Solution of ODE expressed with hyperbolic trigonometric function
 
  
 
  
 
 
 
 
 
 
    
        
        asked 8 years ago  
 
 ljbo gravatar image  
 
 
 
 
 
With Sage 7.3, consider
 
t=var('t')
v=function('v')(t)
m, g, h = var('m g h')
assume(m > 0)
assume(g > 0)
assume(h > 0)
sol = desolve(m*diff(v,t) == m*g - h*v**2, v, ivar=t)
show(sol)
sol.solve(v)
 
The outcomes are
 
−mlog(hv(t)−√ghmhv(t)+√ghm)2√ghm=C+t
 
[v(t)=√ghm(e(2√ghmCm+2√ghmtm)+1)h(e(2√ghmCm+2√ghmtm)−1)]
 
Nevermind desolve came with an implicit solution I had to further solve by hand but then solve missed that the last expression for v(t) is a mere hyperbolic tangent. How would I get the solution in its simpler form?
 
v(t)=√gmh tanh(√ghm(t+mC))
 
Thanks for any hint!
 
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Comments
It doesn't help for the hyperbolic ones, but maxima does have some routines to go back and forth between exponential and trigonometric expressions:
sage: tan(x)._maxima_().exponentialize().sage()
(-I*e^(I*x) + I*e^(-I*x))/(e^(I*x) + e^(-I*x))
sage: tan(x)._maxima_().exponentialize().demoivre().sage()
sin(x)/cos(x)
I don't think these routines are directly exposed in sage and "demoivre" doesn't know about hyperbolic routines.
In general, I would expect computer algebra system would give preference to exponential expressions, because it's easy to define a "normal form" for them, whereas it's not so clear what that should mean for hyperbolic/trigonometric functions in general: there's too much redundancy.
nbruin gravatar imagenbruin (
    
        8 years ago
    )
Thanks a lot for those hints: I had no idea I could get access to Maxima in such a way. As for my problem, from a human point of view, it is natural to want the "shorter" answer. I have just checked that Mathematica does actually produce the formula with tanh. Besides, the math behind exponentialize and demoivre must have an equivalent in hyperbolic trigonometry.
ljbo gravatar imageljbo (
    
        8 years ago
    )
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