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Q: Integrals ( Answered 5 out of 5 stars,   2 Comments )
Subject: Integrals
Category: Science > Math
Asked by: puzzledoldman2-ga
List Price: $25.00
Posted: 12 Feb 2004 15:24 PST
Expires: 13 Mar 2004 15:24 PST
Question ID: 306284
When:     f(T)=0   when T=0
can f(T) be deduced.  If so, what is it?

Request for Question Clarification by mathtalk-ga on 13 Feb 2004 08:25 PST
Hi, puzzledoldman2-ga:

It appears that you have a second-order differential equation, and
only one initial condition, f(0) = 0.  In fact if I'm reading it
correctly, the only way that f(T) appears in the (nonlinear)
differential equation is through the first and second derivatives.

I believe that another "initial condition" such as specifying f'(0)
would be needed in order to determine a unique solution.  Is it
possible that some additional information about the solution is known?

regards, mathtalk-ga

Clarification of Question by puzzledoldman2-ga on 13 Feb 2004 14:50 PST
I'm sorry, that's all the information available.  The .008 is a
constant force pushing a mass of one from its zero rest postion under
the rules of Special Relativity where the resulting acceleration
depends on (1-[v^2/c^2)]^2^.5

Request for Question Clarification by mathtalk-ga on 15 Feb 2004 06:55 PST
Hi, puzzledoldman2-ga:

If the interpretation of f(T) is that of the position of a unit mass
which is "at rest" at time T=0, then I believe the two conditions
implied are:

f(0) = 0

f'(0) = 0

In other words, besides being in the "zero" position, the mass starts
in a rest state of zero velocity.

If this interpretation is agreeable to you, I can post a solution based on it.

regards, mathtalk-ga

Clarification of Question by puzzledoldman2-ga on 15 Feb 2004 11:09 PST
Yes!  Please do!
Subject: Re: Integrals
Answered By: mathtalk-ga on 16 Feb 2004 11:42 PST
Rated:5 out of 5 stars
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

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

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
puzzledoldman2-ga rated this answer:5 out of 5 stars
***** A perfect answer and outstanding explanation!

Subject: Re: Integrals
From: mathtalk-ga on 13 Feb 2004 12:22 PST
The differential equation itself, although nonlinear, is "autonomous"
in the sense that the only dependence on the independent variable T is
implicit in the references to the unknown function itself.  The
"parameters" of the equation are constant, e.g. time-independent and
thus the term autonomous.

As a result any solution to the equation generates "additional"
solutions by translation.  That is, if f(T) satisfies your equation,
so will f(T+c) where c is a constant.  Of course the domain on which
f(T+c) is defined shifts relative to where f(T) is defined, and it
appears to be important in your statement of the problem that f(T) is
defined at T=0.

regards, mathtalk-ga
Subject: Re: Integrals
From: puzzledoldman2-ga on 17 Feb 2004 07:11 PST
For mathtalk-ga
 Your insight is very helpful.  I was using meters per nanosecond for
velocity and .3 m/ns as the value of c.


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