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Science Forum Index » Math - Symbolic Forum » Puzzled by a Non-Linear DDE
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| Patoche |
 Posted: Tue Feb 20, 2007 11:12 am |
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Guest
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Hi there,
Is there an analytical solution to the following Non-Linear Delay
Differential Equation (DDE) ?
(1) b * y(t) - c * t * y'(t) + (1/a) * ( y'(t) )^a - y(t-1) = 0 with
initial conditions y(t) = 0 for all t < 1
and parameters constrained by 0 < a < 1 , b > 0 , c > = 0
I suspect (1) will be difficult to solve analytically. As a first
step, I am therefore investigating the special case c = 0 :
(2) b*y(t) + (1/a) * ( y'(t) )^a - y(t-1) = 0
I am looking for a function y(t) defined on the positive real line and
that solves equation (2).
The case a = 1 is standard in the theory of linear DDEs. I am assuming
a is strictly between 0 and 1.
Assuming y(t) is twice differentiable and that y'(t) is everywhere non-
zero, then (2) implies
(3) b*y'(t) + ( y'(t) )^a * y''(t) / y'(t) - y'(t-1) = 0
Plugging (3) into (2) yields (4), assuming that y''(t) is everywhere
non-zero:
(4) b*y(t) + (1/a) * y'(t)/y''(t) * ( y'(t-1) - b*y'(t) ) -
y(t-1) = 0
Plugging (2) into (3) yields (5) :
(5) b*y'(t) + a * y''(t)/y'(t) * ( y(t-1) - by(t) ) - y'(t-1)
= 0
Provided y(t) and y'(t) are strictly monotonic, then a solution of (2)
should also solve (4) and (5). I was rather hoping that a solution of
(4) and (5) would help in solving (2).
A solution of (4) or (5) may be found, of the following form y(t) =
constant * exp(s*t), where s is equal to s = - ln(b). Equivalently,
y(t) = constant * (1/b)^t
For instance, plugging y(t) = exp(s*t) into (5) yields: b*s +
a*s*( exp(-s)-b ) -s*exp(-s) = 0, or equivalently (1-a)*(b-exp(-s)) =
0, so provided a is strictly greater than 1, then s must satisfy b =
exp(-s), or s = -ln(b), as stated above. The same solution is found
from (4).
However, the function exp(s*t) solves (4) and (5) but does not solve
(2). So here I stop and wonder: I must be wrong to reason that any
solution of (4) or (5) should also be a solution of (2)... Can anyone
spot a mistake somewhere? Does anyone know a way to obtain an
analytical solution of equation (2) ? Of equation (1) ?
Many thanks!
Patrick. |
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