Oscillation Criteria For Even Order Nonlinear Neutral Differential
Equations With Mixed Arguments
1
E. Thandapani,
2S.Padmavathi,
3S. Pinelas
1,2
Ramanuajn Institute for Advanced Study in Mathematics, University of Madras, Chennai 600 005, India.
[email protected]
3
Academia Militar, Departamento de Ciências Exactas e Naturais, Av. Conde Castro Guimarães, 2720-113 Amadora, Portugal.
[email protected]
ABSTRACT
This paper deals with the oscillation criteria for nth order nonlinear neutral mixed type differential equations of the form
(
x
(
t
)
ax
(
t
1)
bx
(
t
2))
(n)=
q
(
t
)
x
(
t
1)
p
(
t
)
x
(
t
2),
(
(
)
(
)
(
))
=
(
)
(
1)
(
)
(
2),
) ( 2 1
ax
t
bx
t
q
t
x
t
p
t
x
t
t
x
n and
(
(
)
(
)
(
))
=
(
)
(
1)
(
)
(
2)
) ( 2 1
ax
t
bx
t
q
t
x
t
p
t
x
t
t
x
nwhere
n
is an even positive integer,a
andb
are nonnegative constants,
1,
2,
1 and
2 are positive real constants,q
(
t
),
p
(
t
)
C
([
t
0,
),
(0,
))
and
,
and
are ratios of odd positive integers with
,
1
. Some examples are provided to illustrate the main results.Keywords:
Oscillation; Even order; Mixed arguments; Nonlinear neutral differential equations.Mathematics Subject Classification 2010:
34C15.Council for Innovative Research
Peer Review Research Publishing System
Journal: Journal of Advances in Mathematics
Vol 5, No. 1
[email protected]
1 INTRODUCTION
In this paper, we study the oscillatory behavior of all solutions of nth order nonlinear neutral differential equations with mixed arguments of the form
(
x
(
t
)
ax
(
t
1)
bx
(
t
2))
(n)=
q
(
t
)
x
(
t
1)
p
(
t
)
x
(
t
2),
(1)
(
(
)
(
)
(
))
=
(
)
(
1)
(
)
(
2),
) ( 2 1
ax
t
bx
t
q
t
x
t
p
t
x
t
t
x
n (2) and
(
(
)
(
)
(
))
=
(
)
(
1)
(
)
(
2)
) ( 2 1
ax
t
bx
t
q
t
x
t
p
t
x
t
t
x
n (3)where
n
is an even positive integer,a
andb
are nonnegative constants,
1,
2,
1 and
2 are positive real constants,q
(
t
),
p
(
t
)
C
([
t
0,
),
(0,
))
and
,
and
are ratios of odd positive integers with
,
1
. As is customary, a solution is called oscillatory if it has arbitrarily large zeros and nonoscillatory if it is eventually positive or eventually negative. Equations (1.1),(1.2) and (1.3) are called oscillatory if all its solutions are oscillatory.Differential equations with delayed and advanced arguments (also called mixed differential equations or equations with mixed arguments) occur in many problems of economy, biology and physics (see for example [2,4,8,9,14]), because differential equations with mixed arguments are much more suitable than delay differential equations for an adequate treatment of dynamic phenomena. The concept of delay is related to a memory of system, the past events are importance for the current behavior, and the concept of advance is related to a potential future events which can be known at the current time which could be useful for decision making. The study of various problems for differential equations with mixed arguments can be seen in [3,7,13,16,19,22]. It is well known that the solutions of these types of equations cannot be obtained in closed form. In the absence of closed form solutions a rewarding alternative is to resort to the qualitative study of the solutions of these types of differential equations. But it is not quite clear how to formulate an initial value problem for such equations and existence and uniqueness of solutions becomes a complicated issue. To study the oscillation of solutions of differential equations, we need to assume that there exists a solution of such equation on the half line.
The problem of asymptotic and oscillatory behavior of solutions of nth order delay and neutral type differential equations has received great attention in recent years see for example [1-27], and the references cited therein. However, there are few results regarding the oscillatory properties of neutral differential equations with mixed arguments.
