Quasi-convex Hamilton-Jacobi equations posed on junctions: The multi-dimensional case

A multi-dimensional junction is obtained by identifying the boundaries of a ﬁnite number of copies of an Euclidian half-space. The main contribution of this article is the construction of a multidimensional vertex test function G ( x, y ). First, such a function has to be suﬃ-ciently regular to be used as a test function in the viscosity solution theory for quasi-convex Hamilton-Jacobi equations posed on a multi-dimensional junction. Second, its gradients have to satisfy appropriate compatibility conditions in order to replace the usual quadratic pe-nalization function | x − y | 2 in the proof of strong uniqueness (comparison principle) by the celebrated doubling variable technique. This result extends a construction the authors previously achieved in the network setting. In the multi-dimensional setting, the construction is less explicit and more delicate.


Introduction
This paper is concerned with extending the theory developed for Hamilton-Jacobi (HJ) equations posed on junctions in [3] to the multi-dimensional setting.
A multi-dimensional junction is made of N copies of R d+1 + glued through their boundaries.
We emphasize that the common boundary of the hyperspaces J i is denoted by Γ.For points For a smooth real-valued function u defined on J, ∂ i u(X) denotes the (spatial) derivative of u with respect to x i at X = (x ′ , x i ) ∈ J i and D ′ u(X) denotes the (spatial) gradient of u with respect to x ′ .The "gradient" of u is defined as follows, With such a notation in hand, we consider the following Hamilton-Jacobi equation posed on the multi-dimensional junction J u t + H i (Du) = 0 t > 0, X ∈ J i \ Γ, u t + F (Du) = 0 t > 0, X ∈ Γ (1.3) submitted to the initial condition u(0, X) = u 0 (X) for X ∈ J. (1.4) The second equation in (1.3) is referred to as the junction condition.
The Hamiltonians are supposed to satisfy the following conditions: (1.5) We next define the A-limited flux function F A associated with the multi-dimensional junction J.
In order to do so, we first consider π 0 i (p ′ ) ∈ R such that p i → H i (p ′ , p i ) reaches its minimum at p i = π 0 i (p ′ ) and H − i is defined by So-called flux-limiter functions A : R d → R are always assumed to be continuous and, in some important cases, to satisfy the following condition, A : R d → R is continuous and quasi-convex. (1.6) The function F A is defined for p = (p 1 , . . ., p N ) and P = (p ′ , p) as F A (P ) = max A(p ′ ), max i=1,...,N H − i (p ′ , p i ) . (1.7) We now consider the following important special case of (1.3), We point out that A could be replaced with max(A, A 0 ) where (1.9) We notice (see Lemma A.1 in Appendix) that the functions A i , i = 0, . . ., N are quasi-convex, continuous and coercive.
As far as general junction conditions are concerned, we assume that the junction function and, in some important cases, (Quasi-convexity) ∀λ, {F ≤ λ} convex. (1.11)

Main results
Theorem 1.1 (General junction conditions reduce to F A ). Let the Hamiltonians satisfy (1.5) and let F : R N → R satisfy (1.10).There exists a unique coercive continuous function A F : R d → R such that every relaxed viscosity sub-solution (resp.super-solution) of (1.3) is a A F -flux limited sub-solution (resp.super-solution) of (1.8).Moreover, if . Theorem 1.3 (Comparison principle on a multi-dimensional junction).Assume that the Hamiltonians satisfy (1.5), the junction function satisfies (1.10) and (1.11) and that the initial datum u 0 is uniformly continuous.Then for all (relaxed) sub-solution u and (relaxed) super-solution v of (1.3)-(1.4)satisfying for some T > 0 and C T > 0,

