Hi, puzzledoldman2-ga:
Based on your Clarifications this ODE already involves an
approximation for the speed of light c. Please allow me to simplify
the notation slightly by this approximation:
11.1111 ~ 100/9 = (10/3)^2
Then your initial value problem (IVP) becomes:
f(0) = 0
f'(0) = 0
f"(t) = SQRT(1 - (f'(t)/0.3)^2)/125
The method of solution section will explain how to solve the problem
with a more precisely specified value of c, should that be required.
Apart from referring to the independent variable t as "time" (with its
implied identification of f(t) as position, f'(t) as velocity, and
f"(t) as acceleration), I've tried to focus my Answer on the
mathematics of the solution. No doubt the interpretation of this
problem in your underlying context of the theory of Special Relativity
would afford further discussion.
* * * * * * * * * * * * * * * * * * * * * * * * * * *
The Solution
------------
For "time" t = 0 to 75*pi/4, the following function f(t) is the unique
solution to the nonlinear IVP posed above:
f(t) = (45/4) * (1 - cos(2t/75))
Verifying that this function satisfies the initial conditions as well
as the ordinary differential equation portion of the problem is
routine. Note:
f'(t) = 0.3 sin(2t/75)
so that f(0) = f'(0) = 0. Furthermore:
f"(t) = cos(2t/75)/125
and it follows for all t in [0,75*pi/4] that:
125 f"(t) = cos(2t/75)
= SQRT( 1 - sin^2(2t/75) )
= SQRT( 1 - (f'(t)/0.3)^2 )
from which it is clear the nonlinear differential equation is satisfied.
In this verification of the solution we make an essential use of the
restriction 0 <= t <= 75*pi/4, so that:
0 <= 2t/75 <= pi/2 [first quadrant]
in order to claim:
cos(2t/75) = SQRT( 1 - sin^2(2t/75) )
Beyond the right endpoint t = 75*pi/4, the second derivative of f(t)
becomes negative. Since the formalism:
f"(t) = SQRT(1 - (f'(t)/0.3)^2)/125
requires a nonnegative (principal) square root, we cannot extend our
claim of this solution beyond that point (where the square root
function is singular).
Note however that the solution does extend backward in time, to t =
-75*pi/4, in a mathematically sensible fashion. The symmetry of
motions prior to t=0 and after could be interpreted as first
"decelerating" from a velocity of -0.3 to rest, then accelerating to
velocity +0.3, each taking place over finite time intervals.
* * * * * * * * * * * * * * * * * * * * * * * * * * *
Method of Solution
------------------
To find the solution f(t), it may clarify matters to think first of a
somewhat simpler differential equation:
g'(t) = SQRT( 1 - (g(t))^2 )
With the initial condition g(0) = 0, we would have a unique solution:
g(t) = sin(t)
on the interval containing the origin t = 0, [-pi/2,+pi/2].
This simpler problem is so closely related to the one you posed that
we can develop your solution by introducing two or three constant
parameters.
Comparing the equation above for g(t) = sin(t) with this one:
f"(t) = SQRT( 1 - (f'(t)/c)^2 )/125
for f'(t), we could make an educated guess that f'(t) has the form:
f'(t) = c sin(bt)
where c is the same as the constant specified in the ODE, and b is an
additional one (available to satisfy our "force" condition). This
would be an especially good guess considering that both g(0) = 0 and
f'(0) = 0.
The "guess" would then lead to an expression for f(t):
f(t) = a - (c/b) cos(bt)
just as was given by our solution. This method could be called
"solution by unknown parameters" as we have only constants a,b,c to
determine.
If we differentiate f'(t) once:
f"(t) = bc cos(bt)
and substitute for both f'(t), f"(t) in your equation:
bc cos(bt) = SQRT( 1 - (c sin(bt)/c)^2 )/125
= SQRT( 1 - sin^2(bt) )/125
then we see that setting b = 1/(125c) will give us a solution valid in
the "first quadrant", i.e. where bt belongs to [0,pi/2].
Thus in the solution offered above, where I assumed:
c = 0.3
we needed to have:
b = 1/(125c) = 2/75
and finally, to satisfy f(0) = 0, we require:
a = c/b = 45/4
Note that some inhomogeneous initial conditions could be matched by
"guessing" a solution with one additional parameter:
f(t) = a - (c/b) cos(bt + d)
This slight additional generality was not required here, but it ties
in nevertheless with my earlier Comment about solutions to
"autonomous" differential equations. The new parameter d acts to
"translate" the solution with respect to time.
regards, mathtalk-ga |