In[10] the authors established some oscillation criteria for the following mixed neutral equations
x
(
t
)
cx
(
t
h
)
c
*x
(
t
h
*)
(n)=
qx
(
t
g
)
px
(
t
g
*),
(4) and
x
(
t
)
cx
(
t
h
)
c
*x
(
t
h
*)
(n)=
qx
(
t
g
)
px
(
t
g
*),
(5) wheren
is an even positive integer,c
,
c
*,
h
,
h
*,
p
andq
are real numbers andg
andg
* are positive constants. In[5] the author established some oscillation results for the following mixed neutral equation
x
(
t
)
p
1x
(
t
1)
p
2x
(
t
2)
=
q
1x
(
t
1)
q
2x
(
t
2),
t
t
0,
(6) withq
1 andq
2 are nonnegative real valued functions.In[25] the authors established some oscillation theorems for the following second order mixed neutral differential equation
(
x
(
t
)
p
1x
(
t
1)
p
2x
(
t
2))
=
q
1(
t
)
x
(
t
1)
q
2(
t
)
x
(
t
2),
t
t
0,
(7)
where
,
and
are ratios of odd positive integers,p
i,
i and
i,i
=
1,2
are positive constants and1,2.
=
)),
[0,
),
,
([
t
0i
C
q
i
In[26] the authors established some oscillation criteria for the following second order mixed neutral differential equation
(
x
(
t
)
ax
(
t
1)
bx
(
t
2))
=
q
(
t
)
x
(
t
1)
p
(
t
)
x
(
t
2),
t
t
0,
(8) and
(
x
(
t
)
ax
(
t
1)
bx
(
t
2))
=
q
(
t
)
x
(
t
1)
p
(
t
)
x
(
t
2),
t
t
0,
(9) where
and
are ratios of odd positive integers with
1
,a
,
b
,
1,
2,
1,
2 are positive constants and1,2.
=
)),
,
[
),
,
([
)
(
),
(
t
p
t
C
t
0t
0i
q
Clearly equations (1.4) and (1.5) with
=
=
=
1
andq
(
t
)
=
q
,
p
(
t
)
=
p
are special cases of equations (1.1) and (1.3) and equation (1.6) withn
=
2
and
=
=
=
1,
q
(
t
)
=
q
1,
p
(
t
)
=
q
2 is a special case of equation (1.3) respectively. Moreover equations (1.7), (1.8) and (1.9) withn
=
2
are special cases of of equations (1.1), (1.2) and (1.3) respectively. Motivated by the above observation in this paper we study the oscillatory behavior of equations (1.1),(1.2) and (1.3) for different values of
1
and
1
. Therefore our results generalize and extend those of [5,10,25,26].In Section 2, we present some sufficient conditions for the oscillation of all solutions of equations (1.1),(1.2) and (1.3). Examples are provided in Section 3 to illustrate the main results.
2 SOME PRELIMINARY LEMMAS
In this section we shall obtain some sufficient conditions for the oscillation of all solutions of the equations (1.1),(1.2) and (1.3). Before proving the main results we state the following lemmas which are essential in the proofs of our oscillation theorems.
Lemma 2.1 Let
A
0,
B
0
and
1.
Then.
)
(
2
1
1 B
A
B
A
IfA
B
,
then.
)
(
B
A
B
A
The proof may be found in [23].
Lemma 2.2 ([21]) Let
([
0,
),
).
C
t
R
u
n Ifu
(n)(
t
)
is eventually of one sign for all larget
,
then there exists a,
>
t
1t
x for somet
1> t
0 , and an integerl
,0
l
n
,
withn
l
even foru
(n)(
t
)
0
orn
l
odd for0
)
(
) (
t
u
n such thatl
>
0
implies thatu
(k)(
t
)
>
0
fort
>
t
x,
k
=
0,1,...,
l
1,
andl
n
1
, implies that0
>
)
(
1)
(
lku
(k)t
fort
>
t
x,
k
=
l
,
l
1,...,
n
1.
Lemma 2.3 ([1]) Let
u
be as in Lemma 2.2. Assume thatu
(n)(
t
)
is not identically zero on any interval[
t
0,
),
andthere exists a
t
1
t
0 such thatu
(n1)(
t
)
u
(n)(
t
)
0
for allt
t
1.
Iflim
(
)
0,
u
t
t
then for every
,
0
<
<
1,
there existsT
t
1,
such that for allt
T
,
).
(
1)!
