DSolve doesn't give all the solutions

Question

I'm trying to solve the following ODE

$\frac{df}{dx}=-\sqrt{2}\left(\frac{1}{4}-2f(x)^{2}\right)$

In the text where it appears, the authors make a distinction into three cases: $f=-1/(2\sqrt{2})$, $f<-1/(2\sqrt{2})$ and $f>-1/(2\sqrt{2})$ depending on the sign of the right hand side of the equation. Further, $f(x)$ is a negative function.

When I give it into Mathematica with DSolve as

 DSolve[D[f[x],x]==-Sqrt[2]*(1/4-2*f[x]^2),f[x],x]

it only gives me the $\tanh$ solution, which according to the text corresponds to the case $f>-1/(2\sqrt{2})$. For the case $f<-1/(2\sqrt{2})$ I'm supposed to be getting a $\coth$ solution but don't see how.

Where are two other cases? Thank you for any help.

Answer 1

EDIT

Sorry. I have made a mistake in the choice of the initial conditions which is corrected below. This error does not change the general staetements.

Corrected exposition

As far as I can see there is no problem with the completeness of the solutions provided by Mathematica.

$Version

(* Out[16]= "10.1.0  for Microsoft Windows (64-bit) (March 24, 2015)" *)

The general solution to the ODE is found by Mathematica to be

sol = DSolve[f'[x] == -Sqrt[2] (1/4 - 2 f[x]^2), f[x], x]

(* Out[2]= {{f[x] -> -((-1 + E^(2 x + 4 Sqrt[2] C[1]))/(
    2 Sqrt[2] (1 + E^(2 x + 4 Sqrt[2] C[1]))))}} *)

As expected, the solution contains one arbitrary constant C[1].

The two branches of the solution correspond just to different choices of C[1].

Notice also that C[1] can be a complex number.

Let us have closer look on the situation.

The critical value of the function is

fc = -(1/(2 Sqrt[2]));

Now we complete the problem by imposing an initial condition on f.

And we distinguish two cases (here I am correcting my mistake of not imposing the condition of on the module)

(1) Abs[f[0]] < Abs[fc]:

ff1[x_] = f[x] /. 
  DSolve[f'[x] == -Sqrt[2] (1/4 - 2 f[x]^2) && f[0] == 1/4, f[x], x][[1]]

During evaluation of In[12]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>

    (* Out[14] = -((1 - 3 E^(2 x) + 2 Sqrt[2] E^(2 x))/(
 2 Sqrt[2] (-1 - 3 E^(2 x) + 2 Sqrt[2] E^(2 x))))    *)

(2) Abs[f[0]] > Abs[fc]:

ff2[x_] = f[x] /. 
  DSolve[f'[x] == -Sqrt[2] (1/4 - 2 f[x]^2) && f[0] == 2, f[x], x][[1]]

During evaluation of In[14]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>

(* Out[15] = -((-7 - 9 E^(2 x) + 4 Sqrt[2] E^(2 x))/(
 2 Sqrt[2] (7 - 9 E^(2 x) + 4 Sqrt[2] E^(2 x)))) *)

and compare the results in a plot:

Plot[{ff1[x], ff2[x]}, {x, 0, 3}, 
 PlotLabel -> "Solution of ODE\nblue curve = ff1, yellow curve = ff2",
  PlotRange -> {-2, 2}]

Summarizing: Mathematica in fact provides the general solution. This includes automatically the cases mentioned in the OP which correspond just to different initial conditions.

Answer 2

If we solve for a general initial condition at x == 0, we get the following:

dsol = DSolve[{D[f[x], x] == -Sqrt[2]*(1/4 - 2*f[x]^2), f[0] == F0}, f, x];
{dsol} = dsol /. 
  HoldPattern@Function[v_, body_] :> Function @@ {v, FullSimplify[body]}

Solve::ifun: Inverse functions are being used.... >>

(*
  {{f -> Function[{x}, (4 F0 Cosh[x] - Sqrt[2] Sinh[x]) / 
     (4 Cosh[x] - 8 Sqrt[2] F0 Sinh[x])]}}
*)

You can see the solution contains all three cases (for a negative f) as follows:

Reduce[F0 < 0 && f[x] < 0 && y == f[x] /. dsol, {F0, x, y}, Reals, 
    Backsubstitution -> True] /. Or | And -> List /. 
  x > _ :> Sequence[] // Grid

Whether to write the nontrivial solutions in terms of Tanh or Coth, or even as above, seems a matter of taste.

-Sqrt[2]/4 Tanh[x - ArcTanh[4 F0/Sqrt[2]]] - f[x] /. dsol // TrigExpand // Simplify
-Sqrt[2]/4 Coth[x - ArcCoth[4 F0/Sqrt[2]]] - f[x] /. dsol // TrigExpand // Simplify
(*  0  *)