MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

The generalized Korteweg-de Vries equation is

$u_t + u_{xxx} + (u^p)_x=0$

for integer $p$. (The original Korteweg-de Vries equation is the case $p=2$.) I need to understand solutions for $p=1$, but I haven't been able to find this case addressed in the literature despite extensive searching. Is there a reason why $p=1$ isn't interesting? Does it reduce to a simpler form? Is it intractable?

If not, what the best way to learn about this case?


share|cite|improve this question
Naively, I'd say that $p=1$ being linear does not exhibit any of the interesting behaviour that the KdV equation exhibits: solitons,... – José Figueroa-O'Farrill May 3 '10 at 11:22
up vote 8 down vote accepted

For $p=1$ your equation indeed reduces to a simpler form. Put $X=x-t, T=t$. Then $u_t=u_T-u_X$, $u_x=u_X$, and hence in the new independent variables $X,T$ your equation (with $p=1$) becomes a well-studied linear third-order equation $$ u_{T}+u_{XXX}=0, $$ which is, inter alia, a linearized version of the "standard" KdV equation with $p=2$, see e.g. the discussion in this book by Ablowitz and Segur (and cf. also Willie's answer).

Note that the case of $p=3$ (the modified KdV equation) is also quite interesting. It is integrable by the inverse scattering transform just as the "usual" KdV and, in fact, is related to it through the Miura transformation (see e.g. the above book for details).

share|cite|improve this answer
Thanks so much. – Jess Riedel May 4 '10 at 15:12

José is correct in his comment. Just to elaborate: in the linear case, one can easily study the equation using Fourier methods. Let $\tilde{u}$ denote the space-time Fourier transform and $\hat{u}$ denote the spatial Fourier transform, the equation with $p=1$ can be written as $$ (\tau - \xi^3 + \xi)\tilde{u} = 0$$ or $$ \partial_t\hat{u} = i(\xi^3 - \xi)\hat{u} $$

The first formulation tells you that the space-time Fourier transform of a solution is a measure supported on the curve $\tau = \xi^3 - \xi$ in frequency space. That the frequency support has curvature implies that the solution should decay in time in physical space, justifying José's comment. (Look up Fourier restriction theorems or Strichartz estimates in the literature for more details.)

We can also see the temporal decay directly from the second formulation using oscillatory integral techniques. The solution can be written as $$ \hat{u}(t,\xi) = e^{i(\xi^3 - \xi)t}\hat{u}_0(\xi) $$ It is then a standard exercise to show that, given initial data of sufficient decay in frequency space (say, the frequency has compact support), taking the inverse Fourier transform of the above solution gives you something with decay in $L^\infty$.

If a solution decays in $L^\infty$, it cannot be a soliton.

We can also see the lack of solitons by posing the traveling wave ansatz $u(t,x) = f(\omega t + x)$. A simple computation shows that the function $f$ must solve $$ -(\omega + 1)f' = f''' \implies f = \exp [ i \sqrt{\omega + 1}(\omega t + x) ] $$ so that traveling waves cannot be spatially localized. In fact, the traveling wave ansatz gives us a dispersion relation for this equation: that waves of spatial frequency $\beta$ travels with velocity $\omega = \beta^2 - 1$. The fact that different frequency components of the wave tend to travel at different velocities illustrates why, starting with a pulsed wave packet, the solution will become wider and wider while its height gets smaller and smaller.

In short, I think the reason why this equation is not heavily studied in the literature is because that as a linear PDE in (1+1) dimensions, it is not really all that interesting to look at.

share|cite|improve this answer
>"That the frequency support has curvature implies that the solution should decay in time in physical space, justifying José's comment." Wow, that's really interesting. Thanks so much – Jess Riedel May 4 '10 at 15:16

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.