The nonlinear pde $$ \partial_t^2\phi-\partial_x^2\phi+\lambda\phi^3=0 $$ has the exact solution $$ \phi(x,t)=\mu\left(\frac{2}{\lambda}\right)^\frac{1}{4}{\rm sn}(p_0t-p\cdot x+\varphi,i) $$ with $\mu$ and $\varphi$ two integration constants and sn the snoidal Jacobi function, provided the dispersion relation holds $$ p^2_0=p^2+\mu^2\sqrt{\frac{\lambda}{2}}. $$ If I interpret $p_0$ as the energy, it seems that is finite. Computing the integral $$ E=\int d^Dx\left[\frac{1}{2}(\partial_t\phi)^2+\frac{1}{2}(\partial_x\phi)^2+\frac{\lambda}{4}\phi^4\right] $$ and extending the volume to infinity, this is divergent. Can one find a sound mathematical explanation for this? Is there a way to "regularize" this integral?


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    $\begingroup$ The exact same phenomenon occurs already when $\lambda=0$ and any plane wave solution. It has infinite energy because it does not decay at infinity. The relation between frequency and energy is more subtle than the motivation for this question suggests. $\endgroup$ – Igor Khavkine May 22 '13 at 8:38
  • $\begingroup$ @IgorKhavkine: Thanks for the comment. Yes, I am aware of the similarity with the plane waves but there one can circumvent the problem (at least looking at standard textbooks). Here the situation seems more awkward. Could you expand on your last sentence? $\endgroup$ – Jon May 22 '13 at 8:48

regularization of the Klein-Gordon equation proceeds in the same way as for the Schrödinger equation; you restrict $x$ to the interval $(0,L)$ and impose periodic boundary conditions $\phi(0,t)=\phi(L,t)$; this quantizes the wave vector $p=p_n(t)$, $n\in\mathbb{Z}$ --- for the linear Schrödinger equation the quantization would be time independent, now it depends on time; in the limit $L\rightarrow\infty$ you would ignore the quantization, calculate the energy $E_L$ by integrating $x$ from $0$ to $L$, and then obtain a finite limit $L^{-1}E_L$; this just means that the energy density per unit length is finite, even though the entire energy content of the infinite system is infinite (as it should be).

  • $\begingroup$ Carlo, thanks a lot. It appeared like this idea does not apply to the nonlinear case by simply redefining the solution with the trick of the $\frac{1}{\sqrt{V}}$ normalization for plane waves. This is my main concern. $\endgroup$ – Jon May 22 '13 at 13:53

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