K-surfaces with free boundaries Hayk Aleksanyan KTH Royal Institute - PowerPoint PPT Presentation

Can we leave part of the boundary free? Fix Γ = { Γ 1 , ..., Γ m } a collection of disjoint ( d − 1)-dimensional closed smooth embedded submanifolds of R d +1 , and let T 0 be a smooth embedded submanifold in R d +1 of codimension 1. Fix also an angle θ > 0. K -surfaces with (Bernoulli) free boundary What conditions should be imposed on Γ, T 0 , and θ in order to get a K -surface spanning Γ and hitting T 0 at an angle θ ? A model case: (The boundary) Take Γ = ∂ Ω where Ω ⊂ R d × { h 0 } is strictly convex, and h 0 > 0. 5/25 Hayk Aleksanyan K-surfaces with free boundaries

Can we leave part of the boundary free? Fix Γ = { Γ 1 , ..., Γ m } a collection of disjoint ( d − 1)-dimensional closed smooth embedded submanifolds of R d +1 , and let T 0 be a smooth embedded submanifold in R d +1 of codimension 1. Fix also an angle θ > 0. K -surfaces with (Bernoulli) free boundary What conditions should be imposed on Γ, T 0 , and θ in order to get a K -surface spanning Γ and hitting T 0 at an angle θ ? A model case: (The boundary) Take Γ = ∂ Ω where Ω ⊂ R d × { h 0 } is strictly convex, and h 0 > 0. (The target manifold) T 0 = R d × { 0 } . 5/25 Hayk Aleksanyan K-surfaces with free boundaries

Can we leave part of the boundary free? Fix Γ = { Γ 1 , ..., Γ m } a collection of disjoint ( d − 1)-dimensional closed smooth embedded submanifolds of R d +1 , and let T 0 be a smooth embedded submanifold in R d +1 of codimension 1. Fix also an angle θ > 0. K -surfaces with (Bernoulli) free boundary What conditions should be imposed on Γ, T 0 , and θ in order to get a K -surface spanning Γ and hitting T 0 at an angle θ ? A model case: (The boundary) Take Γ = ∂ Ω where Ω ⊂ R d × { h 0 } is strictly convex, and h 0 > 0. (The target manifold) T 0 = R d × { 0 } . (The hitting angle) θ = arccos(1 + λ 0 ) − 1 / 2 , for some λ 0 > 0. 5/25 Hayk Aleksanyan K-surfaces with free boundaries

Our setting: formal statement For a convex domain Ω ⊂ R d × { 0 } and parameters h 0 , λ 0 > 0, K 0 ≥ 0, find a concave function u : R d × { 0 } → R + such that  det D 2 ( − u ) = K 0 ψ ( |∇ u | ) ,  in { u > 0 } \ Ω ,  u = h 0 , on ∂ Ω ,   |∇ u | = λ 0 , on Γ u where ψ > 0 is a prescribed real-valued C ∞ function, Γ u = ∂ { u > 0 } \ Ω. 6/25 Hayk Aleksanyan K-surfaces with free boundaries

Our setting: formal statement For a convex domain Ω ⊂ R d × { 0 } and parameters h 0 , λ 0 > 0, K 0 ≥ 0, find a concave function u : R d × { 0 } → R + such that  det D 2 ( − u ) = K 0 ψ ( |∇ u | ) ,  in { u > 0 } \ Ω ,  u = h 0 , on ∂ Ω ,   |∇ u | = λ 0 , on Γ u where ψ > 0 is a prescribed real-valued C ∞ function, Γ u = ∂ { u > 0 } \ Ω. Geometry relevant choices: (the homogeneous equation) K 0 = 0, 6/25 Hayk Aleksanyan K-surfaces with free boundaries

Our setting: formal statement For a convex domain Ω ⊂ R d × { 0 } and parameters h 0 , λ 0 > 0, K 0 ≥ 0, find a concave function u : R d × { 0 } → R + such that  det D 2 ( − u ) = K 0 ψ ( |∇ u | ) ,  in { u > 0 } \ Ω ,  u = h 0 , on ∂ Ω ,   |∇ u | = λ 0 , on Γ u where ψ > 0 is a prescribed real-valued C ∞ function, Γ u = ∂ { u > 0 } \ Ω. Geometry relevant choices: (the homogeneous equation) K 0 = 0, (constant curvature measure) ψ ≡ 1, 6/25 Hayk Aleksanyan K-surfaces with free boundaries

Our setting: formal statement For a convex domain Ω ⊂ R d × { 0 } and parameters h 0 , λ 0 > 0, K 0 ≥ 0, find a concave function u : R d × { 0 } → R + such that  det D 2 ( − u ) = K 0 ψ ( |∇ u | ) ,  in { u > 0 } \ Ω ,  u = h 0 , on ∂ Ω ,   |∇ u | = λ 0 , on Γ u where ψ > 0 is a prescribed real-valued C ∞ function, Γ u = ∂ { u > 0 } \ Ω. Geometry relevant choices: (the homogeneous equation) K 0 = 0, (constant curvature measure) ψ ≡ 1, (constant Gauss curvature) ψ ( t ) = (1 + t 2 ) ( d +2) / 2 6/25 Hayk Aleksanyan K-surfaces with free boundaries

