Geometric properties of the flow



n_∂Γ1(t)·n_∂Γ2(t) = cosθ₃, n_∂Γ2(t)·n_∂Γ3(t) = cosθ₁, n_∂Γ3(t)·n_∂Γ1(t) = cosθ2.

(4.7)

Due to the conditionθ₁+θ₂+θ₃ = 2π and to the claim (4.3) that the three hypersurfaces meet at one triple line, two of the above angle conditions already imply the third one.

4.1.1 Geometric properties of the flow

In this subsection we want to give some basic properties of the flow (4.5). These properties will be an equivalence between Young’s law and a balance of forces, a property of the normals and conormals of the arising hypersurfaces, decreasing of area and properties of stationary states.

The first point is to show an equivalence of the angle conditions (4.7) and Young’s law (4.6) to a balance of forces given by

γ₁n_∂Γ1(t)+γ₂n_∂Γ2(t)+γ₃n_∂Γ3(t) = 0 on L(t). (4.8) Therefore it is important to observe that the three vectors n_∂Γ1(t),n_∂Γ2(t) and n_∂Γ3(t) all lie in a two-dimensional space, namely the orthogonal complement of the tangent space of L(t), i.e.

n_∂Γi(t)(p)∈(T_pL(t))^⊥ .

SinceL(t) is an (n−1)-dimensional submanifold ofRⁿ⁺¹, this orthogonal complement is in fact a two-dimensional space.

Lemma 4.1. Let θ1, θ2 andθ3 be given contact angles for the conormals n_∂Γi(t) as in (4.7) with 0< θ_i < π and θ₁+θ₂+θ₃ = 2π. Then there holds an equivalence between Young’s law (4.6) and the balance of forces (4.8).

Proof. First, let Young’s law be fulfilled. From the angle conditionn_∂Γ1(t)·n_∂Γ2(t)= cosθ₃ 6=±1, we see thatn_∂Γ1(t) andn_∂Γ2(t) are linearly independent vectors and hence are a basis in the two-dimensional space (T_pL(t))^⊥. Since the three vectors n_∂Γi(t), i= 1,2,3 all lie in this space, we can show instead of the identity (4.8) the two scalar identities

γ₁ n_∂Γ1(t)·n_∂Γ1(t)

+γ₂ n_∂Γ2(t)·n_∂Γ1(t)

+γ₃ n_∂Γ3(t)·n_∂Γ1(t)

= 0 and

γ₁ n_∂Γ1(t)·n_∂Γ2(t)

+γ₂ n_∂Γ2(t)·n_∂Γ2(t)

+γ₃ n_∂Γ3(t)·n_∂Γ2(t)

= 0.

CHAPTER 4. TRIPLE LINES WITH BOUNDARY CONTACT

With the help of the angle conditions (4.7) this reads also as







−^γ_γ¹₃ +

−_γ^γ2₃

cosθ₃ = cosθ₂ and −^γ_γ¹₃

cosθ₃+

−^γ_γ²₃

= cosθ₁. Now Young’s law gives the identities ^γ_γ¹

3 = ^sinθ_sinθ¹

3 and ^γ_γ²

3 = ^sin_sin^θ_θ²

3, so that we have to show







−^sinθ_sinθ¹₃ +

−_sin^sinθ_θ²₃

cosθ₃ = cosθ₂ and −^sinθ_sinθ¹₃

cosθ₃+

−^sinθ_sinθ²₃

= cosθ₁. Multiplying with sinθ3 and rearranging the terms gives

−sinθ₁ = cosθ₂sinθ₃+ sinθ₂cosθ₃ and

−sinθ₂ = cosθ₁sinθ₃+ sinθ₁cosθ₃.

Finally, we observe−sinθ₁ =−sin(2π−θ₂−θ₃) =−sin(−(θ₂+θ₃)) = sin(θ₂+θ₃) (analogously for−sinθ₂), and the above identities are addition theorems, which hold true.

On the other hand, let the balance of forces (4.8) be true. Since the three vectors n_∂Γi(t), i= 1,2,3 lie in a two-dimensional space, (4.8) implies that the three vectorsγ₁n_∂Γ1(t),γ₂n_∂Γ2(t) and γ₃n_∂Γ3(t) can be arranged to give a triangle in this two-dimensional space. The angles in this triangle are labelled through (4.7) and the law of sines gives exactly Young’s law (4.6).

With the help of the following construction we choose a direction of the normals n_i(t) of Γ_i(t).

Remark 4.2. With the same argument as above for the outer unit conormals, we see that at the triple line L(t) even the vectors

±n1(t), n_∂Γ1(t), ±n2(t), n_∂Γ2(t), ±n3(t), n_∂Γ3(t)

all lie in a two-dimensional space, namely the orthogonal complement of the tangent space of L(t).

We choose unit normalsn_j(t) of Γ_j(t) in an appropriate direction through the requirement that the angle betweenn_∂Γi(t) and n_j(t) increases byπ/2 compared to the angle between n_∂Γi(t) and n_∂Γj(t), i.e. we have the following formulas

n_i(t)·n_j(t) = cosθ_k, (4.9)

n_∂Γi(t)·n_∂Γj(t) = cosθ_k, (4.10)

n_∂Γi(t)·n_j(t) = cos(θ_k+π

2) =−sinθ_k, (4.11)

each on L(t) and for (i, j, k) = (1,2,3),(2,3,1) and (3,1,2). To be precise we require formula (4.11) at a fixed point of L(t), extend the normals by continuity to all of Γ_j(t) and observe

4.1. MEAN CURVATURE FLOW

Γ₁ Γ2

Γ₃ n₁

n2

n₃

Figure 4.1: The choice of the normals.

the validity of (4.11) on all of L(t) again by continuity. See Figure 4.1 for a sketch in the two-dimensional situation for curves near the triple line.

