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How can I solve this equation:

((a^2 - 1)*((a^2 - 1)*ArcTanh[a] - a))/(2*Sqrt[2*(a^2 - 1)^4]) == C*x

I am looking a solution for $a$ in function of $x$. I tried Solve and similar methods but I got unevaluated expression.

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One way: Derive the differential equation of the family (solve for the constant and differentiate); and use DSolve to solve it.

D[(1/x)((a^2 - 1)*((a^2 - 1)*ArcTanh[a] - a)) /
     (2*Sqrt[2*(a^2 - 1)^4]) /. a -> a[x], x] // Together // Numerator;
{dsol} = DSolve[% == 0, a, x] /. C[1] -> c
(*
  {{a -> Function[{x}, 
      InverseFunction[-(1/2) Log[1 - #1^2] + 
          1/2 Log[-ArcTanh[#1] - #1 + ArcTanh[#1] #1^2] &][c + Log[x]/2]]}}
*)

Evaluating an InverseFunction on exact input can sometimes take a long time. Be sure to use approximate machine reals when plotting or doing other numerical work. (This is accomplished by N below. Alternatively one could use the iterator {x, -1., 1.} instead of {x, -1, 1} to specify the plot domain.)

Block[{c = 1},
 Plot[a[N@x] /. dsol, {x, -1, 1}]
 ]

Mathematica graphics

While InverseFunction can be inconvenient at times, the solution dsol does present a as a function of x and the parameter c.


Update: I guess I should point out that the inverse function is basically just the solution to the equation written in the form of an inverse function.

This simpler solution is equivalent to the produced by dsol:

{a -> Function[{x}, 
   InverseFunction[1/(1 - #1^2)*(-ArcTanh[#1] - #1 + ArcTanh[#1] #1^2) &][Exp[2 c] x]]}

It is equivalent to the OP's original equation via Exp[2 c] == 2 Sqrt[2] C. (Please note that capital C is a Protected Mathematica symbol.)

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  • $\begingroup$ thank you for your answer. Your graph (dsol) presents solution to this equation and it looks like Tanh[Constant*x], where Constant is related to c. For different values of c the shape of Tanh[] will be different. So, how can I find this Constant? For example, to have solution in form of Tanh[x]. $\endgroup$ – John M. Apr 17 '16 at 22:53
  • $\begingroup$ @JohnM. You're welcome. I think you can see from this image that, while dsol produces a sigmoid-like plot similar to Tanh, it is not in fact the same. $\endgroup$ – Michael E2 Apr 17 '16 at 23:23
  • $\begingroup$ Thank you again. This was useful. I noticed that I can not represent dsol with Tanh[x], they are similar but not the same. However, is there a way to represent this curve (sigmoid function) with some function (approximate solution) in term of x? $\endgroup$ – John M. Apr 18 '16 at 0:38
  • $\begingroup$ @Michael E2 Why you have another functional course? $\endgroup$ – user36273 Apr 18 '16 at 9:37
  • 1
    $\begingroup$ @JohnM. Over a finite interval, one can construct an interpolating function with NDSolve or FunctionInterpolation, which would evaluate faster than InverseFunction. Or one could use another approximation method, such as a Chebyshev series expansion I used here. One might try fitting a formula (see FindFit, LinearModelFit, and the whole *Fit family). $\endgroup$ – Michael E2 Apr 18 '16 at 9:59
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If it may also be a graphic solution? Consider:

FunctionDomain[ArcTanh[a], a]
(* -1 < a < 1 *)


((a^2 - 1)*((a^2 - 1)*ArcTanh[a] - a))/(2*Sqrt[2*(a^2 - 1)^4]) == c*x;
x[a_] = 1/c*((a^2 - 1)*((a^2 - 1)*ArcTanh[a] - a))/(2*Sqrt[2*(a^2 - 1)^4]);

With c as parameter

p = Plot[x[a] /. c -> 1, {a, -1, 1}, AxesLabel -> Automatic, GridLines -> Automatic]

enter image description here

p1 = Join @@ Cases[Normal@p, Line[x1__] :> x1, Infinity];
a = Interpolation@Thread@{Last /@ p1, First /@ p1}

Plot[a[x], {x, -3, 3}, GridLines -> Automatic, AxesLabel -> Automatic]

enter image description here

Edit

The easiest way to find the function a(x) is to build the inverse of f(a).

f = 1/c*((#^2 - 1)*((#^2 - 1)*ArcTanh[#] - #))/(2*Sqrt[2*(#^2 - 1)^4]) &;
a = InverseFunction@f

enter image description here

This function can be evaluated numerically with c as parameter, e.g.

c = 1;
Table[a[x], {x, -2, 2}] // N
(* {-0.888998, -0.76291, 0., 0.76291, 0.888998} *)

Plot[{f[x], a[x]}, {x, -2, 2}, GridLines -> Automatic, 
 PlotLegends -> {"f[x]", "a[x]"}, AspectRatio -> 0.8]

enter image description here

Edit 2

Please forgive me, I have a problem with the solutions. I follow here Michel E2's method.

f = ((a^2 - 1)*((a^2 - 1)*ArcTanh[a] - a))/(2*Sqrt[2*(a^2 - 1)^4]) - c x /. a -> a[x];
df = D[f, x] // Simplify

enter image description here

sol = DSolve[df == 0, a, x] /. C[1] -> 0

enter image description here

c = 1;
Plot[a[x] /. sol, {x, -2, 2}, GridLines -> Automatic]

enter image description here

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  • $\begingroup$ thank you for your solution. Using @MichaelE2's and your method I solved my problem. $\endgroup$ – John M. Apr 18 '16 at 22:07
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I am going to give a generic answer which work in such cases. The main idea is to go numerical. You solve it for some parameter values and get an idea of the function which might be the answer.

dat = Table[{c, x, 
           a /. NSolve[((a^2 - 1)*((a^2 - 1)*ArcTanh[a] - a)) /(2*Sqrt[2*(a^2 - 1)^4])
       == c x && 0 < a < 10, a][[1]]}, {c, 0.1, 1, .1}, {x, 0.1, 1., .1}]

dat1 = Flatten[dat, 1];
f = Interpolation[dat1];
Plot3D[f[c, x], {c, 0.1, 1}, {x, 0.1, 1}]

enter image description here

Ignore the Solve::rantz error message. Note that I use [[1]] and a range 0 < a < 10. This comes handy if you have multiple root and also gives faster result because it looks for root only within that region.

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