Existence of hypercylinder expanders of the inverse mean curvature flow
aa r X i v : . [ m a t h . A P ] D ec Existence of hypercylinder expanders of theinverse mean curvature flow
Kin Ming HuiInstitute of Mathematics, Academia SinicaTaipei, Taiwan, R. O. C.Dec 14, 2019
Abstract
We will give a new proof of the existence of hypercylinder expander of the inversemean curvature flow which is a radially symmetric homothetic soliton of the inversemean curvature flow in R n × R , n ≥
2, of the form ( r , y ( r )) or ( r ( y ) , y ) where r = | x | , x ∈ R n , is the radially symmetric coordinate and y ∈ R . More precisely for any λ > n − and µ >
0, we will give a new proof of the existence of a unique evensolution r ( y ) of the equation r ′′ ( y )1 + r ′ ( y ) = n − r ( y ) − + r ′ ( y ) λ ( r ( y ) − yr ′ ( y )) in R which satisfies r (0) = µ , r ′ (0) = r ( y ) > yr ′ ( y ) > y ∈ R . We will prove that lim y →∞ r ( y ) = ∞ and a : = lim y →∞ r ′ ( y ) exists with 0 ≤ a < ∞ . We will also give a new proof of theexistence of a constant y > r ′′ ( y ) = r ′′ ( y ) > < y < y and r ′′ ( y ) < y > y . Key words: inverse mean curvature flow, hypercylinder expander solution, existence,asymptotic behaviourAMS 2010 Mathematics Subject Classification: Primary 35K67, 35J75 Secondary 53C44
Consider a family of immersions F : M n × [0 , T ) → R n + of n -dimensional hypersurfaces in R n + . We say that M t = F t ( M n ), F t ( x ) = F ( x , t ), moves by the inverse mean curvature flowif ∂∂ t F ( x , t ) = − ν H ∀ x ∈ M n , < t < T where H ( x , t ) > ν are the mean curvature and unit interior normal of the surface F t at the point F ( x , t ). Recently there are a lot of study on the inverse mean curvature flowby P. Daskalopoulos, C. Gerhardt, K.M. Hui [H], G. Huisken, T. Ilmanen, K. Smoczyk,J. Urbas and others [DH], [G], [HI1], [HI2], [HI3], [S], [U]. Although there are a lot of1tudy on the inverse mean curvature flow on the compact case, there are not many resultsfor the non-compact case.Recall that [DLW] a n -dimensional submanifold Σ of R n + with immersion X : Σ → R n + and non-vanishing mean curvature H is called a homothetic soliton for the inverse meancurvature flow if there exists a constant λ , − ν ( p ) H ( p ) = λ X ( p ) ⊥ ∀ p ∈ Σ (1.1)where X ( p ) ⊥ is the component of X ( p ) that is normal to the tangent space T X ( p ) ( X ( Σ )) at X ( p ). As proved by G. Drugan, H. Lee and G. Wheeler in [DLW] (1.1) is equivalent to − < H ν, X > = λ ⇔ − < ∆ g X , X > = λ ∀ X ∈ Σ (1.2)where g is the induced metric of the immersion X : Σ → R n + . If the homothetic soliton ofthe inverse mean curvature flow is a radially symmetric solution in R n × R , n ≥
2, of theform ( r , y ( r )) or ( r ( y ) , y ) where r = | x | , x ∈ R n , is the radially symmetric coordinate, y ∈ R ,then by (1.2) and a direct computation r ( y ) satisfies the equation r ′′ ( y )1 + r ′ ( y ) = n − r ( y ) − + r ′ ( y ) λ ( r ( y ) − yr ′ ( y )) , r ( y ) > y ( r ) satisfies the equation, y rr + n − r · (1 + y r ) y r − (1 + y r ) λ ( ry r − y ) = r ′ ( y ) = drdy , r ′′ ( y ) = d rdy and y r ( r ) = dydr , y rr ( r ) = d ydr etc. In the paper [DLW] G. Drugan,H. Lee and G. Wheeler stated the existence and asymptotic behaviour of hypercylinderexpanders which are homothetic soliton for the inverse mean curvature flow with λ > / n .However there are no proof of the existence result in that paper except for the case λ = n − and the proof of the asymptotic behaviour of hypercylinder expanders there are verysketchy. In this paper I will give a new proof of the existence of hypercylinder expandersfor the inverse mean curvature flow with λ > n − . We will also give a new proof of theasymptotic behaviour of these hypercylinder expanders.More precisely I will prove the following main results. Theorem 1.1.
