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Long time regularity of solutions of the Hele–Shaw problem
Nonlinear Analysis 64 (2006) 2817 – 2831 www.elsevier.com/locate/na
Long time regularity of solutions of the Hele–Shaw problem Inwon Kim1 Department of Mathematics, UCLA, USA Received 27 March 2005; accepted 20 September 2005
Abstract In this paper we study the long-time behavior of solutions of the one phase Hele–Shaw problem without surface tension. We show that after a finite time solutions of the Hele–Shaw problem become starshaped and Lipschitz continuous in space. Based on this observation we then prove that the free boundary become smooth in space and time with nondegenerate free boundary speed. © 2005 Published by Elsevier Ltd. Keywords: Viscosity solutions; Free boundary; Hele–Show problem; Long time behavior
1. Introduction Let K ={x ∈ Rn : |x|=1} and suppose that a bounded domain contains K and let 0 =−K and 0 = j (see Fig. 1). Note that j0 = 0 ∪ jK. Let u0 be the harmonic function in 0 with u0 = f ≡ 1 > 0 on K and zero on 0 . In addition we suppose u0 satisfies |Du0 | > 0
on 0 .
(I)
The classical Hele–Shaw problem models an incompressible viscous fluid which occupies part of the space between two parallel, narrowly placed plates. In this case u0 denotes the initial pressure of the fluid and f denotes the pressure from jK into Rn − K. As more fluid is injected through a fixed boundary, the region occupied by the fluid will grow as time increases. Assuming
1 Supported by NSF grant DMS 0244991, 0401436.
E-mail address: [email protected]. 0362-546X/$ - see front matter © 2005 Published by Elsevier Ltd. doi:10.1016/j.na.2005.09.021
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Ω0
u0 = 0 u0 > 0
K
Γ0
Fig. 1.
no surface tension, then the pressure of the fluid u(x, t) solves the following problem: −u = 0 in {u > 0} ∩ Q, on j{u > 0} ∩ Q, ut − |Du|2 = 0 u(x, 0) = u0 (x), u(x, t) = 1 for x ∈ jK,
(HS)
where Q = (Rn − K) × (0, ∞). We define (u) = {(x, t) : u(x, t) > 0}, t (u) = {x : u(x, t) > 0}, (u) = j{(x, t) : u(x, t) = 0}, t (u) = j{x : u(x, t) = 0}. We refer to t (u) and t (u), respectively, as the positive phase and the free boundary of u at time t. Note that the second equation in (HS) indicates that the normal velocity of the free boundary is V = ut /|Du| = |Du| if the solution is smooth. The short-time existence of classical solutions when 0 is C 2+ was proved by Escher and Simonett [9]. When n = 2, Elliot and Janovsky [8] showed the existence and uniqueness of weak solutions formulated by a parabolic variational inequality in H 1 (Q). For our investigation we use the notion of viscosity solutions, which have been recently introduced in [10] (also see Section 2). In general one cannot expect the free boundary to be smooth, mostly due to the collision between parts of the free boundary. In fact one cannot even expect the continuity of solution u in time (see [10] for a counterexample). Nevertheless it is observed (Theorem 3.2) that after t > T0 , where T0 is the time when the positive phase overflows the smallest ball where it is initially is contained, the free boundary becomes starshaped and thus there would be no more collision of the free boundaries. Based on this observation and inspired by the work of Caffarelli et al. (see for example [2,3]) which suggests that Lipschitz, nondegenerate free boundaries of general twophase problems are smooth, we then show that after some time the solution u become Lipschitz continuous and the spatial gradient of u is nondegenerate up to the free boundary (Theorem 4.5) Based on these properties, further regularity analysis in [11] yields that u is indeed a classical solution and the free boundary is smooth after t > T0 (Corollary 4.6). We also mention that alltime smoothness results for solutions of free boundary problems has been obtained with proper restrictions on initial data, see for example [5–7]. Our analysis in this paper is motivated by and closely follows [4], where it is proven that the free boundaries of viscosity solutions of the porous medium equation ut − (um ) = 0,
u0,
(PME)
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become starshaped, Lipschitz continuous and nondegenerate. For (HS) our arguments is simpler because our solution is harmonic at each time and therefore the solution is determined solely by the geometry of the free boundary at a fixed time and the fixed boundary data. On the other hand our viscosity solutions are very weak in the sense that they are in general only lower semicontinuous (Definition 2.5), and therefore delicate estimates are necessary in the beginning of our analysis to prove the continuity of u after t > T0 . We remark that for (PME) the uniform convexity of the free boundary for t > T0 in appropriate scale and convergence to the Barenblatt profile up to any order of derivative have been recently proved in [12], and we expect and plan to prove in the future work that similar phenomenon holds for (HS). In Section 2 we recall the definition and some properties of viscosity solutions of (HS). In Section 3 we prove the Lipschitz continuity of the free boundary in space and time. In particular it follows that u is continuous for t > T0 . In Section 4 we use the results from Section 3 and the free boundary motion law V = |Du| to derive the upper and lower bounds for |Du| up to the free boundary of u. Here we also state the further regularity result obtained in [11]. Remarks. 1. If f is nonconstant or if K is not starshaped, then it is not clear to the author whether or not the free boundary will be starshaped even for large times, since different injection rate from different parts of jK or the uneven geometry of K may affect the free boundary movement. If f = f (t) > 0 is a smooth function of time, then one can use the scaling t u(x, ¯ t) := a(t)u x, a 2 (s) ds ,
a(t) = 1/f (t),
0
to obtain corresponding results. Our arguments and results in this paper also extends to general starshaped K. 2. Condition (I) is to guarantee the lower bound of |Du| on 0 (u). Recently it is proved in ¯ (u) with any positive > 0 when the initial [5] that a lower bound of |Du| can be obtained in positive phase is Lipschitz with small Lipschitz constant. On the other hand we will show later (Theorem 3.2) that for any initial data u0 the positive phase of u become Lipschitz in space with small Lipschitz constant after a finite amount of time T0 . Therefore, it follows that our main result on the large time behavior of u (Theorem 4.5) indeed holds for general initial data: we just need to start the clock at t = T0 + instead of t = 0 by considering u(x, ˜ t) := u(x, t + T0 + ) instead of u(x, t) and by previous arguments we will have (I) for u(x, ˜ 0).
2. Preliminaries For vectors , ∈ Rn we define cos(, ) =
· . ||||
Let us denote by W (, ; x0 ) the cone in Rn with angle , axis in the direction of a unit vector in Rn and vertex at x0 , i.e., W (, ; x0 ) = {y ∈ Rn : cos(y − x0 , ) cos },
0 < .
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Let us recall the notion of viscosity solutions of (HS) defined in [10]. Roughly speaking, viscosity sub and supersolutions are defined by comparison with local (smooth) super and subsolutions. In particular classical solutions of (HS) are also viscosity sub and supersolutions of (HS). Recall that Q = (Rn − K) × (0, ∞) and let be a cylindrical domain D × (a, b) ⊂ Rn × R, where D is an open subset of Rn . Definition 2.1. A nonnegative upper semicontinuous function u defined in is a viscosity subsolution of (HS) if (a) for each a < T < b the set (u) ∩ {t T } is bounded; and (b) for every ∈ C 2,1 ( ) such that u − has a local maximum in (u) ∩ {t t0 } ∩ at (x0 , t0 ), (i) − (x0 , t0 )0 if u(x0 , t0 ) > 0. (ii) ( t − |D |2 )(x0 , t0 ) 0 if (x0 , t0 ) ∈ (u) if − (x0 , t0 ) > 0. Note that because u is only upper lowercontinuous there may be points of (u) at which u is positive. Definition 2.2. A nonnegative lower semicontinuous function v defined in is a viscosity supersolution of (HS) if for every ∈ C 2,1 ( ) such that v − has a local minimum in ∩ {t t0 } at (x0 , t0 ), (i) − (x0 , t0 )0 if v(x0 , t0 ) > 0, (ii) If (z0 , t0 ) ∈ (v), |D |(x0 , t0 ) = 0 and − (x0 , t0 ) < 0, then ( t − |D |2 )(x0 , t0 ) 0. Definition 2.3. u is a viscosity subsolution of (HS) with initial data u0 and fixed boundary data f > 0 if ¯ (a) u is a viscosity subsolution in Q, (b) u = u0 at t = 0; u f on jK; (c) (u) ∩ {t = 0} = (u0 ); Definition 2.4. v is a viscosity supersolution of (HS) with initial data v0 and fixed boundary data ¯ with v = v0 at t = 0 and v f on jK. f if v is a viscosity supersolution in Q For a nonnegative real valued function u(x, t) defined in a cylindrical domain D × (a, b), u∗ (x, t) =
lim sup
u( , s).
