Interior Regularity of Minimizer

and Z

Ω

|φ−un−1|² = Z

Ω\S

|u_n−un−1|²+ Z

S

|ku₀k_∞−un−1|²

= Z

Ω

|u_n−un−1|²+ Z

S

ku₀k²_∞−u²_n−2un−1(ku₀k_∞−u_n)

= Z

Ω

|u_n−un−1|²+ Z

S

(ku₀k_∞−u_n) (ku₀k_∞+u_n−2un−1)

<

Z

Ω

|u_n−un−1|²,

since in setS, we haveku₀k_∞−u_n<0 andku₀k_∞+u_n−2un−1 >2(ku₀k_∞−un−1)≥0 which follows from (7.10). Hence,F_n^h(φ)<F_n^h(un). A contradiction!

We end this section with some estimates regarding the minimizer, which we shall utilize in the succeeding arguments.

Lemma 7.3. If un minimizes F_n^h(n= 1, . . . , M) for h, λ >0, then (i.) k∇u_nk_L2(Ω)≤ k∇u₀k_L2(Ω),

(ii.) ku_n−un−1k_L2(Ω)≤√

hk∇u₀k_L2(Ω), (iii.) f(|{u_n>0}|)≤λ⁻¹k∇u₀k²_L2(Ω). Proof. Denote

J(u) :=

Z

Ω

|∇u|²+λf(|{u >0}|).

Then,F_n^h(u_n)≤ F_n^h(un−1) =J(un−1) implies J(un)≤

Z

Ω

|u_n−un−1|²

h +J(un)≤ J(un−1),

which implies that J(un)≤ J(un−1)≤ · · · ≤ J(u0). Since |{u₀ >0}|=α, then F_n^h(un)≤ F_n^h(un−1) =J(un−1)≤ J(u0) =k∇u₀k²_L2(Ω),

which yields the desired results.

Proof. Consider an arbitrary ball BR such thatBR⊂Ω. Define function φ=η²(un−k)₊,

where η ∈ C₀^∞(Ω; [0,1]) such that η ≡ 1 in a smaller concentric ball B_r and |∇η| ≤ c/(R−r) for some constant c >1. Fix k ∈ R and denote A⁺_R := BR∩ {u_n > k}. By Theorem 7.2, we can assume thatk≤ ku₀k_∞. Since f is 1-Lipschitz continuous, then

f(|{u_n−φ >0}|)−f(|{u_n>0}|) ≤ ||{u_n−φ >0}| − |{u_n>0}||

≤

{u_n−φ >0} ∩A⁺_R +

{u_n>0} ∩A⁺_R

≤ 2 A⁺_R

. Then,F_n^h(u_n)≤ F_n^h(u_n−φ) implies that

0≤ −2 Z

A⁺_R

un−un−1

h φ+

Z

A⁺_R

φ² h −2

Z

A⁺_R

∇u_n· ∇φ+ Z

A⁺_R

|∇φ|²+ 2λ|A⁺_R|. (7.11) Since u_n is bounded byku₀k_∞ almost everywhere, then

− Z

A⁺_R

(u_n−un−1)φ = Z

A⁺_R

(un−1−u_n)η²(u_n−k)

≤ Z

A⁺_R

|un−1−k−(un−k)|(u_n−k)