(
)
(
1 ( 1)t
u
t
n
t
u
n n
Lemma 2.4 ([15])Suppose
q
:
[
t
0,
)
R
is a continuous and eventually nonnegative function, and
is a positive real number. Then the following hold.(I) If
1,
>
)
(
1)!
(
!
)
(
)
(
limsup
1ds
s
q
i
n
i
s
t
t
s
i n i t t t
hold for some
i
=
0,1,...,
n
1,
then the inequality)
(
)
(
)
(
) (
t
y
t
q
t
y
n(II) If
1,
>
)
(
1)!
(
!
)
(
)
(
limsup
1ds
s
q
i
n
i
t
s
s
t
i n i t t t
hold for some
i
=
0,1,...,
n
1,
then the inequality)
(
)
(
)
(
1)
(
nz
(n)t
q
t
z
t
has no eventually positive solution
z
(t
)
which satisfies(
1)
jz
(j)(
t
)
>
0
eventually,j
=
0,1,...,
n
.
Lemma 2.5 ([24]) Assume that for large
t
],
,
[
0
)
(
s
for
all
s
t
t
*q
where
t
* satisfies
(
t
*)
=
t
.
Then,
0,
=
))]
(
(
)[
(
)
(
t
q
t
x
t
t
t
0x
has an eventually positive solution if and only if the corresponding inequality
,
0,
))]
(
(
)[
(
)
(
t
q
t
x
t
t
t
0x
has an eventually positive solution.
In [6,12,18,27], the authors investigated the oscillatory behavior of the following equation
,
0,
=
))]
(
(
)[
(
)
(
t
q
t
x
t
t
t
0x
(10) where([
0,
),
)
C
t
R
q
,
C
([
t
0,
),
R
)
,
(
t <
)
t
,
(
)
=
lim
t
t
and
(0,
)
is a ratio of odd positive integers.Let
=
1
. Then equation (2.1) reduces to the linear delay differential equation,
0,
=
))
(
(
)
(
)
(
t
q
t
x
t
t
t
0x
(11)and it is shown that every solution of equation (2.2) oscillates if
.
1
>
)
(
liminf
) (e
ds
s
q
t t t
(12)3 Oscillation Results
First we study the oscillation of all solutions of equation (1.1).
Theorem 3.1 Let
i>
i for)
>
0,
2
(1
,
1,1
,
1,2,
=
1
a
b
b
a
i
andq
(t
)
andp
(t
)
arenonincreasing functions for
t
t
0.
Assume that the differential inequalities0,
)
(
)
(
)
(
)
(
)
(
2 2 / 1 2 1 / 1 ) (
t
y
t
p
c
t
y
t
q
c
t
y
n (13)0,
)
(
)
(1
2
)
(
)
(
1 1 / / 1 ) (
y
t
a
t
q
t
y
n (14) and0,
)
(
)
2
(1
2
)
(
)
(
/ 2 2 / 1 1 ) (
t
y
b
a
t
p
t
y
n (15)where
(
2
)
},
2
1
,
)
2
(
2
1
,
1
,
1
{
min
=
/ 1 1 / 1 1 1 b
b
b
b
c
have no eventually positive solution, no eventually
positive decreasing solution and no eventually positive increasing solution respectively. Then every solution of equation (1.1) is oscillatory.
Proof: Let
x
(t
)
be a nonoscillatory solution of equation (1.1). Without loss of generality we may assume thatx
(t
)
is eventually positive,i.e., there exists at
1
t
0 such thatx
(t
)
>
0
fort
t
1.
Set.
))
(
)
(
)
(
(
=
)
(
1 2
ax
t
bx
t
t
x
t
z
Then.
0
>
)
(
)
(
)
(
)
(
=
)
(
1 2 1 0 ) (t
t
t
all
for
t
x
t
p
t
x
t
q
t
z
n
(16)Thus
z
(i)(
t
),
i
=
0,1,...,
n
,
are of one sign on[
t
2,
);
t
2
t
1.
As a result we have two cases:(a)z
(t
)
<
0
for,
2
t
t
(b)z
(t
)
>
0
fort
t
2.
Case(a):
z
(t
)
<
0
fort
t
2.
In this case, we let).