For any n ≥ , λ > n − and µ > , there exists a unique even solution r ( y ) ∈ C ( R ) of the equation r ′′ ( y )1 + r ′ ( y ) = n − r ( y ) − + r ′ ( y ) λ ( r ( y ) − yr ′ ( y )) , r ( y ) > , ∀ y ∈ R r (0) = µ, r ′ (0) = which satisfies r ( y ) > yr ′ ( y ) ∀ y ∈ R (1.5)2 nd r ′′ (0) = (cid:18) n − − λ (cid:19) µ . (1.6) Theorem 1.2. (cf. Theorem 20 of [DLW]) Let n ≥ , λ > n − , µ > , and r ( y ) ∈ C ( R ) be theunique solution of (1.4) . Then r ′ ( y ) > ∀ y > , (1.7) a : = lim y →∞ r ′ ( y ) exists and ≤ a < ∞ , (1.8) and lim y →±∞ r ( y ) = ∞ . (1.9) Moreover there exists a constant y > such that r ′′ ( y ) > ∀ < y < y r ′′ ( y ) < ∀ y > y r ′′ ( y ) = . (1.10)Since (1.4) is invariant under reflection y → − y , by uniqueness of solution of ODEthe solution of (1.4) is an even function and Theorem 1.1 is equivalent to the followingtheorem. Theorem 1.3.
For any n ≥ , λ > n − and µ > , there exists a unique solution r ( y ) ∈ C ([0 , ∞ )) of the equation r ′′ + r ′ = n − r − + r ′ λ ( r − yr ′ ) , r ( y ) > , ∀ y > r (0) = µ, r ′ (0) = which satisfies r ( y ) > yr ′ ( y ) ∀ y > and (1.6) . In this section we willl prove Theorem 1.2 and Theorem 1.3. We first start with a lemma.
Lemma 2.1.
For any n ≥ , λ , and µ > , there exists a constant y > such that the equation r ′′ + r ′ = n − r − + r ′ λ ( r − yr ′ ) , r ( y ) > , in [0 , y ) r (0) = µ, r ′ (0) = has a unique solution r ( y ) ∈ C ([0 , y )) which satisfiesr ( y ) > yr ′ ( y ) in [0 , y ) (2.2) Moreover (1.6) holds. roof : Uniqueness of solution of (2.1) follows from standard ODE theory. Hence we onlyneed to prove existence of solution of (2.1). We will use a modification of the fixed pointargument of the proof of Lemma 2.1 of [H] to prove the existence of solution of (2.1). Let0 < ε <
1. We now define the Banach space X ε : = (cid:8) ( g , h ) : g , h ∈ C ([0 , ε ]; R ) (cid:9) with a norm given by || ( g , h ) || X ε = max n k g k L ∞ (0 ,ε ) , k h ( s ) k L ∞ (0 ,ε ) o . For any ( g , h ) ∈ X ε , we define Φ ( g , h ) : = (cid:0) Φ ( g , h ) , Φ ( g , h ) (cid:1) , where for any 0 < y ≤ ε, Φ ( g , h )( y ) : = µ + Z y h ( s ) ds Φ ( g , h )( y ) : = Z y (1 + h ( s ) ) n − g ( s ) − + h ( s ) λ ( g ( s ) − sh ( s )) ! ds . (2.3)For any 0 < η ≤ µ/
4, let D ε,η : = (cid:8) ( g , h ) ∈ X ε : || ( g , h ) − ( µ, || X ε ≤ η (cid:9) . Note that D ε,η is a closed subspace of X ε . We will show that if ε ∈ (0 ,
1) is su ffi cientlysmall, the map ( g , h ) Φ ( g , h ) will have a unique fixed point in D ε,η .We first prove that Φ ( D ε,η ) ⊂ D ε,η if ε ∈ (0 ,
