I attacked this problem by making the substitution $x = e^{-t}$. When subbing for the independent variable, we need to make careful use of the chain rule to express $u(x), u'(x), u''(x)$ in terms of $u(t), u'(t), u''(t)$.

ode = u''[x]+(u'[x]/x)-(u[x]/(x^2))+u[x]-u[x]^3 == 0;
sub = Exp[-t];

newode = ode /. {
    u[x] -> u[t],
    u'[x] -> u'[t]/D[sub, t],
    u''[x] -> (u''[t] - D[sub, {t, 2}] u'[t]/D[sub, t])/D[sub, t]^2,
    x -> sub
}

u[t] - E^(2 t) u[t] - u[t]^3 - E^(2 t) u'[t] + E^(2 t) (u'[t] + u''[t]) == 0

sol = NDSolveValue[{newode, u[1] == 0, u[0] == 1}, u[t], t] /. t -> Exp[-x];
sol /. x -> 0

0.

sol /. x -> 10^15

1.

Plot[Evaluate[sol], {x, 0, 10}, PlotRange -> All]

Here's my issue though... If I plug this solution into the original ODE, I don't get zero, so I think I did something wrong but I can't figure out where. In fact I am having the same issue with Hans Dolhaine's solution he posted above (his is sol2 in the plot below).

Maybe someone here can figure out where I went wrong? I feel it's a simple oversight.

Plot[Evaluate[{
   D[sol, {x, 2}] + (D[sol, x]/x) - (sol/x^2) + sol - sol^3,
   D[sol2, {x, 2}] + (D[sol2, x]/x) - (sol2/x^2) + sol2 - sol2^3
 }], {x, 4, 20}, PlotRange -> All]

It is very easy to show (and everyone knows) that Nonlinear Schrodinger in the simplest case has Tanh kink solution:

f[x_] := Tanh[x/Sqrt[2]];
u''[x] + u[x] - u[x]^3 /. u -> f // FullSimplify

Out[] = 0

That is your equation at asymptotic 'x -> + Infinity' because

Limit[u'[x]/x - u[x]/x^2,x->0] == 0

And it is clear that solution around zero can be linearized. Now that's being said Micahel Trott had offered a proper way of finding this solution that you can see below in his reply to tis comment.

Redoing my math actually in Mathematica, I found I had one term wrong. The final approximation should be:

x/(1/2 (1 + Sqrt[5]) + x)

Attached my notebook

Attachments:

If you assume that the solution is u = x^(Sqrt[5]/2 - 1/2)/(B + x^(Sqrt[5]/2 - 1/2))

and the equation is exact at x=1 then B -> (-3 + Sqrt[5] - Sqrt[38 - 14 Sqrt[5]])/(2 (-3 + Sqrt[5])) or about 2.19353 or approximation u=x^0.618/(2.1935+x^0.618)

which satisfies your boundary conditions

Hello Mattia,

as it is a diff.eq. of second order we need to have information about the first derivative as well. And the division by x leads to problems at x == 0. But we could try to start in the vicinity of 0:

eq = u''[x] + (u'[x]/x) - (u[x]/(x^2)) + u[x] - u[x]^3 == 0;
eps = 10^-4;
sol = u /. 
  NDSolve[{eq, u[eps] == 0, u'[eps] == .2}, u, {x, eps, 5}][[1]]
Plot[sol[x], {x, eps, 3}]

Dear Hans, unfortunately the behavior of the first derivative is unknown. This is a typical Schroedinger problem, in which only BC's are known. Your solution seems to be correct near 0, but the function is monotone and most reach asymptotically 1 at +inf.

Dear Mattia,

a really interesting problem, and I see the point, but what are BC's? And I am somewhat astonished that a quantum-mechanical function does not vanish for x -> infinity.

But besides that I tried a "brute force" solution:

Your original equation was

eq1 = u''[x] + (u'[x]/x) - (u[x]/(x^2)) + u[x] - u[x]^3;

Now I looked for a function (hopefully) converging towards 1 for x -> Infinity. This should workk if v does not grow faster than exp.

uu[x_] := 1 - v[x] Exp[-x]

(Later I tried Exp[ - x^2 ] as well )

This gave a modified differential equation for v[ x ]

 In[3]:= eq2 = x^2 FullSimplify[eq1 /. u -> uu]

  Out[3]= E^(-3 x) (E^(2 x) (1 + x + x^2) v[x] - 3 E^x x^2 v[x]^2 + 
     x^2 v[x]^3 - E^(2 x) (E^x + (1 - 2 x) x 
  \!\(\*SuperscriptBox["v", "\[Prime]",
  MultilineFunction->None]\)[x] + x^2 
  \!\(\*SuperscriptBox["v", "\[Prime]\[Prime]",
  MultilineFunction->None]\)[x]))

Which I tried to solve numerically

eps = 10^-3;
sol = v /. 
  NDSolve[{eq2 == 0, v[eps] == 1, v'[eps] == 0}, v, {x, eps, 5}, 
    Method -> "ImplicitRungeKutta"][[1]]

But even after about 50 minutes of evaluation no result was given back.

