Thanks the replies. So is there no way Mathematica can solve this equation for the boundary condition and give me the tanh(x/sqrt(2)) ? Do I have to guess all my solutions for specific boundary problems ? ;)

and the boundary condition at x^2+y^2+z^2==inf being {x,y,z}/Sqrt[x^2+y^2+z^2])

Why don't you adopt a radial symmetric condition? By the way possibly the above condition is for a scalar function $u(x,y,z)$ contradictory e.g. for {x, y, z} = {0, -Infinity, Infinity}. Not sure how this works for a vector valued function $\vec{u} = u_1(x,y,z) \vec{e_1} + u_2(x,y,z) \vec{e_2} + u_3(x,y,z) \vec{e_3}$.

with the 2- and 3-dimensional case (where u^2 becomes u.u ...)

In 2- and 3-D the equation $\begin{equation}\frac{d^2u(x)}{dx^2}+(1-u(x)^2) u(x)=0\end{equation}$ no obvious generalization for a vector valued function $\vec{u}(\vec{x})$ exists because the Laplace operator $\Delta=\nabla\cdot\nabla$ as a gradient of the divergence of a scalar function $u(\vec{x})$ will not happen to be applicable. If one applies the divergence to a vector valued function, a matrix valued function appears ...

I was remembering only the scalar Laplacian but there is also the Vector Laplacian, you're right.

hmm, it doesn't work for me for some reason

In[5]:= First[FullSimplify[ExpToTrig[u[y] /. DSolve[{D[u[x], {x, 2}] + (1 - u[x]^2) u[x] == 0, u[0] == 0, u'[0] == 1/Sqrt[2]}, u, x]]]]
During evaluation of In[5]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>
During evaluation of In[5]:= DSolve::bvfail: For some branches of the general solution, unable to solve the conditions. >>
During evaluation of In[5]:= DSolve::bvfail: For some branches of the general solution, unable to solve the conditions. >>
Out[5]= y

In[6]:= DSolve[{D[u[x], {x, 2}] + (1 - u[x]^2) u[x] == 0, u[0] == 0, u'[0] == 1/Sqrt[2]}, u, x]
During evaluation of In[6]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>
During evaluation of In[6]:= DSolve::bvfail: For some branches of the general solution, unable to solve the conditions. >>
During evaluation of In[6]:= DSolve::bvfail: For some branches of the general solution, unable to solve the conditions. >>
Out[6]= {}

Any idea why that is ?

Also I don't understand how you got the new boundary condition from the original one.

Ah okay, but that's somewhat cheating as I have to know the solution in advance already. ;) If I wouldn't have had that lucky guess of tanh(x/sqrt(2)) in the first place there would be no way to derive the solution, i guess ?

(and i'm still on mathematica 9)

u[x_] := Tanh[x/Sqrt[2]]
In[8]:= D[u[x], {x, 2}] + (1 - u[x]^2)*u[x]
Out[8]= -Sech[x/Sqrt[2]]^2 Tanh[x/Sqrt[2]] + 
 Tanh[x/Sqrt[2]] (1 - Tanh[x/Sqrt[2]]^2)  
In[9]:= Simplify[%] 
Out[9]= 0

Noisy graph just shows computational roundoff errors

For the boundary condition one gets an error:

DSolve[{D[u[x], {x, 2}] + (1 - u[x]^2) u[x] == 0, u[-Infinity] == -1, u[Infinity] == 1}, u, x]
Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>
DSolve::bvlim: For some branches of the general solution, unable to compute the limit at the given points. Some of the solutions may be lost. >>
Solve::incnst: Inconsistent or redundant transcendental equation. After reduction, the bad equation is 1-2 C[1] == 0. >>

the equation has a solution:

In[21]:= DSolve[D[u[x], {x, 2}] + (1 - u[x]^2) u[x] == 0, u, x]

During evaluation of In[21]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>

Out[21]= {{u -> 
   Function[{x}, 
    I Sqrt[-(1/(1 - Sqrt[1 - 2 C[1]]))]
       JacobiSN[Sqrt[
       x^2 + x^2 Sqrt[1 - 2 C[1]] + 2 x C[2] + 
        2 x Sqrt[1 - 2 C[1]] C[2] + C[2]^2 + Sqrt[1 - 2 C[1]] C[2]^2]/
       Sqrt[2], (1 - Sqrt[1 - 2 C[1]])/(1 + Sqrt[1 - 2 C[1]])] - 
     I Sqrt[-(1/(1 - Sqrt[1 - 2 C[1]]))] Sqrt[1 - 2 C[1]]
       JacobiSN[Sqrt[
       x^2 + x^2 Sqrt[1 - 2 C[1]] + 2 x C[2] + 
        2 x Sqrt[1 - 2 C[1]] C[2] + C[2]^2 + Sqrt[1 - 2 C[1]] C[2]^2]/
       Sqrt[2], (1 - Sqrt[1 - 2 C[1]])/(
       1 + Sqrt[1 - 2 C[1]])]]}, {u -> 
   Function[{x}, -I Sqrt[-(1/(1 - Sqrt[1 - 2 C[1]]))]
       JacobiSN[Sqrt[
       x^2 + x^2 Sqrt[1 - 2 C[1]] + 2 x C[2] + 
        2 x Sqrt[1 - 2 C[1]] C[2] + C[2]^2 + Sqrt[1 - 2 C[1]] C[2]^2]/
       Sqrt[2], (1 - Sqrt[1 - 2 C[1]])/(1 + Sqrt[1 - 2 C[1]])] + 
     I Sqrt[-(1/(1 - Sqrt[1 - 2 C[1]]))] Sqrt[1 - 2 C[1]]
       JacobiSN[Sqrt[
       x^2 + x^2 Sqrt[1 - 2 C[1]] + 2 x C[2] + 
        2 x Sqrt[1 - 2 C[1]] C[2] + C[2]^2 + Sqrt[1 - 2 C[1]] C[2]^2]/
       Sqrt[2], (1 - Sqrt[1 - 2 C[1]])/(1 + Sqrt[1 - 2 C[1]])]]}}

The printing stuff below and around the $MachinePrecision means that it is not meaningful to print because it might or might not be zero, for Chop it's clearly zero.