( ,s)∈D×(a,b)→(x,t)
Note that the lim sup permits times in the future of t, s > t. Definition 2.5. u is a viscosity solution of (HS) (with boundary data u0 and f ) if u is a viscosity supersolution and u∗ is a viscosity subsolution of (HS) (with boundary data u0 and f ).
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Definition 2.6. We say that a pair of functions u0 , v0 : D¯ → [0, ∞) are (strictly) separated (denoted by u0 ≺ v0 ) in D ⊂ Rn if (i) the support of u0 , supp(u0 ) = {u0 > 0} restricted in D¯ is compact and (ii) in supp(u0 ) ∩ D¯ the functions are strictly ordered: u0 (x) < v0 (x). The following properties of viscosity solutions, proven in [10], are frequently used in our paper. Theorem 2.7 (Comparison principle). Let u, v be, respectively, viscosity sub- and supersolu¯ tions in D × (0, T ) ⊂ Q with initial data u0 ≺ v0 in D. If u v on jD and u < v on jD ∩ (u) for 0 t < T , then u(·, t) ≺ v(·, t) in D for t ∈ [0, T ). Theorem 2.8. There exists a unique viscosity solution v(x, t) of (HS) in Q with initial data u0 with condition (I). Moreover the positive set t (v) strictly expands in time. More precisely, s (v) is a compact subset of t (v) for 0 s < t. For the rest of our paper we denote u(x, t) as the unique viscosity solution of (HS) with fixed boundary data 1 and the initial data u0 satisfying (I). Lemma 2.9. (i) u(x, t) → u0 (x) as t → 0+ . Moreover u∗ (x, t) = u(x, t) = 1 for x ∈ jK and u(x, t) = 0 for (x, t) ∈ (u). (ii) u is superharmonic in (u) and u∗ is subharmonic in Q. ¯ ¯ ∗ ) and (u) = (u∗ ). (iii) (u) = (u (iv) u is harmonic in (u). Indeed u(x, t) = ht (x) at each time t > 0, where ht (x) = inf{(x) : −0 in t (u); = 1 on jK; 0 in t (u)}. Proof. 1. (i)–(ii) follows from the definition of u. To prove (iii), observe that for any > 0 the function u(x, ˜ t) := (1 + )u(x, t + ) is a supersolution of (HS) in Q with u∗ ≺ u˜ by the last statement of Theorem 2.8. Hence we apply Theorem 2.7 to u∗ and u˜ to obtain u∗ (x, t)(1 + )u(x, t + ) for any , > 0
(2.1)
In particular u(x, t)u∗ (x, t) u(x, t + ) for any > 0, and (iii) follows. 2. To prove (iv), Let h(x, t) = ht (x) be defined as in (iv) for t 0. It can be checked that ht ¯ t (u) with ht = 1 on jK (see for example [13, Chapter 1.3]). is nonnegative and harmonic in Since u(·, t) is superharmonic in t (u) with u = 1 on jK, by definition h(x, t) u(x, t). On the other hand by (2.1) u∗ (x, t) u(x, t + ) and in particular u∗ (·, t − ) = 0 on t (u) for t > . Moreover u∗ (·, t − ) is subharmonic in (u) with boundary data 1 on jK. Since ht is harmonic and nonnegative in t (u), by maximum principle for harmonic functions we obtain u∗ (x, t − )h(x, t) for any small > 0. Now we obtain u(x, t) lim u∗ (x, t − ) h(x, t) →0
for t > 0,
where the first inequality is due to the lower semicontinuity of u in time. Hence uhu, i.e., u = h for t 0.
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3. Lipschitz continuity of the free boundary For x ∈ Rn , we denote Br (x) := {y ∈ Rn : |x − y| < r}. Since 0 is bounded, we may assume that 0 ⊂ BR0 for some R0 > 0. In this section we would like to show that the free boundary (u) becomes Lipschitz continuous in space and time once it is out of BR0 . First we state the following lemma, which can be proven by a reflection argument and by Theorem 2.7. For detailed proof we refer to [1, Proposition 2.1, Lemma 2.2], where a parallel result is proven for a solution u of the porous medium equation. ¯ Lemma 3.1. For |x0 |, |x1 | > R0 with (x1 , t) ∈ (u) ∩ Q the following holds: If cos(x1 − x0 , x0 ) R0 /|x0 | then u∗ (x1 , t) < u(x0 , t + ) for > 0. Let us define T0 = t (u0 ) = inf{t > 0 : t (u) ⊃ B¯ R0 } > 0.
(3.1)
By comparing u with a radially symmetric solution of (HS) with initial data less than u0 , one can easily show that T0 < ∞. Theorem 3.2. For T0 < t t t¯ < ∞ the free boundary (u) is representable in spherical coordinates of the form r = g(, t), where g is Lipschitz continuous in and t. Moreover g(·, t) is uniformly Lipschitz continuous with Lipschitz constant L = L(t) in for t t with L → 0 as t → ∞. Proof. 1. First we show that t (u) is starshaped and Lipschitz continuous at each time T0 < t < t. For T0 < T , it follows from Lemma 3.1 that x ∗ ,T + for x ∈ T (u∗ ) u (x, T ) < u 1+ for any > 0 and small 0 < < 0 , where 0 depends on T . Also u∗ (x, T ) < 1 = u(x/(1 + ), T + )
on {x : |x| = 1 + }.
Then the maximum principle for harmonic functions yields that u∗ (x, T ) ≺ u(x/(1 + ), T + ) 2. Since
in |x|1 + .
t −T x , +T + w(x, t) = u 1+ 1+
is a viscosity supersolution of (HS) in {x : |x|1+}×[T , ∞) with w =1 u on {x : |x|=1+}, we have u∗ (x, t) ≺ w(x, t) for {x : |x|1 + } ∩ {t T }.
(3.2)
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For given > 0 and > 0, we can choose = /(1 + ) such that (t − T )/(1 + ) + T + = t at t = T + and thus due to (3.2) it follows u∗ (x, T + ) ≺ u(x/(1 + ), T + )
for any , > 0.
(3.3)
In particular (3.3) implies that for t > T0 our solution u(·, t) is decreasing in radial directions and thus t (u) is representable by spherical coordinates r = g(, t) = g(, t) > R0 . 3. Due to Lemma 3.1 for any direction p ∈ Rn such that cos(p, x)R0 /|x| we have u∗ (x + p, t) < u(x, t + ) for , > 0 ¯ if |x + p|, |x| > R0 and (x + p, t) ∈ (u). Hence for x ∈ t (u), T0 < t t there exists 0 > 0 such that for 0 < 0 and > 0 u∗ (x + p, t − ) < u(x, t)
if (x + p, t − ) ∈ (u).
(3.4)
This implies that u(x + p, t − ) = 0 for above , p, given as in (3.4) and thus u(x + p, t) = 0 by the lower semicontinuity of u. In other words at each point x ∈ t (u) there is 0 = (|x|) = cos−1 (R0 /|x|) such that u(x, t) = 0
for y ∈ W (0 , x/|x|; x) ∩ B0 (x).
On the other hand by similar argument it can be easily seen that u(z, t) > 0
for z ∈ W (0 , −x/|x|; x) ∩ B0 (x).
This leads to the Lipschitz continuity g with respect to . Observe that the Lipschitz constant of g(·, t) only depends on 0 , which can be chosen as 0 = (t) = cos−1 [R0 /r(t)], where r(t) = inf{|x| : x ∈ t (u), t = t} > R0 . Since r(t) → ∞ as t → ∞, it follows that 0 → /2 as t → ∞. 4. It remains to show that g is Lipschitz locally in time. Due to (3.2), if (x0 , t0 ) ∈ (u) and t0 > T0 then u∗ ((1 + )x0 , (1 + )(t0 − T0 ) + T0 ) u(x0 , t0 ) = 0 for > 0 and therefore u((1 + )x0 , (1 + )(t0 − T0 ) + T0 ) = 0. If we set t = (1 + )(t0 − T0 ) + T0 and = x0 /|x0 |, above estimates imply g(, t0 ) = x0 and g(, t) (1 + )|x0 |. This and the fact that the u is increasing in time, we obtain |g(, t) − g(, t0 )| g(, t0 ) |t − t0 | t0 − T 0
if T0 < t0 < t.
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Due to the Lipschitz continuity of the free boundary, we are now able to show the following properties of u: Proposition 3.3. (i) u(x, t) is continuous for t > T0 : (ii) If |x0 |, |x1 | > R0 and cos(x1 − x0 , x0 ) R0 /|x0 | then u(x1 , t)u(x0 , t)
for > 0.
