# $L^p$ norms of Fourier transform of solutions of hyperbolic Burger's equation at the time of first blow-up

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.

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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.

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

where

$b_n(t) = -2 J_n(nt)/nt$ (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.

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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