Our setting: formal statement For a convex domain Ω ⊂ R d × { 0 } and parameters h 0 , λ 0 > 0, K 0 ≥ 0, find a concave function u : R d × { 0 } → R + such that  det D 2 ( − u ) = K 0 ψ ( |∇ u | ) ,  in { u > 0 } \ Ω ,  u = h 0 , on ∂ Ω ,   |∇ u | = λ 0 , on Γ u where ψ > 0 is a prescribed real-valued C ∞ function, Γ u = ∂ { u > 0 } \ Ω. Geometry relevant choices: (the homogeneous equation) K 0 = 0, (constant curvature measure) ψ ≡ 1, (constant Gauss curvature) ψ ( t ) = (1 + t 2 ) ( d +2) / 2 For p -Laplace equation, see [A. Henrot and H. Shahgholian; J. Reine und Angew. Math 2000], although methods and motivation are entirely different here. 6/25 Hayk Aleksanyan K-surfaces with free boundaries

Weak solutions (` a la A.D. Aleksandrov) 7/25 Hayk Aleksanyan K-surfaces with free boundaries

Weak solutions (` a la A.D. Aleksandrov) The gradient mapping Let u : Ω → R be convex, x 0 ∈ Ω. 7/25 Hayk Aleksanyan K-surfaces with free boundaries

Weak solutions (` a la A.D. Aleksandrov) The gradient mapping Let u : Ω → R be convex, x 0 ∈ Ω. The set of slopes ω x 0 ( u ) = { p ∈ R d : u ( x ) ≥ u ( x 0 ) + p · ( x − x 0 ) , ∀ x ∈ Ω } is called the gradient mapping of u at x 0 . 7/25 Hayk Aleksanyan K-surfaces with free boundaries

Weak solutions (` a la A.D. Aleksandrov) The gradient mapping Let u : Ω → R be convex, x 0 ∈ Ω. The set of slopes ω x 0 ( u ) = { p ∈ R d : u ( x ) ≥ u ( x 0 ) + p · ( x − x 0 ) , ∀ x ∈ Ω } is called the gradient mapping of u at x 0 . For a set E ⊂ Ω set ω E ( u ) = � ω x ( u ). x ∈ E 7/25 Hayk Aleksanyan K-surfaces with free boundaries

Weak solutions (` a la A.D. Aleksandrov) The gradient mapping Let u : Ω → R be convex, x 0 ∈ Ω. The set of slopes ω x 0 ( u ) = { p ∈ R d : u ( x ) ≥ u ( x 0 ) + p · ( x − x 0 ) , ∀ x ∈ Ω } is called the gradient mapping of u at x 0 . For a set E ⊂ Ω set ω E ( u ) = � ω x ( u ). x ∈ E The Monge-Amp` ere measure We call a convex u : E → R a solution to det D 2 u = K 0 ψ ( |∇ u | ) on � d ξ E , if for any Borel set B ⊂ E one has ψ ( | ξ | ) = K 0 | B | . ω B ( u ) 7/25 Hayk Aleksanyan K-surfaces with free boundaries

Weak solutions (` a la A.D. Aleksandrov) The gradient mapping Let u : Ω → R be convex, x 0 ∈ Ω. The set of slopes ω x 0 ( u ) = { p ∈ R d : u ( x ) ≥ u ( x 0 ) + p · ( x − x 0 ) , ∀ x ∈ Ω } is called the gradient mapping of u at x 0 . For a set E ⊂ Ω set ω E ( u ) = � ω x ( u ). x ∈ E The Monge-Amp` ere measure We call a convex u : E → R a solution to det D 2 u = K 0 ψ ( |∇ u | ) on � d ξ E , if for any Borel set B ⊂ E one has ψ ( | ξ | ) = K 0 | B | . ω B ( u ) The l.h.s. is called the Monge-Amp` ere measure. 7/25 Hayk Aleksanyan K-surfaces with free boundaries

Weak solutions (` a la A.D. Aleksandrov) The gradient mapping Let u : Ω → R be convex, x 0 ∈ Ω. The set of slopes ω x 0 ( u ) = { p ∈ R d : u ( x ) ≥ u ( x 0 ) + p · ( x − x 0 ) , ∀ x ∈ Ω } is called the gradient mapping of u at x 0 . For a set E ⊂ Ω set ω E ( u ) = � ω x ( u ). x ∈ E The Monge-Amp` ere measure We call a convex u : E → R a solution to det D 2 u = K 0 ψ ( |∇ u | ) on � d ξ E , if for any Borel set B ⊂ E one has ψ ( | ξ | ) = K 0 | B | . ω B ( u ) The l.h.s. is called the Monge-Amp` ere measure. The MA measure is weakly* continuous. 7/25 Hayk Aleksanyan K-surfaces with free boundaries

The free boundary condition 8/25 Hayk Aleksanyan K-surfaces with free boundaries

The free boundary condition Fix any regular point x 0 ∈ Γ u , i.e. Γ u has a well-defined inner normal (call it ν ). 8/25 Hayk Aleksanyan K-surfaces with free boundaries

The free boundary condition Fix any regular point x 0 ∈ Γ u , i.e. Γ u has a well-defined inner normal (call it ν ). The condition |∇ u ( x 0 ) | = λ 0 means ∂ u ∂ν ( x 0 ) = λ 0 , which always exists: 8/25 Hayk Aleksanyan K-surfaces with free boundaries

