
This paper examined rural outdoor activity sites for children in Zhangye, Gansu Province, China. The study analyzed the characteristics of children's groups and their behavioral patterns during outdoor activities in rural areas, and explored the factors affecting children's engagement in outdoor activities. A questionnaire survey was conducted to assess the physical and mental health status of rural children in Gansu Province, the distribution of outdoor activity time, and the influence of activity venues on children. Based on the survey results in rural Zhangye, Gansu Province, activity venues and facilities suitable for rural children in Gansu were identified. Finally, five design principles were proposed (i.e., safety, fun, comfort, accessibility, and guiding) to cater to the activity preferences of rural children in Gansu, China and provide a nurturing environment that fostered their physical and mental growth.
Citation: Weidi Zhang, Runbo Liu. Data analysis and spatial design study of children's activity sites in rural Gansu, China[J]. Urban Resilience and Sustainability, 2023, 1(4): 314-333. doi: 10.3934/urs.2023020
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This paper examined rural outdoor activity sites for children in Zhangye, Gansu Province, China. The study analyzed the characteristics of children's groups and their behavioral patterns during outdoor activities in rural areas, and explored the factors affecting children's engagement in outdoor activities. A questionnaire survey was conducted to assess the physical and mental health status of rural children in Gansu Province, the distribution of outdoor activity time, and the influence of activity venues on children. Based on the survey results in rural Zhangye, Gansu Province, activity venues and facilities suitable for rural children in Gansu were identified. Finally, five design principles were proposed (i.e., safety, fun, comfort, accessibility, and guiding) to cater to the activity preferences of rural children in Gansu, China and provide a nurturing environment that fostered their physical and mental growth.
Sliding friction phenomena are nowadays recognized as the simultaneous occurrence of different physical mechanisms that escape the attempt to reduce them to a simple relation, like a friction law. In fact, entire areas of scientific research, such as Tribology, are specialized in the study of these problems [1,2,3].
From a mathematical (PDEs) perspective, a reasonable goal relies on detecting one or more key features that can enter into a rational model whose analysis reveals significant and new aspects of the phenomenon under examination. It was observed by various authors that in sliding friction, the occurrence of local instabilities emerging at different scales plays a decisive role in this physical manifestation [2]. Following this suggestion, we address here the simplest problem related to a linear wave equation in which a damping effect can occur only at certain spatial scales. This toy model is thought to investigate the effects of multiscale character in the wave propagation and the corresponding long-time behavior.
The reference model we have in mind is a variant of the linearly damped wave equation:
{∂ttu(t,x)−Δu(t,x)+∂tu(t,x)=0,(t,x)∈(0,+∞)×Ω,u(t,x)=0,(t,x)∈(0,+∞)×∂Ω,u(0,x)=u0(x),∂tu(0,x)=v0(x),x∈Ω. | (1.1) |
It is well-known that the solution to this equation converges to zero (in H1 norm) as t→+∞ exponentially fast [4,5]. The damping term ∂tu in the equation is responsible for this decay. We are interested in studying a different scenario in which the damping term does not act on the whole function u but on suitable parts of it. In view of the above physical considerations, we could investigate the case in which damping regards a suitable frequency band of the solution. This is implemented by applying a linear operator P to the damping term in the equation, which then reads
{∂ttu(t,x)−Δu(t,x)+P[∂tu(t,⋅)](x)=0,(t,x)∈(0,+∞)×Ω,u(t,x)=0,(t,x)∈(0,+∞)×∂Ω,u(0,x)=u0(x),∂tu(0,x)=v0(x),x∈Ω. | (1.2) |
We illustrate the simplest model we have in mind, leaving more general assumptions on P to the content of this work. With the aim of capturing different behaviors at different spatial scales, one can consider the special case Ω=Rd, d≥1, and the operator P given by
^P[v](t,ξ)=χ{|ξ|>1}(ξ)ˆv(t,ξ),ξ∈Rd, |
where ˆv(t,ξ) denotes the Fourier transform with respect to the spatial variable x and χ{|ξ|>1}(ξ) is the characteristic function of the complement of the unit ball. In this special case, due to the linearity of the equation, the different behavior of the solution at different scales becomes evident by inspecting it in the frequency variable. Indeed, in such a case, the problem is split into
{∂ttˆu(t,ξ)+|ξ|2ˆu(t,ξ)+∂tˆu(t,ξ)=0,t∈(0,+∞), |ξ|>1,(damped regime)∂ttˆu(t,ξ)+|ξ|2ˆu(t,ξ)=0,t∈(0,+∞), |ξ|≤1,(undamped regime)+ initial conditions. | (1.3) |
These different regimes are reflected in the long-time behavior of the solution. Relying on the structure in Eq (1.3) exhibited in the case Ω=Rd, one expects in the case Ω⊂Rd bounded that for high frequencies (i.e., small spatial scales), the solution behaves like the one to Eq (1.1) and converges exponentially fast to zero (dissipative regime). In contrast, for low frequencies (i.e., large spatial scales) the solution behaves like the one to the undamped wave equation and exhibits no decay (conservative regime). It is worth mentioning the recent results in [6] and the references therein for a detailed analysis of the decay in the case Ω=Rd when χ is localized at high frequencies.