(iii) for T > T0 , 0 < < 0 (R0 ) and |x|R0 u(x, t)u(x/(1 + ), (t − T )/(1 + ) + T )
for t > T .
(3.5)
Proof. 1. (ii) follows from (i) and Lemma 3.1. 2. By Lemma 2.9(iv), at each t > 0 u(x, t) = inf{(x) : −0 in t (u); 1 on jK; 0 on t (u)}. For t > T0 , since t (u) is Lipschitz continuous, Perron’s method (we refer to Chapter 1.3 of [13]) yields that u(·, t) is continuous in t (u) and harmonic in t (u) with u = 1 on jK and u = 0 on t (u). 3. To prove continuity of u in time, let us pick (x, t0 ) ∈ (u) such that t0 > T0 . Due to (2.1) u∗ (x, t0 − )u(x, t0 ) for any small > 0. On the other hand from (3.3) we have u(x, ˜ t) := u∗ ((1 + )x, t0 − )u(x, t0 − ) for any > 0 and 0 < < t0 − T0 . Since u˜ is a viscosity subsolution of (HS), we can apply Theorem 2.7 to u˜ and u in Rn − K × [t0 − , ∞) and obtain u∗ ((1 + )x, (1 + )(t − t0 + ) + t0 − ) u(x, t)
for t t0 − .
Now evaluating above inequality at t = t0 , we obtain u∗ (x , t0 + )u((1 + )−1 x , t0 ), where x = (1 + )x. We let = to conclude that for any (x, t) ∈ Rn − K × (t0 , ∞) lim u(x, t + 2 ) lim u((1 + )−1 x, t),
→0
→0
where the second term equals u(x, t) due to the spatial continuity of u. Thus u(x, t)=lim→0 u(x, t + ). On the other hand, since u is lower semicontinuous and increasing in t, we have u(x, t) = lim u(x, t − ). →0
Thus u is continuous in time and now (3.5) follows from (3.3).
Lemma 3.4. There exist A, B, 0 > 0 depending on u0 such that u (x, t) = for 0 < < 0 .
1 + A (1 + )2
u((1 + )x, (1 + A)t + B) u(x, t)
(3.6)
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Proof. 1. It is easy to check that u also solves (HS). Therefore to apply Theorem 2.7 to u and u in Q and to obtain (3.6), we only have to show that for small > 0 there is A, B > 0 such that u0 (x) ≺ u (x, 0) =
1 + A (1 + )2
u((1 + )x, B)
and 1 u (·, t) for x ∈ jK. 2. After a dilation if necessary, we may assume that 0 contains B4 (0). We set c = inf{|Du0 (x)|/4|x| : x ∈ 0 } (note that c > 0 due to condition (I)). Then w(x, t) = u0 ((1 + ct)−1 x) is a viscosity subsolution of (HS) in the domain B1+ct (0) × [0, 1/c], since w(·, t) is harmonic in t (w) and wt c
|x| |Du0 |(1 + ct)−2 |Du0 |2 = |Dw|2 (1 + ct)
on t (w) × [0, 1/c],
where the second inequality holds due to the definition of c. Since w(x, t) is a subsolution of (HS) in B1+ct (0) × [0, 1/c], for any c2 > 0, w2 (x, t) = c2 w(x, c2 t) is also a subsolution of (HS) in B1+ct (0) × [0, 1/(c2 c)]. Now if we choose c2 such that c2 < u0 (x) on |x| = 2, then w2 u on |x| = 2 for 0 t C = 1/(cc2 ). By Theorem 2.7 then w2 u in the domain {x : |x|2} × (0, C). In particular, if x ∈ 0 and |x| 2 then (1 + C −1 t)x ∈ t (u)
for 0 t C.
(3.7)
3. On the other hand if |x|2 then (1 + C −1 t)|x| ∈ B4 (0) ⊂ 0 ⊂ t (u)
for 0 t C.
(3.8)
In terms of u, (3.7) and (3.8) yields that if we let B = C then u(x, 0) > 0 implies that u((1 + )x, B) > 0 for small = C −1 t > 0, in other words ¯ 0 ⊂ {u (x, 0) > 0}.
(3.9)
Finally let us choose A big enough such that for small > 0 1
1 + A (1 + )2
u0 ((1 + )x)
on jK.
(3.10)
(Such A can be chosen, for example, as the maximum of 2|Du0 | on B2 (0) − K.) With above choice of A and B, (3.9), (3.10) and the maximum principle for harmonic functions yields that u0 (x) ≺ u (x, 0) and 1 u for x ∈ jK. Remark. Heuristically speaking, differentiating (3.5) and (3.6) with respect to at = 0 yields that for t > T0 x · (x, t) ut x · (x, t) At + B |Du| t − T0
on (u),
(3.11)
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where = −Du/|Du|. Since formally the free boundary motion law is given as V = |Du| on (u), (3.11) is expected to yield bounds for |Du| on (u) for t > T0 . In the next section we use barrier arguments based on the geometry (Lipschitz continuity) of t (u) to obtain the expected bounds on |Du|. 4. Gradient estimates First we look for the lower bound of |Du| on (u). We start with the following lemma. Using barrier argument we will show that if |Du| is very small near the free boundary then the positive phase expands too slowly, violating (3.6). Lemma 4.1. Let (x0 , t0 ) ∈ (u) and T0 < t t0 < t¯ < ∞. Then there exist positive constants h0 , C1 , C2 depending only on t, t¯ such that sup{u(x, t) : |x − x0 | < h, 0 t − t0 C2 h} > C1 h
for 0 < h < h0 .
Proof. 1. Let x1 = (1 + h/|x0 |)x0 . By Theorem 2.7 for small h > 0 there is L > 1 depending only on the Lipschitz constant of g(·, t0 ) such that u(·, t0 ) = 0 in the ball Bh/L (x1 ). We will construct a barrier based on this ball (see Fig. 2). 2. Consider (x) such that
− = 0 =0 = C1 h
in := B2h (x1 ) − Bh/L (x1 ), on jBh/L (x1 ), on jBh (x1 ),
where the constant C1 > 0 is to be determined. Notice that (x) = (r), where r = |x − x1 | and |r | = C > 0 on jDh/L (x1 ) where C = kC 1 where k = k(n, L). Consider v(x, t) = (1 + Ct)
−1
L 1 + 8C t r h
Γt
0
h h X0
Ωt
θ
0
Fig. 2.
L X1
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in the domain S = 2 × (0, C2 h],
C2 = 1/(8CL)
and 2 = Bh (x1 ) − Dh/L (x1 ).
Observe that on t () we have r(1 + 8C (L/ h)t) = h/L and vt =
8CLr/ h r , 1 + Ct
|Dv| =
(1 + 8CLt/ h) h r = r . 1 + Ct Lr(1 + Ct)
Therefore on t () vt − |Dv|2 C2 (8 − (h/Lr)3 ) 0 since r = h/L(1 + 8C(L/ h)t) h/2L if 0 t < h/8CL. Since v(·, t) is harmonic in 2 , it follows that v is a supersolution of (HS) in S. 3. Now if we assume that u(x, t) C1 h in jDh (x1 ) × [0, C2 h], then u v on the parabolic boundary of S and thus u v in S by Theorem 2.7. Therefore the point (x1 , t0 + h/8CL) belongs to the interior of the set {u = 0}. But this contradicts Lemma 3.4 if we take C1 such that |x1 − x0 | = |8CL| , At 0 + B |t1 − t0 | where A, B is given as in Lemma 3.4 and = (t) = inf{|x| : x ∈ t (u)} > R0 .
Lemma 4.2. Let g given as in Theorem 3.2. If T0 < t < t1 < t¯ < ∞ and g(x1 /|x1 |, t1 ) − |x1 | = h with 0 < h (|x1 | − R0 )/2 then we have u(x1 , t1 )C3 h, where C3 only depend on the initial data u0 and t, t¯. Proof. 1. We recall that for T > T0 and for any > 0 u(x, t)w(x, t) = u(x/(1 + ), (t − T )/(1 + ) + T ). Let us take T = T0 + (t1 − T0 )/2. Then if u(x1 , t1 )ch
then u((1 + )x1 , (1 + )(t1 − T ) + T ) ch.
(4.1)
2. Now suppose that g(x1 /|x1 |, t1 )−|x1 |=M1 h, where M1 > 0 is to be decided. Let =x1 /|x1 | and x0 = g(, t1 ) ∈ t1 (u) . From Lemma 4.1 sup
|x−x0 | h,0 t−t1 C1 h
u(x, t) > C2 h,
(4.2)
where Ci = Ci (t, t¯). We recall that, due to Lemma 3.1 and continuity of u, for t > t and for |x| > R0 there is an angle t > 0 with which u(y, t)u(x, t)
if y ∈ W (t , x/|x|, x).
