International Journal of Mathematics and Statistics Invention (IJMSI) EISSN: 2321 – 4767 PISSN: 2321  4759 www.ijmsi.org Volume 2 Issue 4 ǁ April. 2014ǁ PP0414
An Exposition on Solution Process and Nature of Solutions of Simple Delay Differential Equations Ukwu Chukwunenye Department of Mathematics University of Jos P.M.B 2084 Jos, Nigeria
ABSTRACT: This article gave an exposition on the solutions of singledelay autonomous linear differential equations with algorithmic presentations of the solution processes of two solution methods, with illustrative examples verifying the consistency of both solution methods. The article went further to examine the nature of solutions, obtaining in the process a complete characterization of all nonoscillatory solutions of a problem instance in the absence of initial functions. In the sequel the article proved that there is no nontrivial oscillatory solution if a constant initial function is specified for the delay differential equation in question and hence for delay differential equations in general.
KEYWORDS: Algorithmic, Continuity, Delay, Nature, Process. I.
INTRODUCTION
The method of steps and forward continuation recursive procedure are prevalent in the literature on functional differential equations. However literature on detailed illustrative examples and robust analyses of problem instances is quite sparse, leaving very little room for enhanced understanding and appreciation of this class of differential equations. This article, which makes a positive contribution in the above regard, is motivated by Driver (1977) and leverages on some given problems there to conduct detailed analyses on the process and structure of solutions and thorough analyses of some problem instances. The algorithmic format used in the two solution methods, the verification of the consistency of both methods, the investigation of appropriate feasibility conditions on the solutions and the lucidity of our presentations are quite novel in this area where presentations are skeletal for the most part, forcing readers to plough through several materials and figure out a lot of things on their own with the attendant opportunity cost in time. See also Hale (1977) for more discussion on method of steps.
II.
PRELIMINARIES
For an arbitrary initial function problem:
x t ax t bx t h
(1)
x(t ) t , t h, 0
(2)
at a, b scalar constants, continuous, it is clear that the integrating factor, I t e
If we denote the solution on the interval J k
k 1 h,
kh , k 1, 2,.... by y k t , then
at on J 1 , (1) becomes x t ax t b t h , from which it is clear the integrating factor, I t e .
Hence y1 t can be obtained from the relation:
y1 t ea t b e a t t h dt C1 .
(3)
y1 t exists, since b e a t t h is integrable, in view of the fact that e a t and are continuous, t h being in h, 0 . The solutions yk t , k 1, 2,..., are given recursively by:
yk t ea t b e a t yk 1 t h dt Ck
(4)
y k 1 guarantees the integrability of be a t y k 1 t h and hence the continuity of y k , assuring the existence of y k , for k 1, 2,
where the continuity of
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An Exposition On Solution Process… Above solutions are unique, arising from the unique determination of the C k s from the relation:
yk 1 k 1 h yk k 1 h
(5)
Note that if j 1, 2,..., k 1 , there is no general expression for the if the initial function in (2) is not constant, it is impossible to express
ck j appearing in (6) below. Moreover
y k , in
k yk (t ) d k d k j t j 1 eat (t ). However, if a 0 and (t ) is a polynomial of degree m, j 1 say, m 0, then the y k ' s take the form: the form,
m k 1 yk t d k j t j ck j t j e at 6 j 0 j0 for some constants d k j , j 0,1, m; ck j , j 0,1, , k 1. In other words, yk t is the sum of some
polynomial of degree m and the product of where y0 t t .
e a t and some polynomial Pk 1 t of degree k 1,
The process of getting the above coefficients is easy, but the computations get rather unwieldy. The computational procedure is set out in the following algorithmic steps, noting that: yk t x t for
t k 1 h,
kh.
(7)
Method 1: Algorithmic Steps: [1] Set k = 1 in (6) [2] Plug in y1 t for x t , [3]
y1 t for x t and t h for x t h in (1).
