# Strichartz Estimates for radial Klein-Gordon equation

I'm trying to prove global wellposedness for the Klein-Gordon-Equation with radial initial data. I'm therefore searching for/trying to prove strichartz estimates of the form: $$||e^{it\langle D\rangle}u||_{L^q_{t}L^r_{x}}\lesssim||\langle D\rangle^lu||_{L^{2}}$$ or $$||P_{L}((e^{it\langle D\rangle}u_{M})(e^{it\langle D\rangle}v_{N}))||_{L_{2}}\lesssim L^lM^mN^n||u_{M}||_{L^{2}}||v_{N}||_{L^{2}}$$ with indices $l,m,n$ as small as possible. Where $P_{L}$ is the Littlewoodpaley projector, $e^{it\langle D\rangle}$ the Propagator of the (mass 1) Klein-Gordon-Equation and $L,M,N$ frequencies.

I'm aware, that you have more admissible strichartz exponentens $(q,r)$ for the first kind of estimate if you have radial data (the result is outlined here: http://wiki.math.toronto.edu/DispersiveWiki/index.php/Strichartz_estimates) but I can't see whether/how that can help me since I will get the same $l$ as with none radial data. Am i missing something here?

Also I don't know how to proof the first inequalitity for non-radial-data. I know it is true for $2\leq r<\infty$, $2/q+n/r=n/2$, $l=1/q-1/r+1/2$ and was told to find the proof in Delort and Fang(2000) but I can't. (I hope the proof will help me understand how I can improve the result for radial functions)

• For the standard estimate (first inequality) I don't think you need to refer to something as high powered as Delort and Fang. The black box technique of Keel-Tao ("Endporint Strichartz estimates") essentially carries over. In the K-T paper they only explicitly treat Wave and Schrodinger; but if you insert the $L^1-L^\infty$ decay for Klein-Gordon (which you can get by stationary phase type estimates) you can interpolate exactly the same way against the energy conservation to get the estimates. Mar 14, 2016 at 13:41
• Note however, there is some caveat in 2 dimensions; see sciencedirect.com/science/article/pii/S0021782410001182 for a proof there. ; I am doubtful that you can get better $l$ in the radial case: the argument that sets the differentiability is essentially a scaling argument. You can do it while preserving the radiality of the data. That the $(q,r)$ exponents can be improved is due to that the obstruction there is due to the Knapp counterexample, which gets weakened in the radial cased. Mar 14, 2016 at 13:46
• 1. For the wellposedness problem of nonlinear equations, enlarging admissible ranges for (q,r) allows treating more types of nonlinearities. For an example see Klainerman-Machedon, "Spacetime estimates for null forms and the local existence theorem", CPAM (1993). In particular pay attention to Theorem 2, Theorem 3, Proposition 4, and Remark 1. Mar 21, 2016 at 1:09
• 2. The Fourier transform of radial functions are radial. The reason that angular cut-off are needed in general is that certain "worst-case scenario estimates" that occur, if you try to estimate naively without doing the cut-off, will give you a larger contribution than is actually there. If you know a priori that the functions are radial, certain of these worst-case scenario cannot occur, and you can get better estimates. Mar 21, 2016 at 1:16
• Another way to say this is that the angular cut-offs are designed to solve a particular problem (isolate certain types of interactions), and this problem is exaggerated when the data are not purely radial. So understanding why this procedure is needed in the paper will help you understand what you can get away with when the data is in fact radial. Mar 21, 2016 at 1:33