(4.3)
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(1 + β) X1
h
X0
Mh X1 Fig. 3.
Choose M2 = M2 (t) > 0 such that if we take M1 = M2 then Dh (x0 ) belongs to W (t , ; x1 ). We choose M1 a bit larger than M2 , that is, we set M1 = M2 +
C1 sup{|x| : x ∈ t¯(u)} t − T0
so that Dh (x0 ) belongs to the cone W (t , ; (1 + )x1 ) for 0 , where = C1 /(t1 − T0 )h (see Fig. 3). 3. Due to (4.1) if u(x1 , t1 ) < C2 h then u((1 + )x1 , t1 + (t1 − T )) < u(x1 , t1 ) < C2 h for h small and 0 . Due to the choice of M1 and (4.3) it follows that u(x, t) < C2 h in the cylinder Bh (x0 ) × [t1 , t1 + C1 h]. This contradicts (4.2) and thus we can conclude by taking C3 = C2 /M1 . Next we proceed to obtain an upper bound for |Du| near (u). As before the main idea is that if |Du| is very big near the free boundary then the positive phase expands too fast in time, violating (3.5). Lemma 4.3. Let g(x/| ¯ x|, ¯ t0 ) − |x| ¯ = h with 0 h (|x| ¯ − R0 )/2 and T0 < t t0 t¯. Then there are constants L depending on t, n and M = M(L, t, t¯, u0 , n) such that ¯ < h/L}Mh. inf{u(x, t0 ) : |x − x| Proof. 1. Let x0 = x¯ + h ∈ t0 (u), where = x/| ¯ x|. ¯ Due to the Lipschitz continuity of (u) in space, there is a constant L = L(t) > 0 such that D2h/L (x) ¯ ∈ t0 . Consider a harmonic function ¯ − Dh/L (x) ¯ such that (x) in D2h/L (x) ¯ (x) = 0 on jD2h/L (x), (x) = Mh on jDh/L (x), ¯ ¯ where M3 > 0 where M > 0 is a constant chosen so that |D| = M3 (1 + M3 )/L on jD2h/L (x), is to be decided. Note that (x) = (r), r = |x − x| ¯ and M only depends on M3 and n. ¯ by zero and let us define Let us extend outside D2h/L (x) −1 h h M3 r− (t − t0 ) + . v(x, t) = 1+ h L L
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in the domain S =(Rn −Dh/L (x))×[t ¯ 0 , t0 +h]. Then for t0 t t0 +h, r −h/L (1+M3 )(h/L) on t (v) and thus v satisfies h M3 / h |r |2 2 vt − |Dv| = | r − | − r L (1 + (M3 / h)(t − t0 ))2 (1 + (M3 / h)(t − t0 ))2 h M3 r− − M3 (1 + M3 )/L 0 C h L for (x, t) ∈ t (v) ∩ {t0 t t0 + h}. Hence v is a supersolution of (HS) in S. 2. Now suppose that Mh < u(·, t0 ) on Dh/L (x). ¯ Then v(x, t0 ) = (x) ≺ u(x, t0 ) in Rn − ¯ × [t0 , t0 + h]. Hence by Dh/L (x). ¯ Moreover, since u increases in time, v = Mh < u on jDh/L (x) applying Theorem 2.7 to v and u in the domain S, it follows that v ≺ u in S. In particular h h ¯ ⊂ (u). x¯ + (2 + M3 ) , t0 + h = x0 + (2 + M3 − L) , t0 + h ∈ (v) L L (4.4) 3. On the other hand, due to (3.5) for any > 0 we have u((1 + s)x0 , t0 + (t0 − T0 )) = 0
if s.
(4.5)
Now (4.4) and (4.5) leads to a contradiction if we choose =h/(t0 −T0 ), s=(2+M3 −L)h/L|x0 | and M3 = L − 2 + L/(t − T0 ), where = (t¯) = sup{|x| : x ∈ t¯(u)}.
Lemma 4.4. For T0 < t t1 t¯ there is h0 = h(t) > 0 satisfying the following: if g(x1 /|x1 |, t1 ) − |x1 | = h with 0 < h < h0 then we have u(x1 , t1 )M h, for M = M (t, t¯, u0 , n) > 0. Proof. 1. Choose h0 = h0 (t) > 0 such that |x| > R0 + 3h0 on t (u). Let 0 < h < h0 , = x1 /|x1 | and let x0 = x1 + h ∈ t1 (u). Let t as given in (4.3). We choose L = L(t) > 2 such that D2h/L (x2 ) ∈ W (t , −; x1 )
where x2 = x1 − h
(see Fig. 4). By the choice of h0 , D2h/L (x2 ) ⊂ {|x| > R0 }. Observe that Lemma 4.3 holds for x¯ = x2 with this choice of L. 2. Assume u(x1 , t1 ) > 2M h, where M is the constant M obtained as in Lemma 4.3 with the choice of L given in step 1. Due to the monotonicity of u along the cone W (t , −, x1 ), we have u(·, t1 )u(x1 , t1 ) > M (2h) in D2h/L (x2 ). This contradicts Lemma 4.3 since g(x2 /|x2 |, t1 ) − |x2 | = 2h. Now we are ready to state our main theorem: Theorem 4.5. Let u be a viscosity solution of (HS) with initial data u0 and fixed boundary data 1 on K. Then (u) is Lipschitz continuous in space and time for t > T0 , where T0 is given
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Ωt
1
2h 2h
L
X2
K2
X1
X0
Γt
1
Fig. 4.
in (3.1). Moreover there is a neighborhood N of (u) in each strip of the form T0 < t t t¯ and constants C1 , C2 > 0 depending only on t, t¯, u0 and n such that C1
ut , |Du|
|Du(x, t)| C2 ,
if (x, t) ∈ (u) ∈ N. Proof. Let us pick t ∈ (t, t¯). We only have to obtain the bounds for |Du|. From the Lipschitz continuity of t (u) (Theorem 3.2) and properties of harmonic functions in a Lipschitz domain (see [3, Lemma 4]), for t t > T0 there are positive constants c1 , c2 depending only on t such that if h is small enough then 0 < c1 |Du|(x − h, t)
u(x − h, t) c2 |Du|(x − h, t), h
where x ∈ t (u) and = x/|x|. Now Lemma 4.3 and 4.4 leads to the conclusion.
Based on above theorem the following result is proved in [11, Section 6]: Corollary 4.6. Let u0 , u be given as in Theorem 4.5. Then after t > T0 u is a classical solution. More precisely u and (u) after t > T are analytic in space and time and u satisfies the free boundary condition V = |Du| in the classical sense. References [1] D.G. Aronson, L.A. Caffarelli, The initial trace of a solution of the porous medium equation, Trans. Am. Math. Soc. 280 (1) (1983) 351–366. [2] I. Athanasopoulos, L. Caffarelli, S. Salsa, Regularity of the free boundary in parabolic phase-transition problems, Acta Math. 176 (1996) 245–282. [3] L. Caffarelli, A Harnack inequality approach to the regularity of free boundaries, Part I: Lipschitz free boundaries are C 1, , Rev. Mat. Iberoam. 3 (2) (1987) 139–162. [4] L.A. Caffarelli, J.L. Vazquez, N.I. Wolanski, Lipschitz continuity of solutions and interfaces of the N-dimensional porous medium equation, Indiana Univ. Math. J. 36 (1987) 373–401. [5] S. Choi, D. Jerison, I. Kim, The regularity and speed of the Hele–Shaw flow, submitted.
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[6] P. Daskalopoulos, R. Hamilton, K. Lee, All time C ∞ -regularity of interface in degenerated diffusion: a geometric approach, Duke Math. J. 108 (2001) 295–327. [7] P. Daskalopoulos, K. Lee, All time smooth solutions of the one-phase Stefan problem and the Hele–Shaw flow, Commun. Partial Differential Equations 12 (2004) 71–89. [8] C.M. Elliot, V. Janovsky, A variational inequality approach to Hele–Shaw flow with a moving boundary, Proc. R. Soc. Edinburgh, Sect. A 88 (1–2) (1981) 93–107. [9] J. Escher, G. Simonett, Classical solutions of multidimensional Hele–Shaw models, SIAM J. Math. Anal. 28 (5) (1997) 1028–1047. [10] I. Kim, Uniqueness and existence result of Hele–Shaw and Stefan problem, Arch. Rat. Mech. Anal. 168 (2003) 299–328. [11] I. Kim, Regularity of the free boundary for the one phase Hele–Shaw problem, J. Differential Equations, to appear. [12] K. Lee, J.L. Vazquez, Geometrical properties of solutions of the porous medium equation for large times, Indiana Math. J. 52 (2003) 991–1016. [13] M. Tsuji, Potential theory in modern function theory, Maruzen Co., LTD., Tokyo, 1959.