Compare the coefficients of the resulting left and right hand expressions to secure the in the relevant ranges for
j , and obtain y1 t . Verify that the condition
ck , j s and d k , j s
y1 0 0 is satisfied,
where y0 t t . [1] [2]
Set k = 2 in (6). Substitute y2 t for x t , y 2 t for x t and y1 t h for x (t h ) in (1)
[3]
Compare the coefficients of the resulting left and right hand expressions to secure the in the relevant ranges for
ck , j s and d k , j s
j , and hence obtain y2 (t ). Check that the condition y2 h y1 h is
satisfied. This implementation process continues, so that at the k
th
stage, y k t is substituted for
x t , y k t for x t and the already secured yk 1 t h substituted for x t h in (1). Then the
coefficients of the resulting left and right hand expressions are compared to yield the
ck , j s and d k j s for
j 0,1, 2,...., m, m 1, and hence y t , for k 3 . Verify that the continuity condition
yk k 1 h yk 1 k 1 h holds. Alternatively, proceed as follows: Method 2: Algorithmic Steps: [1]
Set k = 1 in (76) and then y0 t h t h and evaluate the integral.
[2]
Set y1 (0) 0 y0 0 to secure C1 , following the computation of the integral on the right of (4)
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An Exposition On Solution Process… Having initialized the process in step 1, y k t is obtained by using (4) in conjunction with the condition yk
k 1 h y k 1 h ,
k 2,3, . The integral in (4) can be determined since
k 1
y k 1 t h is already known. The condition
yk
k 1 h
y k 1
k 1 h secures C k .
Example 1 Solve the initial function problem:
x t ax t bx t 1 on [0, 2], a R. x t t 1 t , t 1, 0
Solution: Method 1
h 1, k 1, c1 c10 , d1 d10 y1 t d1 d11 t c1 e a t ; y1 t a y1 t b y0 t h
d11 a c1 e
at
a d1 a d11 t b 1 t 1 a c1 e a t b
d11 a d1 a d11 b t 0 d11 If a 0, then c1 1 d1 1
b a2 ,
a
, d1
1 a
d11
b a2
.
a 1. The condition y1 0 0 d1 c1 1 .
Verification
b b 2 1, a 0. 2 a a b b b at e , if a 0; that is Therefore, y1 t 2 t 1 a a a b b b x t 2 t 1 2 e a t for t 0 1 , if a 0 . a a a c1 d1 1
2
The condition
b 0 precludes degeneracy of (1) to an ordinary differential equation.
y2 t d 2 d 21 t c2 c21 t e
at
, where
d 2 d 2, 0
y 2 t a y2 t b y1 t 1
d 21 ac2eat c21e at ac21t e a t a d 2 a d 21 t ac2e
at
ac21te
at
a t 1 b d1 d11 t 1 c1e
d21 a d2 b d1 d11 b a d21 b d11 t c21 bc1e
We claim that the members of the set 1, t , e
at
a
e
at
0
are linearly independent for a 0.
Proof:
1 t e at at at at at 0 1 ae ae 0, for a 0. This proves the claim. Indeed 1, t , e , and te are linearly 0 0 aeat independent as the Wronskian turns out to be a 0 for a 0. By the above claim it is valid to set each of the above coefficients to zero. Consequently, 4
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An Exposition On Solution Process…
b d11 b b b2 b 2 , c21 c1 b e a b 1 2 e a , a a a a a 2 2 2 d b d11 b d1 1b b b b2 b2 d 2 21 2 2 2 3 2. a aa a a a a
d 21
The requirement y2 (1) y1 (1) must be enforced.