(
))
(
)
(
)
(
(
=
)
(
=
)
(
<
0
u
t
z
t
bx
t
2
x
t
ax
t
1
b
x
t
2 Then.
),
(
1
)
(
2 2 1/t
t
t
u
b
t
x
Using above inequality in equation (1.1), we have
)
(
)
(
)
(
)
(
)
(
=
0
( )
1
2t
x
t
p
t
x
t
q
t
u
n),
(
)
(
)
(
)
(
)
(
/ 1 2 / 2 2 ) (
u
t
b
t
p
t
u
b
t
q
t
u
n or0
)
(
)
(
)
(
)
(
)
(
1 / 1 2 1 / 2 2 ) (
t
u
t
p
c
t
u
t
q
c
t
u
nhas a positive solution, which is a contradiction.
Case(b):
z
(t
)
>
0
fort
t
2.
Now we set.
),
(
2
)
(
)
(
=
)
(
t
z
t
a
z
t
1b
1z
t
2t
t
2y
(17) Then)
(
2
)
(
)
(
=
)
(
( ) ( ) 1 1 ( ) 2 ) (
a
z
t
b
z
t
t
z
t
y
n n n n)
(
)
(
)
(
)
(
=
q
t
x
t
1
p
t
x
t
2
(
1)
(
1
1)
(
1)
(
2
1)
a
q
t
x
t
p
t
x
t
(
)
(
)
(
)
(
)
.
2
1
2
1
2
2
2
2
b
q
t
x
t
p
t
x
t
Using the monotonicity of
q
(t
)
andp
(t
)
,a
,b
1,1
and Lemma 2.1 in the above inequality, we get
(
)
(
)
(
)
2
)
(
)
(
1 1 1 1 1 2 ) (
x
t
ax
t
bx
t
t
q
t
y
n
(
)
(
)
(
)
.
2
)
(
2 2 1 2 2 1
p
t
x
t
ax
t
bx
t
Now using
z
(t
)
>
0
fort
t
2 in the above inequality, we obtain,
0,
>
)
(
2
)
(
)
(
2
)
(
)
(
1 / 1 1 / 2 2 ) (t
t
t
z
t
p
t
z
t
q
t
y
n
(18) which implies that the functiony
(i)(
t
),
i
=
0,1,...,
n
are of one sign. We shall prove thaty
(t
)
>
0
eventually. If not, then).
(
2
)
(
2
)
(
)
(
=
)
(
=
)
(
<
0
1 1
2 1
2
y
t
z
t
a
z
t
b
z
t
b
z
t
t
v
Hence).
(
2
)
(
2 1
v
t
b
t
z
Using the last inequality in (3.6), we obtain
0.
)
(
2
2
)
(
)
(
2
2
)
(
)
(
/ 2 2 / 1 1 2 1 / / 1 1 ) (
v
t
b
t
p
t
v
b
t
q
t
v
n or.
0,
)
(
)
(
)
(
)
(
)
(
1 / 1 2 1 / 2 2 2 ) (t
t
t
v
t
p
c
t
v
t
q
c
t
v
n
Using the procedure of case(a),
v
(t
)
is a positive solution of (3.1), a contradiction. Thus we consider two possible cases:(i):z
(t
)
<
0
eventually,(ii)z
(t
)
>
0
eventually.Case(i): Assume that
z
(t
)
<
0
for all 2.
* 3t
t
t
Then from (3.4) we have(
1)
iz
(i)(
t
)
>
0
fort
t
4
t
3* and1
0,1,...,
=
n
i
. We claim thaty
(t
)
<
0
fort
t
4.
To prove it, assumey
(t
)
>
0
fort
t
4.
Then differentiate (3.5), we have)
(
2
)
(
)
(
=
)
(
<
0
1 1
2
t
z
t
a
z
t
b
z
t
y
or)
(
2
)
(
)
(
=
)
(
>
0
1 1
2
y
t
w
t
a
w
t
b
w
t
where
w
= z
>
0
on[
t
4,
).
Since the functionw
is decreasing on[
t
4,
),
we have,
0,
)
(
)
2
(1
)
(
>
0
1w
t
2for
t
t
4b
a
t
y
a contradiction. Thus
y
(t
)
<
0
fort
t
4 and from (3.6), we have.
0,1,...,
=
0
>
)
(
1)
(
iy
(i)t
for
t
t
4and
i
n
(19) Now, using the monotonicity ofz
(t
)
, we obtain.
