∂(x

∂t +tx∂(x

∂x −(x+tu)∂(x

∂u =−x+x= 0

Thus the operator satisﬁes the invariant condition.

The second linear equation is given as

t2=−du

(x+tu)(63)

The equation (63) simpliﬁes to the ﬁrst order Ordinary

differential equation given by

t+u)+dt

t= 0 (64)

but from equation (62) we have that x

t=λ1, which

we replace in equation (64) and arrive at the equation

(λ1+u)+dt

t= 0.(65)

Solving the equation (65) we obtain the second invari-

ant given by the equation

v=tu +x(66)

Checking the invariant vwe have that

X1(v) = t2∂(tu +x)

∂t (67)

+tx∂(tu +x)

∂x

−(x+tu)∂(tu +x)

∂u

=t2u+tx −t2u−tx

= 0

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Thus the operator satisﬁes the invariant condition. We

designate one of the invariants to be a function of the

other i.e.

v=φ(λ1)

The invariant solution is given by

u=φ(λ1)

t−λ1(68)

We substitute for

ut=1

t2(x−φ(λ1)−λ1φ0(λ1)) (69)

ux=1

t2φ0(λ1)−1

uxx =1

t3φ00(λ1)

into the equation (1) and we obtain that

ut−uxx −uux=φ00(λ1) + φ0(λ1))φ= 0 (70)

which when integrated once yields that

φ0(λ1) + 1

2φ2(λ1) = 1

2C1(71)

This gives that

dφ(λ1)

dλ1

2(C1−φ2(λ1)) (72)

This implies that

Zdφ(λ1)

C1−φ2(λ1)=Z1

2dλ1(73)

which gives that

Zdφ(λ1)

C1−φ2(λ1)=1

2λ1+A(74)

with Aa constant. For C1= 0,we obtain that

φ(λ1) = 2

λ1+ 2A(75)

We conclude that the invariant solution is given by

u(t, x) = 2

x+ 2At −x

t(76)

4 Conclusion

The purpose of the project was to gain an insight into

the method of optimal system a non linear equation

using the simpliﬁcation of a vector. The challenges

were how to simplify the vector used to determine

the optimal system. However the determination of the

rank the coefﬁcients matrix of operators helped solve

the problem. From the present project, the method of

ﬁnding optimal systems of one-dimensional subalge-

bras, proved to be effective. We would like to explore

them further and even for higher dimensional subal-

gebras. Future projects would also include extending

on the current one to determine an optimal system of

the invariant solutions for the equation (1).

Acknowledgements: The authors would like to ac-

knowledge the assistance of fellow colleagues in go-

ing through the manuscript and for their invaluable

inputs. The research was supported by the ﬁnancial

assistance from Tshwane University of Technology.

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Contribution of Individual Authors to the

Creation of a Scientific Article (Ghostwriting

Policy)

The authors equally contributed in the present

research, at all stages from the formulation of the

problem to the final findings and solution.

Sources of Funding for Research Presented in a

Scientific Article or Scientific Article Itself

No funding was received for conducting this study.

Conflict of Interest

The authors have no conflicts of interest to declare

that are relevant to the content of this article.

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