Comparison with known results
Our results are related to [1,2] where an optimal control problem in a two-domain setting is studied.The state of the system evolves according to two different dynamics on each side of an hypersurface.Moreover, the two dynamics at the interface corresponding to the maximal and minimal Ishii's discontinuous solutions of the associated Hamilton-Jacobi equation are identified.One of the two value functions is characterized in terms of partial differential equations.We showed in [3] that, in the one-dimensional setting, both can be characterized by using the notion of flux-limited solutions introduced in [3].The result of the present paper indicates that such a connexion holds in the general two-domain setting.Moreover, we can deal with quasi-convex Hamiltonians instead of convex ones.
The reader is also referred to [5,4] for optimal control problems in multi-domains.In particular, the authors impose some transmission conditions.As we already mentioned it in [3], Definition 2.4 is strongly related to these works.
Organization of the article.The paper is organized as follows.In Section 2, the notion of viscosity solution in the setting of multi-dimensional junction is introduced.Section 3 is devoted to the construction of the vertex test function.The proof of a technical lemma is presented in an appendix.

Notation. For a function
2 Viscosity solutions on a multi-dimensional junction 2.1 Definitions

Class of test functions
For T > 0, set J T = (0, T ) × J.The class of test functions on J T is chosen as follows,

Classical viscosity solutions
In order to define classical viscosity solutions, we recall the definition of upper and lower semicontinuous envelopes u * and u * of a (locally bounded) function u defined on [0, T ) × J: with equality at (t 0 , x 0 ) for some t 0 > 0, we have ii) We say that u is a (classical viscosity) sub-solution (resp.super-solution) of ( iii) We say that u is a (classical viscosity) solution if u is both a sub-solution and a super-solution.
Definition 2.2 (Flux-limited solutions).Consider a continuous flux-limiter function A : R d → R.
Then u is a A-flux limited sub-solution (resp.super-solution, solution) of (1.8) if it is a classical sub-solution (resp.super-solution, solution) of (1.3) with F = F A .

Relaxed viscosity solutions
We next introduce relaxed viscosity solutions.
i) We say that u is a relaxed sub-solution (resp.relaxed super-solution) of (1.3) in J T if for all test function ϕ ∈ C 1 (J T ) such that u * ≤ ϕ (resp.u * ≥ ϕ) in a neighborhood of (t 0 , x 0 ) ∈ J T with equality at (t 0 , x 0 ) for some t 0 > 0, we have ii) We say that u is a relaxed (viscosity) solution if u is both a sub-solution and a super-solution.

A reduced set of test functions
Let π ± i : R d × R → R be defined as follows Definition 2.4 (Flux-limited viscosity solutions -again).Assume the Hamiltonians satisfy (1.5) and consider a continuous flux-limiter function Proposition 2.5 (Equivalence of Definitions 2.2 and 2.4).The definitions 2.2 and 2.4 are equivalent.
Proof.It is clear that flux-limited sub-solutions (resp.super-solutions) are reduced sub-solutions (resp.reduced super-solutions).To prove that the converse holds true, we proceed as in [3] by considering critical slopes in x.Precisely, it is enough to prove the following lemmas.
Lemma 2.6 (Critical slopes for super-solutions).Let u be a super-solution of (1.8) away from Γ and let ϕ touch u * from below at P 0 = (t 0 , X 0 ) with X 0 ∈ Γ.Then the "critical slopes" defined as follows Lemma 2.7 (Critical slopes for sub-solutions).Let u be a sub-solution of (1.8) away from Γ and let ϕ touch u * from above at P 0 = (t 0 , X 0 ) with X 0 ∈ Γ.Then the "critical slopes" defined as follows The proofs of these lemmas are straightforward adaptations of the corresponding ones in [3] so we skip them.The remaining of the proof is also analogous but we give some details in the sub-solution case for the reader's convenience.
Let ϕ touch u * from below at P 0 = (t 0 , X 0 ) with X 0 ∈ Γ and let λ denote ϕ t (P 0 ) and P = (p ′ , p 1 , . . ., p N ) denote Dϕ(P 0 ).We want to prove We know from Lemma 2.6 that for all i = 1, . . ., N , for some pi ≤ 0. In particular, We write next Moreover, we have from (2.3) that Hence, we can consider the following test function From the definition of reduced sub-solutions, we thus get which is the desired contradiction.