Regards Hans

Well, not monotone for small values of x, but......

eq1 = u''[x] + (u'[x]/x) - (u[x]/(x^2)) + u[x] - u[x]^3;
uu[x_] := 1 - v[x] Exp[-x^1.3];
eq2 = x^2 FullSimplify[eq1 /. u -> uu];
eq2a = FullSimplify[Solve[eq2 == 0, v''[x]][[1, 1]]] /. Rule -> Equal
eps = 10^-3;
sol = v /. 
  NDSolve[{eq2a, v[eps] == 1, v'[eps] == -.1}, v, {x, eps, 50}, 
    Method -> "ExplicitRungeKutta"][[1]]
Plot[1 - Exp[-x^2] sol[x], {x, eps, 10}, PlotRange -> All]

Hans, Interesting approach. I tried to map the infinite domain to a finite one i.e. something like x1=x/(1+x), but the resulting ode doesn't seem to solve with NDSolve.

Unfortunately there is a mistake in the last plot-command, working with Exp[ - x^2 ] instead of Exp[ - x^1.3 ] which was used to solve the differential equation. With the last setting the approach doesn't work.

But another question: the differentials in the original equation are obviously the laplacian in cylindical coordinates. The function part is

u[x] - u[x]/x^2 - u[x]^3

and this may be written as

-u[x] (u[x] - Sqrt[-1 + x^2]/x) (u[x] + Sqrt[-1 + x^2]/x)

What does that mean for the allowed values of x? If u is real then x > 1 ?

DSolve is unlikely to solve equations with nonlinear terms in it

Dear Vitaliy, this equation gives the vortex solution of the GPE for ultra cold imperfect Bose gases. I have no idea about the behavior of the first derivate but I know that the wave function u[x] is u[x=0]=0 and u[x=+inf]=1, and it's monotone. I don't think that domain has necessary to be infinitive but big enough. The problem is that I don't know how to use NDSolve knowing only BC's. The second problem is that I have to solve a second equation for the modified GPE, which is: $$u?[x]+(u?[x]/x)?(u[x]/(x2))+u[x]?u[x]3?b?(1/x)?D[x?D[u[x]2]?u[x]=0$$ depending on the parameter b, and to compare the two plots.

But can I use NDSolve to solve the equations with u[x=0] = 0 and u[x=1000] = 1 (for example)? How? I'm not an expert in Mathematica. Thanks a lot.

Ah, ok! When deriving the new domain, I had a typo and used $t = e^{-x}$ instead of $x = e^{-t}$...

Vitaly, Your transform is similar to the one I tried t=x/(1+x) yours: t=Sqrt[x]/(1+Sqrt[x]) . Any reason Sqrt[x}? In any case, NDSolve was unable to solve the transformed equation. I think the reason is that the equation becomes very stiff near the boundaries. Although I'm sure there is a way, I didn't really spend a lot of time on it.
Mattia mentioned that the equation was solved before, maybe it is worth looking at some literature to see how the problem has been handled in the past.

Actually, Kay, I like your transform much better. Could you post your code or attach a notebook or the transformed ODE and NDSolve errors? We could try to figure out a custom method in NDSolve that would handle stiffness.

Vitaly, Ok here it is:

In[201]:= ode1[u_] := D[u, {x, 2}] + D[u, x]/x - u/x^2 + u - u^3;

In[202]:= pdc[ode1[u[x]]]

Out[202]//TraditionalForm= -(u(x)/x^2)+\[PartialD]^2u(x)/\[PartialD]x^2-u(x)^3+u(x)+\[PartialD]u(x)/\[PartialD]x/x

transform equation

In[203]:= teq1 = Expand[ode1[u[x/(1 + x)]]] /. x/(1 + x) -> t /. x -> t/(1 - t)

