WELL-POSEDNESS FOR HYPERBOLIC PROBLEMS (0.2)

WELL-POSEDNESS FOR HYPERBOLIC PROBLEMS We will use the familiar Hilbert spaces H = L 2 (Ω) and V = H 1 (Ω). We consider the Cauchy problem (.1) c u = ( 2 t c )u = f L 2 ((, T ) Ω) on [, T ] Ω u() = u H 1 (Ω) u () = u 1 L 2 (Ω). We will denote, as the inner product on L 2 (Ω). If we pair the above equation with a test function φ Cc 1 (, T ; H 1 (Ω)), then we formally obtain (.2) [ 2 t u(t), φ(t) c u(t), φ(t) ]dt = [ c 2 u(t), φ t u(t), t φ(t) ]dt = f(t), φ(t) dt. f(t), φ(t) dt using an application of Green s theorem and integration by parts where any integral over a subset of ([, T ] Ω) vanishes due to support assumptions on φ. We say u is a weak solution of (.1) if it satisfies (.2) for any such test functions φ, u L 2 (, T ; H 1 (Ω)) H 1 (, T ; L 2 (Ω)), and u satisfies the initial conditions in (.1). Hence, it will be useful to define the sesquilinear form a(u, v) := c u, c v, u, v H 1 (Ω). Theorem.1. There exists a unique weak solution u to (.1) such that u L 2 ((, T ); H 1 (Ω)) and u L 2 ((, T ); L 2 (Ω)). Proof. (Existence) Since H 1 (Ω) is separable, let {w j } j=1 H1 (Ω) be a countable basis for H 1 (Ω) in the following sense: for all m, w 1,..., w m are linearly independent, and, Each x H 1 (Ω) is limit of elements of the form finite ξ k w k, for ξ k C. We define the approximate solution u m (t) of order m of the problem in the following way m u m (t) = g km (t)w k, the g km s being determined so that (.3) a(u m (t), w l ) + u m(t), w l = f(t), w l, 1 l m, g km () = ξ km, g km() = η km k=1 1

2 WELL-POSEDNESS FOR HYPERBOLIC PROBLEMS with m ξ km w k u in H 1 (Ω) as m, k=1 m η km w k u 1 in L 2 (Ω) as m. k=1 Thus, the g km s are determined by a second order linear ordinary differential system of the form Aẍ(t)+Cx(t) = b(t) for m m constant matrices A, C and b(t) a vector containing m components, which admits a unique solution since A is invertible. The hope is that the sequences {u m } m=1 and {u m} m=1 have convergent subsequences in L 2 (, T ; H 1 (Ω)) and L 2 (, T ; L 2 (Ω)) respectively that converge to a weak solution of the PDE. Using the Banach-Alaoglu theorem, it will suffice to show that these sequences are uniformly bounded which requires an energy estimate. Energy Estimate Let u C 2 c ([, T ], H 1 (Ω)), such that c u = f weakly in the sense of (.3) (i.e. that a(u(t), v) + u (t), v = f(t), v, v H 1 (Ω) ). It is shown on page 213 of the textbook that a(, ) defines an equivalent inner product and norm on H 1 (Ω), i.e. one has constants A, B > such that for all φ H 1 (Ω) A φ H 1 a(φ, φ) = c φ c φ L2 (Ω) B φ H 1 (Ω). Thus, WLOG we may interchange these two norms in all of our estimates if needed. Define the energy of u at time t by E(u(t)) = a(u(t), u(t)) + u (t) 2. Then using u (t) H 1 (Ω) as a test function, we have d dt E(u(t)) = a(u (t), u(t)) + a(u(t), u (t)) + u (t), u (t) + u (t), u (t) = u (t), f(t) + f(t), u (t) Thus, we obtain = 2R f(t), u (t). E(u(t)) = E(u()) + 2 t If we integrate this expression from to T we obtain u 2 L 2 (,T ;H 1 (Ω)) + u 2 L 2 ((,T ) Ω) T E(u()) + 2 C ( R f(s), u (s) ds. (E(u()) + t R f(s), u (s) dsdt ) f(s), u (s) ds C(E(u()) + ɛ 1 f 2 L 2 (,T ;Ω) ) + ɛ u 2 L 2 ((,T ) Ω)

WELL-POSEDNESS FOR HYPERBOLIC PROBLEMS 3 where we use Cauchy-Schwartz and Young s inequality in the final inequality above, and ɛ >. With possibly a different constant C and making ɛ < 1 we obtain (.4) u 2 L 2 (,T ;H 1 (Ω)) + u 2 L 2 ((,T ) Ω) C(E(u()) + f 2 L 2 ((,T ) Ω) ). In fact, we can get an improved result showing that the map t E(u(t)) is actually Holder continuous with weaker assumptions on u (see Lemma.2). Returning to the proof Next, by Bessel s inequality we have u m () 2 H 1 (Ω) u 2 H 1 (Ω) and u m() L 2 (Ω) u 1 2 L 2 (Ω) m. If we combine these two inequalities and apply the proof of the energy inequality to u m, we obtain u m 2 L 2 (,T ;H 1 (Ω)) + u m 2 L 2 ((,T ) Ω) C(E(u()) + f 2 L 2 (,T ;Ω) ) with a constant independent of m. Next, apply the Banach-Alaoglu theorem to the Hilbert spaces L 2 (, T ; H 1 (Ω)) and L 2 (, T ; L 2 (Ω)) to extract subsequence u mk from u m such that for some u L 2 (, T ; H 1 (Ω)) and v L 2 ((, T ) Ω) we have u mk u m k u in L 2 (, T ; H 1 (Ω)) v in L 2 ((, T ) Ω). Now using continuity of the weak derivative, we obtain from the first convergence that u m k u in a weak distribution sense. But by uniqueness of the weak limit, we must have v = u. Next, we check that u is indeed a weak solution to our PDE. We first show that u satisfies the differential equation. Take r ψ = ψ j (t)w j j=1 with ψ j C 1 c ((, T )). After multiplying equation (.3) by ψ j and summing, we obtain for m k r a(u mk (t), ψ(t)) + u m k (t), ψ(t) = f(t), ψ(t) so after an integration by parts we get [a(u mk (t), ψ(t)) u m k (t), ψ (t) ]dt = Then letting m k above gives [a(u(t), ψ(t)) u (t), ψ (t) ]dt = f(t), ψ(t) dt. f(t), ψ(t) dt. Since the w j form a basis for H 1 (Ω), the set of such ψ is actually dense in the space of functions ψ L 2 (, T ; H 1 (Ω)) such that ψ L 2 ((, T ) Ω), ψ() = ψ(t ) = so u indeed satisfies the differential equation.

