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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?

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up vote 5 down vote accepted

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 $y_0 \to \infty$.

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Thanks, that's very helpful! (I think those numeric parameter values you quote are for $b$ rather than $c$.) –  Greg Egan Sep 23 '12 at 8:58
    
Unless I'm confused, I think it follows that the manifold of solutions will be a countably infinite collection of disjoint copies of $T^n$. For example, in one dimension each value for $b$ that satisfies the boundary conditions gives a solution with a certain number of periods across the torus, say $m$, which can then be translated by any distance $d$ around the torus. So the general solution is: $\phi_{m,d}(x) = a(b_m)\; \text{sn}\left(b_m (x - d), c(b_m)\right)$ –  Greg Egan Sep 23 '12 at 11:37
    
Yes, those are the $b$ values. –  Robert Israel Sep 23 '12 at 19:35
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