Further existence results for elliptic–parabolic and forward–backward parabolic equations

We give an existence result for first order evolution equation of the type Ru′+Au=f\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {R}u' + \mathcal {A}u = f$$\end{document} where R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {R}$$\end{document} may be a function depending also on time assuming positive, null and negative sign, then the equation may be elliptic–parabolic, both forward and backward. The result is given in an abstract setting with Banach spaces depending on time (the functions u are defined in an interval [0, T] and u(t)∈X(t)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$u(t) \in X(t)$$\end{document} for a.e. t) and R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {R}$$\end{document} which is in fact a linear operator. We also extend a previous existence result for the equation (Ru)′+Au=f\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(\mathcal {R}u)' + \mathcal {A}u = f$$\end{document} to the setting of moving Banach spaces. We also give a time regularity result in a particular case and give many examples of different possible choices of R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {R}$$\end{document}.


Introduction
In this paper we consider differential equations of mixed type in abstract form whose concrete model example is where r is a function which may assume positive, null and negative values and consequently this equation may be of elliptic-parabolic type, parabolic both forward and backward. Equations of mixed type have been considered since at least one century ago, since, as far as many authors say, they are mentioned in [7]. Here we recall some simple and more known examples: Communicated The first was considered in [1] in 1968, the second one in [15] in 1971 and they clearly are particular cases of equations like where A is an elliptic operator and r a changing sign function. This type of equations seem to be interesting in many areas and have arisen in connection with many different problems: in the study of some stochastic differential equation, in the kinetic theory, in some physical models (like electron scattering, neutron transport). For these applications we confine to quote the recent paper [10] and for the many others we refer to the references contained in the already quoted paper [15] and in [2,3]. Just in these papers Beals treated equations like (2), but always with simple r . For instance, in [2] the equation is considered, but, as Gevrey seems to suggest and as Beals says, some coefficients like are of interest in some applications. Among papers taking into account equations with some more general r (and A linear) we recall [9,20]. In particular in [9] a coefficient depending also on time is considered and a condition of regularity in time is requested, i.e.
A recent paper where the author consider r = r (x, t) (and A linear) is [8]. Another paper we want to recall is [16] which is more general, even if incomplete. As regards equations like (2) where r ≥ 0, the known results are more general than those regarding forward-backward parabolic equations. We recall the paper [21] and the book (see chapter 3) [22] for some general results. Finally we want to recall [4] for many examples and applications of equations with non-negative coefficients.
All the results cited above are generalized in the present paper and in [17], where it is considered the abstract equation with suitable boundary data, where A is a monotone operator and R is a linear operator depending also on time which can be not invertible. When R is a multiplication operator, i.e. (Ru)(x, t) := r (x, t)u(x, t), also discontinuous and unbounded coefficients, i.e. r ∈ L 1 loc , are admitted without assuming the existence of r t . Just to show an example, an equation included in the result in [17] is The aim of this paper is giving existence results for the equation R and A operators, with suitable boundary data, in a quite general setting and to extend to this general setting also the results about Eq. (3). As regards boundary data, roughly speaking and supposing (Ru)(x, t) := r (x, t)u(x, t), we give the initial datum about u at time 0 where r "is positive", we prescribe a final datum for u at time T where r "is negative", no datum is prescribed where r = 0, both at time t = 0 and t = T . These boundary conditions are coherent with the Fichera conditions for the well-posedness of a boundary value problem of elliptic-parabolic type (see the paper [6], but see also [11] for a more recent paper discussing these conditions).
In the last section we give some examples which could help to clarify that. Since we consider abstract equation we consider functions defined in [0, T ] and valued in a Banach space. But the setting is not standard, we consider a family of triplets where V (t) is a reflexive Banach space which continuously embeds in the Hilbert space H (t), while V (t) denotes the dual space of V (t). In this way our functions will be defined in [0, T ] and, for each t ∈ [0, T ], u(t) will denote an element in In the last section we show with some examples why this setting can be interesting.

