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 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 and f''(T)=.008{1-11.1111[f'(T)]^2}^.5 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:
 ```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```
 puzzledoldman2-ga rated this answer: `***** A perfect answer and outstanding explanation!`

 ```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```
 ```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. PuzzledOldMan```