Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

I am struggling to understand the behavior of the Fourier transform (in the $x$ variable) of initially smooth solutions of the hyperbolic Burger's equation in 1-D,

$ \partial_t u + u~ \partial_x u =0$ .

I start with a smooth and rapidly decaying initial condition $u(x,t)=u_0(x)$ on $\Bbb R$ . This solution evolves in time until it breaks down. At the time of first breakdown $t=T$ I look at the Fourier transform $\hat u(k,T)$ of the solution $u(x,T)$.

In particular, I am trying hard to understand how and why the $L^p$ norms of the Fourier transform $\hat u$ remain finite at the time of first blow-up for $p>1$. I think that if one uses weak (or Lorentz) norms, then this non-blow-up extends even to the weak $L^1$ norm.

The only way I have been able to understand this property is via the convervation law for the $L^\infty$ norm of $u$. For the $\|u \|_{L^\infty} $ norm to be defined at the time of first blow-up, the Fourier transform needs to remain in a weak $L^1$ space. Interpolation explains the rest.

My question is whether there is a way to understand the non-blow-up of the said $L^p$ norms of the Fourier transform $\hat u$ without invoking the conservation law for the $L^\infty$ norm of $u$.

What I seek is some kind of direct Fourier-analytic way to see what is going on. I have reached an impasse.

I will be very grateful for any insight or advice.

share|cite|improve this question
I am starting to wonder whether maybe there is NO known way except to use conservation laws. (The bounty just ended too.) Why I think there may be no answer: the Burger equation has a 2-parameter scale invariance symmetry. Without a conservation law, it may not be possible to "peg" suitable norms to try to avoid the worst case scenario. By worst case scenarios, I mean the usual inequalities such as Hölder's and Young's (convolution) inequalities, etc. – Gandhi Viswanathan Jun 11 '12 at 19:01

This is a fun question! I have started playing with it but it the case of periodic data $u(x,t) = sin(x)$. In this case you can write down the explicit solution.

$u(x,t) = \sum_{n=1}^\infty b_n(t) \sin(nx)$


$b_n(t) = -2 J_n(nt)/nt$ (Bessel function of order n)

From this you can compute some $L^p$ norms explicitly to get a sense of what is happening. This is not a full solution but it is as far as I got before I had to get back to work ...

The above result is from

G.W. Platzman, An exact integral of complete spectral equations for unsteady one- dimensional flow, Tellus, XVI (1964), pp. 422–431.

share|cite|improve this answer
Thanks! The solution $u(.,t)$ becomes progressively more "inclined" until the first derivative $\partial u/\partial x$ becomes infinite at $x= \pm n \pi$. Beyond this time of first breakdown, there no strong solutions, because $u$ becomes multivalued. Of course, weak solutions exist... My question is whether we can understand, purely in terms of Fourier transforms, why the Fourier transform $\hat u$ remains in a weak $L^1$ space at the time of breakdown. I want to know whether there is a way of seeing this without recourse to the conservation of $L^p$ norms of the $u$. – Gandhi Viswanathan Jun 4 '12 at 21:34
Correction: I should have written $x=(2n+1)\pi$... – Gandhi Viswanathan Jun 4 '12 at 22:42

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.