Notations, hypotheses and preliminary results
Consider the following family of evolution triplets where H (t) is a separable Hilbert space, V (t) a reflexive Banach space which continuously and densely embeds in H (t) and V (t) the dual space of V (t), and we suppose there is a constant k which satisfies for every w ∈ H (t), v ∈ V (t) and every t ∈ [0, T ]. We will suppose the existence of a Banach space U such that (for simplicity we consider the constant k as in (7)) and define, for some p ≥ 2, the set Moreover we will suppose that the functions are measurable for every u ∈ U and for the same p ∈ [2, +∞) used to define U we define the spaces as the completion of U with respect to the norms From now on we will suppose that the following holds true: Lemma 2.1 Denote by V the dual space of V and assume (10) holds.
Proof Consider the application On one hand we have immediately that On the other hand, consider f ∈ V * . For every ε > 0 one can choose a subdivision of [0, T ], i.e. t 0 = 0 < t 1 < t 2 < . . . t N = T and a step function Notice that for g ∈ H (t), z ∈ V (t) the duality between g and z satisfies In particular if g, z ∈ U we get that t → g, z V (t)×V (t) is continuous, and in fact uniformly where the last inequality is due to the fact that t → u i , w i V (t)×V (t) is uniformly continuous. By the (uniform) continuity of t → w i V (t) for every i we get that, choosing the points t i suitably near, we can suppose that By (10) we can also choose {t i } in such a way that and then, by (11), Then we have Now we focus our attention on the first term in the right hand side: Summing up we have By the arbitrariness of ε we finally get Then we have shown that T is a linear isometry between V * and V and in particular that which clearly implies that v = 0. By that we finally derive that T (V * ) is dense in V . Being also closed we conclude that T (V * ) = V .
By the assumptions above and supposing (10) by the previous lemma one gets that with continuous and dense embeddings, with Definition 2.2 Consider a family of linear operators R(t) such that R depends on a parameter t ∈ [0, T ] and being L(H (t)) the set of linear and bounded operators from H (t) in itself. We say that R belongs to the class E(C 1 , C 2 ), C 1 , C 2 ≥ 0, if it satisfies what follows for every u, v ∈ U : then we can define the square root of R(t) 2 (see, e.g., Chapter 3 in [14]), which is a positive operator, |R(t)| = R(t) 2 1/2 and then define the two positive operators By this decomposition we can also write and with respect to the norm w Clearly the operation depends on R. In this way = 0 (see, e.g., Theorem 10.37 in [14]) and R + (t) : Given an operator R ∈ E(C 1 , C 2 ) it is possible to define two other linear operators. First we can define the derivative of R which, unlike R, is valued in L(V (t), V (t)), i.e. the set of linear and bounded operators from By the density of U in V (t) we can extend R (t) to V (t).

Remark 2.3
Notice that the last request in Definition 2.2 and (8) imply that Via R, R + , R − and R we can also define which turn out to be linear and bounded by the constant C 1 and, by density of U in V, an operator which turns out to be linear, self-adjoint and bounded by C 2 . As done before we can define, in a way analogous to that done for the spaces (14), with respect to the norm w H = |R| 1/2 w H , where |R| = R + + R − .
Analogously, we define H + and H − and P + and P − the orthogonal projections fromH onto H + and H − respectively. H 0 is the kernel of R and P 0 the projection defined in H onto H 0 .