The free boundary condition Fix any regular point x 0 ∈ Γ u , i.e. Γ u has a well-defined inner normal (call it ν ). The condition |∇ u ( x 0 ) | = λ 0 means ∂ u ∂ν ( x 0 ) = λ 0 , which always exists: by concavity of u , for any t 2 > t 1 > 0 we get �� � � 1 − t 1 x 0 + t 1 ≥ t 1 u ( x 0 + t 1 ν ) = u ( x 0 + t 2 ν ) u ( x 0 + t 2 ν ) . t 2 t 2 t 2 8/25 Hayk Aleksanyan K-surfaces with free boundaries

The free boundary condition Fix any regular point x 0 ∈ Γ u , i.e. Γ u has a well-defined inner normal (call it ν ). The condition |∇ u ( x 0 ) | = λ 0 means ∂ u ∂ν ( x 0 ) = λ 0 , which always exists: by concavity of u , for any t 2 > t 1 > 0 we get �� � � 1 − t 1 x 0 + t 1 ≥ t 1 u ( x 0 + t 1 ν ) = u ( x 0 + t 2 ν ) u ( x 0 + t 2 ν ) . t 2 t 2 t 2 ( Geometric reformulation ) There is a unique support plane G in R d × { 0 } for Γ u at x 0 . 8/25 Hayk Aleksanyan K-surfaces with free boundaries

The free boundary condition Fix any regular point x 0 ∈ Γ u , i.e. Γ u has a well-defined inner normal (call it ν ). The condition |∇ u ( x 0 ) | = λ 0 means ∂ u ∂ν ( x 0 ) = λ 0 , which always exists: by concavity of u , for any t 2 > t 1 > 0 we get �� � � 1 − t 1 x 0 + t 1 ≥ t 1 u ( x 0 + t 1 ν ) = u ( x 0 + t 2 ν ) u ( x 0 + t 2 ν ) . t 2 t 2 t 2 ( Geometric reformulation ) There is a unique support plane G in R d × { 0 } for Γ u at x 0 . Any support hyperplane H to the graph( u ) at ( x 0 , 0) ∈ R d × R , must pass through G . 8/25 Hayk Aleksanyan K-surfaces with free boundaries

The free boundary condition Fix any regular point x 0 ∈ Γ u , i.e. Γ u has a well-defined inner normal (call it ν ). The condition |∇ u ( x 0 ) | = λ 0 means ∂ u ∂ν ( x 0 ) = λ 0 , which always exists: by concavity of u , for any t 2 > t 1 > 0 we get �� � � 1 − t 1 x 0 + t 1 ≥ t 1 u ( x 0 + t 1 ν ) = u ( x 0 + t 2 ν ) u ( x 0 + t 2 ν ) . t 2 t 2 t 2 ( Geometric reformulation ) There is a unique support plane G in R d × { 0 } for Γ u at x 0 . Any support hyperplane H to the graph( u ) at ( x 0 , 0) ∈ R d × R , must pass through G . Hence, there is one free parameter, the slope of H . 8/25 Hayk Aleksanyan K-surfaces with free boundaries

The free boundary condition Fix any regular point x 0 ∈ Γ u , i.e. Γ u has a well-defined inner normal (call it ν ). The condition |∇ u ( x 0 ) | = λ 0 means ∂ u ∂ν ( x 0 ) = λ 0 , which always exists: by concavity of u , for any t 2 > t 1 > 0 we get �� � � 1 − t 1 x 0 + t 1 ≥ t 1 u ( x 0 + t 1 ν ) = u ( x 0 + t 2 ν ) u ( x 0 + t 2 ν ) . t 2 t 2 t 2 ( Geometric reformulation ) There is a unique support plane G in R d × { 0 } for Γ u at x 0 . Any support hyperplane H to the graph( u ) at ( x 0 , 0) ∈ R d × R , must pass through G . Hence, there is one free parameter, the slope of H . The extreme H (i.e. the “most inclined on the graph”) must have slope λ 0 . 8/25 Hayk Aleksanyan K-surfaces with free boundaries

The main results: homogeneous case Theorem A ( K 0 = 0, the homogeneous case) Let K 0 = 0, and Ω ⊂ R d be bounded convex C 1 , 1 -regular domain. Then, there exists a unique weak solution u . 9/25 Hayk Aleksanyan K-surfaces with free boundaries

The main results: homogeneous case Theorem A ( K 0 = 0, the homogeneous case) Let K 0 = 0, and Ω ⊂ R d be bounded convex C 1 , 1 -regular domain. Then, there exists a unique weak solution u . Moreover the graph of u is a ruled surface, u is C 1 , 1 on { u > 0 } \ Ω, the free boundary Γ u is C 1 , 1 , if in addition, Ω is strictly convex, then so is the free boundary. 9/25 Hayk Aleksanyan K-surfaces with free boundaries

