On a unit torus $T^n$ (or equivalently, on $\mathbb{R}^n$ with periodic boundary conditions), the linear Helmholtz equation:


$\nabla^2 \phi + k^2 \phi=0$


will have no non-trivial solutions for generic values of $k$, while for special values of $k$ it will have a finite-dimensional vector space of solutions, with a basis of functions:


$\phi\_m(x)=\exp(2\pi i m\_j x^j)$


for each $n$-tuple of integers $m=(m\_1,m\_2,...m\_n)$ such that:


$m\_1^2+m\_2^2+...+m\_n^2=\left(\frac{k}{2\pi}\right)^2$


Now, consider a non-linear version of this problem, such as:


$\nabla^2 \phi + k^2 \phi + \lambda \phi^3=0$


still on $T^n$. The solutions will no longer comprise a vector space, but rather a manifold.


My question is: how can I determine the cardinality of the dimension of that manifold? Will it again be zero-dimensional generically and finite for some special parameter values, or will the behaviour change?



Space of solutions of nonlinear Helmholtz equation on a torus
Ignoring boundary conditions, the PDE has solutions $\phi = a\; \text{sn}\left(b x, c\right)$ where $\text{sn}$ is the Jacobi SN function (in Maple's parametrization: note that Mathematics uses a different convention), ${a}^{2}=2\;{\dfrac {{b}^{2}-{k}^{2}}{\lambda}}$, ${c}^{2}={\dfrac {{k}^
{2}}{{b}^{2}}}-1$. These are periodic in $x$, with a period depending on $b$ and $c$. For any given $k$ and $\lambda$ (at least in some region) I believe there should be infinitely many values of $a,b,c$ for which the period divides $2 \pi$. For example, with $k=2$ and $\lambda=1$ I get a period dividing $2 \pi$ for $c = 2.722857918$, $3.502129242$, $4.303773851$, $5.116503598$, etc. Of course we can replace $bx$ by $\sum\_j b\_j x^j$ with $\sum\_j b\_j^2 = b^2$. 


EDIT: We can think about it this way. Assume $k, \lambda > 0$.  

By scaling $y$ and time $t$ we can non-dimensionalize the autonomous differential equation $\ddot{y} + k^2 y + \lambda y^3 = 0$ to $\ddot{y} + y + y^3 = 0$.
The phase-plane trajectories of this autonomous differential equation are the closed curves $\dfrac{v^2}{2} + \dfrac{y^2}{2} + \dfrac{y^4}{4} = C$ for $C > 0$, where $v = dy/dt$. The period $P$ is of the orbit through $(y=y\_0, v=0)$ is, by symmetry, $4$ times the time needed to get from $(y\_0,0)$ to $(0,\sqrt{ y\_0^2 + y\_0^4/2})$, and thus
$$ P = 4 \int\_0^{y\_0} \dfrac{dy}{\sqrt{ y\_0^2 - y^2 + y\_0^4/2 - y^4/2}}$$
Under the change of variables $y = s y\_0$ this becomes
$$ P = 4 \sqrt{2} \int\_0^1 \dfrac{ds}{\sqrt{2 - 2 s^2 + y\_0^2 - y\_0^2 s^4}} $$
which goes to $0$ as $y\_0 \to \infty$.