Out[203]= u[t] - ((1 - t)^2 u[t])/t^2 - u[t]^3 + (
 2 t Derivative[1][u][t])/((1 - t) (1 + t/(1 - t))^3) - (
 3 Derivative[1][u][t])/(1 + t/(1 - t))^2 + ((1 - t) Derivative[1][u][t])/(
 t (1 + t/(1 - t))) + (
 t^2 (u^\[Prime]\[Prime])[t])/((1 - t)^2 (1 + t/(1 - t))^4) - (
 2 t (u^\[Prime]\[Prime])[t])/((1 - t) (1 + t/(1 - t))^3) + (
  u^\[Prime]\[Prime])[t]/(1 + t/(1 - t))^2

In[204]:= teq2 = Simplify[%] t^2

Out[204]= (-1 + 2 t) u[t] - 
 t^2 u[t]^3 + (-1 + 
    t)^3 t ((-1 + 2 t) Derivative[1][u][t] + (-1 + t) t (u^\[Prime]\[Prime])[
      t])

In[205]:= Factor[teq2]

Out[205]= -u[t] + 2 t u[t] - t^2 u[t]^3 + t Derivative[1][u][t] - 
 5 t^2 Derivative[1][u][t] + 9 t^3 Derivative[1][u][t] - 
 7 t^4 Derivative[1][u][t] + 2 t^5 Derivative[1][u][t] + 
 t^2 (u^\[Prime]\[Prime])[t] - 4 t^3 (u^\[Prime]\[Prime])[t] + 
 6 t^4 (u^\[Prime]\[Prime])[t] - 4 t^5 (u^\[Prime]\[Prime])[t] + 
 t^6 (u^\[Prime]\[Prime])[t]

In[206]:= eps = .1;

In[207]:= NDSolve[{teq2 == 0, u[0] == 0, u[1.] == 1.}, u[t], {t, eps, 1.}]

During evaluation of In[207]:= Power::infy: Infinite expression 1/0.^2 encountered. >>

During evaluation of In[207]:= Infinity::indet: Indeterminate expression 0. ComplexInfinity encountered. >>

During evaluation of In[207]:= Power::infy: Infinite expression 1/0.^2 encountered. >>

During evaluation of In[207]:= Power::infy: Infinite expression 1/0.^2 encountered. >>

During evaluation of In[207]:= General::stop: Further output of Power::infy will be suppressed during this calculation. >>

During evaluation of In[207]:= Infinity::indet: Indeterminate expression 0. ComplexInfinity encountered. >>

During evaluation of In[207]:= Infinity::indet: Indeterminate expression 0. ComplexInfinity encountered. >>

During evaluation of In[207]:= General::stop: Further output of Infinity::indet will be suppressed during this calculation. >>

During evaluation of In[207]:= NDSolve::ndnum: Encountered non-numerical value for a derivative at t == 0.`. >>

Out[207]= NDSolve[{(-1 + 2 t) u[t] - 
    t^2 u[t]^3 + (-1 + 
       t)^3 t ((-1 + 2 t) Derivative[1][u][t] + (-1 + t) t (
         u^\[Prime]\[Prime])[t]) == 0, u[0] == 0, u[1.] == 1.}, 
 u[t], {t, 0.1, 1.}]

This blows up because of division by zero at the low end. I can fix that by using the low end asymptotic term as t->0 t u'==u:

In[217]:= NDSolve[{teq2 == 0, t u'[eps] == u[eps], u[1.] == 1.}, 
 u[t], {t, eps, 1.}]

During evaluation of In[217]:= NDSolve::ndsz: At t == 0.9999112864801112`, step size is effectively zero; singularity or stiff system suspected. >>

Out[217]= NDSolve[{(-1 + 2 t) u[t] - 
    t^2 u[t]^3 + (-1 + 
       t)^3 t ((-1 + 2 t) Derivative[1][u][t] + (-1 + t) t (
         u^\[Prime]\[Prime])[t]) == 0, 
  t Derivative[1][u][0.1] == u[0.1], u[1.] == 1.}, u[t], {t, 0.1, 1.}]

Now it blows up near t=1 because the differential equation turns into an algebraic equation (u=u^3) via an apparent stiffness at 1 (coefficients of the derivatives start to vanish)

Attachments:

I tried several values for the derivative. Obviously one can improve my choice u'[eps] = 1.11 but I do not think that the improved plot will be very different.

Clearly there is ONLY ONE exact value of u'[eps] to obtain that u[x] to 1 for x \to Infinity, but it is not necessary to find the EXACT value to produce a reasonable plot (for a BSc thesis).

Mattia is working on it. Notice that with s > 1/2 the equartion becomes nasty (with s = 0 one has the familiar Gross-Pitaevskii equation for a quantized vortex).