≤ Z

A⁺_R

|u_n−1−k||u_n−k|+ Z

A⁺_R

(un−k)²

≤ Z

A⁺_R

(|u_n−1|+|k|) (|u_n|+|k|) + Z

A⁺_R

(u_n−k)²

≤ 4ku₀k²_∞|A⁺_R|+ R² (R−r)²

Z

A⁺_R

(un−k)², and

Z

A⁺_R

φ² = Z

A⁺_R

η⁴(un−k)² ≤ R² (R−r)²

Z

A⁺_R

(un−k)². Using Young’s inequality, we have for some ε1 ∈(0,1),

−2 Z

A⁺_R

∇u_n· ∇φ = −2 Z

A⁺_R

∇u_n·

η²∇u_n+ 2(u_n−k)η∇η

≤ (−2 +ε₁) Z

A⁺_R

η²|∇u_n|²+ 4 ε1

Z

A⁺_R

(u_n−k)²|∇η|². Again, by Young’s inequality, we have for some ε2 ∈(0,1−ε1),

Z

A⁺_R

|∇φ|² = Z

A⁺_R

η⁴|∇u_n|²+ 4 Z

A⁺_R

η³(un−k)∇u_n·∇η+ 4 Z

A⁺_R

η²(un−k)²|∇η|²

≤ (1 +ε₂) Z

A⁺_R

η⁴|∇u_n|²+ 4 1 +_ε¹

2

Z

A⁺_R

η²(u_n−k)²|∇η|²

≤ (1 +ε₂) Z

A⁺_R

η²|∇u_n|²+ 4 1 +_ε¹

2

Z

A⁺_R

(u_n−k)²|∇η|².

Thus, (7.11) becomes (1−ε₁−ε₂)

Z

A⁺r

|∇u_n|² ≤ (−1+ε₁+ε₂) Z

A⁺_R\A⁺_r

η²|∇u_n|²+ 4 1+_ε¹

1+_ε¹

2

Z

A⁺_R

(u_n−k)²|∇η|² + 3R²

h(R−r)² Z

A⁺_R

(un−k)²+ 2 max

4ku₀k²_∞ h , λ

|A⁺_R|.

≤ C

(R−r)² Z

A⁺_R

(u_n−k)²+ 2 max

4ku₀k²_∞ h , λ

|A⁺_R|, whereC(h, R) := max

4c(1+_ε¹

1+_ε¹

2),3R²h⁻¹

. Hence, Z

A⁺r

|∇u_n|² ≤ ρ

"

1

(R−r)²max

A⁺_R

(u_n−k)²+ 1

#

|A⁺_R|,

where ρ(1−ε1 −ε2) = max

4c(1 + _ε¹

1 + _ε¹

2),3R²h⁻¹,8ku₀k²_∞h⁻¹,2λ

. Analogously, takingφ=η²(un−k)− yields the above inequality forA⁻_R:=BR∩ {u_n< k}. Moreover, by Theorem 7.2,u_n is bounded with ess supu_n≤ ku₀k_∞. It follows that u_n belongs to Di Giorgi classB₂(Ω,ku₀k∞, ρ,∞,0).

Fix x0 ∈ Ω and consider an open ball BR := B(x0, R) ⊂ Ω for some R ≤1. For any x, y∈BR, denote r =|x−y|< R. By Theorem H.7 [62, Chapter 2, Theorem 6.1], we can find constantsγ <1 and C(γ, N, R)>0 such that

|u_n(x)−un(y)| ≤ sup

Br

un−inf

Br

un≤C|x−y|^γ. Then,u_n is H¨older continuous near x₀.

Next, we show that the minimizer is a subsolution to an elliptic partial differential equation on domain Ω. In fact, on the region where it is strictly positive, the minimizer is a solution and locally aC² function.

Theorem 7.5. If un minimizesF_n^h(n= 1, ..., M) for h, λ >0, then (in the weak sense) (i.) ∆un≥ u_n−un−1

h in Ω,

(ii.) ∆u_n= u_n−un−1

h in the open set {u_n>0}.

Proof. For ε >0 and nonnegative functions φ∈C₀^∞(Ω), considerψ:=un−εφ. Hence, if ψ(x) > 0 for some x ∈ Ω, then clearly we have u_n(x) > εφ(x) > 0. Thus, we get f(|{ψ >0}|)≤f(|{u_n>0}|), which implies that

0 ≤ F_n^h(ψ)− F_n^h(u_n)

≤ −2ε Z

Ω

u_n−un−1

h φ+∇u_n· ∇φ

+ε² Z

Ω

|φ|²

h +|∇φ|²

. Dividing by 2εand takingε→0 yields

Z

Ω

u_n−un−1

h φ+∇u_n· ∇φ≤0, (7.12)

thereby, proving (i).

To prove (ii), it suffices to show the reverse inequality of (7.12). Consider ψ:=u_n+εφ whereφ∈C₀^∞({u_n>0}) with

|ε|< m

maxφ where m:= min

suppφun>0.