b2 b a b a y2 (1) y1 (1) d 2 d 21 (c2 c21 )e d1 d11 c1e 2 e 2 e a a a 2 2 2 2 b b b b b b b c2 2 ea 2 ea 2 3 2 2 b 2 e a . Hence, a a a a a a a 2 2 2 b b b b b b x(t ) y2 (t ) 2 3 2 2 t 2 ea (t 1) e a (t 1) e at 2 e at a a a a a a 2 2 b b b 2 3 ea (t 1) bea (t 1) 2 ea (t 1) b 1 2 tea (t 1) , for t [1, 2]. a a a a
a
b2 b2 b2 b b b a b2 b2 b b b b a e e 2 b b 2 2 ea 2 ea 3 2 2 2 2 3 2 a a a a a a a a a a a a b b b y1 (1) 2 ea 2 ea . Therefore the condition y2 (1) y1 (1) is satisfied. a a a We proceed to show that y2 (t ) satisfies the delay differential equation on [1, 2]. y2 (1) 2
b 2 b a ( t 1) b at 2b 2 a ( t 1) b 2 a ( t 1) a ( t 1) at a ( t 1) e be ae e e abe e be a ( t 1) 2 2 a a a a a 2 2 b b 2 e a (t 1) abte a ( t 1) te a ( t 1) . a a 2 2 b b b2 b a (t 1) b at b 2 a (t 1) a ( t 1) at ay2 (t ) by1 (t 1) 2 2 2 t e be ae e 2 2 e abea (t 1) a a a a a a 2 2 2 2 2 b b b b b b2 e a ( t 1) abte a ( t 1) te a ( t 1) 2 t be a ( t 1) 2 ea ( t 1) . a a a a a a y 2 (t )
Therefore y 2 (t ) ay2 (t ) by1 (t 1). This completes the proof that y2 (t ) solves the initial function problem on the interval
[1, 2].
Method 2: Algorithmic Steps Set k = 1 in (4) and then y0 t h t h , evaluate the integral. Set y1 (0) 0 y0 0 to secure C1 , following the computation of the integral on the right of (4) Having initialized the process in step 1, y k t is obtained by using (4) in conjunction with the condition yk
k 1 h y k 1 h , k 1
k 2,3, . The integral in (4) can be determined since
yk 1 t h is already known. The condition yk k 1 h yk 1 k 1 h secures Ck . Consider the interval J1 [0, 1]. Set y0 (t 1) (t 1) 1 t 1 t , on J 0 [1, 0]. Apply (4) with
b at b at at a t at k 1, h 1 to get y1 t e b e t dt C1 e t e 2 e C1 . a a
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An Exposition On Solution Process…
y1 0 (0) 1 b
y1 t
a
2
a
C1 1 C1 1
2
b
b
t 1
a
b a
y1 t
2
b
t
a
b a
2
1
b at e a2
1 at e . a2
Consider the interval J 2 [1, 2]. Set y2 (1) y1 (1)
b a
2
b a
ea
b 2
e a . Apply (4), with k 2, to get
a b b b y2 t ea t b e a t y1 t 1 dt C2 e a t b e a t 2 (t 1) 1 2 e a ( t 1) dt C2 a a a
2 b2 a t b2 a t b2 a t a a b y2 t e 2 e dt e tdt e dt e b dt e dt C2 2 a a a a 2 2 2 2 b b b b 1 1 y2 t ea t 3 e a t 2 e a t e a bt e a 2 t te a t 2 e a t C2 a a a a a a 2 2 2 2 b b b b b b b 1 1 y2 (1) 2 ea 2 ea ea 3 e a 2 e a e a b e a 2 e a 2 e a C2 a a a a a a a a a at
b a
2
b a
2
C2
b a b a b a
b
ea
a
b
ea
2
e
a
2
a b
2
a
ea
b2 a
ea 2
e
a
3
b
2
a
3
1
b2
a
b
2
a b2
2
b
b a
2
2
b2
b
2
a
3
a
2
e
a
be
2
ea t
b
a
y2 t
b
2
a
3
b
2
a
2
2
e a
b a
e a ( t 1) bt
e a 1 b
2
a
2
a
2
b2 a
3
C2 e a
C2 e a