),
(
)
(1
)
(
2
)
(
)
(
=
)
(
1 1z
t
2a
z
t
1t
t
4b
t
z
a
t
z
t
y
,
0,
)
(
)
(1
2
)
(
)
(
1 / / 1 1 4 ) (t
t
t
y
a
t
q
t
y
n
(20)has a positive decreasing solution, a contradiction.
Case(ii): Assume that
z
(t
)
>
0
for allt
t
3
t
2.
Now we consider the following two subcases:Subcase(i): Assume that
y
(t
)
<
0
for allt
t
3.
Proceeding as in Case(i) and using the monotonicity ofz
(t
),
we obtain).
(
)
(1
)
(
1
1
a
z
t
t
y
Using the last inequality in (3.6) and the monotonicity of
y
(t
),
we obtain)
(
2
)
(
)
(
1 / 1 ) (
q
t
z
t
t
y
n)
(
)
(1
2
)
(
1 / / 1
y
t
a
t
q
).
(
)
(1
2
)
(
1 1 / / 1
y
t
a
t
q
Thus once again
y
(t
)
is a positive decreasing solution of the equation (3.2), which is a contradiction.Subcase(ii): Assume that
y
(t
)
>
0
for allt
t
3.
Then we have)
(
)
2
(1
)
(
2
)
(
)
(
=
)
(
1 1
2 1
2
a
z
t
b
z
t
a
b
z
t
t
z
t
y
and this with (3.6) implies
,
0,
)
(
)
2
(1
2
)
(
)
(
/ 2 2 3 / 1 1 ) (t
t
t
y
b
a
t
p
t
y
n
(21)has a positive increasing solution which satisfies
,
,
1,2,...,
=
0
>
)
(
3 ) (t
t
and
n
i
for
t
y
i
(22)a contradiction. This completes the proof.
Corollary 3.1 Let
i>
i for)
>
0
2
(1
1,
,
1,2,
=
1 b
a
b
a
i
and
=
=
1
andq
(t
)
andp
(t
)
are nonincreasing functions fort
t
0.
If1
<
<
0
,
1)!
(
>
))
(
)
(
(
)
(
liminf
1 2 1 2 1
e
n
b
ds
s
p
s
q
s
n t t t
(23)1
0,1,...,
=
),
(1
2
>
)
(
1)!
(
!
)
(
)
(
limsup
1 1 1 1 1 1
i
n
i
q
s
ds
a
i
n
t
s
s
t
i n i t t t
(24) and),
2
(1
2
>
)
(
1)!
(
!
)
(
)
(
limsup
1 1 1 2 2 2 2
b
a
ds
s
p
i
n
i
s
t
t
s
i n i t t t (25)where
i
=
0,1,...,
n
1,
then every solution of equation (1.1) is oscillatory.Proof: Let
y
(t
)
be a positive solution of (3.1), fort
t
1
t
0.
Then we have(
)
0
)(
t
y
n for allt
t
1.
Further0
>
)
(
1) (t
y
n for allt
t
1,
otherwisey
(t
)
ast
.
Hence we havey
(
t
)
>
0,
y
(n1)(
t
)
>
0
and0,
)
(
) (
t
y
n fort
t
1.
Then by Lemma 2.2 and Lemma 2.3 we obtain.
(0,1),
),
(
1)!
(
)
(
t
1y
( 1)t
t
t
2t
1n
t
y
n n
From (3.1) and the monotonicity of
y
(t
),
we have.
0,
)
(
)
(
)
(
)
(
1 2 2 ) (t
t
t
y
b
t
p
t
q
t
y
n
Combining the last two inequalities, we obtain
.
0,
)
(
)
(
1)!
(
))
(
)
(
(
)
(
1 2 2 1) ( 1 2 1 ) (t
t
t
y
t
n
b
t
p
t
q
t
y
n n n
Let
(
t
)
=
y
(n1)(
t
).
Then we see that
(t
)
is a positive solution of equation2 2 1 1 2 1
)
(
)
0,
(
1)!