Long time regularity of solutions of the Hele–Shaw problem Inwon Kim1 Department of Mathematics, UCLA, USA Received 27 March 2005; accepted 20 September 2005
Abstract In this paper we study the long-time behavior of solutions of the one phase Hele–Shaw problem without surface tension. We show that after a finite time solutions of the Hele–Shaw problem become starshaped and Lipschitz continuous in space. Based on this observation we then prove that the free boundary become smooth in space and time with nondegenerate free boundary speed. © 2005 Published by Elsevier Ltd. Keywords: Viscosity solutions; Free boundary; Hele–Show problem; Long time behavior
1. Introduction Let K ={x ∈ Rn : |x|=1} and suppose that a bounded domain contains K and let 0 =−K and 0 = j (see Fig. 1). Note that j0 = 0 ∪ jK. Let u0 be the harmonic function in 0 with u0 = f ≡ 1 > 0 on K and zero on 0 . In addition we suppose u0 satisfies |Du0 | > 0
on 0 .
(I)
The classical Hele–Shaw problem models an incompressible viscous fluid which occupies part of the space between two parallel, narrowly placed plates. In this case u0 denotes the initial pressure of the fluid and f denotes the pressure from jK into Rn − K. As more fluid is injected through a fixed boundary, the region occupied by the fluid will grow as time increases. Assuming
1 Supported by NSF grant DMS 0244991, 0401436.
E-mail address: [email protected]. 0362-546X/$ - see front matter © 2005 Published by Elsevier Ltd. doi:10.1016/j.na.2005.09.021
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Ω0
u0 = 0 u0 > 0
K
Γ0
Fig. 1.
no surface tension, then the pressure of the fluid u(x, t) solves the following problem: −u = 0 in {u > 0} ∩ Q, on j{u > 0} ∩ Q, ut − |Du|2 = 0 u(x, 0) = u0 (x), u(x, t) = 1 for x ∈ jK,
(HS)
where Q = (Rn − K) × (0, ∞). We define (u) = {(x, t) : u(x, t) > 0}, t (u) = {x : u(x, t) > 0}, (u) = j{(x, t) : u(x, t) = 0}, t (u) = j{x : u(x, t) = 0}. We refer to t (u) and t (u), respectively, as the positive phase and the free boundary of u at time t. Note that the second equation in (HS) indicates that the normal velocity of the free boundary is V = ut /|Du| = |Du| if the solution is smooth. The short-time existence of classical solutions when 0 is C 2+ was proved by Escher and Simonett [9]. When n = 2, Elliot and Janovsky [8] showed the existence and uniqueness of weak solutions formulated by a parabolic variational inequality in H 1 (Q). For our investigation we use the notion of viscosity solutions, which have been recently introduced in [10] (also see Section 2). In general one cannot expect the free boundary to be smooth, mostly due to the collision between parts of the free boundary. In fact one cannot even expect the continuity of solution u in time (see [10] for a counterexample). Nevertheless it is observed (Theorem 3.2) that after t > T0 , where T0 is the time when the positive phase overflows the smallest ball where it is initially is contained, the free boundary becomes starshaped and thus there would be no more collision of the free boundaries. Based on this observation and inspired by the work of Caffarelli et al. (see for example [2,3]) which suggests that Lipschitz, nondegenerate free boundaries of general twophase problems are smooth, we then show that after some time the solution u become Lipschitz continuous and the spatial gradient of u is nondegenerate up to the free boundary (Theorem 4.5) Based on these properties, further regularity analysis in [11] yields that u is indeed a classical solution and the free boundary is smooth after t > T0 (Corollary 4.6). We also mention that alltime smoothness results for solutions of free boundary problems has been obtained with proper restrictions on initial data, see for example [5–7]. Our analysis in this paper is motivated by and closely follows [4], where it is proven that the free boundaries of viscosity solutions of the porous medium equation ut − (um ) = 0,
u0,
(PME)
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become starshaped, Lipschitz continuous and nondegenerate. For (HS) our arguments is simpler because our solution is harmonic at each time and therefore the solution is determined solely by the geometry of the free boundary at a fixed time and the fixed boundary data. On the other hand our viscosity solutions are very weak in the sense that they are in general only lower semicontinuous (Definition 2.5), and therefore delicate estimates are necessary in the beginning of our analysis to prove the continuity of u after t > T0 . We remark that for (PME) the uniform convexity of the free boundary for t > T0 in appropriate scale and convergence to the Barenblatt profile up to any order of derivative have been recently proved in [12], and we expect and plan to prove in the future work that similar phenomenon holds for (HS). In Section 2 we recall the definition and some properties of viscosity solutions of (HS). In Section 3 we prove the Lipschitz continuity of the free boundary in space and time. In particular it follows that u is continuous for t > T0 . In Section 4 we use the results from Section 3 and the free boundary motion law V = |Du| to derive the upper and lower bounds for |Du| up to the free boundary of u. Here we also state the further regularity result obtained in [11]. Remarks. 1. If f is nonconstant or if K is not starshaped, then it is not clear to the author whether or not the free boundary will be starshaped even for large times, since different injection rate from different parts of jK or the uneven geometry of K may affect the free boundary movement. If f = f (t) > 0 is a smooth function of time, then one can use the scaling t u(x, ¯ t) := a(t)u x, a 2 (s) ds ,
a(t) = 1/f (t),
0
to obtain corresponding results. Our arguments and results in this paper also extends to general starshaped K. 2. Condition (I) is to guarantee the lower bound of |Du| on 0 (u). Recently it is proved in ¯ (u) with any positive > 0 when the initial [5] that a lower bound of |Du| can be obtained in positive phase is Lipschitz with small Lipschitz constant. On the other hand we will show later (Theorem 3.2) that for any initial data u0 the positive phase of u become Lipschitz in space with small Lipschitz constant after a finite amount of time T0 . Therefore, it follows that our main result on the large time behavior of u (Theorem 4.5) indeed holds for general initial data: we just need to start the clock at t = T0 + instead of t = 0 by considering u(x, ˜ t) := u(x, t + T0 + ) instead of u(x, t) and by previous arguments we will have (I) for u(x, ˜ 0).
2. Preliminaries For vectors , ∈ Rn we define cos(, ) =
· . ||||
Let us denote by W (, ; x0 ) the cone in Rn with angle , axis in the direction of a unit vector in Rn and vertex at x0 , i.e., W (, ; x0 ) = {y ∈ Rn : cos(y − x0 , ) cos },
0 < .
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Let us recall the notion of viscosity solutions of (HS) defined in [10]. Roughly speaking, viscosity sub and supersolutions are defined by comparison with local (smooth) super and subsolutions. In particular classical solutions of (HS) are also viscosity sub and supersolutions of (HS). Recall that Q = (Rn − K) × (0, ∞) and let be a cylindrical domain D × (a, b) ⊂ Rn × R, where D is an open subset of Rn . Definition 2.1. A nonnegative upper semicontinuous function u defined in is a viscosity subsolution of (HS) if (a) for each a < T < b the set (u) ∩ {t T } is bounded; and (b) for every ∈ C 2,1 ( ) such that u − has a local maximum in (u) ∩ {t t0 } ∩ at (x0 , t0 ), (i) − (x0 , t0 )0 if u(x0 , t0 ) > 0. (ii) ( t − |D |2 )(x0 , t0 ) 0 if (x0 , t0 ) ∈ (u) if − (x0 , t0 ) > 0. Note that because u is only upper lowercontinuous there may be points of (u) at which u is positive. Definition 2.2. A nonnegative lower semicontinuous function v defined in is a viscosity supersolution of (HS) if for every ∈ C 2,1 ( ) such that v − has a local minimum in ∩ {t t0 } at (x0 , t0 ), (i) − (x0 , t0 )0 if v(x0 , t0 ) > 0, (ii) If (z0 , t0 ) ∈ (v), |D |(x0 , t0 ) = 0 and − (x0 , t0 ) < 0, then ( t − |D |2 )(x0 , t0 ) 0. Definition 2.3. u is a viscosity subsolution of (HS) with initial data u0 and fixed boundary data f > 0 if ¯ (a) u is a viscosity subsolution in Q, (b) u = u0 at t = 0; u f on jK; (c) (u) ∩ {t = 0} = (u0 ); Definition 2.4. v is a viscosity supersolution of (HS) with initial data v0 and fixed boundary data ¯ with v = v0 at t = 0 and v f on jK. f if v is a viscosity supersolution in Q For a nonnegative real valued function u(x, t) defined in a cylindrical domain D × (a, b), u∗ (x, t) =
lim sup
u( , s).