With an analogue version of Lemma 4.1 we can write instead of (4.8) also

γ₁n₁(t) +γ₂n₂(t) +γ₃n₃(t) = 0 on L(t). (4.12) In the next lemma we show a decreasing of the weighted total area.

Lemma 4.3. For evolving hypersurfaces which move according to weighted mean curvature flow and fulfill the angle conditions from (4.5), the weighted total area

A(t) = X3

i=1

γ_iA_i(t), (4.13)

with A_i(t) = R

Γi(t)

1dHⁿ, decreases in time t.

Proof. With the proof of the formula for the derivative of area from Lemma 2.46, we see d

dtA_i(t) = − Z

Γi(t)

V_iH_idHⁿ+ Z

Si(t)

v_∂Γi

|{z}=0

dHⁿ⁻¹+ Z

L(t)

v_∂ΓidHⁿ⁻¹,

where the normal boundary velocity v_∂Γi at the outer boundaryS_i(t) vanishes due to the right angle condition as in Lemma 2.46. Therefore we observe for the weighted total area

d

dtA(t) = − X3

i=1

γ_i Z

Γi(t)

V_iH_idHⁿ+ X3 i=1

γ_i Z

L(t)

v_∂ΓidHⁿ⁻¹.

For the normal boundary velocities at the triple line, we observe v_∂Γi(t, p) =n_∂Γi(t, p)· d

dτc_i(τ)

τ=t

for p∈∂Γ_i(t),

CHAPTER 4. TRIPLE LINES WITH BOUNDARY CONTACT

where c_i : (t−ε, t+ε)→ Rⁿ⁺¹ are curves with c_i(τ) ∈∂Γ_i(τ) and c_i(t) = p. Since we require L(t) = ∂Γ₁(t) = ∂Γ₂(t) = ∂Γ₃(t), and since the definition of normal boundary velocity is independent of the curve, we can take one curve for all three hypersurfaces Γi, that is we get

v_∂Γi(t, p) =n_∂Γi(t, p)· d

Plugging this into the derivative of the weighted total area and usingV_i =γ_iH_i, we can deduce d

This shows the lemma.

As in the case for one hypersurface we want to describe Γ_i(t) with the help of functions ρ_i : [0, T)×Γ^∗_i →Ras graphs over some fixed stationary solution of (4.5). This means we fix three hypersurfaces Γ^∗_i, which lie in Ω, and the boundary has a decomposition ∂Γ^∗_i = L^∗_i ∪S_i^∗, such that the three hypersurfaces meet at a triple line L^∗ =L^∗₁ =L^∗₂ =L^∗₃ and the other parts are sections with the outer fixed boundary, i.e. S_i^∗ =∂Γ^∗_i ∩∂Ω.

These hypersurfaces shall fulfill the angle conditions and the mean curvature equations from (4.5) withV_i ≡0. As above we can show that the outer unit conormalsn_∂Γ^∗ For these stationary solutions the following lemma holds.

Lemma 4.4. Stationary hypersurfaces as above are minimal hypersurfaces, i.e. they fulfill H_i^∗ ≡0, and the identity Remember thatσ_i^∗ is our notation for the second fundamental form ofΓ^∗_i with respect to the unit normal n^∗_i.

4.1. MEAN CURVATURE FLOW

Proof. The fact H_i^∗≡0 follows directly from the mean curvature equations with V_i ≡0.

Forq ∈L^∗, we can decompose the tangent space T_qΓ^∗_i with the help of the outer unit conormal n_∂Γ^∗_i of Γ^∗_i at L^∗ into

TqΓ^∗_i = TqL^∗∪span{n_∂Γ^∗_i}. Therefore we can complete n_∂Γ^∗

i to an orthonormal basis {n_∂Γ^∗

i, t₁, . . . , t_n−1} of T_qΓ^∗_i with the help of suitable vectors t₁, . . . , t_n−1 ∈ T_qL^∗. Note that we choose for every i= 1,2,3 the same set of vectorst₁, . . . , t_n−1. Since the mean curvatureH_i^∗ is the trace of the Weingarten map, see Definition 2.19, we obtain the identity

γ_iH_i^∗ = γ_iσ_i^∗(n_∂Γ^∗

i, n_∂Γ^∗

i) +γ_i

n−1X

j=1

σ_i^∗(t_j, t_j). (4.14) With the above result about Γ^∗_i being a minimal hypersurface and with our notation of normal curvature, we can write this as

0 = γ_iκ_n∂Γ∗ i +γ_i

n−1X

j=1

σ^∗_i(t_j, t_j). Summing over i= 1,2,3 gives for the second term

X3 i=1

γ_i

n−1X

j=1

σ^∗_i(t_j, t_j) =−

n−1X

j=1

X3 i=1

γ_i∂_tjn^∗_i ·t_j =−

n−1X

j=1

∂_tj X3 i=1

γ_in^∗_i

!

| {z }

=0 on L^∗

·t_j = 0,

where the last zero appears due to the fact that t_j is a tangent vector of L^∗. For the normal curvatures in direction n_∂Γ^∗_i this gives finally

γ₁κ_n∂Γ∗

1 +γ₂κ_n∂Γ∗

2 +γ₃κ_n∂Γ∗

3 = 0 on L^∗

and we finished the proof.

Im Dokument Stability Analysis of Geometric Evolution Equations with Triple Lines and Boundary Contact (Seite 105-109)