Hyperbolic equations of Von Karman type

We report some recent results on weak and semi-strong solutions to a coupled hyperbolic-elliptic system of Von Karman type on ${IR}^{2m}$, $m \in {IN}_{\geq 2}$.

space IR 2m , m ∈ IN ≥2 , and present two recent results on this system, namely the existence of weak global solutions, and the existence and uniqueness of a local semistrong solution in the case m = 2 (which turns out to be the most complicated), under a smallness assumption on the initial values. We only give an indication of the main steps of the proofs of these results; for a full proof, we refer to our forthcoming memoir [6].
1.1. The equations. All functions we consider are real valued. Let m ∈ IN ≥2 , and u 1 , . . . , u m , u m+1 ∈ C ∞ (IR 2m ). We define the m-linear form where we adopt the usual summation convention for repeated indices, and δ i1 ··· im Arguing formally for the moment, we assume that the elliptic equation where ∆ := − ∇ j j u, can be uniquely solved in a suitable functional framework in terms of u, thereby defining a map u → f (u). Let T > 0. We consider the Cauchy problem of hyperbolic type, consisting in the determination of a function u on [0, T ] × IR 2m , satisfying the equation u tt + ∆ m u = N (f (u), u, . . . , u) , (4) and subject to the initial conditions where u 0 and u 1 are given functions defined on IR 2m . We refer to problem (3)+(4)+ (5) as "problem (VKH)", and we look for solutions to this problem in a suitable scale of Banach spaces, depending on the regularity of the data of the problem. We recall that in the original von Karman system on IR 2 , the equation under consideration is (4), with f (u) given by (3), all written for m = 2 instead of m = 1. In this model, the unknown function u represents the vertical displacement of the plate, and the corresponding term f (u) represents the so-called "Airy stress function", which is related to the internal elastic forces acting on the plate, and depends on its deformation, as measured by u.
1.2. Function spaces. 1. For 1 ≤ p ≤ +∞, we set L p := L p (IR 2m ), denote its norm by | · | p , and by · , · the scalar product in L 2 . For k ≥ 0 we denote by H k := H k (IR 2m ) the usual Sobolev space on IR 2m , equipped with the equivalent norm where We also introduce the spaceH k , defined as the completion of H k with respect to the norm (7);H k is a Hilbert space, with corresponding scalar product The main properties of the spacesH k and the functions N and I introduced in (1) and (2) are described in [3] and [4], where we proved the following where C p → +∞ as p → ∞.
2) The functions N and I are completely symmetric in all of their arguments; in addition, the identity (obtained from (2 by integration by parts), holds. Consequently, by Hölder's inequality and (9, (10, it follows that, if u 1 , . . . u m , u m+1 ∈H m , with C independent of u 1 , . . . , u m+1 .
with C independent of u 1 , . . . , u m .
The function f satisfies the estimate with C independent of u.
2) If in addition u ∈H m+1 , then f ∈H m+h for 0 < h ≤ m, and with C independent of u.
Proof. The first claim follows from part (3) of proposition 1 and the Lax-Milgram theorem; (18) follows from (17), taking ϕ = f and using (14). If u ∈H m+1 , part (4) of proposition 1 implies that ∆ m f = − N (u, . . . , u) ∈ L 2 ; in addition, by (16),  (3) does indeed define f := f (u) in problem (VKH). Using Hardy spaces techniques, it is possible to show that f ∈ L ∞ ; note that this conclusion does not follow from the mere fact that f ∈H m , which is not imbedded in L ∞ (recall (11)). On the other hand, part (1) of proposition 1 implies that 3. Given T > 0 and a Banach space X, we denote by: erywhere defined, bounded and weakly continuous; the latter meaning that for all ψ ∈ X , the scalar function [0, T ] t → u(t), ψ X×X , where the brackets denote the duality pairing between X and its dual X , is continuous. When there is no chance of confusion, we shall drop the reference to X × X in duality pairings. Finally, for h ∈ IN and T > 0 we introduce the anisotropic Sobolev spaces endowed with their natural norms We shall need the following results on the time-dependent spaces introduced above; for a proof, see e.g. Lions -Magenes, [11, ch. 1] and Lions, [10, ch. 1].
Proposition 2. Let X and Y be Banach spaces, with X reflexive and X → Y . Then: To define the type of solutions to problem (VKH) we wish to consider, we note that proposition 1 allows us to prove . This allows us to give the following definitions.
In general, τ may depend on h; if τ = T , we call u a global solution. In addition, we distinguish between semi-strong and regular solutions, corresponding, respectively, to the cases 0 < h < m and h > m.
The distinction between different types of solutions in definition 1.3 is motivated by the fact that for semi-strong solutions, as for weak ones, both u tt and ∆ m u are distribution-valued functions of t (in H h−m ) (in contrast, for h > m the solutions are actually classical, due to the strong Sobolev imbeddings). On the other hand, parts (3) and (4) of proposition 1 imply that the right sides of (3) and (4) are functions, either in L 1 if h = 0, or at least in L 2 if h > 0. Also, note that if u 0 = u 1 = 0, the function u ≡ 0 is a strong (thus, also weak) solution of problem (VKH); thus, we assume u 0 = 0 or u 1 = 0.
2. The existence of global weak solutions to problem (VKH) can be established by a straightforward generalization of J. L. Lions' result mentioned earlier for the case m = 1; likewise, the existence and uniqueness of local semi-strong solutions can be established by methods similar to those we used for strong and regular solutions in chapter 7, sct. 2, of our book [5]. We claim: Theorem 1.4. Let T > 0, u 0 ∈ H m and u 1 ∈ L 2 . Then: 1. There exists at least one weak solution u ∈ X w m0 (T ) to problem (VKH ). 2. Any weak solution u ∈ X w m0 (T ) to problem (VKH ) is continuous at t = 0, in the sense that 3. If for each u 0 and u 1 there is only one weak solution u ∈ X w m0 (T ) to problem (VKH ), then u ∈ X s m0 (T ). Theorem 1.5. Let m = 2, h = 1, u 0 ∈ H 3 and u 1 ∈ H 1 . There exists δ > 0 such that if then for some τ 1 > 0 there exists a unique semi-strong solution u ∈ X s 2,1 (τ 1 ) to problem (VKH ).
In the next two sections we outline the main steps required for the proofs of theorems 1.4 and 1.5; we plan to provide a fully detailed proof in [6], where we also prove, with similar methods, the following general result: Theorem 1.6. Let m > 2, 1 ≤ h < m, u 0 ∈ H m+h and u 1 ∈ H h . There exist τ h > 0 and a unique semi-strong solution u ∈ X s mh (τ h ) to problem (VKH ). In addition, inf 1≤h<m τ h ≥ τ 1 .