The main results: homogeneous case Theorem A ( K 0 = 0, the homogeneous case) Let K 0 = 0, and Ω ⊂ R d be bounded convex C 1 , 1 -regular domain. Then, there exists a unique weak solution u . Moreover the graph of u is a ruled surface, u is C 1 , 1 on { u > 0 } \ Ω, the free boundary Γ u is C 1 , 1 , if in addition, Ω is strictly convex, then so is the free boundary. Example (truncated cone) Take Ω = B ( x 0 , r ) in R d ( d ≥ 2). Fix λ 0 > 0 and h 0 = 1. Then � � u ( x ) = 1 + λ 0 − λ 0 1 + 1 r | x − x 0 | , r ≤ | x − x 0 | ≤ r λ 0 is the solution, with free boundary | x − x 0 | = r (1 + 1 /λ 0 ). 9/25 Hayk Aleksanyan K-surfaces with free boundaries

The main results: elliptic case Theorem B ( K 0 > 0, the strictly convex (elliptic) case) Let K 0 > 0, and Ω ⊂ R d be bounded strictly convex smooth domain. Let also ψ : R + → (0 , ∞ ) be non-decreasing and smooth. 10/25 Hayk Aleksanyan K-surfaces with free boundaries

The main results: elliptic case Theorem B ( K 0 > 0, the strictly convex (elliptic) case) Let K 0 > 0, and Ω ⊂ R d be bounded strictly convex smooth domain. Let also ψ : R + → (0 , ∞ ) be non-decreasing and smooth. Then, there exists a small constant K = K (Ω , ψ, λ 0 ) > 0, such that for any K 0 ∈ (0 , K ) there exists a weak solution u , which is C ∞ on { u > 0 } \ Ω and the free boundary Γ u is C ∞ as well. 10/25 Hayk Aleksanyan K-surfaces with free boundaries

The main results: elliptic case Theorem B ( K 0 > 0, the strictly convex (elliptic) case) Let K 0 > 0, and Ω ⊂ R d be bounded strictly convex smooth domain. Let also ψ : R + → (0 , ∞ ) be non-decreasing and smooth. Then, there exists a small constant K = K (Ω , ψ, λ 0 ) > 0, such that for any K 0 ∈ (0 , K ) there exists a weak solution u , which is C ∞ on { u > 0 } \ Ω and the free boundary Γ u is C ∞ as well. On NON-existence The smallness of K 0 cannot be eliminated entirely! 10/25 Hayk Aleksanyan K-surfaces with free boundaries

The main results: elliptic case Theorem B ( K 0 > 0, the strictly convex (elliptic) case) Let K 0 > 0, and Ω ⊂ R d be bounded strictly convex smooth domain. Let also ψ : R + → (0 , ∞ ) be non-decreasing and smooth. Then, there exists a small constant K = K (Ω , ψ, λ 0 ) > 0, such that for any K 0 ∈ (0 , K ) there exists a weak solution u , which is C ∞ on { u > 0 } \ Ω and the free boundary Γ u is C ∞ as well. On NON-existence The smallness of K 0 cannot be eliminated entirely! Work out the case of radial solutions (when Ω is a ball) by hand. 10/25 Hayk Aleksanyan K-surfaces with free boundaries

Some ideas of the proofs: the homogeneous case H X 0 X 0 � Ω Ω Y 0 R d × { 0 } A scematic view for the homogeneous case. 11/25 Hayk Aleksanyan K-surfaces with free boundaries

Convex polygonal domains Let Ω ⊂ R d × { 0 } be a convex polygon , and let F 1 , ..., F n be the facets of � Ω := Ω × { h 0 } . 12/25 Hayk Aleksanyan K-surfaces with free boundaries

Convex polygonal domains Let Ω ⊂ R d × { 0 } be a convex polygon , and let F 1 , ..., F n be the facets of � Ω := Ω × { h 0 } . For each 1 ≤ i ≤ n let H i be the hyperplane in R d +1 passing through F i and having slope λ 0 . Identify each H i with the linear function. 12/25 Hayk Aleksanyan K-surfaces with free boundaries

Convex polygonal domains Let Ω ⊂ R d × { 0 } be a convex polygon , and let F 1 , ..., F n be the facets of � Ω := Ω × { h 0 } . For each 1 ≤ i ≤ n let H i be the hyperplane in R d +1 passing through F i and having slope λ 0 . Identify each H i with the linear function. Then u ( x ) = inf 1 ≤ i ≤ n H i ( x ), x ∈ R d , solves the homogeneous problem. 12/25 Hayk Aleksanyan K-surfaces with free boundaries

Convex polygonal domains Let Ω ⊂ R d × { 0 } be a convex polygon , and let F 1 , ..., F n be the facets of � Ω := Ω × { h 0 } . For each 1 ≤ i ≤ n let H i be the hyperplane in R d +1 passing through F i and having slope λ 0 . Identify each H i with the linear function. Then u ( x ) = inf 1 ≤ i ≤ n H i ( x ), x ∈ R d , solves the homogeneous problem. The most delicate part is to show that there is no X ∈ R d +1 in the strip 0 < x d +1 < h 0 where more than d planes meet, i.e. the graph of u has NO vertex (a geometric proof). 12/25 Hayk Aleksanyan K-surfaces with free boundaries

Approximation by polygons: existence Let Ω be bounded, convex and C 1 . For each X 0 ∈ � Ω there is a support hyperplane H X 0 in R d +1 through X 0 and having slope λ 0 . 13/25 Hayk Aleksanyan K-surfaces with free boundaries