It follows that the penalization ofψ is equal to that ofu_n, and so, we get 0 ≤ F_n^h(ψ)− F_n^h(u_n)

≤ 2ε Z

{un>0}

un−un−1

h φ+∇u_n· ∇φ

+ε² Z

{un>0}

|φ|²

h +|∇φ|²

. Dividing by 2ε and takingε→0 gives the desired result.

Corollary 7.6. Ifu_nminimizesF_n^h(n= 1, ..., M)forh, λ >0, thenu_n∈C_loc² ({u_n>0}).

Proof. Fix x₀ ∈ {u_n > 0}. Consider function φ ∈ C₀^∞({u_n > 0}) such that φ ≡ 1 in B_r :=B(x₀, r)⊂ {u_n >0} for some r >0. Let w= ψ∗g where where ψ denotes the fundamental solution of Laplace equation and functiong is defined by

g := u_n−un−1

h φ.

By Theorem 7.4 and [49, Lemma 4.2], we see thatw∈C²(Ω) and ∆w=gin Ω. Hence, by Theorem 7.5, we get

∆(u_n−w) = u_n−un−1

h (1−φ) = 0 inB_r,

which implies that u_n−w is harmonic (hence,C^∞) in B_r. Thus, u_n isC² nearx₀. The following remark will be utilized in the succeeding arguments.

Remark. Fix a ball Br⊂Ω. Define functional I_n,Br(u) :=

Z

Br

|u−un−1|²

h +|∇u|²

dx.

Then, there exists a unique minimizer v ∈ A := {v ∈ H¹(Br) : v = un on ∂Br} such that

I_n,Br(v) = min

A I_n,Br(·).

Moreover, in the weak sense,

∆v= v−un−1

h inBr, (7.13)

which follows from the proof of Theorem 7.5 by simply dropping the penalty arguments.

Existence. A similar argument as in the proof of Theorem 7.1 shows that we can find a minimizing sequence {u^k} ⊂ A that is uniformly bounded in H¹(Br) and whose

subsequence converges weakly inH¹(Br) to somev∈H¹(Br) and I_n,Br(v)≤lim inf

k→∞ I_n,Br(u^k).

Note that there exists a trace operator T :H¹(Br) → L²(∂Br). SinceT is a compact operator, then we can find a subsequence (still denoted by u^k) such that

T(u^k)→T(v) strongly inL²(∂Br).

Noting that T(u^k) = un on ∂Br, we see v = un on ∂Br in the sense of trace, and so, v∈A.

Uniqueness. Suppose there exist minimizersu, v∈A, that is,I_n,Br(u) =I_n,Br(v) =m.

Considerw:= ¹₂(u+v). By the parallelogram law, we have kw−un−1k²_L2(Br) =

¹₂(u−un−1) +¹₂(v−un−1)

2 L²(Br)

= ¹₂ku−un−1k²_L2(Br)+ ¹₂kv−un−1k²_L2(Br)−

u−v 2

2

L²(Br)

k∇wk²_L2(Br) = ¹₂k∇uk²_L2(Br)+¹₂k∇vk²_L2(Br)−

∇u− ∇v 2

2

L²(Br)

Hence,

m≤ I_n,Br(w) = ¹₂I_n,Br(u) +¹₂I_n,Br(v)−1 4

Z

Br

|u−v|²

h −1

4 Z

Br

|∇(u−v)|²

≤ m− 1

4hku−vk²_L2(Br),

which implies that ku−vk_L2(Br)= 0, and so, u=v almost everywhere inB_r.

Using the above remark, we now show that asλ→ ∞, the set of positive values of the minimizer un preserves the prescribed measure. Later, we shall establish that for large enough λ > 0, this prescribed measure can be attained without having to take λ to infinity.

Theorem 7.7. If un minimizesF_n^h(n= 1, . . . , M), then we have

α≤ |{u_n>0}| ≤α+λ⁻¹k∇u₀k²_L2(Ω). (7.14) Proof. Assume that |{u_n >0}|< α. We will deduce a contradiction by constructing a suitable perturbationv of the minimizerun.