(
))
(
)
(
(
)
(
t
t
t
t
n
b
t
p
t
q
t
n
(26)But according to the Lemma 2.5 and the condition (2.3), condition (3.11) guarantees that inequality (3.14) has no positive solution, which is a contradiction. Hence (3.1) has no eventually positive solution. Moreover in view of Lemma 2.4 (II) and the condition (3.12), inequality (3.8) has no eventually positive solution which satisfies (3.7), which is a contradiction. Hence (3.2) has no eventually positive decreasing solution. Also in view of Lemma 2.4 (I) and the condition (3.13), inequality (3.9) has no eventually positive decreasing solution which satisfies (3.10), which is a contradiction. Hence (3.3) has no eventually positive increasing solution.
Next we consider the equation (1.2) and present sufficient conditions for the oscillation of all solutions.
Theorem 3.2 Let
i>
i for)
>
0,
,
1,
2
(1
1,2,
=
a
1
b
a
b
i
,
1
andq
(t
)
andp
(t
)
are nondecreasing functions fort
t
0.
Assume that the differential inequalities0,
)
(
)
(
)
(
)
(
)
(
2 / 1 1 2 / 2 1 ) (
t
y
t
p
c
t
y
t
q
c
t
y
n (27)0,
)
(
)
(1
2
)
(
)
(
1 / / 1 1 ) (
y
t
b
t
q
t
y
n (28) and0,
)
(
)
(1
2
)
(
)
(
1 / / 2 2 ) (
y
t
b
t
p
t
y
n (29) where(
2
)
},
2
1
,
)
2
(
2
1
,
1
,
1
{
min
=
/ 1 1 / 1 1 2 a
a
a
a
c
have no eventually positive solution, no eventually
positive decreasing solution and no eventually positive increasing solution respectively. Then every solution of equation (1.2) is oscillatory.
Proof: Let
x
(t
)
be a nonoscillatory solution of equation (1.2). Without loss of generality we may assume thatx
(t
)
is eventually positive,i.e., there exists at
1
t
0 such thatx
(t
)
>
0
fort
t
1.
Set
)
(
))
(
)
(
(
=
)
(
1 2 1t
x
t
ax
t
bx
t
z
and proceeding as in the proof of Theorem 3.1, we see that the function
z
1(i)(
t
),
i
=
0,1,...,
n
,
are of one sign on.
);
,
[
t
2
t
2
t
1 As a result we have two cases:(a)z
1(
t
)
<
0
fort
t
2,
(b)z
1(
t
)
>
0
fort
t
2.
Case(a):z
1(
t
)
<
0
fort
t
2.
In this case, we let).
(
))
(
)
(
)
(
(
=
)
(
=
)
(
<
0
u
1t
z
1t
x
t
ax
t
1
bx
t
2
a
x
t
1 Then.
),
(
1
)
(
1/ 1 2 1t
t
t
u
a
t
x
Using above inequality in equation (1.2), we have
)
(
)
(
)
(
)
(
)
(
=
0
u
1(n)t
q
t
x
t
1
p
t
x
t
2),
(
)
(
)
(
)
(
)
(
1/ 1 1 1/ 2 1 ) ( 1
u
t
a
t
p
t
u
a
t
q
t
u
n or0
)
(
)
(
)
(
)
(
)
(
2 1 / 1 2 1 1 / 1 2 ) ( 1
t
u
t
p
c
t
u
t
q
c
t
u
nhas a positive solution, which is a contradiction. Case(b):
z
1(
t
)
>
0
fort
t
2.
Now we set.
),
(
)
(
2
)
(
=
)
(
1 1 1 1 1 2 2 1z
t
b
z
t
t
t
a
t
z
t
y
(30) Then)
(
)
(
2
)
(
=
)
(
1( ) 1 1( ) 1 1( ) 2 ) ( 1
a
z
t
b
z
t
t
z
t
y
n n n n)
(
)
(
)
(
)
(
=
1
2
t
x
t
p
t
x
t
q
(
)
(
)
(
)
(
)
2
1
1
1
1
1
2
1
a
q
t
x
t
p
t
x
t
(
2)
(
1
2)
(
2)
(
2
2)
.
b
q
t
x
t
p
t
x
t
Using the monotonicity of
q
(t
)
andp
(
t
),
a
,
b
1,1
and Lemma 2.1 in the above inequality, we get
(
)
(
)
(
)
2
)
(
)
(
1 1 1 1 1 2 ) ( 1
x
t
ax
t
bx
t
t
q
t
y
n
(
)
(
)
(
)
.
2
)
(
2 2 1 2 2 1
p
t
x
t
ax
t
bx
t
Now using