Approximation by polygons: existence Let Ω be bounded, convex and C 1 . For each X 0 ∈ � Ω there is a support hyperplane H X 0 in R d +1 through X 0 and having slope λ 0 . Define x ∈ R d . h ∗ ( x ) = inf H X 0 ( x ) , X 0 ∈ ∂ � Ω The infimum does not collapse due to the uniform bound on the slopes. 13/25 Hayk Aleksanyan K-surfaces with free boundaries

Approximation by polygons: existence Let Ω be bounded, convex and C 1 . For each X 0 ∈ � Ω there is a support hyperplane H X 0 in R d +1 through X 0 and having slope λ 0 . Define x ∈ R d . h ∗ ( x ) = inf H X 0 ( x ) , X 0 ∈ ∂ � Ω The infimum does not collapse due to the uniform bound on the slopes. Approximate Ω by polygonal domains, and for each polygon take the solution constructed above. Then, the limit will converge to h ∗ and will give a weak solution for the homogeneous problem (uses the weak* continuity of MA measure). 13/25 Hayk Aleksanyan K-surfaces with free boundaries

Every weak solution is a ruled surface Proposition (Line segments on the graph) Let u be any weak solution, and assume X 0 is on the graph of u . Then, there is a line segment though X 0 joining the free boundary with R d × { h 0 } and lying entirely on the graph of u . 14/25 Hayk Aleksanyan K-surfaces with free boundaries

Every weak solution is a ruled surface Proposition (Line segments on the graph) Let u be any weak solution, and assume X 0 is on the graph of u . Then, there is a line segment though X 0 joining the free boundary with R d × { h 0 } and lying entirely on the graph of u . Proof. For a weak solution u fix X 0 in the interior of M := graph( u ). 14/25 Hayk Aleksanyan K-surfaces with free boundaries

Every weak solution is a ruled surface Proposition (Line segments on the graph) Let u be any weak solution, and assume X 0 is on the graph of u . Then, there is a line segment though X 0 joining the free boundary with R d × { h 0 } and lying entirely on the graph of u . Proof. For a weak solution u fix X 0 in the interior of M := graph( u ). Fix a support hyperplane Π to M through X 0 , and define X := Hull (Π ∩ M ); we need to see that X intersects the h 0 - and 0-level surfaces of u . 14/25 Hayk Aleksanyan K-surfaces with free boundaries

Every weak solution is a ruled surface Proposition (Line segments on the graph) Let u be any weak solution, and assume X 0 is on the graph of u . Then, there is a line segment though X 0 joining the free boundary with R d × { h 0 } and lying entirely on the graph of u . Proof. For a weak solution u fix X 0 in the interior of M := graph( u ). Fix a support hyperplane Π to M through X 0 , and define X := Hull (Π ∩ M ); we need to see that X intersects the h 0 - and 0-level surfaces of u . Assume NOT, then we can squeeze a strictly convex surface “between” Π and M (using “smoothing of polytopes” after M. Ghomi), violating the condition det D 2 u = 0. 14/25 Hayk Aleksanyan K-surfaces with free boundaries

Every weak solution is a ruled surface Proposition (Line segments on the graph) Let u be any weak solution, and assume X 0 is on the graph of u . Then, there is a line segment though X 0 joining the free boundary with R d × { h 0 } and lying entirely on the graph of u . Proof. For a weak solution u fix X 0 in the interior of M := graph( u ). Fix a support hyperplane Π to M through X 0 , and define X := Hull (Π ∩ M ); we need to see that X intersects the h 0 - and 0-level surfaces of u . Assume NOT, then we can squeeze a strictly convex surface “between” Π and M (using “smoothing of polytopes” after M. Ghomi), violating the condition det D 2 u = 0. The case when X 0 ∈ ∂ � Ω ∪ Γ u follows by approximation. � 14/25 Hayk Aleksanyan K-surfaces with free boundaries

Comparison principle Proposition Let Ω 1 ⊂ Ω 2 be convex domains, and let a concave function u i be a weak solutions for Ω i , i = 1 , 2. Define ω i := Hull (Γ i ). Then 15/25 Hayk Aleksanyan K-surfaces with free boundaries

Comparison principle Proposition Let Ω 1 ⊂ Ω 2 be convex domains, and let a concave function u i be a weak solutions for Ω i , i = 1 , 2. Define ω i := Hull (Γ i ). Then if either of Γ i is C 1 , then ω 1 ⊂ ω 2 , 15/25 Hayk Aleksanyan K-surfaces with free boundaries

Comparison principle Proposition Let Ω 1 ⊂ Ω 2 be convex domains, and let a concave function u i be a weak solutions for Ω i , i = 1 , 2. Define ω i := Hull (Γ i ). Then if either of Γ i is C 1 , then ω 1 ⊂ ω 2 , if either of u i is C 1 in { u i > 0 } \ Ω, then U 1 ≤ U 2 , where U i is the extension of u i into Ω i as identically h 0 . 15/25 Hayk Aleksanyan K-surfaces with free boundaries

Comparison principle Proposition Let Ω 1 ⊂ Ω 2 be convex domains, and let a concave function u i be a weak solutions for Ω i , i = 1 , 2. Define ω i := Hull (Γ i ). Then if either of Γ i is C 1 , then ω 1 ⊂ ω 2 , if either of u i is C 1 in { u i > 0 } \ Ω, then U 1 ≤ U 2 , where U i is the extension of u i into Ω i as identically h 0 . Proof. Argue by contradiction, and use the existence of line segments on the graphs. � . 15/25 Hayk Aleksanyan K-surfaces with free boundaries