The existence result
In this section we will give one of the main results of the paper. We will consider a function R such that, given two non-negative constants, and all the spaces we introduced in (6), (8), (9), (14) and from now on we will assume (10) in such a way that Lemma 2.1 holds. Our goal is to give an existence result for an abstract equation like for some suitable operator A we will specify below. We want to stress that, despite of the fact that no derivative of R appears in the equation, we require R to be differentiable, i.e. R ∈ E(C 1 , C 2 ). This fact will be needed to get the existence of a solution to the previous equation and we will also show (see example 6 in the last section) that without this assumption at least uniqueness of the solution fails. Anyway to require that R is diffierentiable is not so restrictive (as shown in the examples in the last section) because if, for instance, R is a multiplication operator, i.e. Ru = r (x, t)u(x, t) for some function r , R could be differentiable even if r is discontinuous.
We will use this assumption about R to split properly the operator u → Ru + Au as indicated in (33) to give the existence of a solution.
We will use the operator R to define Ru in an apparently involute way. First for a function u ∈ V ⊂ H we consider the generalized derivative of Ru and require that it belongs to V , where the generalized derivative is defined as a function w ∈ V such that Notice that by (8) we have than we can define in a classical way the generalized derivative of Ru in L p (0, T ; U ) (here p denotes p/( p − 1)), and then to require that (Ru) ∈ V . For a function u ∈ V for which Ru admits generalized derivative in V we define With this definition in mind we now define the space Remark 3.1 Notice the space u ∈ V | (Ru) ∈ V endowed with the norm · W R coincides with W R . Indeed, in view of the definition of Ru given in (19), R u belong to V .
Moreover one can endow this space both with the norm · W R and with the norm u = u V + (Ru) V since they are equivalent.

Proposition 3.2 Under assumption (18) we have that for every u, v ∈ W R the function t → (R(t)u(t), v(t)) H (t) is absolutely continuous and the following hold:
and there exists a constant c, which depends only on T , such that In By the density of C 1 ([0, T ]; U ) one gets the first part of the thesis. Now we show the estimate.
and similarly one extends by which one concludes.

Remark 3.3 By the previous result we get that
In an analogous way, taking is continuous.
Now we prove a compactness result and, following the analogous in [13] (precisely Lemma 5.1 and Theorem 5.1), we divide the proof in a preliminary lemma and the compactness result. We prove both the results because in our situation things are very different from the standard case being the spaces defined in (5) depending on a parameter, while in [13] the spaces are fixed.
The proof of Theorem 3.5 follows the ideas of the analogous one of Lions, but it is not an immediate adaptation. To prove it we will use the following lemma.
Proof If, by contradiction, we suppose that the thesis is false we have that there exist a value of η, sayη, such that for each h ∈ N there is u h ∈ U for which Then, since (k defined in (7)) Moreover, since u h V = 1, we also get (up to select a subsequence still denoted by (u h ) h ) the existence ofū ∈ V such that u h →ū in V -weak.
In particular, up to a subsequence, u h → u weakly in V; for simplicity we can suppose that u = 0 and that We conclude if we show that Notice that, by Proposition 3.2, we have that for some positive constantc. Now define, for each t ∈ [0, T ], Then we get Being R(t)u h (t) V (t) equibounded, to get (24) it will be sufficient to show that the sequence } h converges to 0 pointwise. To show that we will confine to t = 0, being the choice of t irrelevant. Fix δ > 0, then consider φ ∈ U such that Notice that the function in particulat in σ = 0. Consider λ ∈ (0, 1] which will be fixed later and define the two spaces H λ and V λ as the completion of U with respect to the norms, respectively, Consider ϕ : [0, T ] → [0, 1] such that ϕ(0) = −1 and ϕ(T ) = 0: we have that Define a h the first addend, b h the second one. Notice that Then, after fixing ε > 0, we can choose λ small enough such that For the term a h we have Then, from (26), we derive that We can repeat the same argument for every t ∈ [0, T ] and get that for every δ > 0 lim sup By the arbitrariness of δ we derive that Before to state the existence result we recall some definitions and a classical result, for which we refer to [23] (see Section 32.4).
such that if we define the abstract operator this turns out to be A monotone, bounded, coercive and hemicontinuous.
We denote where R is defined in (15) and satisfies (18). Thanks to Remark 3.3 it makes sense to consider the two quantities for every w + ∈ U ∩ (H + (0) ⊕ H 0 (0)) and w − ∈ U ∩ (H − (T ) ⊕ H 0 (T )) and then to define Now we consider the three operators and state the following result.