Stability
In the following proposition, we assert that, for the special junction functions F A , the junction condition is in fact always satisfied in the classical sense, that is to say in the sense of Definition 2.1.Proof.We treat successively the super-solution case and the sub-solution case.
Case 1: the super-solution case.Let u be a relaxed super-solution and let us assume by contradiction that there exists a test function ϕ touching u from below at P 0 = (t 0 , X 0 ) for some t 0 ∈ (0, T ) and X 0 ∈ Γ, such that Consider next the test function φ satisfying φ ≤ ϕ in a neighborhood of P 0 , with equality at P 0 such that and Using the fact that ) at P 0 , we deduce a contradiction with (2.4) using the viscosity inequality satisfied by ϕ for some i ∈ {1, . . ., N }.
Case 2: the sub-solution case.Let now u be a relaxed sub-solution and let us assume by contradiction that there exists a test function ϕ touching u from above at P 0 = (t 0 , X 0 ) for some t 0 ∈ (0, T ) and X 0 ∈ Γ, such that ϕ t + F A (Dϕ) > 0 at P 0 . (2.5)

Let us define
where we have used the fact that H i (D ′ ϕ(P 0 ), +∞) = +∞.Then we can construct a test function φ satisfying φ ≥ ϕ in a neighborhood of P 0 , with equality at P 0 , such that and Using the fact that ) at P 0 for all i, we deduce a contradiction with (2.5) using the viscosity inequality for ϕ for some i ∈ {1, . . ., N }.

General junction conditions reduce to flux-limited ones
We first prove Theorem 1.1.
Proof of Theorem 1.1.With the notation of Remark 1.2 in hand, we first remark that there exists only one λ ≥ A 0 (p ′ ) such that there exists p + = (p + 1 , . . ., p + N ) with p + i ≥ p 0 i such that The coercivity of A F is a direct consequence of the fact that A F ≥ A 0 .We thus prove next that A F is continuous.Consider a sequence (p ′ n ) n converging towards p ′ and let p + n = (p + 1,n , . . ., p + N,n ) with p + i,n ≥ p 0 i be such that We first claim that (p + i,n ) n is bounded.Indeed, if not, then A n → +∞ and, for n large enough, F (p ′ n , p 0 ) ≥ A n which is impossible.The claim also implies that (A n ) n is also bounded.Consider now to converging subsequences, still denoted by (p + n ) n and (A n ) n , and let p + and A be their limits.We can pass to the limit in (2.6) and get We only do the proof for super-solutions since the proof for sub-solutions follows along the same lines.Let ϕ be a test function touching u * from below at P 0 = (t 0 , X 0 ).We only need to consider the case where X 0 ∈ Γ.We can also assume that In view of the definition of A F , we get The proof is now complete.
We now turn to the following useful proposition.The proof of this proposition is postponed and can be found in Appendix.

Existence
Using Perron's method as in [3], we easily get existence of relaxed viscosity solutions for general junction functions F satisfying (1.10).

Vertex test function
This section is devoted to the construction of the vertex test function to be used in the proof of the comparison principle.We will use below the following shorthand notation In particular, keeping in mind the definition of Du (see (1.2)), Problem (1.8) on the junction can be rewritten as follows u t + H(X, Du) = 0 for all (t, X) ∈ (0, +∞) × J.
Then our key result is the following one.
iii) (Compatibility condition on the diagonal) For all X ∈ J, iv) (Compatibility condition on the gradients) For all (X, Y ) ∈ J 2 , vi) (Gradient bounds) For all K > 0, there exists We now assert that Theorem 1.3 is a direct consequence of Theorem 3.1.

The case of smooth convex Hamiltonians
Assume that the Hamiltonians H i satisfy the following assumptions for i = 1, ..., N , and the flux limiter It is useful to associate with each H i satisfying (3.6) its partial inverse functions π ± i : where we recall that The regularity of π ± can be derived thanks to the inverse function theorem.As far as the concavity of π + i is concerned, we can drop the subscript i and we do so for clarity.let (p ′ , λ), (q ′ , µ) ∈ epi A and t ∈ (0, 1).Then which is the desired result.The monotonicity of π + is easy to derive from the convexity of H.The proof of the lemma is now complete.
We next define the function G 0 for X ∈ J i , Y ∈ J j , i, j = 1, ..., N , as follows, where with A ≥ A 0 and A quasi-convex.