1) is su ffi ciently small. Let ( g , h ) ∈ D ε,η .Then | h ( s ) | ≤ η ≤ µ/ µ ≤ g ( s ) ≤ µ ∀ ≤ s ≤ ε (2.4) ⇒ g ( s ) − sh ( s ) ≥ µ > ∀ ≤ s ≤ ε (2.5)and | Φ ( g , h )( y ) − µ | ≤ Z y | h ( s ) | ds ≤ ηε ≤ η ∀ ≤ y ≤ ε. (2.6)Hence by (2.3), (2.4) and (2.5), | Φ ( g , h )( y ) | ≤ (1 + η ) n − µ + + η ) | λ | µ ! ε ≤ η ∀ ≤ y ≤ ε (2.7)if 0 < ε ≤ ε where ε = min , η (1 + η ) − n − µ + + η ) | λ | µ ! − . Φ ( D ε,η ) ⊂ D ε,η for any 0 < ε ≤ ε .We now let 0 < ε ≤ ε . Let ( g , h ) , ( g , h ) ∈ D ε,η and δ : = || ( g , h ) − ( g , h ) || X ε . Then by(2.4) and (2.5), | h i ( s ) | ≤ η ≤ µ/ , µ ≤ g i ( s ) ≤ µ g i ( s ) − sh i ( s ) ≥ µ > ∀ ≤ s ≤ ε, i = , . (2.8)Hence by (2.8), we have | Φ ( g , h )( y ) − Φ ( g , h )( y ) | ≤ Z y | h ( s ) − h ( s ) | ds ≤ δε ≤ δ ∀ ≤ y ≤ ε (2.9)and | Φ ( g , h )( y ) − Φ ( g , h )( y ) |≤ Z y | h ( s ) − h ( s ) | | h ( s ) + h ( s ) | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n − g ( s ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + h ( s ) λ ( g ( s ) − sh ( s )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)! ds + ( n − Z y (1 + h ( s ) ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) g ( s ) − g ( s ) g ( s ) g ( s ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ds + Z y (1 + h ( s ) ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + h ( s ) λ ( g ( s ) − sh ( s )) − + h ( s ) λ ( g ( s ) − sh ( s )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ds ∀ ≤ y ≤ ε ≤ c δε + (1 + η ) Z r (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + h ( s ) λ ( g ( s ) − sh ( s )) − + h ( s ) λ ( g ( s ) − sh ( s )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ds ∀ ≤ y ≤ ε (2.10)where c = η n − µ + + η ) | λ | µ ! + n − + η )9 µ . Now by (2.8), (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + h ( s ) g ( s ) − sh ( s ) − + h ( s ) g ( s ) − sh ( s ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12) (1 + h ( s ) )( g ( s ) − sh ( s )) − (1 + h ( s ) )( g ( s ) − sh ( s )) (cid:12)(cid:12)(cid:12) µ ∀ ≤ s ≤ ε (2.11)and (cid:12)(cid:12)(cid:12) (1 + h ( s ) )( g ( s ) − sh ( s )) − (1 + h ( s ) )( g ( s ) − sh ( s )) (cid:12)(cid:12)(cid:12) ≤| h ( s ) − h ( s ) || h ( s ) + h ( s ) || g ( s ) − sh ( s ) | + (1 + h ( s ) ) | g ( s ) − g ( s ) + sh ( s ) − sh ( s ) |≤ c δ ∀ ≤ s ≤ ε (2.12) c = η µ + η ! + + η ! .
5y (2.10), (2.11) and (2.12), | Φ ( g , h )( y ) − Φ ( g , h )( y ) | ≤ c + + η ) c | λ | µ ! δε ∀ ≤ y ≤ ε. (2.13)We now let ε = min ε , c + + η ) c | λ | µ ! − and 0 < ε ≤ ε . Then by (2.9) and (2.13), k Φ ( g , h ) − Φ ( g , h ) k X ε ≤ k ( g , h ) − ( g , h ) k X ε ∀ ( g , h ) , ( g , h ) ∈ D ε,η . Hence Φ is a contraction map on D ε,η . Then by the Banach fix point theorem the map Φ has a unique fix point. Let ( g , h ) ∈ D ε,η be the unique fixed point of the map Φ . Then Φ ( g , h ) = ( g , h ) . Hence g ( y ) = µ + Z y h ( s ) ds ⇒ g ′ ( y ) = h ( y ) ∀ < y < ε and g (0) = µ (2.14)and h ( y ) = Z y (1 + h ( s ) ) n − g ( s ) − + h ( s ) λ ( g ( s ) − sh ( s )) ! ds ∀ < y < ε ⇒ h ′ ( y ) = (1 + h ( y ) ) n − g ( y ) − + h ( y ) λ ( g ( y ) − sh ( y )) ! ∀ < y < ε. (2.15)By (2.5), (2.14) and (2.15), g ∈ C ([0 , ε )) ∩ C (0 , ε ) satisfies (2.1) and (2.2) with y = ε . Letting y → g ∈ C ([0 , ε )) and the lemma follows. (cid:3) By an argument similar to the proof of Lemma 2.1 we have the following lemma.