( ,s)∈D×(a,b)→(x,t)
Note that the lim sup permits times in the future of t, s > t. Definition 2.5. u is a viscosity solution of (HS) (with boundary data u0 and f ) if u is a viscosity supersolution and u∗ is a viscosity subsolution of (HS) (with boundary data u0 and f ).
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Definition 2.6. We say that a pair of functions u0 , v0 : D¯ → [0, ∞) are (strictly) separated (denoted by u0 ≺ v0 ) in D ⊂ Rn if (i) the support of u0 , supp(u0 ) = {u0 > 0} restricted in D¯ is compact and (ii) in supp(u0 ) ∩ D¯ the functions are strictly ordered: u0 (x) < v0 (x). The following properties of viscosity solutions, proven in [10], are frequently used in our paper. Theorem 2.7 (Comparison principle). Let u, v be, respectively, viscosity sub- and supersolu¯ tions in D × (0, T ) ⊂ Q with initial data u0 ≺ v0 in D. If u v on jD and u < v on jD ∩ (u) for 0 t < T , then u(·, t) ≺ v(·, t) in D for t ∈ [0, T ). Theorem 2.8. There exists a unique viscosity solution v(x, t) of (HS) in Q with initial data u0 with condition (I). Moreover the positive set t (v) strictly expands in time. More precisely, s (v) is a compact subset of t (v) for 0 s < t. For the rest of our paper we denote u(x, t) as the unique viscosity solution of (HS) with fixed boundary data 1 and the initial data u0 satisfying (I). Lemma 2.9. (i) u(x, t) → u0 (x) as t → 0+ . Moreover u∗ (x, t) = u(x, t) = 1 for x ∈ jK and u(x, t) = 0 for (x, t) ∈ (u). (ii) u is superharmonic in (u) and u∗ is subharmonic in Q. ¯ ¯ ∗ ) and (u) = (u∗ ). (iii) (u) = (u (iv) u is harmonic in (u). Indeed u(x, t) = ht (x) at each time t > 0, where ht (x) = inf{(x) : −0 in t (u); = 1 on jK; 0 in t (u)}. Proof. 1. (i)–(ii) follows from the definition of u. To prove (iii), observe that for any > 0 the function u(x, ˜ t) := (1 + )u(x, t + ) is a supersolution of (HS) in Q with u∗ ≺ u˜ by the last statement of Theorem 2.8. Hence we apply Theorem 2.7 to u∗ and u˜ to obtain u∗ (x, t)(1 + )u(x, t + ) for any , > 0
(2.1)
In particular u(x, t)u∗ (x, t) u(x, t + ) for any > 0, and (iii) follows. 2. To prove (iv), Let h(x, t) = ht (x) be defined as in (iv) for t 0. It can be checked that ht ¯ t (u) with ht = 1 on jK (see for example [13, Chapter 1.3]). is nonnegative and harmonic in Since u(·, t) is superharmonic in t (u) with u = 1 on jK, by definition h(x, t) u(x, t). On the other hand by (2.1) u∗ (x, t) u(x, t + ) and in particular u∗ (·, t − ) = 0 on t (u) for t > . Moreover u∗ (·, t − ) is subharmonic in (u) with boundary data 1 on jK. Since ht is harmonic and nonnegative in t (u), by maximum principle for harmonic functions we obtain u∗ (x, t − )h(x, t) for any small > 0. Now we obtain u(x, t) lim u∗ (x, t − ) h(x, t) →0
for t > 0,
where the first inequality is due to the lower semicontinuity of u in time. Hence uhu, i.e., u = h for t 0.
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3. Lipschitz continuity of the free boundary For x ∈ Rn , we denote Br (x) := {y ∈ Rn : |x − y| < r}. Since 0 is bounded, we may assume that 0 ⊂ BR0 for some R0 > 0. In this section we would like to show that the free boundary (u) becomes Lipschitz continuous in space and time once it is out of BR0 . First we state the following lemma, which can be proven by a reflection argument and by Theorem 2.7. For detailed proof we refer to [1, Proposition 2.1, Lemma 2.2], where a parallel result is proven for a solution u of the porous medium equation. ¯ Lemma 3.1. For |x0 |, |x1 | > R0 with (x1 , t) ∈ (u) ∩ Q the following holds: If cos(x1 − x0 , x0 ) R0 /|x0 | then u∗ (x1 , t) < u(x0 , t + ) for > 0. Let us define T0 = t (u0 ) = inf{t > 0 : t (u) ⊃ B¯ R0 } > 0.
(3.1)
By comparing u with a radially symmetric solution of (HS) with initial data less than u0 , one can easily show that T0 < ∞. Theorem 3.2. For T0 < t t t¯ < ∞ the free boundary (u) is representable in spherical coordinates of the form r = g(, t), where g is Lipschitz continuous in and t. Moreover g(·, t) is uniformly Lipschitz continuous with Lipschitz constant L = L(t) in for t t with L → 0 as t → ∞. Proof. 1. First we show that t (u) is starshaped and Lipschitz continuous at each time T0 < t < t. For T0 < T , it follows from Lemma 3.1 that x ∗ ,T + for x ∈ T (u∗ ) u (x, T ) < u 1+ for any > 0 and small 0 < < 0 , where 0 depends on T . Also u∗ (x, T ) < 1 = u(x/(1 + ), T + )
on {x : |x| = 1 + }.
Then the maximum principle for harmonic functions yields that u∗ (x, T ) ≺ u(x/(1 + ), T + ) 2. Since
in |x|1 + .
t −T x , +T + w(x, t) = u 1+ 1+
is a viscosity supersolution of (HS) in {x : |x|1+}×[T , ∞) with w =1 u on {x : |x|=1+}, we have u∗ (x, t) ≺ w(x, t) for {x : |x|1 + } ∩ {t T }.
(3.2)
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For given > 0 and > 0, we can choose = /(1 + ) such that (t − T )/(1 + ) + T + = t at t = T + and thus due to (3.2) it follows u∗ (x, T + ) ≺ u(x/(1 + ), T + )
for any , > 0.
(3.3)
In particular (3.3) implies that for t > T0 our solution u(·, t) is decreasing in radial directions and thus t (u) is representable by spherical coordinates r = g(, t) = g(, t) > R0 . 3. Due to Lemma 3.1 for any direction p ∈ Rn such that cos(p, x)R0 /|x| we have u∗ (x + p, t) < u(x, t + ) for , > 0 ¯ if |x + p|, |x| > R0 and (x + p, t) ∈ (u). Hence for x ∈ t (u), T0 < t t there exists 0 > 0 such that for 0 < 0 and > 0 u∗ (x + p, t − ) < u(x, t)
if (x + p, t − ) ∈ (u).
(3.4)
This implies that u(x + p, t − ) = 0 for above , p, given as in (3.4) and thus u(x + p, t) = 0 by the lower semicontinuity of u. In other words at each point x ∈ t (u) there is 0 = (|x|) = cos−1 (R0 /|x|) such that u(x, t) = 0
for y ∈ W (0 , x/|x|; x) ∩ B0 (x).
On the other hand by similar argument it can be easily seen that u(z, t) > 0
for z ∈ W (0 , −x/|x|; x) ∩ B0 (x).
This leads to the Lipschitz continuity g with respect to . Observe that the Lipschitz constant of g(·, t) only depends on 0 , which can be chosen as 0 = (t) = cos−1 [R0 /r(t)], where r(t) = inf{|x| : x ∈ t (u), t = t} > R0 . Since r(t) → ∞ as t → ∞, it follows that 0 → /2 as t → ∞. 4. It remains to show that g is Lipschitz locally in time. Due to (3.2), if (x0 , t0 ) ∈ (u) and t0 > T0 then u∗ ((1 + )x0 , (1 + )(t0 − T0 ) + T0 ) u(x0 , t0 ) = 0 for > 0 and therefore u((1 + )x0 , (1 + )(t0 − T0 ) + T0 ) = 0. If we set t = (1 + )(t0 − T0 ) + T0 and = x0 /|x0 |, above estimates imply g(, t0 ) = x0 and g(, t) (1 + )|x0 |. This and the fact that the u is increasing in time, we obtain |g(, t) − g(, t0 )| g(, t0 ) |t − t0 | t0 − T 0
if T0 < t0 < t.
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Due to the Lipschitz continuity of the free boundary, we are now able to show the following properties of u: Proposition 3.3. (i) u(x, t) is continuous for t > T0 : (ii) If |x0 |, |x1 | > R0 and cos(x1 − x0 , x0 ) R0 /|x0 | then u(x1 , t)u(x0 , t)
for > 0.
(iii) for T > T0 , 0 < < 0 (R0 ) and |x|R0 u(x, t)u(x/(1 + ), (t − T )/(1 + ) + T )
for t > T .
