Dimensional estimates for singular sets in geometric variational problems with free boundaries

If f:Rn→Rn is a Lipschitz diffeomorphism with det⁡(∇⁡f)>0 on Rn, then f⁢(E) is a set of finite perimeter in f⁢(A), with f⁢(∂*⁡E)=Hn-1∂*⁡(f⁢(E)) and

where for any invertible linear map L:Rn→Rn one defines cof⁡L=(det⁡L)⁢(L-1)*. Moreover, for every G⊂A, one has

If X,Y:Rn→Rn are linear maps, then

and thus

If g∈C2⁢(A), then

and

Proof.

Step one. We notice the validity of the following formula: if S∈Cc1⁢(A;ℝn) and E⊂Ω, then

Indeed, if S∈Cc2⁢(A;ℝn), then the assertion follows by the Divergence Theorem and by the identity

The case when S∈Cc1⁢(A;ℝn) is then obtained by approximation.

Step two. Since ft⁢(A)=A, we find ft⁢(E)∩A=ft⁢(E∩A). Hence by the area formula,

By (A.3), (A.7) and the Taylor expansion of g, we get

Inasmuch, div⁡(g⁢T)=∇⁡g⋅T+g⁢div⁡T and div⁡(g⁢Z)=∇⁡g⋅Z+g⁢div⁡Z, by step one and by (A.4), one finds (A.8) and

We now complete the proof of (A.9) by showing that

Indeed, by differentiating (A.4) along T one finds 0=∇⁡T⁢[T]⋅νΩ+T⋅IIΩ⁢[T], and then we conclude by (A.5). ∎

We have

and

Proof.

By (A.3), Lemma 2, and by the Taylor expansion of Φ at νE, we get

where we have also used Φ⁢(νE)=∇⁡Φ⁢(νE)⋅νE and ∇2⁡Φ⁢(νE)⁢[νE]=0. By equation (A.6) and by ft⁢(A)=A, we find (A.10) and (A.11). ∎

Let Φ∈E*⁢(λ), g∈C2⁢(Rn), A be a bounded open set, H be an open half-space and E be a minimizer of Φ+∫g on (A,H). Then

for every function ζ∈Cc1⁢(A) with the property that spt⁡ζ∩ΣA⁢(E)=∅. Moreover, there exists a constant C=C⁢(n,λ,Lip⁡(g)) such that

whenever ζ∈Cc1⁢(A) with spt⁡ζ∩ΣA⁢(E)=∅.

Proof.

As proved in [1, Section 2.4] we have

If ζ∈Cc1⁢(A∖ΣA⁢(E)), then there exists an N∈C1⁢(ℝn;ℝn) such that

We set T=ζ⁢∇⁡Φ⁢(N)∈Cc1⁢(A;ℝn) and we note that, by (A.15), ft⁢(x)=x+t⁢T⁢(x) defines a family of admissible variations for |t|≤ε0 and ε0 suitably small. Since ft is affine in t, by (A.2), one has Z=0. In particular, by Lemma 1, Lemma 2, and by minimality of E,

where, setting for simplicity Φ=Φ⁢(νE), ∇⁡Φ=∇⁡Φ⁢(νE), and ∇2⁡Φ=∇2⁡Φ⁢(νE), one has

We start by noticing that (A.14) gives

where IIE⁢(x) is extended to be zero on (Tx⁢RA⁢(E))⊥ and a:RA⁢(E)→ℝn is a continuous vector field. Hence

By ∇2⁡Φ⁢[νE]=0 and the symmetry of ∇2⁡Φ, one finds tr⁡(∇2⁡Φ⁢[a]⊗νE)=0, so that

where we have set

Moreover, by ∇⁡Φ⋅νE=Φ and again by ∇2⁡Φ⁢[νE]=0, we find

so that (A.16) gives

The validity of this condition for every ζ∈Cc1⁢(A∖ΣA⁢(E)) gives the well-know stationarity condition

We now compute Γ1. By ∇⁡Φ⋅νE=Φ, we find

so that, by IIEΦ⁢[∇⁡Φ]⋅νE=0 and by ∇2⁡Φ⁢[a]⋅νE=0 (which follow by the symmetry of ∇2⁡Φ and by ∇2⁡Φ⁢[ν]=0), we find

By (A.17), (A.18) and a simple computation, one gets

We now start computing Γ2. By (A.17), we have

at the same time, writing ∇⁡T=X+Y where X=∇⁡Φ⊗∇⁡ζ+ζ⁢IIEΦ and Y=ζ⁢∇2⁡Φ⁢[a]⊗νE, and noticing that Y2=0, while

we find that,

Hence,

We now compute Γ3. By (A.17) and (∇⁡T)*⁢[νE]=Φ⁢∇⁡ζ, we find

At the same time, writing ∇⁡T=X+Y with X and Y as above, we find

By taking into account the fact that (Y*)2=0 (as Y2=0) and by exploiting once more that ∇2⁡Φ⁢[νE]=0 and IIEΦ⁢[νE]=0, we find that

so that

In conclusion,

so that

On noticing that Γ4=Φ2⁢∇2⁡Φ⁢[∇⁡ζ]⋅∇⁡ζ, we conclude the proof of (A.12). By (1.1), one has

and thus

Hence, (A.12) implies (A.13). ∎