The vertex test function in
In order to prove Proposition 3.3, we first need to study G 0 for X ∈ J i and Y ∈ J j with i = j.Then, one can write with G ij A is defined in (3.10).Remark that for X ∈ J i and Y ∈ J j , we have We also consider the simplex Lemma 3.5 (Necessary conditions for the maximiser).Given Z ∈ Q, the supremum defining G ij is reached for some (P, λ) ∈ G ij A and there exists (α i , α j , α 0 ) ∈ T such that Proof.G ij (Z) is defined by maximizing a linear function under a equality constraint and an inequality constraint.Constraints are qualified if When constraints are qualified, Karush-Kuhn-Tucker theorem asserts that there exists α j ∈ R and α 0 ≥ 0 such that If one sets In particular, (α i , α j , α 0 ) ∈ T .Hence, the result is proved when constraints are qualified.This is in particular true if and the simple fact that A i (p ′ ) ≤ A 0 (p ′ ) imply in particular that A(p ′ ) = A 0 (p ′ ).We arrive at the same conclusion if ∂ j H j (p ′ , p j ) ≥ 0. In other words, Constraints are qualified as soon as ∀p ′ , A(p ′ ) > A 0 (p ′ ). (3.11) In particular, the result of the lemma holds true under this latter condition: where From the previous case, we know that there exists P ε and λ ε such that We can extract a subsequence such that α ε → α.Moreover, P ε • Z − λ ε is bounded from above and λ ε = H i (p ′ε , p ε i ) = H j (p ′ε , p ε j ).Since H i and H j are assumed to be superlinear, we conclude that we can also extract a converging subsequence from P ε .This achieves the proof of the lemma.Lemma 3.6 (Uniqueness of P ).Let Z ∈ Q.If there exists α, P and β, Q such that Then P = Q.
Proof.We consider the function Ψ : R d+2 × T → R defined as follows By assumption, we have If P denotes Q − P and ᾱ denotes β − α, then 0 = Taking the scalar product with P yields 0 = with T i ≥ 0, i = 1, 2 and Indeed, keeping in mind that Hence, we get 0 = 1 0 We distinguish three cases.We will use several times the fact that ).We will also use the corresponding property for p j : p j = π − j (p ′ , p j ).
-In the first subcase, α i = 1, we get p ′ = q ′ and p i = q i and Z = ∇ P H i (P ) and 0 = (p j − q j )z j = (P − Q) We conclude by remarking that p j = π − j (p ′ , λ) = q j .The second subcase is similar.
In this third case, this implies that two components of a = α + θ ᾱ = (a i , a j , a 0 ) are not 0.
-If a 0 = 0 then p ′ = q ′ and p i = q i and p j = q j , i.e.P = Q.
-If a i = 0 then p ′ = q ′ and p j = q j and z i = 0 and λ = µ and p i = π + (p ′ , λ) = q i .The third subcase a j = 0 is similar to the second one.
The proof of the lemma is now complete.
The two previous lemmas imply the following one.
Lemma 3.7 (Gradients of G 0 ij ).The function G 0 ij is C 1 in J i × J j , up to the boundary, and where (p ′ , λ) = (P(X, Y ), L(X, Y )) are uniquely determined by the relation In particular, the maps P and L are continuous in J i × J j .
The following lemma is elementary but it will be used below.
Lemma 3.8 (G 0 ij at the boundary).The restriction of G ij to {z i = 0} and {z j = 0} equals respectively (H i ∨ A) * and (H j ∨ A) * .