Weak solutions.
In this section we report the main steps of the proof of theorem 1.4, based on a Galerkin approximation scheme.
1. The Discretized System. We choose a complete orthonormal basis W = (w j ) j≥1 of H m (for instance, via Hermite functions); for n ∈ IN ≥1 we set W n := span{w 1 , . . . , w n }, and denote by P n : L 2 → W n the corresponding orthogonal projection in H m , which has the property that if u ∈ H r , r ≥ 0, then P n u → u in H r . We consider then the approximate initial value problem u n tt + ∆ m u n = P n N (f n , u n , . . . , u n ) =: P n (R n ) , ∆ m f n = − N (u n , u n , . . . , u n ) ; (31) By Carathéodory's theorem, this problem has a local solution u n ∈ C([0, t n ]; W n ), with u n t ∈ AC([0, t n ]; W n ), for some t n ∈ [0, T ]. 2. Crucial a priori estimate. Proposition 3. There exists R 0 ≥ 1, independent of n and t n , such that for all t ∈ [0, t n ], Proof. [J. L. Lions.] Multiplying (30) by u n t in L 2 yields d dt |u n t | 2 2 + |∇ m u n | 2 2 + |u n | 2 2 = 2 R n + u n , u n t .

Convergence.
Since R 0 is independent of t n , we can extend each u n to all of [0, T ]. Since R 0 is also independent of n, we deduce the weak convergence of subsequences, still denoted by (u n ) n≥1 and (f n ) n≥1 to limits u and f ; more precisely, In particular, u ∈ L 2 (0, T ; H m ) and u t ∈ L 2 (0, T ; L 2 ); thus, by the trace theorem (part (3) (3) and (4) of proposition 2), we deduce that (41) These facts allow us to prove that (42) thus, from (40) we deduce that f solves equation (3), which means that f = f (u). In turn, (42) also allows us to prove that 4. Continuity. We now show that if problem (VKH) admits a unique solution u ∈ X w m0 (T ), then in fact u ∈ X s m0 (T ). By the weak continuity of u, u t and f into H m , L 2 andH m , respectively, it is sufficient to show the continuity of the norms |u t (·)| 2 , |∇ m u(·)| 2 and |∇ m f (·)| 2 , which in turn is implied by the continuity of the (square of the) norm 1 By this we mean the use of a cut-off function ζ of sufficiently large support so that, in the decomposition where (g n ) n≥1 is a bounded sequence of L 1 , the term Bn is as small as desired, uniformly in n, and the term An can be estimated on the compact set Ω := supp(ζ).
We argue as in Majda, [12, ch. 2, sct. 1]. Since equation (4) is reversible in time, it is sufficient to prove the right continuity of u and u t at any t 0 ∈ [0, T [. The assumed uniqueness of weak solutions allows us to limit ourselves to the case t 0 = 0, because on any interval [t 0 , T ], u would coincide with the solutionũ of problem (VKH) on the interval [t 0 , T ], with initial valuesũ(t 0 ) = u(t 0 ) ∈ H m andũ t (t 0 ) = u t (t 0 ) ∈ L 2 at t = t 0 (recall that if u ∈ X w m0 (T ), then u(t), u t (t) are, for each t ∈ [0, T ], welldefined elements of, respectively, H m and L 2 ). Finally, the weak continuity of u, u t and f implies thatẼ thus, we only need to prove that Canceling the term 2 u n , u n t from (37) we deduce that d dtẼ 0 (u n (t)) = 0 ; =Ẽ 0 (u(0)) , from which (48) follows. Finally, we note that this argument shows that, even in absence of uniqueness, u and u t are continuous at any t 0 such that either E 0 (u(t 0 )) =Ẽ 0 (u(0)) orẼ 0 (u(t 0 )) = lim n→∞Ẽ 0 (u n (t 0 )) .
In particular, u is continuous at t 0 = 0 (in the sense of (28)), where both conditions of (51) hold. This concludes the proof of theorem 1.4.