Since u_n≥0, then settingE :={u_n= 0}, we have|E|>|Ω| −α >0. Takex∈∂E\∂Ω such that

|B(x, r)∩E|>0, (7.15) for all r >0. Fix r, sufficiently small such thatB_r:=B(x, r)⊂Ω and

0<|B_r| ≤α− |{u_n>0}|. (7.16)

Let v ∈ H¹(Ω) be the unique minimizer of I_n,Br in Br and equal to un outside Br. Then, we must have I_n,Br(v)≤ I_n,Br(u_n), that is,

Z

Br

|v−un−1|²

h +|∇v|² ≤ Z

Br

|u_n−un−1|²

h +|∇u_n|². (7.17)

Note thatf(|{u_n>0}|) = 0. Also, by (7.16), we have

|{v >0}| = |{u_n>0}\B_r|+|{v >0} ∩Br|

≤ |{u_n>0}|+|B_r| ≤α, which implies that f(|{v >0}|) = 0.

On the other hand, (7.17) givesF_n^h(v)≤ F_n^h(u_n). Sinceu_n minimizes F_n^h, then we get F_n^h(v) = F_n^h(un). Equivalently, I_n,Br(v) = I_n,Br(un), which follows from the previous remark thatv =un. By (7.15), we see thatv is a nonconstant function. Note that

∆v−v

h =−un−1

h ≤0 inB_r.

By [49, Theorem 3.5], v cannot achieve a nonpositive minimum in B_r, that is, v >0 in Br. However, since |B(x, r)∩E|>0, then v6=un. A contradiction!

Lastly, the second inequality of (7.14) follows from the first inequality and Lemma 7.3, as follows:

|{u_n>0}| −α=f(|{u_n>0}| −α)≤λ⁻¹k∇u₀k²_L2(Ω), which gives the desired result.

Following the arguments in [77], we show that the minimizerunis locally Lipschitz con-tinuous. To start, we establish the following lemma that guarantees Lipschitz continuity at the free boundary.

Lemma 7.8. (cf. [77, Lemma 4.1]) If un minimizes F_n^h(n= 1, . . . , M) for h, λ > 0, then there exists constants r₀ >0 and C(r₀)>0 such that ifx∈Ω satisfies

r(x) :=dist(x,{u_n= 0})<min

dist(x, ∂Ω) 2 , r0

, (7.18)

thenun(x)≤Cr√

λ. Here, constant C does not depend onN and λ.

Proof. Letx∈Ω such that (7.18) holds. Assume thatu_n(x)>0, that is,r :=r(x)>0.

(The case when u_n(x) = 0 is trivial.) We claim that for some M >0, we haveu_n(x)≤ M r. Suppose not. Then for any M >0, there existsx0∈Ω satisfying (7.18) such that

u_n(x₀)> M r. (7.19)

By Theorem 7.2, we see that for any n, Z

Ω

|u_n|^p ≤ |Ω|ku₀k^p_∞<∞, ∀p≥1, and so, sup

n≥1

kun−1k_Lp(Ω) <+∞.

It follows from [49, Theorem 8.17 and 8.18] that u_n(x₀) ≤ C₁ inf

B(x0,³₄r)

u_n(x) +k(r),

!

(7.20) whereC₁ =C₁(r, h, q)>0 with the property limr↓0C₁(r, h, q)>0,q > N, and

k(r) =r^2(1−N/q)sup

n≥1

un−1

h

L^q/2(Ω)=o(r) as r↓0. (7.21) Lety∈∂B(x0, r)∩ {u_n= 0} and consider a functionv∈H¹(B(y, r)) such that

I_n,B(y,r)(v) = min

A I_n,B(y,r)(·),

where A :={v ∈ H¹(B(y, r)) :v =un on∂B(y, r)}. By the previous remark, we have for any φ∈C₀^∞(B(y, r)),

Z

B(y,r)

v−un−1

h φ+∇v· ∇φ= 0. (7.22)

By Theorem 7.5, we have (in the weak sense)

∆(u_n−v) ≥ u_n−v

h , inB(y, r), and so, [49, Theorem 8.1] gives

sup

B(y,r)

(un−v)≤ sup

∂B(y,r)

(un−v) = 0.