The vertex test function in
In view of the definition of G 0 , see (3.9), we have In particular, we derive from Lemma 3.8 the following one.
Lemma 3.9 (Continuity of G 0 ).The function G 0 is continuous in J × J.
We now turn to the regularity of G 0 ii .
Proof.Since G 0 ii is convex, it is C 1 if and only if there exists one and only one subgradient.Consider two distinct subgradients (p ′ , p i ) and (q ′ , q i ) of (max(H i , A)) * at some point (z ′ , z i ) = (x ′ − y ′ , x i − y i ).This implies that Since H i is strictly convex, this can only happen when H i = A and since (p ′ , p i ) = (q ′ , q i ), this implies z i = 0. Lemma 3.11 (Gradients of G 0 ii ).For (X, Y ) ∈ J i × J i such that x i = y i , we have which holds true for some α i ∈ [0, 1].In particular, the maps P and L are continuous in Proof.Lemma 3.10 implies that P = (p ′ , p i , −p i ) is unique.Hence λ is unique too.Moreover for some α i ∈ [0, 1].Lemma 3.6 allows us to conclude.

Proof of Proposition 3.3
We now turn to the proof of Proposition 3.3.
Proof of Proposition 3.3.The proof proceeds in several steps.
Step 1: Regularity.We already noticed in Lemma 3.9 that G 0 ∈ C(J 2 ) and Lemmas 3.7 and 3.10 imply that G 0 ∈ C 1 (R) for each region R given by .12) Step 2: Computation of the gradients.For each R given by (3.12) and for all (X, Y ) ∈ R ⊂ J i × J j , Lemmas 3.7 and 3.11 imply that Notice in particular that P and L are continuous in J × J except on ∪ N i=1 {x i = y i = 0}.
Step 3: Checking the compatibility condition on the gradients.Let us consider (x, y) ∈ J 2 with x = y = 0 or x = y.We have ) and H − j (p ′ , p j ) = λ, we get the result.If now (X, Y ) = (0, 0), then p i = π + i (p ′ , λ) and λ = A(p ′ ).Hence, we get (3.14) in this case too.One can derive (3.15) in the same way.
Step 4: Superlinearity.In view of the definition of G 0 , we deduce from (3.13) that for all R > 0 and λ From the definition (3.8) of π ± i and the assumption (3.6) on the Hamiltonians, we deduce that π 0 (R) → +∞ as R → +∞ which implies that for any K ≥ 0, there exists a constant C K ≥ 0 such that Therefore we get (3.4) with g 0 (a) = sup K≥0 (Ka − C K ).

The general case
Let us consider a slightly stronger assumption than (1.5), namely      (3.17) Proof.In view of (3.16), it is easy to check that (β • H i ) ′′ > 0 if and only if we have Because D 2 H ′′ i (p 0 i ) > 0, we see that the right hand side is negative for λ close enough to H i (p 0 i ).Then it is easy to choose a function β satisfying (3.18) and (3.17).Finally, compositing β with another convex increasing function which is superlinear at +∞ if necessary, we can ensure that β • H i superlinear.Proof.We assume that the Hamiltonians H i satisfy (3.16).Let β be the function given by Lemma 3.12.If u solves (1.8) on J T , then u is also a viscosity solution of β(u t ) + Ĥi (Du) = 0 for t ∈ (0, T ) and x ∈ J * i , β(u t ) + F Â(Du) = 0 for t ∈ (0, T ) and x = 0 This implies Because of the lower bound on β ′ given by Lemma 3.12, we get |(β −1 ) ′ | L ∞ (R) ≤ 1/δ which yields the compatibility condition (3.3) with γ = γ/δ arbitrarily small.
We are now in position to prove Theorem 3.1 in the general case.

Proposition 2 .
8 (F A junction conditions are always satisfied in the classical sense).Assume the Hamiltonians satisfy (1.5) and consider a continuous flux-limiter function A. If F = F A , then relaxed viscosity solutions in the sense of Definition 2.3 coincide with viscosity solutions in the sense of Definition 2.1.

Theorem 3 . 1 (
The vertex test function).Let A : R d → R be quasi-convex and γ > 0. Assume the Hamiltonians satisfy (1.5).Then there exists a function G : J 2 → R enjoying the following properties.i) (Regularity)