Lemma 2.2.
For any n ≥ , λ , , µ > , M > , δ > , r , r ∈ R , satisfying δ ≤ r ≤ M , | r | ≤ M , r − y r ≥ δ , there exists a constant δ ∈ (0 , y / depending on λ , δ , y and M such that for any y / < y < y the equation r ′′ + r ′ = n − r − + r ′ λ ( r − yr ′ ) , r ( y ) > , in [ y , y + δ ) r ( y ) = r , r ′ ( y ) = r (2.16) has a unique solution r ( y ) ∈ C ([ y , y + δ )) which satisfiesr ( y ) > yr ′ ( y ) in [ y , y + δ ) . (2.17)6 emma 2.3. Let n ≥ , < λ , n − , µ > and y > . Suppose r ( y ) ∈ C ([0 , y )) is the solutionof (2.1) which satisfies (2.2) . Then the following holds.(i) If λ > n − , then r ′ ( y ) > ∀ < y < y . (ii) If < λ < n − , then r ′ ( y ) < ∀ < y < y . Proof : By Lemma 2.1, (1.6) holds. We divide the proof into two cases:
Case 1 : λ > n − By (1.6), r ′′ (0) >
0. Hence there exists a constant δ > r ′ ( s ) > < s < δ .Let (0 , a ), δ ≤ a ≤ y , be the maximal interval such that r ′ ( s ) > ∀ < s < a . Suppose a < y . Then r ′ ( a ) = r ′′ ( a ) ≤
0. On the other hand by (2.1), r ′′ ( a ) = (cid:18) n − − λ (cid:19) r ( a ) > a = y and (i) follows. Case 2 : 0 < λ < n − By (1.6), r ′′ (0) <
0. Hence there exists a constant δ > r ′ ( s ) < < s < δ .Let (0 , a ), δ ≤ a ≤ y , be the maximal interval such that r ′ ( s ) < ∀ < s < a . Suppose a < y . Then r ′ ( a ) = r ′′ ( a ) ≥
0. On the other hand by (2.1), r ′′ ( a ) = (cid:18) n − − λ (cid:19) r ( a ) < a = y and (ii) follows. (cid:3) Lemma 2.4.
Let n ≥ , λ > n − , µ > and y > . Suppose r ( y ) ∈ C ([0 , y )) is the solution of (2.1) which satisfies (2.2) . Then there exist a constant δ > such thatr ( y ) − yr ′ ( y ) ≥ δ ∀ < y < y . (2.18) Proof : Let w ( y ) = r ( y ) − yr ′ ( y ), a = min ≤ y ≤ y / w ( y ), a = µλ ( n − and a = min( a , a ). Then a > a >
0. By Lemma 2.3, r ( y ) ≥ µ ∀ < y < y . (2.19)Suppose there exists y ∈ ( y / , y ) such that w ( y ) < a . Let ( a , b ) be the maximal intervalcontaining y such that w ( y ) < a for any y ∈ ( a , b ). Then a > y / w ( a ) = a and w ( y ) < µ λ ( n − ∀ a < y < b . (2.20)7y (2.1), (2.19), (2.20), and a direct computation, w ′ ( y ) = y (1 + r ′ ( y ) ) + r ′ ( y ) λ w ( y ) − n − r ( y ) ! ∀ < y < y ≥ y (1 + r ′ ( y ) ) λ w ( y ) + λ w ( y ) − n − µ !! ∀ a < y < b ≥ y λ w ( y ) > ∀ a < y < b . Hence w ( y ) > w ( a ) = a ∀ a < y < b and contradiction arises. Hence no such y exists and w ( y ) ≥ a for any y ∈ (0 , y ). Thus(2.18) holds with δ = a . (cid:3) Lemma 2.5.