Proposition 3.8 The operator L
Proof The proof is the same of the analogous in [17]. The only difference is that one has to consider first u ∈ C 1 ([0, T ]; U ) ∩ W 0 R and then use the definition of V, V * and Lemma 22.

Definition 3.9
We say u is a solution of the problem ⎧ ⎨ and the two conditions P + (0)u(0) = ϕ and P − (T )u(T ) = ψ are satisfied. We say u is a solution of the problem We start now giving an existence result for the following problem The idea now is to apply the previous proposition and Theorem 3.7 to the equation Ru +Au = f adding and subtracting a term involving the derivative of R and, in this way, to get the sum of two operators satisfying assumption of Theorem 3.7. We will see with an example (see example 6 in the last section) that, even if the derivative of R is not involved in the equation, the lack of regularity in time for R can cause problems, at least lack of uniqueness of solutions.
With this is mind we write Theorem 3.10 Suppose R satisfies assumptions (18). Suppose true one of the following: Then problems (32) admits a solution for every f ∈ V . If moreover M is strictly monotone the solution is unique. Suppose true one of the following: iii ) M = A + 1 2 R is monotone, bounded, coercive and hemicontinuous. iv ) M = A is monotone, bounded, coercive, hemicontinuous and R u, u V ×V ≥ 0 for every u ∈ V.
Then problems (31) with ϕ = ψ = 0 admits a solution for every f ∈ V . If moreover M is strictly monotone the solution is unique.
Proof The proof follows from Theorem 3.7 and Proposition 3.8, taking in Theorem 3.7 M = M and Lu = Ru + 1 2 R u in points i ) and iii ), Lu = Ru in point ii ), Lu = (Ru) in point iv ).

Remark 3.11 In fact we obtain an existence result also for the Cauchy problem
Now we want to consider the Cauchy-Dirichlet problem with non-zero "initial" data ⎧ ⎨ To do that we have to add some assumptions, both on A and R. Assumptions on A are explicitly given in the theorem which follows. Assumptions on R are more implicit and are hidden in (35). Consider the following spaces: (see (14) for the definition ofH − ,H 0 ,H + ). Then we suppose This assumptions indirectly involves the operator R, as we show with an example at the end of the paper (see example 2 in the last section).

Remark 3.12
We want to stress that we remove the assumption H + (0)∩ H − (T ) = {0} which, on the contrary, was made in [17].
Then the following theorem holds. (15), R ∈ E(C 1 , C 2 ) and A : V → V is hemicontinuous. Suppose that there exist two constants α, β > 0 such that
Proof First we show the existence result: consider , , ϕ, ψ ∈ U with P + (0) = ϕ, P − (T ) = ψ and define Now solve the following problem (one can easily verify that the operatorÃv := A(v + η) is bounded, coercive, strongly monotone and hemicontinuous) and consider u = v + η; then the function u solves (34) with ϕ, ψ ∈ U . Before concluding we come to the estimates. Notice that, since Pu = Ru + Au and, for p = 2, Pu = Ru + 1 2 R u + Au − 1 2 R u and by Proposition 3.2 we get for p > 2
Similarly as done above one can obtain the estimate for the difference u n − u m for every n, m ∈ N, and then there is a function u ∈ W such that u n → u in V and Ru n → Ru in V .
Up to select a subsequence we also get that Au n weakly converge to some b ∈ V and then Au n , u n V ×V → b, u V ×V . Since A is monotone and hemicontinuous A is type M, then we can conclude that b = Au.
Similarly one proves the following theorem, which is an extension to the case of Banach spaces depending on the parameter t of the result stated in [17].

Remark 3.15
If A is linear we also have the corresponding existence results for the problems ⎧ ⎨ for every λ ∈ R. It is sufficient indeed to consider the change of variable which has a unique solution v. Then u(t) = v(t)e −λt solves the original problem.

A time regularity result
Here we give a regularity result for the solution of (34) and (31) only in the following special case: For a much more detailed study about regularity, not only in t, we refer to [19].