4 WELL-POSEDNESS FOR HYPERBOLIC PROBLEMS We now show that u satisfies the first initial condition. space of functions Define the following C T = {φ φ C ([, T ]), φ(t ) = φ (T ) = }, and consider the functions of the form By partial integration we have ψ = φ v, v H 1 (Ω), φ C T ([, T ]). [ u m k, ψ + u mk, ψ ]dt = u mk (), ψ(). The right hand side converges to u, ψ() if k. The left hand side converges to Hence [ u, ψ + u, ψ ]dt = u(), ψ(). u(), ψ() = u, ψ(). It follows that u satisfies the first initial condition. Using the equation (.3) for u mk tested with ψ, one has an extra term due to the partial integration: Letting k, we obtain Now let φ k C c [a(u mk, ψ) u m k, ψ ]dt = [a(u(t), ψ(t)) u (t), ψ (t) ]dt = ((, T ]) such that f, ψ dt + u m k (), ψ(). φ k H( t) pointwise, f(t), ψ(t) dt + u 1, ψ(). φ k () = 1 for each k, and φ k δ() in distribution (H(t) here denotes the Heaviside function). So replacing ψ above with ψ k = φ k v instead and letting k means the first term on the left goes to as well as the first term on the right (since the support of φ k shrinks to inside (, T )). But u (t), ψ k (t) dt u (), v. We thus obtain Hence, u () = u 1 as desired. Uniqueness u (), v = u 1, v v L 2 (Ω). Suppose u 1 and u 2 are two weak solutions of the PDE (.1). Then u = u 1 u 2 solves c u =, u() = and u () =. If we knew that u L 2 (, T ; L 2 (Ω)), then our energy estimate (.4) goes through as before to show u. Since we do not have this type of regularity in time, we proceed to regularize in time.

WELL-POSEDNESS FOR HYPERBOLIC PROBLEMS 5 Thus, let ρ m, m = 1, 2,... be a regularizing sequence, that is ρ m (t) ρ m (t)dt = 1 suppρ m {}. Let ɛ >. There is an M such that suppρ m ( ɛ, ɛ) for all m > M. Set φ(t) = ρ m (s t)(ρ m u)(s), then, for s [ɛ, T ɛ], we have that φ Cc ([, T ], H 1 (Ω)). Thus, u satisfies (.2) with this choice of v and we denote u m (t) = ρ m u(t). Notice that ρ m(s t)u(t)dt = ρ m u(s), so using (.2) with u paired with φ (f being in (.2)) gives (.5) a(u m (s), u m(s)) + u m(s), u m(s) = ρ m, u m(s) =. Thus, the energy estimate proceeds exactly as before to show E(u m (t))dt C The mollification we have constructed ensures that u m u u m u E(u m ()). in L 2 (, T ; H 1 (Ω)) in L 2 (, T ; L 2 (Ω)). Then letting m and using E(u()) = shows so that u. E(u(t))dt =.1. Hölder-1/2 continuity of the energy function. Lemma.2. Let u L 2 (, T ; H 1 (Ω)), u L 2 (, T ; L 2 (Ω)), f L 2 (, T ; L 2 (Ω)). If u satisfies (.2), then the function t E(u(t)) is Hölder continuous of order 1/2 on a set I that differs from (, T ) by a set of measure. Proof. Using mollifiers, the proof of (.5) shows with the notation in that section that a(u m (s), u m(s)) + u m(s), u m(s) = f m (s), u m(s) and hence d dt E(u m(t)) = 2R f m (t), u m(t). Looking at the right hand side shows the limit as m actually exists in L 1 ((, T )) and we have d dt E(u(t)) = 2R f(t), u (t) in L 1 ((, T )). It follows that de(u(t)) dt is in L 1 ((, T )) so that E(u(t)) is bounded in [, T ]; hence, u(t) H 1 (Ω), u (t) L2 (Ω) are bounded as well. Another consequence

6 WELL-POSEDNESS FOR HYPERBOLIC PROBLEMS is E(u(t)) is absolutely continuous on a set I that differs from [, T ] by a set of measure zero. Integrating this expression from s to t, s < t I, gives t ( t ) 1/2 ( t E(u(t)) E(u(s)) = 2 R f(r), u (r) dr 2 1dr f(r), u (r) 2 dr s s s 2 ( ) 1/2 T t s f(r), u (r) 2 dr 2C t s f L2 (,T ;L 2 (Ω)) where C > is a uniform constant such that u (t) L2 (Ω) C for t [, T ]. ) 1/2