Download Communications in Mathematical Physics - Volume 241 by M. Aizenman (Chief Editor) PDF

By M. Aizenman (Chief Editor)

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119) ι∈G k,G k,G ∈ Hδ,M × Hδ,M , we conclude that ι∗ ( ), ι∈G 1 8 1 8 ˜ ι∗ (φ) = ( , ψ, ζ, σ, h), ι∈G o. Our main existence theorem now follows readily from the above proposition and the Inverse Function Theorem. Acknowledgement. D. advisor Professor Richard Schoen, who suggested this problem and gives me constant directions and encouragement. I also would like to thank Professor Robert Bartnik and Professor Hubert Bray for many stimulating discussions during their visit at the AIM-Stanford workshop on General Relativity in April 2002.

Now it follows from (73) that thus integrating by parts and using the fact that (divϒ)i = O(r −δ−2 ), Dj (divϒi ) = O(r −δ−3 ) and δ > − (divϒ) = 0, 3 2 we see that divϒ ≡ 0 in M. Therefore, (73) and (74) become ϕ =0 ϒ =0 and (75) in M  ϕ − ϒnn = 0    ∂ϕ − div ϒ(n, ·) = 0 ∂n divϒ = 0    ϒ| = wgo | on , (76) which proves Lemma 2. It is easily seen that (ϒ, ϕ) = (go , 1) satisfies both (75) and (76). To eliminate such a trivial solution, we choose δ ∈ (−1, − 21 ] throughout the rest of our discussion.

It is easily seen that is a linear isometry of T (S 2 ) which rotates every tangent vector π 2 clockwise. e. ∇S 2 = 0. 42 P. Miao First we look for Type (I) solutions. Straightforward calculation, though not quite a pleasant thing to do, shows that (75) and (76) are reduced to the following system of coupled ODEs:  [d(r) − 2c(r)] =0 d (r) + 2r d (r) − r22 d(r) + r22 a(r) − L(L+1)   r2   c(r) =0 c (r) + 2r c (r) + r22 c(r) + r43 b(r) − L(L+1) r2 (88) L(L+1) 4 2 2 2  b (r) − r 2 b(r) + r a(r) − r d(r) − r c(r) − r 2 [b(r) − 2rc(r)] = 0    a (r) + 2r a (r) − r42 a(r) + r42 d(r) − L(L+1) [a(r) − 4r b(r) + 2c(r)] = 0 r2 with the boundary condition  co − a(1) =0    co − Lb(1) =0  a (1) + 2a(1) − 2d(1) − L(L + 1)b(1) = 0   b (1) + 2b(1) + d(1) =0   c(1) =0 for L ≥ 2, and  2  r d (r) + 2rd (r) − 4d(r) + 2a(r) − 4r b(r) = 0 rb (r) − 6r b(r) + 2a(r) − 2d(r) =0  2 r a (r) + 2ra (r) − 6a(r) + 4d(r) + 8r b(r) = 0 (89) (90) with the boundary condition    co − a(1) =0 co − b(1) =0 a (1) + 2a(1) − 2d(1) − 2b(1) = 0   b (1) + 2b(1) + d(1) =0 (91) for L = 1.

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