Regularity of a weak solution and free boundary Proposition Let Ω be bounded convex C 1 , 1 -regular domain, and let Ω H X 0 ( x ), x ∈ R d . Then, Γ h ∗ is C 1 , 1 and h ∗ is C 1 , 1 h ∗ ( x ) = inf X 0 ∈ ∂ � in R d \ Ω. 16/25 Hayk Aleksanyan K-surfaces with free boundaries

Regularity of a weak solution and free boundary Proposition Let Ω be bounded convex C 1 , 1 -regular domain, and let Ω H X 0 ( x ), x ∈ R d . Then, Γ h ∗ is C 1 , 1 and h ∗ is C 1 , 1 h ∗ ( x ) = inf X 0 ∈ ∂ � in R d \ Ω. The proof: follow the shared line segment. H X 0 X 0 � Ω Ω Y 0 R d × { 0 } 16/25 Hayk Aleksanyan K-surfaces with free boundaries

Uniqueness and strict convexity 17/25 Hayk Aleksanyan K-surfaces with free boundaries

Uniqueness and strict convexity If Ω is C 1 , 1 , then h ∗ is C 1 , 1 , and has C 1 , 1 free boundary. 17/25 Hayk Aleksanyan K-surfaces with free boundaries

Uniqueness and strict convexity If Ω is C 1 , 1 , then h ∗ is C 1 , 1 , and has C 1 , 1 free boundary. Then, any weak solution can be compared with h ∗ , hence the uniqueness. 17/25 Hayk Aleksanyan K-surfaces with free boundaries

Uniqueness and strict convexity If Ω is C 1 , 1 , then h ∗ is C 1 , 1 , and has C 1 , 1 free boundary. Then, any weak solution can be compared with h ∗ , hence the uniqueness. Strict convexity of h ∗ follows by comparison with conical solutions. A quantitative version of strict convexity follows from Blaschke inclusion principle and comparison of the solution with conical barriers (from above). 17/25 Hayk Aleksanyan K-surfaces with free boundaries

Elliptic case, K 0 > 0, the strategy (The class of super-solutions ) concave functions u ∈ W + ( K 0 , λ 0 , Ω) s.t. u = h 0 on ∂ Ω and det D 2 ( − u ) ≥ K 0 ψ ( |∇ u | ) on { u > 0 }\ Ω and |∇ u | ≤ λ 0 on Γ u . 18/25 Hayk Aleksanyan K-surfaces with free boundaries

Elliptic case, K 0 > 0, the strategy (The class of super-solutions ) concave functions u ∈ W + ( K 0 , λ 0 , Ω) s.t. u = h 0 on ∂ Ω and det D 2 ( − u ) ≥ K 0 ψ ( |∇ u | ) on { u > 0 }\ Ω and |∇ u | ≤ λ 0 on Γ u . Show that W + � = ∅ (by construction, an envelope of certain paraboloids). 18/25 Hayk Aleksanyan K-surfaces with free boundaries

Elliptic case, K 0 > 0, the strategy (The class of super-solutions ) concave functions u ∈ W + ( K 0 , λ 0 , Ω) s.t. u = h 0 on ∂ Ω and det D 2 ( − u ) ≥ K 0 ψ ( |∇ u | ) on { u > 0 }\ Ω and |∇ u | ≤ λ 0 on Γ u . Show that W + � = ∅ (by construction, an envelope of certain paraboloids). (Perron’s method) Show that there is a minimal element in W + , and that it solves the problem. The free boundary condition is the most delicate part (is being handled by a special type of extension, which we named Blaschke extension ). 18/25 Hayk Aleksanyan K-surfaces with free boundaries

Elliptic case, K 0 > 0, the strategy (The class of super-solutions ) concave functions u ∈ W + ( K 0 , λ 0 , Ω) s.t. u = h 0 on ∂ Ω and det D 2 ( − u ) ≥ K 0 ψ ( |∇ u | ) on { u > 0 }\ Ω and |∇ u | ≤ λ 0 on Γ u . Show that W + � = ∅ (by construction, an envelope of certain paraboloids). (Perron’s method) Show that there is a minimal element in W + , and that it solves the problem. The free boundary condition is the most delicate part (is being handled by a special type of extension, which we named Blaschke extension ). (For smoothness of the free boundary) extend the solution beyond the free boundary, to reduce the matters to interior case. 18/25 Hayk Aleksanyan K-surfaces with free boundaries

Construction of super-solution Assumptions: Ω is bounded, strictly convex and C 2 , ψ is non-decreasing (need to adjust the free boundary condition) and smooth. 19/25 Hayk Aleksanyan K-surfaces with free boundaries

Construction of super-solution Assumptions: Ω is bounded, strictly convex and C 2 , ψ is non-decreasing (need to adjust the free boundary condition) and smooth. Let κ 0 > 0 be the smallest principal curvature of ∂ Ω. Then, Ω rolls freely inside a ball of radius r 0 := 1 /κ 0 (W. Blaschke’s rolling ball theorem (2d case), and [J. Rauch, JDG, 1974] for d > 2). ( Intuition : A “more curved” fits inside the “less curved” one). 19/25 Hayk Aleksanyan K-surfaces with free boundaries

