$L^p$ norms of Fourier transform of solutions of hyperbolic Burgers' 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 Burgers' 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.
 A: 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.
A: This question is almost a decade old and my answer is a bit of a misdirect: The $L^p$ conservation can be observed directly.
Indeed, let $1<p<\infty$, then assuming everything is sufficiently smooth, multiply the PDE by $|u|^{p-1}$ and integrate to find
$$
\int\,|u|^{p-1}\partial_t u \,dx = \frac{1}{p}\frac{\mathrm{d}}{\mathrm{d}t}\int |u|^{p}\,dx = -\int\,|u|^{p-1}u\partial_xu\,dx = \frac{1}{p}\int\,\partial_x(|u|^{p+1})\,dx = 0.
$$
If $p=1$, then multiply the PDE by $\text{sgn}\, u$ and integrate to get
$$
\frac{\mathrm{d}}{\mathrm{d}t} \int\,|u|\,dx = -\frac{1}{2}\int\,\partial_x (u^2)\,dx = 0.
$$
Thus $\|u\|_{L^p}=\|u_0\|_{L^p}$ for all $1\leqslant p < \infty$.
So in some sense the $L^\infty$ conservation is the odd one out since you need Lagrangian characteristics rather than energy methods!
Now, for the Fourier transform norms, note that $\|\hat{u}\|_{L^\infty}\leqslant\|u\|_{L^1}$ by definition and $\|\hat{u}\|_{L^2}=\|u\|_{L^2}$ by Plancherel's. Thus we can interpolate between $L^2$ and $L^\infty$ for all $2<p<\infty$ using standard $L^p$ interpolation. For $1<p<\infty$ we can use the Hausdorff-Young inequality which says $\|\hat{u}\|_{L^p}\lesssim \|u\|_{L^q}$ for $p$ and $q$ dual exponents.
One needs more information to conclude that $\|\hat{u}\|_{L^1}$ remains bounded. I would guess you're right and the $L^\infty$ conservation gives $\hat{u}$ in weak-$L^1$ or better.
