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I am new to mathematica and am trying to solve a differential equation. Actually, I am not entirely sure if the system can be called differential equation.

I am interested in finding out the second derivative of the system. The actual function and the first derivative only enter the system at local points, i.e. 1/2, 1, and a constant t which should be uniquely determined by the system.

I have tried Dsolve first but I guess the system is to complicated to get an analytical solution, so a numerical one would be nice to have too.

My code is:

\[Sigma]A = 1/2 + (1 - t) f'[1] - f[1] + f[t] - (t - 1/2) xbB'[t] + xbB[t] 
-xbB[1/2] + 1/2 xaB'[1] - xaB[1] + xaB[1/2] - xaB'[1] + xaB'[1/2] ;

\[Sigma]B = 1/2 - 1/2 xaB'[1] + xaB[1] - xaB[1/2] + f'[1] - f'[1/2] - (1-t) 
f'[1] + f[1] - f[t] + (t - 1/2) xbB'[t] - xbB[t] + xbB[1/2];

c = 1/2;

NDSolve[
 {
 f''[t] == 0,
 xaB''[1] == c, 
 f''[q] == - xbB''[q],

 f'[t] == 0,
 xbB'[1/2] == 0,
 xaB'[1/2] == 0,

 t/(1 - t) == \[Sigma]A / \[Sigma]B,

 ((c - f''[q] )/(c + f''[q]))^2 == ((1 - q)/q)^2 *\[Sigma]A/\[Sigma]B,
 (c - f''[q] )/(c + f''[q]) *(c - xaB''[q])/(c + xaB''[q] )  q/(1-q) == (1 - 
 q)/q}, f''[q], {q, 1/2, 1}]

I have only included the conditions on the functions that I really need for the solution of f'' to be correct. However, I have tried different combinations of starting conditions. As I am only interested in the second derivatives, the constants in the other 2 functions can be chosen freely. or to solve the system for f or f' instead of f'' (I'm not even sure if NDSolve does solve for the derivatives directly?)

As I said I'm new to mathematica and this Forum so if there is any further info I should supply or format I should use, just let me know. I have searched for related questions during hours but none seemed to apply to my problem.

Thank you in advance for any assistance!

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  • $\begingroup$ Finding the concrete value for t might help a lot. NDSolving for {f[q], xaB[q], xbB[q]} and only after that taking the second derivative of f[q] might help. Including relevant conditions for xaB and xbB might help. Can you make progress on any of those? $\endgroup$ – Bill Jun 5 '16 at 18:12
  • $\begingroup$ Thanks for the help. Unfortunately, there are no further conditions I have on t (besides it lying strictly between ½ and 1. Regarding the conditions on xaB and xbB: Since I don't care about the constant terms of these functions I introduced some conditions at one point of each function but that did not help. The only further thing I know is that f'' and xaB'' should be increasing, xbB'' decreasing. is there any way to impose this? I guess only by involving yet another derivative? Many many thanks for your help in any case $\endgroup$ – JKJ Jun 5 '16 at 19:22
  • $\begingroup$ One other idea. You have a gob load of denominators. Can you carefully justify cross multiplying to get rid of all those without dividing by zero? That sometimes helps IF it is valid to do. But it still doesn't seem to be able to find any starting point without any information about xaB or t, other than a rational function of t is equal to two unknown rational functions. $\endgroup$ – Bill Jun 5 '16 at 22:28
  • $\begingroup$ Thanks again (: I have done that already, just in case. Cross multiplying shouldn't pose a problem to the solution. I have also tried several values for xaB[½] or xaB[1] to see if mma finds a solution then, it still states "cannot find a starting value for the variable xaB". Do you know if Mathematica does solve systems of equations were only the derivatives enter as an actual function (i.e. not only with a constant)? Because I tried to give values for t and then Mma says "Input is not an ordinary differential equation" $\endgroup$ – JKJ Jun 6 '16 at 0:22
  • $\begingroup$ I am usually very cautious. I try to use notation as close to the accepted and customary shown in the help pages as possible. That was why I originally suggested that you try to solve for {f[q],xaB[q],xbB[q]} and provide sufficient information for it to be able to find those and if that more likely works and all you need is f'[q] to differentiate the result afterwards. But I think you are running out of luck on this because none of my ideas work and nobody much brighter than either of us is showing us how it really should be done. $\endgroup$ – Bill Jun 6 '16 at 1:29

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