Theorem 3.16
Under the assumption of Theorem 3.13 and given f ∈ V , ϕ ∈H + , ψ ∈H − denote by u the solution of problem (34). Suppose moreover that R ∈ E(C 1 , C 2 ), that A satisfies Assume that f , the generalized derivative of f , belongs to V and suppose there exist u 0 , u T ∈ V such that P + u 0 = ϕ, P − u T = ψ (P + and P − the projections from H ontoH + andH − , respectively) f (0) − Au 0 ∈ Im R, f (T ) − Au T ∈ Im R. Then Denote by v the solution of (41) and consider Clearly w ∈ H 1 (0, T ; V ). Writing Rv as (Rv) − R v and integrating the equation in (41) in [0, t] one obtains Since w = v by the previous equality we can derive that Then we get Notice that by assumption Rv(0) − f (0) + Au 0 = 0 ∈ H and, by (41), but we can choose u 0 in such a way that We can do that, if necessary, modifying u 0 and taking u 0 ∈ V to be the solution of the problem Clearly in this way the initial condition in (41) is maintained. Therefore, since u 0 satisfies (43), we have that the function w − u solves the problem Now consider the functioñ Similarly as done for w, but now integrating between T and t ∈ (0, T ), we get that the functionw − u solves the problem Now define the function z := w −w and notice that is independent of t and in particular z is the solution of the (elliptic) problem Then z = 0, i.e. w −w = 0, i.e.
By that we derive that the solution of problem (44) is 0, i.e.
By that, being w the solution of (41), we also get the estimate of point ii ).

Examples
In  , a x, t, Du(x) + b(x, t, u) with a : for every ξ ∈ R n and for some positive λ o , o and some M ≥ 0. Then A will be defined as in (27). We fix now our attention on the operator R. Consider a function and Finally for every t ∈ [0, T ] we denote and (see also (14)) Then consider the problem, for some f ∈ L 2 (0, T ; 1. Clearly Theorem 3.13 includes the "standard" equations. If r ≡ 1 we have the forward parabolic equation if r ≡ −1 we have the backward parabolic equation if r ≡ 0 we have a family of elliptic equations 2. Suppose r = r (x), r ∈ L ∞ ( ). As long as r ≥ 0 every function is admitted, even, for example, 3. Suppose r = r (t). Assumptions (18) are satisfied if r ∈ W 1,∞ (0, T ), therefore every r ∈ W 1,∞ (0, T ) is admitted. Two interesting situations are the following: the first when r (0) ≥ 0 and r (T ) ≤ 0 leads to the problem where a datum is given in the whole both at time 0 and at time T ; the second where r (0) ≤ 0 and r (T ) ≥ 0, which leads to the problem where no information is given in the whole both at time 0 and at time T . 4. More interesting is the case when r = r (x, t). As long as the situation is very similar to that analyzed in example 3, so every r such that r , r t ∈ L ∞ ( × (0, T )) is admitted, provided that assumption (35) is satisfied. Suppose now r does not admit a partial derivative with respect to time.
Well, assumption (18) could be satisfied anyway. To show it we consider a very simple example: suppose n = 1, = (a, b), T > 0, consider a function and define the sets and the function r To verify that R ∈ E we consider w 1 , w 2 ∈ H 1 0 (a, b) and evaluate Moreover notice that, as long as γ is decreasing so that γ ≤ 0, assumption (39) in Theorem 3.13 is easily satisfied, since turns out to be bounded thanks to the fact that γ is bounded, and In Figure 1.a below two possible admissible configurations are shown, the first one referring to a situation when ω + and ω 0 are like those defined in (50), the second one refers to a possible configuration with Notice that there is a constant C such that for every Then the first assumption required in (39) in Theorem 3.13 is satisfied if and then γ can be also increasing, provided that A possible configuration is shown in Figure 2, where γ is not necessarily negative, but it has to satisfy (52).
Analogous considerations can be made if n ≥ 2: taking the simplest example where for each t ∈ [0.T ] we have = + (t) ∪ 0 (t) and In this case we need that that for w 1 , w 2 ∈ H 1 0 ( ) the following hold: These hold if + (t) is open and the interface separating + (t) and 0 (t) is Lipschitz continuous (see, e. g., Proposition 3, section 3.4.4, in [5]). Moreover, since u, v ∈ H 1 0 ( ), it makes sense to consider the trace on this interface (see, e. g., Theorem 1, section 4.3, in [5] As regards the operator A, consider, for instance, it is like (46), but with a and b independent of t, i.e.
To consider a simper example suppose b(x, u) = b(x)u with b ∈ L ∞ ( ). Now consider f ∈ H 1 (0, T ; H −1 ( )) and the two functions u 0 and u T solutions respectively of the two problems and consider ϕ the restriction to + (0) (see (47)) of u 0 and ψ the restriction to − (T ) of u T . Then the solution u of (49) belongs to H 1 (0, T ; H 1 0 ( )). 6 Then we can fix η ∈ H 1 0 ( ) and solve separately the two problems and call u 1 the solution of the first problem, u 2 the solution of the second problem. Notice that the function u η (t) = u 1 (t) for t ∈ [0, T /2], u η (t) = u 2 (t) for t ∈ [T /2, T ] solves problem and this is true for every η ∈ H 1 0 ( ), and so we have infinite different solutions. Notice that if r depends only on t, r (0) < 0, r (T ) > 0, r increasing and continuous the problem above has a unique solution, even if there are no initial and final data.
In the following example we modify (46) and consider for A a monotone operator whose growth is more than linear. We consider the simple example where p > 2 (one could also consider p > 2n/(2 + n) is such a way that W 1, p ⊂ L 2 , but for simplicity we confined to for every ξ ∈ R n and for some positive λ o , o . Then we can consider the functions considered in example 4, but we have to confine to non-increasing γ in the first example and to some + (t) such that d dt