This implies that

0≤u_n≤v inB(y, r). (7.23)

Now, extend functionv by un outside the ball, and define

vb:=

(

v, on B(y, r) u_n, on Ω\B(y, r).

By the minimality ofu_n and takingφ=v−u_n in (7.22), we get Z

B(y,r)

|v−u_n|²

h +|∇(v−u_n)|² = F_n(u_n)− F_n(v) +λ(f(|{bv >0}|)−f(|{u_n>0}|)) + 2

Z

B(y,r)

(v−un−1)(v−u_n)

h +∇v· ∇(v−u_n)

≤ λ(f(|{bv >0}|)−f(|{u_n>0}|)) (7.24) By (7.23), we see that{u_n>0} ∩B(y, r)⊆ {v >0} ∩B(y, r). It follows from Theorem 7.7 that

|{bv >0}| = |{v >0} ∩B(y, r)|+|{u_n>0}\B(y, r)|

≥ |{u_n>0} ∩B(y, r)|+|{u_n>0}\B(y, r)|

= |{u_n>0}| ≥α.

Thus, (7.24) becomes Z

B(y,r)

|∇(v−u_n)|² ≤ λ(|{v >b 0}| − |{u_n>0}|)

= λ|{bv >0} ∩ {u_n= 0} ∩B(y, r)|. (7.25) By (7.21), there exists small ε1 1 such that forr < ε1, we have

k(r)≤r inf

B(x0,³₄r)

un(x).

Also, by (7.23), we get inf

B(x0,³₄r)

u_n(x)≤ inf

B(x0,³₄r)∩B(y,r)

u_n(x)≤ inf

B(x0,³₄r)∩B(y,r)

v(x).

Hence, (7.19) and (7.20) implies

M r < u_n(x₀)≤C₁(1 +r) inf

B(x0,³₄r)∩B(y,r)

v(x), that is,

v(x) ≥ M r 2C1

, inB(x₀,³₄r)∩B(y, r).

Invoking this inequality in [49, Theorem 8.18] yields the following for 1≤p < N/(N−2), inf

B(y,¹₂r)

v(x) +k(r) ≥ C₂(r, h)r^−N/pkvk_Lp(B(y,r))

≥ C₂r^−N/pkvk_Lp(B(x0,³₄r)∩B(y,r))

≥ C₂r^−N/p|B(x₀,³₄r)∩B(y, r)|^1/pM r 2C1

=: 2C₃M r.

where lim

r↓0 C₃(C₂, C₁⁻¹) >0. Again, by (7.21), we can find sufficiently small ε₂ 1 so that forr < ε2, we have

k(r)≤r inf

B(y,¹₂r)

v(x)< inf

B(y,¹₂r)

v(x), which implies that

inf

B(y,¹₂r)

v(x)≥C₃M r. (7.26)

Now, define a function

w(x) = C₃M r

exp

−µρ²(x) r²

−exp(−µ)

, whereρ(x) = dist(x, y) and

µ ≥ N +

r

N²+r₀²

h. (7.27)

Here, we take r0 = min(ε1, ε2)1. Then, we have

∆w−w−un−1

h ≥ ∆w− w

h

= C₃M rexp

−µρ² r²

4µ²ρ²

r⁴ −2µN r² − 1

h

+C₃M r

h exp(−µ)

≥ 0 inB(y, r)\B(y,¹₂r). (7.28)

(Indeed, ifx∈B(y, r)\B(y,¹₂r), then ¹₂r ≤ρ(x)≤r. Also from (7.27), we get (µ−N)² ≥ N²+r₀²

h. Then, we have

4µ²ρ²

r⁴ −2µN r² − 1

h ≥ µ²

r² −2µN r² −1

h = (µ−N)² r² −N²

r² − 1 h

≥ r²

h r²₀−r²

≥0.)

Meanwhile, we note that for any x ∈ ∂B(y, r), we have v(x) = un(x) ≥ 0, and so, w(x) = 0≤v(x). Also, (7.26) implies that if ρ(x) = ¹₂r, then

w(x) = C3M r

exp −¹₄µ

−exp(−µ) ≤C3M r

≤ inf

B(y,¹₂r)

v(x)

≤ v(x) inB(y,¹₂r).