Remark 2.
By part (2) of proposition 2, the strong continuity of u, u t and f from [0, T ] into, respectively, H m , L 2 andH m , would follow if u satisfied the same identity (49) satisfied by its Galerkin approximants u n ; that is, if However, identity (52) is formally obtained from (3) via a multiplication by 2 u t in L 2 , and none of the individual terms of (3) need be in L 2 if u ∈ X w m0 (T ) only. Of course, the difficulty lies in the nonlinear term N (f, u, . . . , u), which we can only prove to be bounded from [0, T ] into L 1 , as we see from the estimate which follows from (9). Thus, we are not able to determine whether (52) holds or not, and the problem of whether there exists a weak solution u ∈ X s m0 (T ) to problem (VKH) remains open.
3. Semi-strong solutions, m = 2. In this section we report the main steps of the proof of theorem 1.5. We first show the existence of a solution u ∈ X w 2,1 (τ 1 ) to problem (VKH), then the uniqueness of solutions in this space whose initial data satisfy the smallness assumption (29), and, finally, that this solution is in fact in X s 2,1 (τ 1 ). 1.1. We consider the same Galerkin approximants defined in (30) and (31), with m = 2, but with instead of (32) and (33). For τ ∈]0, T ], u ∈ X 2 2,1 (τ ), and t ∈ [0, τ ], we set We know from the a priori estimate (34) that for all n ≥ 1 and all t ∈ [0, T ], thus, defining e.g. τ 1 := 1 2 C M1 , we conclude from (73) that each u n satisfies, on the common interval [0, τ 1 ], the uniform bound We can then proceed as in the proof of theorem 1.4 to show the existence of a function u ∈ X w 2,1 (τ 1 ), which is a local solution of problem (VKH) on the interval [0, τ 1 ].
Then, z solves the problem wheref := f (ũ). We can show that both terms of (77) are in H −1 , pointwise in t; thus, we can multiply (77) by 2 z t ∈ H 1 , to obtain, as in (59) and (64), 2.2. Acting as in the first part of this proof, in particular using (78) and proceeding as in (66) to estimate |∇ 2f t | 2 , we arrive at the estimate Thus, integrating (80) and recalling that Ψ 0 (z(0)) = 0, we obtain that As in (69), under a condition similar to (70), thus, we deduce from (82) that for all t ∈ [0, τ ], By Gronwall's inequality we conclude then that E 0 (z(t)) ≡ 0, which implies the asserted uniqueness.
3. The proof of the strong continuity of u and u t into H 3 and H 1 is similar to that of the continuity claim in theorem 1.4, whereby it is sufficient to prove the right continuity of E 1 (u(·)) at t = 0, for which it suffices to show that Recalling the definition of Ψ 1 in (64), we write E 1 (u(t)) = Ψ 1 (u(t)) + D(u(t)) , D(u) := N (∇u, ∇u), f (u) .
4. Concluding Remarks. The smallness assumption (29) could be removed if the Galerkin approximants of u satisfied a bound of the type Indeed, in this case we could estimate the last term of (59) as |B n | ≤ 2 C |∇ 2 f n | ∞ |∇ 3 u n | 2 |∇u n t | 2 ≤ 2 C Λ (E 1 (u n )) 2 , and we would obtain from (59) that which yields an inequality similar to (72). However, the only bound on f n we know so far is (61), which implies a bound on |∇ 2 f n (t)| p for all p ≥ 2, but not for p = ∞ (recall (11)). This difficulty disappears if m > 2 or if m = 2 and h ≥ 2 (strong solutions). We illustrate this in the case m > 2, h = 1; that is, for solutions in X s m1 (τ 1 ). In this case, omitting the dependence on the variables n and t, we have u ∈ H m+1 , so that ∇ 2 u ∈ H m−1 → L 2m and, therefore, ∆ m f ∈ L 2 . Thus, f ∈H 2m ∩H m , which implies that ∇ 2 f ∈H 2m−2 ∩H m−2 →H m+1 ∩ L m → L ∞ . In fact, using proposition 1, (18) and (34), we arrive at the estimate in accord with (91). However, at this stage we do not know if assumption (29) is actually necessary. ♦