Construction of super-solution Assumptions: Ω is bounded, strictly convex and C 2 , ψ is non-decreasing (need to adjust the free boundary condition) and smooth. Let κ 0 > 0 be the smallest principal curvature of ∂ Ω. Then, Ω rolls freely inside a ball of radius r 0 := 1 /κ 0 (W. Blaschke’s rolling ball theorem (2d case), and [J. Rauch, JDG, 1974] for d > 2). ( Intuition : A “more curved” fits inside the “less curved” one). If x 0 ∈ ∂ Ω is fixed, and the ball B = B ( z 0 , r 0 ) touches Ω at x 0 and Ω ⊂ B , then the paraboloid P ( x ) = h 0 + α r 2 0 − α | x − z 0 | 2 (with a properly chosen α > 0) satisfies 19/25 Hayk Aleksanyan K-surfaces with free boundaries

Construction of super-solution Assumptions: Ω is bounded, strictly convex and C 2 , ψ is non-decreasing (need to adjust the free boundary condition) and smooth. Let κ 0 > 0 be the smallest principal curvature of ∂ Ω. Then, Ω rolls freely inside a ball of radius r 0 := 1 /κ 0 (W. Blaschke’s rolling ball theorem (2d case), and [J. Rauch, JDG, 1974] for d > 2). ( Intuition : A “more curved” fits inside the “less curved” one). If x 0 ∈ ∂ Ω is fixed, and the ball B = B ( z 0 , r 0 ) touches Ω at x 0 and Ω ⊂ B , then the paraboloid P ( x ) = h 0 + α r 2 0 − α | x − z 0 | 2 (with a properly chosen α > 0) satisfies P ( x 0 ) = h 0 and P ( x ) ≥ h 0 on Ω, 19/25 Hayk Aleksanyan K-surfaces with free boundaries

Construction of super-solution Assumptions: Ω is bounded, strictly convex and C 2 , ψ is non-decreasing (need to adjust the free boundary condition) and smooth. Let κ 0 > 0 be the smallest principal curvature of ∂ Ω. Then, Ω rolls freely inside a ball of radius r 0 := 1 /κ 0 (W. Blaschke’s rolling ball theorem (2d case), and [J. Rauch, JDG, 1974] for d > 2). ( Intuition : A “more curved” fits inside the “less curved” one). If x 0 ∈ ∂ Ω is fixed, and the ball B = B ( z 0 , r 0 ) touches Ω at x 0 and Ω ⊂ B , then the paraboloid P ( x ) = h 0 + α r 2 0 − α | x − z 0 | 2 (with a properly chosen α > 0) satisfies P ( x 0 ) = h 0 and P ( x ) ≥ h 0 on Ω, det D 2 ( − P ) ≥ K 0 ψ ( |∇ P | ) on { P > 0 } . 19/25 Hayk Aleksanyan K-surfaces with free boundaries

Construction of super-solution Assumptions: Ω is bounded, strictly convex and C 2 , ψ is non-decreasing (need to adjust the free boundary condition) and smooth. Let κ 0 > 0 be the smallest principal curvature of ∂ Ω. Then, Ω rolls freely inside a ball of radius r 0 := 1 /κ 0 (W. Blaschke’s rolling ball theorem (2d case), and [J. Rauch, JDG, 1974] for d > 2). ( Intuition : A “more curved” fits inside the “less curved” one). If x 0 ∈ ∂ Ω is fixed, and the ball B = B ( z 0 , r 0 ) touches Ω at x 0 and Ω ⊂ B , then the paraboloid P ( x ) = h 0 + α r 2 0 − α | x − z 0 | 2 (with a properly chosen α > 0) satisfies P ( x 0 ) = h 0 and P ( x ) ≥ h 0 on Ω, det D 2 ( − P ) ≥ K 0 ψ ( |∇ P | ) on { P > 0 } . |∇ P | ≤ λ 0 on ∂ { P > 0 } . 19/25 Hayk Aleksanyan K-surfaces with free boundaries

Construction of super-solution Assumptions: Ω is bounded, strictly convex and C 2 , ψ is non-decreasing (need to adjust the free boundary condition) and smooth. Let κ 0 > 0 be the smallest principal curvature of ∂ Ω. Then, Ω rolls freely inside a ball of radius r 0 := 1 /κ 0 (W. Blaschke’s rolling ball theorem (2d case), and [J. Rauch, JDG, 1974] for d > 2). ( Intuition : A “more curved” fits inside the “less curved” one). If x 0 ∈ ∂ Ω is fixed, and the ball B = B ( z 0 , r 0 ) touches Ω at x 0 and Ω ⊂ B , then the paraboloid P ( x ) = h 0 + α r 2 0 − α | x − z 0 | 2 (with a properly chosen α > 0) satisfies P ( x 0 ) = h 0 and P ( x ) ≥ h 0 on Ω, det D 2 ( − P ) ≥ K 0 ψ ( |∇ P | ) on { P > 0 } . |∇ P | ≤ λ 0 on ∂ { P > 0 } . Do this for a dense set of points, and take the infimum: gives an element of W + . 19/25 Hayk Aleksanyan K-surfaces with free boundaries

Perron in action Any element of W + is larger than the solution to the homogeneous equation. Hence, w ∈ W + w ( x ) , x ∈ R d , u ∗ ( x ) := inf does not collapse. 20/25 Hayk Aleksanyan K-surfaces with free boundaries