Hence, we have

w(x) ≤ v(x) on∂(B(y, r)\B(y,¹₂r)).

Note that from (7.28), we get

∆w− w

h ≥ −un−1

h = ∆v− v

h inB(y, r)\B(y,¹₂r).

Then, the maximum principle [49, Theorem 8.1] gives sup

B(y,r)\B(y,¹₂r)

(w−v) ≤ sup

∂(B(y,r)\B(y,¹₂r))

(w−v)≤0, that is,

w(x) ≤ v(x) inB(y, r)\B(y,¹₂r).

There exists constant C4>0 such thatw(x)≥C4(r−ρ) in B(y, r)\B(y,¹₂r). (Indeed, if x ∈ B(y, r)\B(y,¹₂r), then ¹₂r ≤ ρ(x) < r, that is, r−ρ ≤ ¹₂r. Moreover, since the exponential function is nonincreasing, then for someδ >0,

w(x) ≥ ¹₂δC₃M r≥εC₃M(r−ρ).

Take C₄ :=δC₃M >0.) Combining this with (7.26) implies that

v(x)≥C5M(r−ρ) inB(y, r), (7.29)

for some constantC5>0.

Take two disjoint balls B(y_i,¹₈r) ⊂B(y,¹₂r) for i= 1,2. For ξ ∈ ∂B(y, r), let z_i(ξ) be the point on the line segmentξyisuch that the length of segment`i(ξ) :=ξzi(ξ) becomes the largest with z_i(ξ)∈/ B(y_i, δr) and u(z_i(ξ)) = 0. In caseu(x)6= 0 for all x∈ξy_i, we set zi(ξ) = ξ. For x ∈ B(y, r)\y_i, let ξi(x) ∈ ∂B(y, r) be a point such that x ∈ yiξi. Then by (7.29), we have for someC6 >0

v(x)≥C₅M(r−ρ)≥ψ_i(x) :=C₆M rdist(y_i, ξ_i)−dist(y_i, x) dist(y_i, ξ_i) . Hence,

Z zi

ξ

d d`i

ψ_id`_i =:ψ_i(z_i) ≤ v(z_i) :=

Z zi

ξ

d d`i

(v−u_n)d`_i, that is,

Z zi

ξ

C5M r

dist(yi, ξ)d`i ≤ Z zi

ξ

|∇(v−un)|d`i.

Integrating with respect to ξ ∈ ∂B(y, r), summing up with respect to i, noting that r/dist(y_i, ξ) =O(1) asr ↓0, squaring both sides, and finally, invoking Schwarz’ inequal-ity yields

C₆²M²|V| ≤ Z

V

|∇(v−u_n)|², (7.30)

where

V :=V₁∪V₂, V_i = [

ξ∈∂B(y,r)

`_i(ξ).

Combining this with (7.25) gives

C₆²M²|V| ≤λ|{v >0} ∩ {u_n= 0} ∩B(y, r)| ≤λ|{u_n= 0} ∩B(y, r)|, (7.31) thereby, contradicting the arbitrariness of M.

Finally, we note that {u_n= 0} ∩B(y, r)⊂V. By (7.31), we get

C₆²M²|{u_n= 0} ∩B(y, r)| ≤C₆M|V| ≤λ|{u_n= 0} ∩B(y, r)|.

Hence, invoking the above claim givesu_n(x)≤rC₆⁻¹√

λfor anyx∈Ω satisfying (7.18).

Using the above lemma, we can show that the minimizer u_n is locally Lipschitz in Ω.

Theorem 7.9. If un minimizes F_n^h(n= 1, . . . , M) for h, λ >0, then un∈C_loc^0,1(Ω). In particular, the (local) Lipschitz coefficients of u_n do not depend on n.

Proof. A similar argument as in the proof of [77, Proposition 4.1] whereQ_max=√ λ.

ドキュメント内 Numerical Analysis of Multiphase Curvature-driven Interface Evolution with Volume Constraint (ページ 99-109)