(3.5)
Proof. 1. (ii) follows from (i) and Lemma 3.1. 2. By Lemma 2.9(iv), at each t > 0 u(x, t) = inf{(x) : −0 in t (u); 1 on jK; 0 on t (u)}. For t > T0 , since t (u) is Lipschitz continuous, Perron’s method (we refer to Chapter 1.3 of [13]) yields that u(·, t) is continuous in t (u) and harmonic in t (u) with u = 1 on jK and u = 0 on t (u). 3. To prove continuity of u in time, let us pick (x, t0 ) ∈ (u) such that t0 > T0 . Due to (2.1) u∗ (x, t0 − )u(x, t0 ) for any small > 0. On the other hand from (3.3) we have u(x, ˜ t) := u∗ ((1 + )x, t0 − )u(x, t0 − ) for any > 0 and 0 < < t0 − T0 . Since u˜ is a viscosity subsolution of (HS), we can apply Theorem 2.7 to u˜ and u in Rn − K × [t0 − , ∞) and obtain u∗ ((1 + )x, (1 + )(t − t0 + ) + t0 − ) u(x, t)
for t t0 − .
Now evaluating above inequality at t = t0 , we obtain u∗ (x , t0 + )u((1 + )−1 x , t0 ), where x = (1 + )x. We let = to conclude that for any (x, t) ∈ Rn − K × (t0 , ∞) lim u(x, t + 2 ) lim u((1 + )−1 x, t),
→0
→0
where the second term equals u(x, t) due to the spatial continuity of u. Thus u(x, t)=lim→0 u(x, t + ). On the other hand, since u is lower semicontinuous and increasing in t, we have u(x, t) = lim u(x, t − ). →0
Thus u is continuous in time and now (3.5) follows from (3.3).
Lemma 3.4. There exist A, B, 0 > 0 depending on u0 such that u (x, t) = for 0 < < 0 .
1 + A (1 + )2
u((1 + )x, (1 + A)t + B) u(x, t)
(3.6)
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Proof. 1. It is easy to check that u also solves (HS). Therefore to apply Theorem 2.7 to u and u in Q and to obtain (3.6), we only have to show that for small > 0 there is A, B > 0 such that u0 (x) ≺ u (x, 0) =
1 + A (1 + )2
u((1 + )x, B)
and 1 u (·, t) for x ∈ jK. 2. After a dilation if necessary, we may assume that 0 contains B4 (0). We set c = inf{|Du0 (x)|/4|x| : x ∈ 0 } (note that c > 0 due to condition (I)). Then w(x, t) = u0 ((1 + ct)−1 x) is a viscosity subsolution of (HS) in the domain B1+ct (0) × [0, 1/c], since w(·, t) is harmonic in t (w) and wt c
|x| |Du0 |(1 + ct)−2 |Du0 |2 = |Dw|2 (1 + ct)
on t (w) × [0, 1/c],
where the second inequality holds due to the definition of c. Since w(x, t) is a subsolution of (HS) in B1+ct (0) × [0, 1/c], for any c2 > 0, w2 (x, t) = c2 w(x, c2 t) is also a subsolution of (HS) in B1+ct (0) × [0, 1/(c2 c)]. Now if we choose c2 such that c2 < u0 (x) on |x| = 2, then w2 u on |x| = 2 for 0 t C = 1/(cc2 ). By Theorem 2.7 then w2 u in the domain {x : |x|2} × (0, C). In particular, if x ∈ 0 and |x| 2 then (1 + C −1 t)x ∈ t (u)
for 0 t C.
(3.7)
3. On the other hand if |x|2 then (1 + C −1 t)|x| ∈ B4 (0) ⊂ 0 ⊂ t (u)
for 0 t C.
(3.8)
In terms of u, (3.7) and (3.8) yields that if we let B = C then u(x, 0) > 0 implies that u((1 + )x, B) > 0 for small = C −1 t > 0, in other words ¯ 0 ⊂ {u (x, 0) > 0}.
(3.9)
Finally let us choose A big enough such that for small > 0 1
1 + A (1 + )2
u0 ((1 + )x)
on jK.
(3.10)
(Such A can be chosen, for example, as the maximum of 2|Du0 | on B2 (0) − K.) With above choice of A and B, (3.9), (3.10) and the maximum principle for harmonic functions yields that u0 (x) ≺ u (x, 0) and 1 u for x ∈ jK. Remark. Heuristically speaking, differentiating (3.5) and (3.6) with respect to at = 0 yields that for t > T0 x · (x, t) ut x · (x, t) At + B |Du| t − T0
on (u),
(3.11)
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where = −Du/|Du|. Since formally the free boundary motion law is given as V = |Du| on (u), (3.11) is expected to yield bounds for |Du| on (u) for t > T0 . In the next section we use barrier arguments based on the geometry (Lipschitz continuity) of t (u) to obtain the expected bounds on |Du|. 4. Gradient estimates First we look for the lower bound of |Du| on (u). We start with the following lemma. Using barrier argument we will show that if |Du| is very small near the free boundary then the positive phase expands too slowly, violating (3.6). Lemma 4.1. Let (x0 , t0 ) ∈ (u) and T0 < t t0 < t¯ < ∞. Then there exist positive constants h0 , C1 , C2 depending only on t, t¯ such that sup{u(x, t) : |x − x0 | < h, 0 t − t0 C2 h} > C1 h
for 0 < h < h0 .
Proof. 1. Let x1 = (1 + h/|x0 |)x0 . By Theorem 2.7 for small h > 0 there is L > 1 depending only on the Lipschitz constant of g(·, t0 ) such that u(·, t0 ) = 0 in the ball Bh/L (x1 ). We will construct a barrier based on this ball (see Fig. 2). 2. Consider (x) such that
− = 0 =0 = C1 h
in := B2h (x1 ) − Bh/L (x1 ), on jBh/L (x1 ), on jBh (x1 ),
where the constant C1 > 0 is to be determined. Notice that (x) = (r), where r = |x − x1 | and |r | = C > 0 on jDh/L (x1 ) where C = kC 1 where k = k(n, L). Consider v(x, t) = (1 + Ct)
−1
L 1 + 8C t r h
Γt
0
h h X0
Ωt
θ
0
Fig. 2.
L X1
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in the domain S = 2 × (0, C2 h],
C2 = 1/(8CL)
and 2 = Bh (x1 ) − Dh/L (x1 ).
Observe that on t () we have r(1 + 8C (L/ h)t) = h/L and vt =
8CLr/ h r , 1 + Ct
|Dv| =
(1 + 8CLt/ h) h r = r . 1 + Ct Lr(1 + Ct)
Therefore on t () vt − |Dv|2 C2 (8 − (h/Lr)3 ) 0 since r = h/L(1 + 8C(L/ h)t) h/2L if 0 t < h/8CL. Since v(·, t) is harmonic in 2 , it follows that v is a supersolution of (HS) in S. 3. Now if we assume that u(x, t) C1 h in jDh (x1 ) × [0, C2 h], then u v on the parabolic boundary of S and thus u v in S by Theorem 2.7. Therefore the point (x1 , t0 + h/8CL) belongs to the interior of the set {u = 0}. But this contradicts Lemma 3.4 if we take C1 such that |x1 − x0 | = |8CL| , At 0 + B |t1 − t0 | where A, B is given as in Lemma 3.4 and = (t) = inf{|x| : x ∈ t (u)} > R0 .
Lemma 4.2. Let g given as in Theorem 3.2. If T0 < t < t1 < t¯ < ∞ and g(x1 /|x1 |, t1 ) − |x1 | = h with 0 < h (|x1 | − R0 )/2 then we have u(x1 , t1 )C3 h, where C3 only depend on the initial data u0 and t, t¯. Proof. 1. We recall that for T > T0 and for any > 0 u(x, t)w(x, t) = u(x/(1 + ), (t − T )/(1 + ) + T ). Let us take T = T0 + (t1 − T0 )/2. Then if u(x1 , t1 )ch
then u((1 + )x1 , (1 + )(t1 − T ) + T ) ch.
(4.1)
2. Now suppose that g(x1 /|x1 |, t1 )−|x1 |=M1 h, where M1 > 0 is to be decided. Let =x1 /|x1 | and x0 = g(, t1 ) ∈ t1 (u) . From Lemma 4.1 sup
|x−x0 | h,0 t−t1 C1 h
u(x, t) > C2 h,
(4.2)
where Ci = Ci (t, t¯). We recall that, due to Lemma 3.1 and continuity of u, for t > t and for |x| > R0 there is an angle t > 0 with which u(y, t)u(x, t)
if y ∈ W (t , x/|x|, x).
(4.3)
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(1 + β) X1
h
X0
Mh X1 Fig. 3.