With
in the more general case. So in this case examples like those shown in Figure 1 are admissible, but that in Figure 2 is not. 8. Now consider the following R : [0, T ] → L(L 2 ( )). For a fixed function r ∈ L ∞ × × [0, T ] we define R(t)u (x):= r (x, y, t)u(y) dy u ∈ L 2 ( ) .
Clearly r could be a convolution kernel, i.e. r (x, y, t) = r (x − y, t) (suitable extended to zero outside of × (0, T )). If assumptions (35) Notice that Ru in (34) belongs, a priori, to V , but this is well defined since we recall that Ru = (Ru) − R u.
In this case (see again [18]) one has that there is q > p such that W 1,q 0 ( ) is dense in V (t) for every t ∈ [0, T ]. Then we consider U = W 1,q 0 ( ), V (t) and H (t) as above , for every ξ ∈ R n and for some L ≥ 1.
Consider the spaces and the operator just introduced and once defined In this way R(t) turns out to be bounded for every t even if μ is unbounded and we will need (see (13)) that the following function is absolutely continuous (and differentiable) for every u, v ∈ W 1,q 0 ( ): Then for every ϕ ∈ L 2 + (0), μ + (·, 0) , ψ ∈ L 2 − (T ), μ − (·, T ) and f ∈ V the problem ⎧ ⎪ ⎪ ⎨ has a unique solution. where μ ∈ L 1 ( × × (0, T )) and A, for instance, as in (54). 11. Another example of varying spaces is the following: consider first a function q : → [1, +∞) L q(·) ( ):= u ∈ L 1 loc ( ) |u(x)| q(x) dx < +∞ endowed with the norm (see, for instance, [12] for definitions and properties of these spaces) Clearly W 1,q(·) 0 ( ) is defined as the space Acknowledgements Open access funding provided by Università degli Studi di Padova within the CRUI-CARE Agreement. The author is very grateful to the referee. Her or his doubts and questions forced the author to improve the paper and avoid a mistake. Finally the author is grateful to Annalisa Cesaroni for drawing my attention to the Fichera conditions, which I did not know.
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