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## A NONLINEAR PARTIAL

A NONLINEAR PARTIAL DIFFERENTIAL EQUATION FOR THE VOLUME PRESERVING MEAN CURVATURE FLOW5 Proof. By definition it holds that thus (2.1) gives that ˜ρ(x 1 , · · · , x n+1 , t) = 1 + ρ ∗( x |x| , t ) , (3.2) S t = {x ∈ R n+1 ∗ : φ(x, t) = 0}, where for t fixed, φ(·, t) : R n+1 ∗ → R is given by (3.3) φ(x, t) := |x| − ˜ρ(x, t). Obviously any x ∈ S t is a root of φ(x, t). In what follows, we use that ˜ρ is a function defined on any x ∈ R n+1 ∗ , independent of |x|. The derivatives in formulae are applied in the space R n+1 in cartesians and at the end we consider x ∈ S t ⊂ R n+1 in order to compute the exact values on the hypersurface S t . For the velocity and mean curvature of S t we have respectively the formulae, [4] V = − ∂ tφ |∇ x φ| | x∈S t , H = 1 ( n div ∇x φ ) x | x∈St . |∇ x φ| Using now the velocity formula in (1.1) we arrive at (3.4) ∂ t φ = (H − h)|∇ x φ| on S t . From (3.3) we compute (3.5) ∂ t φ(x, t) = −∂ t˜ρ(x, t), ∇ x φ = while for any 0 ≤ i ≤ n + 1 it follows that ∂|x| = 1 n+1 ∂x i 2 ( ∑ x 2 j) −1/2 2x i = x i |x| . j=1 Therefore, the second equality of (3.5) gives (3.6) ∇ x φ = x |x| − ∇ x˜ρ. ( ∂|x| ∂|x| ) − ˜ρ x1 , · · · , − ˜ρ xn+1 , ∂x 1 ∂x n+1 In (3.4) we replace ∂ t φ by (3.5) and use (3.6) to obtain finally (3.7) ∂ t˜ρ = (h − H) ∣ x |x| − ∇ ∣ x˜ρ ∣. Let us consider λ > 0, then the next equality easily follows Hence, for any λ > 0, we obtain where ˜ρ(λx, t) = 1 + ρ ∗ ( λx |λx| , t) = 1 + ρ∗ ( x , t) = ˜ρ(x, t). |x| 0 = ∂ λ (˜ρ(x, t)) = ∂ λ (˜ρ(λx, t)) = ˜ρ(λx, t) = ˜ρ(y, t), n+1 ∑ i=1 and y := λx. So, x∇ y ˜ρ(y, t) = 0 and therefore λ > 0 yields y∇ y ˜ρ(y, t) = λx∇ y ˜ρ(y, t) = 0. ˜ρ yi (y, t) ∂y i ∂λ ,

6 DIMITRA ANTONOPOULOU, GEORGIA KARALI Setting λ := 1 we get that and thus (3.8) x |x| ⊥∇ x˜ρ(x, t), ∣ x (∣ |x| − ∇ ∣∣ x x˜ρ ∣ = ∣ 2 + |∇ x˜ρ| 2) 1/2 √ = 1 + |∇x˜ρ| |x| 2 . Replacing (3.8) in (3.7) we obtain (3.9) ∂ t˜ρ = (h − H) √ 1 + |∇ x˜ρ| 2 . The next step is to calculate the mean curvature H in terms of ˜ρ. By (3.6), (3.8) and the definition (3.3) of φ it follows that ( ∇x φ ) ( nH := div x = 1 + |∇ x˜ρ| 2) − 1 ( 2 n ) (3.10) |∇ x φ| |x| − ∆ x˜ρ ( − 1 + |∇ x˜ρ| 2) − 3 n+1 ∑ 2 |∇ x˜ρ| ( x j |x| − ˜ρ x j ) ∂ (|∇ x˜ρ|). ∂ xj Further, ∂ ∂ xj (|∇ x˜ρ|) = so replacing in (3.10) we get j=1 ∂ (( n+1 ∑ ) 1 ) n+1 ∑ ˜ρ 2 2 x ∂x i = |∇ x˜ρ| −1 ˜ρ xi ˜ρ xix j , j i=1 (3.11) nH = A − for and But we note that A := n+1 ∑ B := ( 1 + |∇ x˜ρ| 2) − 3 2 B, i=1 ( 1 + |∇ x˜ρ| 2) − 1 ( 2 n ) |x| − ∆ x˜ρ j=1 ( x j |x| − ˜ρ x j ) ( n+1 ∑ i=1 ˜ρ xi ˜ρ xix j ). n+1 ∑ n+1 ∑ (3.12) ∇ x˜ρ Hess x (˜ρ) ∇ x˜ρ T = ˜ρ xi ˜ρ xix j ˜ρ xj = thus replacing in B we get (3.13) B = n+1 ∑ j=1 x j |x| ( n+1 ∑ i=1 j=1 i=1 n+1 ∑ j=1 n+1 ∑ ˜ρ xj i=1 ˜ρ xi ˜ρ xix j ) − ∇ x˜ρ Hess x (˜ρ) ∇ x˜ρ T , where ∇ x˜ρ T is the transpose of ∇ x˜ρ. Since x⊥∇ x˜ρ we deduce that n+1 ∑ x j ˜ρ xj = 0. j=1 By differentiation with respect to x i we obtain n+1 ∑ x j ˜ρ xjx i + ˜ρ xi = 0 j=1 ˜ρ xi ˜ρ xix j ,

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