Choose M2 = M2 (t) > 0 such that if we take M1 = M2 then Dh (x0 ) belongs to W (t , ; x1 ). We choose M1 a bit larger than M2 , that is, we set M1 = M2 +
C1 sup{|x| : x ∈ t¯(u)} t − T0
so that Dh (x0 ) belongs to the cone W (t , ; (1 + )x1 ) for 0 , where = C1 /(t1 − T0 )h (see Fig. 3). 3. Due to (4.1) if u(x1 , t1 ) < C2 h then u((1 + )x1 , t1 + (t1 − T )) < u(x1 , t1 ) < C2 h for h small and 0 . Due to the choice of M1 and (4.3) it follows that u(x, t) < C2 h in the cylinder Bh (x0 ) × [t1 , t1 + C1 h]. This contradicts (4.2) and thus we can conclude by taking C3 = C2 /M1 . Next we proceed to obtain an upper bound for |Du| near (u). As before the main idea is that if |Du| is very big near the free boundary then the positive phase expands too fast in time, violating (3.5). Lemma 4.3. Let g(x/| ¯ x|, ¯ t0 ) − |x| ¯ = h with 0 h (|x| ¯ − R0 )/2 and T0 < t t0 t¯. Then there are constants L depending on t, n and M = M(L, t, t¯, u0 , n) such that ¯ < h/L}Mh. inf{u(x, t0 ) : |x − x| Proof. 1. Let x0 = x¯ + h ∈ t0 (u), where = x/| ¯ x|. ¯ Due to the Lipschitz continuity of (u) in space, there is a constant L = L(t) > 0 such that D2h/L (x) ¯ ∈ t0 . Consider a harmonic function ¯ − Dh/L (x) ¯ such that (x) in D2h/L (x) ¯ (x) = 0 on jD2h/L (x), (x) = Mh on jDh/L (x), ¯ ¯ where M3 > 0 where M > 0 is a constant chosen so that |D| = M3 (1 + M3 )/L on jD2h/L (x), is to be decided. Note that (x) = (r), r = |x − x| ¯ and M only depends on M3 and n. ¯ by zero and let us define Let us extend outside D2h/L (x) −1 h h M3 r− (t − t0 ) + . v(x, t) = 1+ h L L
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in the domain S =(Rn −Dh/L (x))×[t ¯ 0 , t0 +h]. Then for t0 t t0 +h, r −h/L (1+M3 )(h/L) on t (v) and thus v satisfies h M3 / h |r |2 2 vt − |Dv| = | r − | − r L (1 + (M3 / h)(t − t0 ))2 (1 + (M3 / h)(t − t0 ))2 h M3 r− − M3 (1 + M3 )/L 0 C h L for (x, t) ∈ t (v) ∩ {t0 t t0 + h}. Hence v is a supersolution of (HS) in S. 2. Now suppose that Mh < u(·, t0 ) on Dh/L (x). ¯ Then v(x, t0 ) = (x) ≺ u(x, t0 ) in Rn − ¯ × [t0 , t0 + h]. Hence by Dh/L (x). ¯ Moreover, since u increases in time, v = Mh < u on jDh/L (x) applying Theorem 2.7 to v and u in the domain S, it follows that v ≺ u in S. In particular h h ¯ ⊂ (u). x¯ + (2 + M3 ) , t0 + h = x0 + (2 + M3 − L) , t0 + h ∈ (v) L L (4.4) 3. On the other hand, due to (3.5) for any > 0 we have u((1 + s)x0 , t0 + (t0 − T0 )) = 0
if s.
(4.5)
Now (4.4) and (4.5) leads to a contradiction if we choose =h/(t0 −T0 ), s=(2+M3 −L)h/L|x0 | and M3 = L − 2 + L/(t − T0 ), where = (t¯) = sup{|x| : x ∈ t¯(u)}.
Lemma 4.4. For T0 < t t1 t¯ there is h0 = h(t) > 0 satisfying the following: if g(x1 /|x1 |, t1 ) − |x1 | = h with 0 < h < h0 then we have u(x1 , t1 )M h, for M = M (t, t¯, u0 , n) > 0. Proof. 1. Choose h0 = h0 (t) > 0 such that |x| > R0 + 3h0 on t (u). Let 0 < h < h0 , = x1 /|x1 | and let x0 = x1 + h ∈ t1 (u). Let t as given in (4.3). We choose L = L(t) > 2 such that D2h/L (x2 ) ∈ W (t , −; x1 )
where x2 = x1 − h
(see Fig. 4). By the choice of h0 , D2h/L (x2 ) ⊂ {|x| > R0 }. Observe that Lemma 4.3 holds for x¯ = x2 with this choice of L. 2. Assume u(x1 , t1 ) > 2M h, where M is the constant M obtained as in Lemma 4.3 with the choice of L given in step 1. Due to the monotonicity of u along the cone W (t , −, x1 ), we have u(·, t1 )u(x1 , t1 ) > M (2h) in D2h/L (x2 ). This contradicts Lemma 4.3 since g(x2 /|x2 |, t1 ) − |x2 | = 2h. Now we are ready to state our main theorem: Theorem 4.5. Let u be a viscosity solution of (HS) with initial data u0 and fixed boundary data 1 on K. Then (u) is Lipschitz continuous in space and time for t > T0 , where T0 is given
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Ωt
1
2h 2h
L
X2
K2
X1
X0
Γt
1
Fig. 4.
in (3.1). Moreover there is a neighborhood N of (u) in each strip of the form T0 < t t t¯ and constants C1 , C2 > 0 depending only on t, t¯, u0 and n such that C1
ut , |Du|
|Du(x, t)| C2 ,
if (x, t) ∈ (u) ∈ N. Proof. Let us pick t ∈ (t, t¯). We only have to obtain the bounds for |Du|. From the Lipschitz continuity of t (u) (Theorem 3.2) and properties of harmonic functions in a Lipschitz domain (see [3, Lemma 4]), for t t > T0 there are positive constants c1 , c2 depending only on t such that if h is small enough then 0 < c1 |Du|(x − h, t)
u(x − h, t) c2 |Du|(x − h, t), h
where x ∈ t (u) and = x/|x|. Now Lemma 4.3 and 4.4 leads to the conclusion.
Based on above theorem the following result is proved in [11, Section 6]: Corollary 4.6. Let u0 , u be given as in Theorem 4.5. Then after t > T0 u is a classical solution. More precisely u and (u) after t > T are analytic in space and time and u satisfies the free boundary condition V = |Du| in the classical sense. References [1] D.G. Aronson, L.A. Caffarelli, The initial trace of a solution of the porous medium equation, Trans. Am. Math. Soc. 280 (1) (1983) 351–366. [2] I. Athanasopoulos, L. Caffarelli, S. Salsa, Regularity of the free boundary in parabolic phase-transition problems, Acta Math. 176 (1996) 245–282. [3] L. Caffarelli, A Harnack inequality approach to the regularity of free boundaries, Part I: Lipschitz free boundaries are C 1, , Rev. Mat. Iberoam. 3 (2) (1987) 139–162. [4] L.A. Caffarelli, J.L. Vazquez, N.I. Wolanski, Lipschitz continuity of solutions and interfaces of the N-dimensional porous medium equation, Indiana Univ. Math. J. 36 (1987) 373–401. [5] S. Choi, D. Jerison, I. Kim, The regularity and speed of the Hele–Shaw flow, submitted.
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[6] P. Daskalopoulos, R. Hamilton, K. Lee, All time C ∞ -regularity of interface in degenerated diffusion: a geometric approach, Duke Math. J. 108 (2001) 295–327. [7] P. Daskalopoulos, K. Lee, All time smooth solutions of the one-phase Stefan problem and the Hele–Shaw flow, Commun. Partial Differential Equations 12 (2004) 71–89. [8] C.M. Elliot, V. Janovsky, A variational inequality approach to Hele–Shaw flow with a moving boundary, Proc. R. Soc. Edinburgh, Sect. A 88 (1–2) (1981) 93–107. [9] J. Escher, G. Simonett, Classical solutions of multidimensional Hele–Shaw models, SIAM J. Math. Anal. 28 (5) (1997) 1028–1047. [10] I. Kim, Uniqueness and existence result of Hele–Shaw and Stefan problem, Arch. Rat. Mech. Anal. 168 (2003) 299–328. [11] I. Kim, Regularity of the free boundary for the one phase Hele–Shaw problem, J. Differential Equations, to appear. [12] K. Lee, J.L. Vazquez, Geometrical properties of solutions of the porous medium equation for large times, Indiana Math. J. 52 (2003) 991–1016. [13] M. Tsuji, Potential theory in modern function theory, Maruzen Co., LTD., Tokyo, 1959.