Let n ≥ , λ > n − , µ > and y > . Suppose r ( y ) ∈ C ([0 , y )) is the solution of (2.1) which satisfies (2.2) . Then there exists a constant M > such that < r ′ ( y ) ≤ M ∀ < y < y (2.21) and µ ≤ r ( y ) ≤ µ + M y ∀ < y < y . (2.22) Proof : By (2.1), (2.2) and Lemma 2.3, r ′′ + r ′ ≤ n − r ≤ n − µ ∀ < y < y . (2.23)Integrating (2.23) over (0 , y ),tan − ( r ′ ( y )) ≤ ( n − y µ ∀ < y < y . (2.24)By Lemma 2.3 and (2.24), (2.21) holds with M = tan ( n − y µ ! . By (2.21) we get (2.22) and the lemma follows. (cid:3)
Lemma 2.6.
Let n ≥ , λ > n − , µ > and y > . Suppose r ( y ) ∈ C ([0 , y )) is the solution of (2.1) which satisfies (2.2) . Then eitherr ′′ ( y ) > ∀ < y < y (2.25) or there exists a constant y ∈ (0 , y ) such that r ′′ ( y ) = and ( r ′′ ( y ) > ∀ < y < y r ′′ ( y ) < ∀ y < y < y (2.26)8 roof : We will use a modification of the proof of Lemma 15 of [DLW] to prove this lemma.By (1.6), r ′′ (0) >
0. Hence there exists a constant δ > r ′′ ( s ) > < s < δ .Let (0 , y ), δ ≤ y ≤ y , be the maximal interval such that r ′′ ( s ) > ∀ < s < y . If y = y , then (2.25) holds. If y < y , then r ′′ ( y ) =
0. By Lemma 2.3 and (2.1), r ′′′ ( y )1 + r ′ ( y ) = r ′ ( y ) r ′′ ( y ) (1 + r ′ ( y ) ) − n − r ( y ) r ′ ( y ) − r ′ ( y ) r ′′ ( y ) λ ( r ( y ) − yr ′ ( y )) − y (1 + r ′ ( y ) ) r ′′ ( y ) λ ( r ( y ) − yr ′ ( y )) ∀ < y < y (2.27) ⇒ r ′′′ ( y )1 + r ′ ( y ) = − ( n − r ′ ( y ) r ( y ) < < δ ′ < y − y such that r ′′ ( y ) < y < y < y + δ ′ .Let ( y , z ) be the maximal interval such that r ′′ ( s ) < ∀ y < s < z . If z < y , then r ′′ ( z ) = r ′′ ( z ) ≥
0. On the other hand by Lemma 2.3 and (2.27), r ′′′ ( z )1 + r ′ ( z ) = − ( n − r ′ ( z ) r ( z ) < z = y and (2.26) follows. (cid:3) We are now ready to prove Theorem 1.3.
Proof of Theorem 1.3 : By Lemma 2.1 there exists a constant y ′ > r ( y ) ∈ C ([0 , y ′ )) which satisfies (1.6) and (2.2) in (0 , y ′ ). Let (0 , y ) bethe maximal interval of existence of solution r ( y ) ∈ C ([0 , y )) of (2.1) which satisfies (2.2)and (1.6). Suppose y < ∞ . By Lemma 2.2, Lemma 2.4 and Lemma 2.5, there existsa constant δ ∈ (0 , y ) such that for any y / < y < y there exists a unique solution r ( y ) ∈ C ([ y , y + δ )) of (2.16) which satisfies (2.17) with r = r ( y ) and r = r ′ ( y ). Let y ∈ (cid:16) y − δ , y (cid:17) and let r ( y ) ∈ C ([ y , y + δ )) be the unique solution of (2.16) given byLemma 2.2 which satisfies (2.17) with r = r ( y ) and r = r ′ ( y ). We then extend r ( y ) toa solution of (1.11) in (0 , y + δ ) by setting r ( y ) = r ( y ) for any y ≤ y < y + δ . Since y + δ > y , this contradicts the maximality of the interval (0 , y ). Hence y = ∞ and thereexists a unique solution r ( y ) ∈ C ([0 , ∞ )) of the equation (1.11) which satisfies (1.12) and(1.6) and the theorem follows. (cid:3) Proof of Theorem 1.2 : We will give a simple proof di ff erent from the sketchy proof of thisresult in [DLW] here. By (i) of Lemma 2.3 (1.7) holds. By Lemma 2.6 either r ′′ ( y ) > ∀ y > y > a : = lim y →∞ r ′ ( y ) exists (2.29)9nd a >