Perron in action Any element of W + is larger than the solution to the homogeneous equation. Hence, w ∈ W + w ( x ) , x ∈ R d , u ∗ ( x ) := inf does not collapse. Show an existence of a minimizing sequence, and hence u ∗ ∈ W + (plus strict concavity of u ∗ ). 20/25 Hayk Aleksanyan K-surfaces with free boundaries

Perron in action Any element of W + is larger than the solution to the homogeneous equation. Hence, w ∈ W + w ( x ) , x ∈ R d , u ∗ ( x ) := inf does not collapse. Show an existence of a minimizing sequence, and hence u ∗ ∈ W + (plus strict concavity of u ∗ ). Solving the Dirichlet problem for affine boundary data, and using strong comparison principle, show u ∗ solves the equation in { u ∗ > 0 } \ Ω. 20/25 Hayk Aleksanyan K-surfaces with free boundaries

Perron in action Any element of W + is larger than the solution to the homogeneous equation. Hence, w ∈ W + w ( x ) , x ∈ R d , u ∗ ( x ) := inf does not collapse. Show an existence of a minimizing sequence, and hence u ∗ ∈ W + (plus strict concavity of u ∗ ). Solving the Dirichlet problem for affine boundary data, and using strong comparison principle, show u ∗ solves the equation in { u ∗ > 0 } \ Ω. Still need to show that |∇ u ∗ | = λ 0 on the free boundary (we have only ≤ everywhere by construction). 20/25 Hayk Aleksanyan K-surfaces with free boundaries

Blaschke extension and the free boundary condition H x 0 H ⊥ x 0 R d × { 0 } x 0 Reflection of a surface at a single point on the free boundary. 21/25 Hayk Aleksanyan K-surfaces with free boundaries

...details 22/25 Hayk Aleksanyan K-surfaces with free boundaries

...details Define a convex body S + ∗ bounded by the graph( u ∗ ) if 0 < x d +1 < h 0 , Ω × { h 0 } if x d +1 = h 0 , and when x d +1 < h 0 take the intersection of all extreme halfspaces at Γ u ∗ . 22/25 Hayk Aleksanyan K-surfaces with free boundaries

...details Define a convex body S + ∗ bounded by the graph( u ∗ ) if 0 < x d +1 < h 0 , Ω × { h 0 } if x d +1 = h 0 , and when x d +1 < h 0 take the intersection of all extreme halfspaces at Γ u ∗ . For each x ∈ Γ u ∗ , if H x is an extreme supporting hyperplane x passing through H x ∩ ( R d × { 0 } ) and to the graph, define H ⊥ the normal to H x . 22/25 Hayk Aleksanyan K-surfaces with free boundaries

...details Define a convex body S + ∗ bounded by the graph( u ∗ ) if 0 < x d +1 < h 0 , Ω × { h 0 } if x d +1 = h 0 , and when x d +1 < h 0 take the intersection of all extreme halfspaces at Γ u ∗ . For each x ∈ Γ u ∗ , if H x is an extreme supporting hyperplane x passing through H x ∩ ( R d × { 0 } ) and to the graph, define H ⊥ the normal to H x . define S − x as the mirror reflection of S + x with respect to H ⊥ x . 22/25 Hayk Aleksanyan K-surfaces with free boundaries

...details Define a convex body S + ∗ bounded by the graph( u ∗ ) if 0 < x d +1 < h 0 , Ω × { h 0 } if x d +1 = h 0 , and when x d +1 < h 0 take the intersection of all extreme halfspaces at Γ u ∗ . For each x ∈ Γ u ∗ , if H x is an extreme supporting hyperplane x passing through H x ∩ ( R d × { 0 } ) and to the graph, define H ⊥ the normal to H x . define S − x as the mirror reflection of S + x with respect to H ⊥ x . Fix x 0 ∈ Γ u ∗ , and take a dense sequence x j ⊂ Γ u ∗ near x 0 . Define a nested sequence of convex bodies m � S m = S + S − ∗ ∩ x j , j =1 and take a limit as m → ∞ . Call the limit convex body S B the Blaschke reflection body. 22/25 Hayk Aleksanyan K-surfaces with free boundaries

...details Show that the boundary of S B is a graph over R d close to x 0 . 23/25 Hayk Aleksanyan K-surfaces with free boundaries

...details Show that the boundary of S B is a graph over R d close to x 0 . Assume at x 0 ∈ Γ u ∗ we have |∇ u ∗ ( x 0 ) | = λ < λ 0 . Then a slight tilt of the extreme support plane H x 0 , say H , will intersect a cap from S B . 23/25 Hayk Aleksanyan K-surfaces with free boundaries

...details Show that the boundary of S B is a graph over R d close to x 0 . Assume at x 0 ∈ Γ u ∗ we have |∇ u ∗ ( x 0 ) | = λ < λ 0 . Then a slight tilt of the extreme support plane H x 0 , say H , will intersect a cap from S B . Slightly translate H parallel towards Ω, to H δ , and in a slab between H and H δ replace the boundary of S B by an exact solution. For δ > 0 small enough, this will violate the minimality of u ∗ . 23/25 Hayk Aleksanyan K-surfaces with free boundaries