On the Einstein-Euler Equations

On the Einstein-Euler Equations Tetu Makino (Yamaguchi U, Japan) November 10, 2015 / Int l Workshop on the Multi-Phase Flow at Waseda U 1

1 Introduction. Einstein-Euler equations: (A. Einstein, Nov. 25, 1915) R µν 1 2 g µν(g αβ R αβ ) = 8πG c 4 T µν (1) T µν = (c 2 ρ + P )U µ U ν P g µν (2) (A0): P is supposed to be a given analytic function of ρ > 0 such that 0 < P, 0 < dp/dρ < c 2 and P = Aρ γ (1 + [ρ γ 1 /c 2 ] 1 ) (3) as ρ +0 with constants A > 0, 1 < γ < 2. 2

Look for a co-moving spherically symmetric metric ds 2 = e 2F (t,r) c 2 dt 2 e 2H(t,r) dr 2 R(t, r) 2 (dθ 2 + sin 2 θdϕ 2 ) (4) such that U ct = e F, U r = U θ = U ϕ = 0 with spherically symmetric density distribution ρ(t, r). Equations turn out to be F R e t = V F V ( m e t = GR R 3 + 4πP ) c 2 (1 + V 2 c 2 2Gm ) P c 2 R R (ρ + P/c 2 ) F ρ ( V e t = (ρ + P/c2 ) R + 2V ) R (C. W. Misner and D. H. Sharp, 1964) 3 (5a) (5b) (5c)

Here X = X/ r, and r m = 4π ρr 2 R dr; (6) 0 e F = [ κ + exp u ] c 2, (7) ρ dp ( Aγ ) where u = = 0 ρ + P/c 2 γ 1 ργ 1 (1 + [ρ γ 1 /c 2 ] 1 ), (8) e 2H = (1 + V 2 c 2 2Gm ) 1(R c 2 ) 2. (9) R 4

2 Equilibria. The Tolman-Oppenheimer-Volkoff eq s: dm dr = 4πr2 ρ, dp dr = (ρ + P/c2 ) G(m + 4πr3 P/c 2 ) r 2 (1 2Gm/c 2 r) (10a) (10b) (J. P. Oppenheimer and G. M. Volkoff, 1939) 5

Fix ρ = ρ(r) such that m = 4π 3 ρ cr 3 + O(r 5 ), P = P c (ρ c + P c /c 2 )4πG(ρ c /3 + P c /c 2 ) r2 2 + O(r4 ) (11) as r 0 and ρ(r) 0 as r r + (< ). Put m + = 4π r+ 0 ρ(r)r 2 dr, κ + = 1 2Gm + c 2 r + > 0 (12) 6

Lemma 1. ū(r) = Gm + r+κ 2 (r + r)(1 + [r + r, (r + r) 1 γ 1 ]1 ), (13) + ( (γ 1) Gm ) 1 + γ 1 ρ(r) = Aγ r+κ 2 (r + r) 1 γ 1 (1 + [r+ r, (r + r) 1 γ 1 ]1 ). + (14) Here (A1) : 1 γ 1 [X 1, X 2 ] q = k 1 +k 2 q a k1 k 2 X k 1 1 Xk 2 2 is supposed to be an integer. 7

3. Perturbation from the equilibrium. satisfy R = r(1 + y(t, r)), V = rv(t, r) (15) F y e t = (1 + P/c2 ρ)v, (16a) F v e t = 1 c 2 (1 + P ρ y)2 v r (rv) 1 G ( r 3 (1 + y) 2 m + 4π c 2 P r3 (1 + y) 3) + (1 + r2 v 2 c 2 2Gm )(1 c 2 + P/c 2 1 (1 + y)2 P ρ) (16b) r(1 + y) r ρ r Here m = m(r) is given and ρ is the function of r, y, ry given by ρ = ρ(r)(1 + y) 2( 1 + y + r y ) 1 (17) r 8

4 Linearization. Linearization of (15) at y = v = 0: 2 y + Ly = 0 (18) t2 Here Ly = 1 b a = d dr ΓP r4 ( a dy ) dr + Qy, 1 + P/c 2 ρ ef +H, b = (1 + P/c 2 ρ) 1 ρr 4 F +3H e with Γ := ρ P dp dρ. 9

Lemma 2. L is a self-adjoint operator in L 2 ((0, r + ); b(r)dr) whose spectrum consists of simple eigenvalues λ 1 < λ 2 < < λ ν < +. 10

5 Pulsating stars. Fix a positive eigenvalue λ of L and its eigenfunction ψ(r) such that ψ(r + ) = 1. Put y 1 = sin( λt + Θ 0 )ψ(r) (19) Look for a solution of (16a)(16b) of the form: Here y = εy 1 + ε 2 ỹ, v = εv 1 + ε 2 ṽ (20) v 1 = e F (1 + P/c 2 ρ) 1 y 1 t Theorem 1. Given T > 0, there exists a positive ϵ 0 (T ) such that for ε ϵ 0 (T ) there is a solution (y, v) C ([0, T ] [0, r + ]) of (16a)(16b) of the form (20) such that sup j t r k (ỹ, ṽ) L j+k n 11 C(n)

Note R(t, r + ) = r + (1 + ε sin( λt + Θ 0 ) + O(ε 2 )) (21) and ρ(t, r) = {C(t)(r + r) 1 γ 1 (1 + O(r+ r)) (0 r < r + ) 0 (r + r) (22) with a smooth function C(t) of t such that C(t) = ( γ 1 Aγ Gm ) 1 + γ 1 r+κ 2 + + O(ε) 12