0. We now divide the proof into two cases.
Case 1 : a = ∞ Then there exists y > r ′ ( y ) > p n − λ ∀ y > y (2.30)By (1.11) and (2.30), r ′′ ( y )1 + r ′ ( y ) ≤ r ( y ) n − − + r ′ ( y ) λ ! ∀ y > ≤ r ( y ) n − − + n − λλ ! < ∀ y > y which contradicts (2.28). Hence a , ∞ . Case 2 : a < ∞ By (1.12), 0 < yr ′ ( y ) r ( y ) < ∀ y > . (2.31)Now by (2.29) and the l’Hosiptal rule,lim y →∞ r ( y ) y = lim y →∞ r ′ ( y ) = a ⇒ lim y →∞ yr ′ ( y ) r ( y ) = lim y →∞ r ′ ( y )lim y →∞ r ( y ) / y = . (2.32)By (1.11), (2.29), (2.31), (2.32) and the l’Hosiptal rule,lim y →∞ r ( y ) r ′′ ( y )1 + a = lim y →∞ r ( y ) r ′′ ( y )1 + r ′ ( y ) = n − − + a λ · y →∞ (cid:16) − yr ′ ( y ) r ( y ) (cid:17) = − ∞ which contradicts (2.28). Hence a < ∞ does not hold. Thus by Case 1 and Case 2 (2.28)cannot hold. Hence there exists y > a : = lim y →∞ r ( y ) ∈ ( µ, ∞ ] exists . Since by (1.10) ( r ( y ) − yr ′ ( y )) ′ = − yr ′′ ( y ) > y > y , a : = lim y →∞ ( r ( y ) − yr ′ ( y )) ∈ ( r ( y ) − y r ′ ( y ) , ∞ ] exists . (2.33)Suppose a ∈ ( µ, ∞ ) . (2.34)10hen a = . (2.35)By (1.8), (2.33) and (2.34), a : = lim y →∞ yr ′ ( y ) = a − a ∈ [0 , a − r ( y ) + y r ′ ( y )) exists . Suppose a >
0. Then there exists y > y such that yr ′ ( y ) ≥ a / ∀ y ≥ y ⇒ r ( y ) ≥ r ( y ) + a y / y ) ∀ y ≥ y ⇒ a = ∞ which contradicts (2.34). Hence a = . (2.36)Letting y → ∞ in (1.11), by (2.35) and (2.36),lim y →∞ r ′′ ( y ) = (cid:18) n − − λ (cid:19) a > a = ∞ . Thus (1.9) holds and thetheorem follows. (cid:3) References [DH] P. Daskalopoulos and G. Huisken,
Inverse mean curvature flow of entire graphs ,arXiv:1709.06665v1.[DLW] G. Drugan, H. Lee and G. Wheeler,
Solitons for the inverse mean curvature flow ,Pacific J. Math. 284 (2016), no. 2, 309–326.[G] C. Gerhardt,
Flow of nonconvex hypersurfaces into spheres , J. Di ff erential Geom. 32 (1990),no. 1, 299–314.[H] K.M. Hui, Existence of self-similar solution of the inverse mean curvature flow , DiscreteContin. Dyn. Syst. Series A (to appear), arXiv:1801.08250.[HI1] G. Huisken and T. Ilmanen,
The Riemannian Penrose inequality , Internat. Math. Res.Notices (1997), no. 20, 1045–1058.[HI2] G. Huisken and T. Ilmanen,
The inverse mean curvature flow and the Riemannian Penroseinequality , J. Di ff erential Geom. 59 (2001), no. 3, 353–437.11HI3] G. Huisken and T. Illmanen, Higher regularity of the inverse mean curvature flow , J.Di ff erential Geom. 80 (2008), no. 3, 433–451.[S] K. Smoczyk, Remarks on the inverse mean curvature flow , Asian J. Math. 4 (2000), no. 2,331–335.[U] J. Urbas,