b b y2 t ea t 3 e a t 2 e a t e a bt e a a a 2
b2
b a
2
te a ( t 1)
a
b2 2
e a
a b b2 1 a t 1 a t t te 2 e a2 a a a 2
2 b
2
a
2
b2 a
3
t
e a be a b
b2 a
2
e a
2
a3
b b b b2 b2 eat 2 e a e a 1 2 2 3 e a be a 2 e a a a a a a 2 2 2 2 2 b b b b b y2 t 3 2 e a ( t 1)bt 2 te a (t 1) 2 t 3 a a a a a b b b b2 b2 2 e a (t 1) e a (t 1) e a t 2 e a t 2 3 e a (t 1) be a (t 1) 2 e a (t 1) a a a a a 2 2 2 2 2 b b b b b y2 t 3 2 e a ( t 1)bt 2 te a ( t 1) 2 t 3 a a a a a b b b b2 b2 2 e a (t 1) e a (t 1) ea t 2 ea t 2 3 ea (t 1) bea (t 1) 2 ea (t 1) a a a a a 2 2 2 2 2 b b b b b b b b b2 2 3 2 2 t 1 2 e at b 2 2 2 3 e a (t 1) b 2 te a ( t 1) a a a a a a a a a www.ijmsi.org
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An Exposition On Solution Process…
x(t ) 2
b2 b2 b2 b at b b b2 b 2 a (t 1) b 2 a (t 1) t 1 e b 2 e b te 2 a3 a 2 a 2 a a2 a2 a3 a2 a
is the solution of the initial function problem on the interval using Method 1. The case
[1, 2]. This is consistent with the result obtained
a 0 x (t ) bx(t 1), t 0; x(t ) (t ) 1 t , t [1, 0]. Hence
x (t ) bt x(t )
1 2
bt 2 d1 , t [0,1]. The continuity condition x(0) (0) 1
1 d1 1 x(t ) y1 (t ) bt 2 1 on [0,1]. 2 1 2 1 2 2 3 On (1,2), x (t ) b (t 1) b x(t ) y2 (t ) b (t 1) bt d 2 . 2 6 1 1 2 b 3 The continuity condition y2 (1) y1 (1) d 2 1 b x(t ) y2 (t ) b (t 1) bt 1. 2 6 2 at a t Using the relation yk t e b e yk 1 t h dt Ck , with a 0, h 1 and k 1 yields
t2 y 1 t by0 t h dt C1 b t 1 dt C1 btdt C1 b C1 ; y 1 0 (0) 1 C1 1 2 2 t y 1 t x(t ) b 1, t [0,1]. 2 y2 t by1 t h dt C2 b
y2 1 y 1 1 b C2
1 2
1 2 1 2 3 2 b(t 1) 1 dt C2 6 b (t 1) bt C2 ;
b 1 C2 1
1 2
b y2 t x (t )
1 6
b 2 (t 1) 3 bt 1
1 2
b, t [1, 2].
This also agrees with the preceding result. Nature of nontrivial oscillatory solutions One can appreciate the tedium involved in extending the solutions beyond the interval [1, 2]. Equation (1) is a first order scalar, linear, homogeneous delay differential equation with real coefficients. Solutions obtained thus far have been real nonoscillatory. In general oscillatory solutions of (2) can be obtained in the absence of initial function specification by assuming solutions of the form:
x t e t ,
(8)
to obtain the transcendental equation:
a de h 0
(9)
This transcendental equation is analyzed completely to determine the nature of nontrivial solutions of oscillatory type. We assert that in general real solutions do not exist in the form
x t et
.
Our analysis and conclusion in the next example will validate the above assertion. Example 2 Prove that:
x(t ) x t 2
(10)
admits nontrivial oscillating solutions and determine the nature of such solutions.
