MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:


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?

share|cite|improve this question
up vote 6 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 $y_0 \to \infty$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.