The problem can be solved analytically.
First we transform the equation a bit. Integrate the ODE once we obtain
neweq = Integrate[D[Sign[x] u[x], x] + D[u[x]^2 D[u[x], x], x], x] == c
(* Sign[x] u[x] + u[x]^2 Derivative[1][u][x] == c *)
Then it's not hard to notice Sign[x] u[x] + u[x]^2 Derivative[1][u][x]
is an odd function. We can analyze it manually, but here I'll use DChange
to make the post a bit more interesting:
(* Definition of DChange isn't included in this post,
please find it in the link above. *)
DChange[Sign[x] u[x] + u[x]^2 u'[x], x == -X, x, X, u[x]] // Expand
(* -Sign[X] u[X] - u[X]^2 Derivative[1][u][X] *)
Thus Sign[x] u[x] + u[x]^2 u'[x] == 0
at x == 0
. Since c
is a constant, we conclude c == 0
.
Next we write it as an ODE of $x(u)$ for convenience of subsequent discussion:
neweqreverse = neweq /. c -> 0 /. {u[x] -> u, x -> x[u], u'[x] -> 1/x'[u]}
(* u Sign[x[u]] + u^2/Derivative[1][x][u] == 0 *)
Solve the ODE for $x>0$ and $x<0$ separately:
{eqR, eqL} = Simplify[neweqreverse, #] & /@ {x[u] > 0, x[u] < 0}
(* {u + u^2/Derivative[1][x][u] == 0, u (-1 + u/Derivative[1][x][u]) == 0} *)
solR = DSolveValue[{eqR, x[top] == 0}, x[u], u] // Simplify
(* 1/2 (top^2 - u^2) *)
solL = DSolveValue[{eqL, x[top] == 0}, x[u], u] // Simplify
(* 1/2 (-top^2 + u^2) *)
Notice here top
is the value of $u(0)$.
For $u(-5)=u(5)=0$, the graphic of solutions can be obtained with e.g.
ParametricPlot[{#, u}, {u, -5, 5}, PlotRange -> All,
RegionFunction -> Function[{x}, x < 0], AspectRatio -> 1/GoldenRatio]~Show~
ParametricPlot[{#2, u}, {u, -5, 5},
RegionFunction -> Function[{x}, x > 0]] & @@ ({solL, solR} /. c -> 0 /.
Solve[solR == 5 /. c -> 0 /. u -> 0, top][[1]])

As we can see, there exist 2 non-trivial solutions.
BTW it's easy to notice that $u = 0$ only if $x=\pm \frac{\text{top}^2}{2}$, so b.c.s like $u(-5)=u(6)=0$ don't form a well posed problem.
Remark
Solution for $m=\frac{1}{2}$ case i.e.
D[Sign[x] u[x], x] + D[u[x]^(1/2) D[u[x], x], x] == 0
can be discussed in the same manner. Solution for $u(-6)=u(6)=0$ when $m=\frac{1}{2}$ can be plotted with e.g.
ParametricPlot[{#, u}, {u, -10, 10}, PlotRange -> All,
RegionFunction -> Function[{x}, x < 0], AspectRatio -> 1/GoldenRatio]~Show~
ParametricPlot[{#2, u}, {u, -10, 10}, RegionFunction -> Function[{x}, x > 0],
PlotRange -> All] & @@ ({solL, solR} /. c -> 0 /.
Solve[solR == 6 /. c -> 0 /. u -> 0, top][[1]])

As illustrated, there's only one non-trivial solution when $m=\frac{1}{2}$.
One can directly solve neweq /. c -> 0
with DSolve
. A warning will be generated then, but the results are correct.
u[x]==u[-x]
symmetric? $\endgroup$