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An Exposition On Solution Process… Prove also that there are no nontrivial oscillatory solutions if a constant initial function is specified for the delay differential equation (10). Proof Our proof will rely partly on the following result which we proceed to establish:
x ln 1 x x 1
(11)
To establish this result, observe that the result is equivalent to:
ex 1 x
(12)
Set:
f x ex 1 x
(13)
e x 1 0, x 0 f x 0, x 0
Then,
f 0 0, f ( x ) 0 iff x 0 . Thus (0, 0) is the only stationary point of f(x).
Now f ( x ) e , x . In particular f 0 0 0, 0 is a global minimum of f x . x
Hence f x 0 x; that is e 1 x, x ; this translates to x ln 1 x . x
Assume a solution of the form x (t ) e
t
in the example. Then e
a 0, b 1 and h
2
2
, noting that
.
The equation:
e
2
(14) 2
e 0 if is real. Therefore if is real, then 0, so /2 ln / 2 that e But ln 1 0 , by an appeal to (11). is not satisfied for any real 0 , since
Therefore,
1 / 2 , or 1 1 . Since 2
2
1 0, we conclude that
2
2
0
must be negative. Therefore, the equation e 2 cannot be satisfied t by any real . In other words, any solution of the form x t e must yield i , when and are real numbers such that 0 . Plugging this into the transcendental equation yields:
This contradicts the fact that
i e
2
cos 2
i sin 2
,
(15)
from which we infer that:
e
2
cos 2
(16)
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An Exposition On Solution Process… and:
e If
sin 2
2
17
0 , the above relations imply that:
cos 0 2 and:
(18)
2
sin
(19)
From (18) we infer that:
2
2k 1
2 k 1
2
, k integer (20)
is an odd integer, positive or negative.
19
the relation
Therefore:
1 is infeasible.
1
(21)
The relations (20) and (21) imply that
1 , thus 1 or 1 ; hence if 1, then 1 or 1 and
x t c1 cos t c2 sin t ,
(22)
for real constants c1 and c2. (22) can be rewritten in the form:
x t c12 c22 sin t for some real δ with 0
(23) . Clearly if c1 and c2 do not vanish at the same time, then
2
c12 c2 2 0 , so
that (23) represents nontrivial oscillatory solutions, being sinusoidal. Remarks If 0, If
8
e
0 since e
2
2
0.
0 , then 0 , since real solutions do not exist in the form (8) 3 4 k If 0 , then cos 0 1 4 k 2 2 2 2 0 2
1 4 k 3 4 k , for any integer k. If 0 , then cos
4 k 1 4 k (1st quadrant) or 3 4 k 4 4 k (4th quadrant for any integer k). (16) and (17) imply that e
cos and e 2 sin 2 2
2
2 e 2 e 1 2 2 e 1 2 2 e 2 e 2 www.ijmsi.org
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An Exposition On Solution Process…
e 2 solutions to the transcendental equation (9) are given by i and the set
, :
e 2 , 0 . There is an infinity of oscillatory solutions. These solutions are
given by:
x t e t c1 cos t c2 sin t ,
,
for all real
for which
2 e 2
and e
24
2 . These solutions are nontrivial if
c12 c2 2 0 . Furthermore, if 0 then 0 and tan 2 . If 0 , then 1 4 k 3 4 k and if 0 , then, 4 k 1 4 k , for any integer k. If 0 , then 0
and e
1
If 0, the relation
2 2
2
j
j!
j 3
2 e 2
2
2 2 . Therefore 2 0 if 0.
2
is infeasible for 1, since e 2
2
feasible for 0 0.45445 . In other words, 0 , for 0 0.45445 . Clearly 2
1 . In fact it is
e 1 for
0 . In particular 0 2 1 for 0 0.45445 . Hence, 0, 1 1
for this set of

values. From the preceding analysis, we have obtained a complete characterization of all nonoscillatory solutions of (10) in the absence of initial functions. Next we will prove that there is no nontrivial oscillatory solution if a constant initial function is specified for the delay differential equation (10). Suppose
Then (24) x 0 c1 0 . If
0 , so that 1,
x t c1 cos t c2 sin t x
2
x t t , t 2, 0 with respect to (10).
is a continuous initial function specified by
then
c 2 and x 4 2
2 2
c1 c2 4 .
