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Hi I'm new to mathematica so I may have a somewhat trivial question. my code below solves and plots the results of a first order nonlinear ODE. The solutions mathematica gives me contains & and #1 which I don't know what those mean. 

Clear[m, z, p, h, ld, to, td, mo, hc, mb, cb, t]
Clear[mm, sol, sol20, tt, mmm]

sol = DSolve[{m'[
 t] == -((m[t]^(2/3))/(p^(2/3)))*(h/ld (to - td + ((mo - m[t])*hc)/(mb*cb))), m[0] == mo}, m[t], t]
mm[t_] := Part[sol, 1, 1, 2];
mmm[tt_] := 
mm[t] /. {p -> 900, hc -> -17000*1000, ld -> -400*1000, 
mo -> 60*0.453592, cb -> 0.49*1000, mb -> 2000*0.453592, 
td -> 553, to -> 490, h -> 70, t -> tt}
mmm[20]
Plot[mmm[x], {x, 0, 7500}]

Im wondering if this solution is an analytical solution and if so how can I extract a simplified form (for perhaps a specific range or only non-imaginary range) or something more presentable. OR is this using numerical techniques to solve it and if so which method is being used.

here is the output solution

{{m[t] -> 
   InverseFunction[((-1)^(2/3) (2 Sqrt[3] ArcTan[(-1 + (2 (-1)^(1/3) hc^(1/3) #1^(
         1/3))/(hc mo + cb mb (-td + to))^(1/3))/Sqrt[3]] + 
     2 Log[(hc mo + cb mb (-td + to))^(1/3) + (-1)^(1/3) hc^(1/3) #1^(1/3)] - 
     Log[(hc mo + cb mb (-td + to))^(
       2/3) - (-1)^(1/3) hc^(1/3) (hc mo + cb mb (-td + to))^(
        1/3) #1^(1/3) + (-1)^(2/3) hc^(2/3) #1^(2/3)]))/(
  2 hc^(1/3) (hc mo + cb mb (-td + to))^(2/3)) &][(h t)/(cb ld mb p^(2/3)) + ((-1)^(
  2/3) (2 Sqrt[3]ArcTan[(-1 + (
       2 (-1)^(1/3) hc^(1/3) mo^(1/3))/(hc mo + cb mb (-td + to))^(1/3))/Sqrt[3]] + 
    2 Log[(-1)^(1/3) hc^(1/3) mo^(1/3) + (hc mo + cb mb (-td + to))^(1/3)] - 
    Log[(-1)^(2/3) hc^(2/3) mo^(2/3) - (-1)^(1/3) hc^(1/3) mo^(
       1/3) (hc mo + cb mb (-td + to))^(1/3) + (hc mo + cb mb (-td + to))^(2/3)]))/(
     2 hc^(1/3) (hc mo - cb mb td + cb mb to)^(2/3))]}}
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    – user9660
    Commented Nov 26, 2014 at 16:04

1 Answer 1

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This is a symbolic solution obtained by separating variables and integrating, like this:

Integrate[
 1/(-((m[t]^(2/3))/(p^(2/3)))*
      (h/ld (to - td + ((mo - m[t])*hc)/(mb*cb))) /. m[t] -> m), m]
(*
((-1)^(2/3) cb ld mb p^(
   2/3) (2 Sqrt[3]
       ArcTan[(-1 + (
        2 (-1)^(1/3) hc^(1/3) m^(1/3))/(hc mo + cb mb (-td + to))^(
        1/3))/Sqrt[3]] + 
     2 Log[(-1)^(1/3) hc^(1/3) m^(1/3) + (hc mo + cb mb (-td + to))^(
        1/3)] - Log[(-1)^(2/3) hc^(2/3) m^(2/3) - (-1)^(1/3) hc^(1/3)
         m^(1/3) (hc mo + cb mb (-td + to))^(
        1/3) + (hc mo + cb mb (-td + to))^(2/3)]))/(2 h hc^(
   1/3) (hc mo + cb mb (-td + to))^(2/3))
*)

(Evaluate this antiderivative at m -> mo and m -> m[t], subtract and set equal to t, try solve for m[t]. It looks like inverse functions had to be used. I had no success simplifying this further.)

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  • $\begingroup$ Hi Mike that is helpful, I was confused as to how to interpret the &, #1 symbols and thought something numerical might be happening.. If I wanted to plot this in matlab how could I interpret those symbols? $\endgroup$
    – JPR984
    Commented Dec 1, 2014 at 19:15
  • $\begingroup$ The #1 (or simply #) and & are part of the "pure function" notation. See this answer for links to the documentation. The & binds #1 to the first argument of a function. For instance g = f[#1] & is roughly equivalent to g[x_] := f[x]. I don't know Matlab. The equations x == InverseFunction[f[#] &][y] and y == f[x] are equivalent, assuming f is 1-1 of course. HTH $\endgroup$
    – Michael E2
    Commented Dec 2, 2014 at 14:12

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