By J. Prolla
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After a scaling we can assume Q = (0, 1)n . Moreover, it suffices to consider u ∈ C 1 (Q, R). 17) where xi = (˜ x1 , . . , x˜i , xi+1 , . . , xn ). Squaring this equation and using Cauchy– Schwarz on the right hand side we obtain n 2 2 u(x) − 2u(x)u(˜ x) + u(˜ x) 2 1 ≤ |∂i u|dxi i=1 n 0 n ≤n |∂i u|dxi i=1 0 1 (∂i u)2 dxi . 19) Q and finishes the proof. Now we are ready to show Rellich’s compactness theorem. ✷ 48 Chapter 4. 4 (Rellich’s compactness theorem) Let U be a bounded open subset of Rn .
After a scaling we can assume Q = (0, 1)n . Moreover, it suffices to consider u ∈ C 1 (Q, R). 17) where xi = (˜ x1 , . . , x˜i , xi+1 , . . , xn ). Squaring this equation and using Cauchy– Schwarz on the right hand side we obtain n 2 2 u(x) − 2u(x)u(˜ x) + u(˜ x) 2 1 ≤ |∂i u|dxi i=1 n 0 n ≤n |∂i u|dxi i=1 0 1 (∂i u)2 dxi . 19) Q and finishes the proof. Now we are ready to show Rellich’s compactness theorem. ✷ 48 Chapter 4. 4 (Rellich’s compactness theorem) Let U be a bounded open subset of Rn .
5). We pick as underlying Hilbert space H01 (U, R3 ) with scalar product u, v = (∂j ui )(∂j vi )dx. 30) U Let X be the closure of X in H01 (U, R3 ), that is, X = {v ∈ C 2 (U , R3 )|∂j vj = 0 and v|∂U = 0} = {v ∈ H01 (U, R3 )|∂j vj = 0}. 5) by w ∈ X and integrate over U η∂k ∂k vj − (vk ∂k )vj + Kj wj dx = U (∂j p)wj dx = 0. 32) U Using integration by parts this can be rewritten as η(∂k vj )(∂k wj ) − vk vj (∂k wj ) − Kj wj dx = 0. 34) U where a(u, v, w) = uk vj (∂k wj ) dx. 34) represents a necessary solubility condition for the NavierStokes equations.