Hi,
Your f goes to fixed values as x goes to +/- infinity consequently f' needs to go to 0 as x goes to +/- infinity.
I did the integration in my head so I think I was off by a factor of 2. There is an integration constant which I set so that f' vanishes at +/- infinity:
In[14]:= eq2 =
D[f[x], x]^2 == \[Lambda]^2 (-1/2 v^2 f[x]^2 + 1/4 f[x]^4 + 1/4 v^4)
Out[14]= Derivative[1][f][
x]^2 == \[Lambda]^2 (v^4/4 - 1/2 v^2 f[x]^2 + f[x]^4/4)
In[15]:= D[#, x] & /@ eq2 //
Simplify Collect[Solve[%, f''[x]], \[Lambda]]
Out[15]= {}[
2 Derivative[1][f][x] (f^\[Prime]\[Prime])[
x] == \[Lambda]^2 (-v^2 f[x] Derivative[1][f][x] +
f[x]^3 Derivative[1][f][x])]
Which gives your equation 1
I then take the square root of each side picking the correct branch
In[74]:= eq3 =
f'[x] == -\[Lambda] Sqrt[-1/2 v^2 f[x]^2 + 1/4 f[x]^4 + 1/4 v^4]
Out[74]= Derivative[1][f][x] == -\[Lambda] Sqrt[
v^4/4 - 1/2 v^2 f[x]^2 + f[x]^4/4]
and then I integrated to find x:
In[268]:= Integrate[-1/(\[Lambda] Sqrt[-1/2 v^2 f^2 + 1/4 f^4 +
1/4 v^4]), f]
Out[268]= (2 (f^2 - v^2) ArcTanh[f/
v])/(v Sqrt[(f^2 - v^2)^2] \[Lambda])
now simplify and you find:
In[273]:= Clear[v, \[Lambda]]
In[274]:= f[x_] := -v Tanh[\[Lambda] v x /2]
ok now I check if I did my algebra right this time:
In[275]:= eq1
Out[275]=
1/2 v^3 \[Lambda]^2 Sech[(v x \[Lambda])/2]^2 Tanh[(v x \[Lambda])/
2] == 1/2 v^3 \[Lambda]^2 Tanh[(v x \[Lambda])/2] -
1/2 v^3 \[Lambda]^2 Tanh[(v x \[Lambda])/2]^3
In[276]:= Simplify[%]
Out[276]= True
eureka!