6. Cauchy problem. Consider (CP): F y v e = (16a), e F t t = y t=0 = ψ 0 (x), v t=0 = ψ 1 (x) (16b), Theorem 2. For given T there is a positive δ(t ) such that, if ψ 0, ψ 1 C ([0, 1]) satisfy max k K ( Dk (ψ 0, ψ 1 ) L δ(t ), then the problem (CP) has a solution in C ([0, T ] [0, 1]). Here K is a number depending on γ. 13

7 Metric on the vacuum exterior. The Schwarzschild metric: where ds 2 = K (cdt ) 2 1 K (dr ) 2 (R ) 2 (dθ 2 + sin 2 θdϕ 2 ), (23) Patched metric: K := 1 2Gm + c 2 R, t = t (t, r), R = R (t, r), (r r + ) ds 2 = g 00 (cdt) 2 + 2g 01 cdtdr + g 11 dr 2 + g 22 (dθ 2 + ). 14

Theorem 3. There are t, R C ([0, T ] [r +, + )) such that the patched g µν are of C 1 ([0, T ] [0, + )). But then 2 R r+ r 2 2 R ( r+ R ) 2, +0 r 2 = A 0 r A = V 2 ( Gm+ c 2 c 2 R 2 + 1 1 V )(1 κ+ c 2 + V 2 t c 2 2Gm ) 2 + c 2 R r+ 0 does not vanish if V 0. 15

8 Analysis of the structure of the equation system. Introduce the variable x instead of r by x = tan2 θ 1 + tan 2 θ, θ = π 2ξ + r 0 ρ ΓP e F +H dr (24) Then L = x(1 x) d2 dx 2 ( 5 2 (1 x) N 2 x ) d dx + + L 1 (x)x(1 x) d dx + L 0(x) (25) Here N = 2γ γ 1 (26) and L 1, L 0 are analytic on a neighborhood of [0, 1]. 16

r = C 0 x(1 + [x]1 ) as x 0 (27) r + r = C 1 (1 x)(1 + [1 x] 1 ) as x 1 (28) The equations (16a)(16b) turn out to be y t Jv = 0, v t + H 1Ly + H 2 = 0 (29) Here J, H 1, H 2 are analytic functions of x, y, z(= ry r ), v, w(= rv r ), y r, y rr on a neighborhood of [0, 1] {(0,, 0)}. and J = e F (1 + P/c 2 ρ)(1 + O(y, z, )), H 1 = e F (1 + P/c 2 ρ) 1 (1 + O(y, z, )), H 2 = O( (y, z, ) 2 ) 17

Apply the Nash-Moser theory... As x 1: Why? L (1 x) d2 dx 2 + N 2 d dx = d2 dξ 2 N 1 ξ d dξ (= ξ), for 1 x = ξ2 4, but z = r y r Const. y x 1 ξ y ξ y 2 ξ 2 Loss of regularity: usual Picard iteration doens t work; Dionne s trick to reduce the problem to a quasi-linear one doesn t work... 18

Cf: Y. Shibata and Y. Tsutsumi / Nonlin. Anal., 11(1987), 335-365 2 u ( ) t 2 u + F t, x, { j t x k u} j+ k 2 = 0 on [0, T ] Ω, u Ω = 0, u u t=0 = ψ 0 (x), = ψ 1 (x). t t=0 Ω R N, F C, F (t, x, 0) = 0, { } F (t, x, 0) = 0, L [N/2] + 8, ψ 0 [N/2]+8 + ψ 1 [N/2]+7 B + compatibility of order L 1 T = T (N, Ω, F, B), sol. u L l=0 Cl ([0, T ], H L l (0) ) 19

9 General-purpose frame. The system of equations to be considered be y J(x, y, z)v = 0, t v t + H 1(x, y, z, v)ly + H 2 (x, y, z, v, w) = 0, where x runs over [0, 1], z stands for y/ x, w stands for v/ x and L = x(1 x) d2 dx 2 ( N0 2 (1 x) N 1 2 x ) d dx + + L 1 (x)x(1 x) d dx + L 0(x) with N 0, N 1 > 2. L 0 (x), L 1 (x) are analytic functions of x on a neighborhood of [0, 1]. 20

J, H 1, H 2 are of class A 0, A 0, A 2, where A q is the set of all analytic functions of x, Y 1, Y 2, on a neighborhood of [0, 1] {(0, 0, )} of the form k 1 +k 2 + q a k1 k 2 (x)y k 1 1 Y k 2 2. Fix T > 0 and y, v C ([0, T ] [0, 1]) small. Look for a solution (y, v) C ([0, T ] [0, 1]) of the form y = y + ỹ v = v + ṽ such that ỹ t=0 = 0, ṽ t=0 = 0. 21

Assumptions: (B0): H 1 (x, 0, 0, 0) = J(x, 0, 0) 1 and C > 1 such that 1 C < J(x, 0, 0) < C. (B1): As x 0 and as x 1 we have z J 0, ( z H 1 )Ly + z H 2 0, w H 2 0. Here a function f(x, y, z, v, w, y, y ) on a neighborhood of (x 0, 0, 0, ) is said to be 0 as x x 0 iff f(x 0, y, z, ) = 0 y, z,.,that is, analytic Ω such that f(x, y, z, ) = (x x 0 )Ω(x, y, z, ). 22

Conclusion: δ(t ) > 0, K such that if max j+k K j t x(y k, v ) L δ(t ) then solution (y, v) C ([0, T ] [0, 1]). Application to pulsating stars: y = εy 1, v = εv 1 Application to Cauchy problem: y = ψ 0 (x) + tj(x, 0, 0)ψ 1 (x), v = ψ 1 (x). Proof is due to Nash-Moser(-Hamilton) theory: R. Hamilton / Bull. AMS., 7(1982), pp. 65-222. 23

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