2 c c2 2 1
Suppose
t 0 ,
c1
2 c1 c1 c1 2 c1 0 c2 0 x t 0 t 2 trivial solution. 2
Suppose
a constant. Then, c1 c2
0 , so that 4 k 1 4 k t
Recall that x(t ) e
Let us evaluate
or
3 4 k 4 4 k , k integer. , ; x(t ) 0 , t , 0 . 2 2
c1 cos( t ) c2 sin( t ) , t
. Since x(t ) for all t such that t , t , 4
2
4
2
t , , these requirements are simultaneously satisfied only by 1. Therefore, 2 , 0 . From the preceding results, , 0 and for feasibility take k 0 such that 4 2 2 2 www.ijmsi.org
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An Exposition On Solution Process…
2 4 2 x 0 c2 0 and x e c1 c2 0 e 2 2 4 2 2 2 4 4 4 4 x 0 c c e c e c 1 e e But c2 0, 1 0 2 2 2 2 2 2 2 4 2 4 4 4 c2 0 or 1 e e 0. But 1 e e 0 if and only if 2 2
e
4
e
4
2 . 2
0, if v 0 2 dw v v Let us examine the function w e e . Clearly, e e 0, if v 0 2 dv 0, if v 0 v
v
Therefore w is decreasing for v < 0 and increasing for v > 0; w has global minimum at v = 0, with minimum value 2
2 0.585786. (Note: w (v) 0 v R ). We deduce that w is never less that
2 4 4 2 2. This proves that 1 e e 0, proving that c2 0 ; hence c1 0. 2 , which implies that only the trivial solution exists. We conclude that x(t ) 0, t 2 For the case 0, we earlier proved that 0 0.45445 and 1 1, for feasibility.
We wish to evaluate
4
2
x(t ) for all t such that t ,
t
4
,
2
. Since
, , these requirements are simultaneously satisfied only by 1. Since 1 1, the case 2 0 is infeasible. These prove that there are no nontrivial oscillatory solutions if a constant initial function is t
specified for the delay differential equation (10). Remarks: One could also reason along the same lines as in Driver (1977) to establish the existence of infinitely many complex roots for the equation
z
1
e
2.
To achieve this, set
h z
, w( z ) 0, where w( z ) 1 ze , and h is an arbitrary delay. Clearly w( z ) 1, for z 0. But
w( z ) has an isolated singularity at z 0. Thus by Picard’s theorem, in every neighbourhood of z 0, w( z ) takes on every value infinitely often with the exception of 1. Hence w( z) 0 infinitely often and so
the equation e 2 must have infinitely many complex roots: i; , real. However the subsequent analyses undertaken above are imperative for solving the problem.
III.
CONCLUSION
This article gave an exposition on how singledelay scalar differential equations could be solved using two methods presented in algorithmic formats. The results, which relied on the continuity of solutions, verified the consistency of both methods using the same problem, as well as gave a clue on the associated level of difficulty. It went on to conduct detailed and rigorous analyses on the nature, characterization and flavor of solutions for some problem instances. The article revealed that functional differential equations are difficult to
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An Exposition On Solution Processâ€Ś solve, even for the most innocent of problemssuch as the autonomous linear delay type with further simplifying restrictions; this difficulty is largely attributable to the many issues that must be dealt with, two such issues being the transcendental nature of the associated characteristic equations and the prescribed class of initial functions.
REFERENCE [1] [2]
Driver, R. D. (1977). Ordinary and Delay Differential Equations. Applied Mathematical Sciences 20, SpringerVerlag, New York. Hale, J. K. (1977). Theory of functional differential equations. Applied Mathematical Science, Vol. 3, SpringerVerlag, New York.
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