We describe the results obtained in this paper concerning the problem (1.2). In Section 2, under suitable assumptions on the operator P, we prove well-posedness of the problem (1.2) and we study qualitative decay properties of the energy. Section 3 is devoted to an explicit analysis for special choices of the operator P in the one-dimensional case. Finally, in Section 4, we show that, in special cases, solutions to problem (1.2) split into a dissipative and a conservative part.
Let us emphasize that the linear framework addressed in the present work constitutes the simplest setting in which we aim to investigate the role of the spatial scales in dissipative phenomena. We are aware that a correct scenario to grasp the very nature of this physics should include nonlinear terms. More complex behaviors may occur when damping is coupled with adhesion [7,8,9], peeling [10,11,12,13,14], and in more general nonlinear settings [15,16]. We plan to address these issues in future works.
In this section we introduce the model, we prove well-posedness of the problem (1.2), and we study some qualitative properties of the energy.
Let d≥1 and assume that
(Ω1) Ω⊂Rd is a bounded, connected, open set with boundary of class C2.*
*These assumptions are enough to apply Poincaré's inequality in H10(Ω) and the elliptic regularity theory up to the boundary.
We study the initial boundary value problem
{∂ttu(t,x)−Δu(t,x)+P[∂tu(t,⋅)](x)=0,(t,x)∈(0,+∞)×Ω,u(t,x)=0,(t,x)∈(0,+∞)×∂Ω,u(0,x)=u0(x),∂tu(0,x)=v0(x),x∈Ω, | (2.1) |
where P:L2(Ω)→L2(Ω) satisfies the following properties:
(P1) P is a bounded linear operator;
(P2) P is monotone, i.e., ⟨P[u],u⟩L2(Ω)≥0 for all u∈L2(Ω).
Some general comments on the properties (P1) and (P2) of the operator P are in order. For more explicit examples of operators P satisfying the properties (P1) and (P2), we refer to Section 3.
Remark 2.1. If P=IdL2(Ω), we recover the classical wave equation with damping, i.e., Eq (2.1) reduces to
{∂ttu(t,x)−Δu(t,x)+∂tu(t,x)=0,(t,x)∈(0,+∞)×Ω,u(t,x)=0,(t,x)∈(0,+∞)×∂Ω,u(0,x)=u0(x),∂tu(0,x)=v0(x),x∈Ω. | (2.2) |
To explain in which sense the term P[∂tu(t,⋅)] may model a frequency-dependent damping, we point out the following example.
Example 2.2. If P:L2(Ω)→L2(Ω) is an orthogonal projection, then the properties (P1) and (P2) are satisfied. Indeed, orthogonal projections in a Hilbert space are characterized as linear self-adjoint idempotent operators. Continuity follows from the fact that
‖P[v]‖2L2(Ω)=⟨P[v],P[v]⟩L2(Ω)=⟨P2[v],v⟩L2(Ω)=⟨P[v],v⟩L2(Ω)≤‖P[v]‖L2(Ω)‖v‖L2(Ω). |
From the inequality above, we also obtain that P is monotone, since
⟨P[v],v⟩L2(Ω)=‖P[v]‖2L2(Ω)≥0. |
If an orthogonal projection is interpreted as a filter that selects some of the frequencies in the wave u(t,⋅), then the term P[∂tu(t,⋅)] can be seen as a damping term acting on those selected frequencies. For more details on this interpretation, see Section 3.
Remark 2.3. Since P is acting only on the spatial variable, it commutes with the differentiation with respect to time. We explain this fact in a more rigorous way. Given φ∈C∞c(R×Ω), we have that t↦P[φ] belongs to C∞(R;L2(Ω)) and
∂tP[φ(t,⋅)](x)=P[∂tφ(t,⋅)](x),for all t∈R, x∈Ω. |
This follows from the fact that, by linearity of P, for all t∈R, h≠0, x∈Ω, we have that
P[φ(t+h,⋅)](x)−P[φ(t,⋅)](x)h=P[φ(t+h,⋅)−φ(t,⋅)h](x), |
Exploiting boundedness of P and passing to the limit as h→0, we deduce that t∈R↦P[φ(t,⋅)]∈L2(Ω) is differentiable and we obtain the claimed identity (2.3). We iterate the argument for higher-order derivatives to conclude that t↦P[φ(t,⋅)] belongs to C∞(R;L2(Ω)).
We recast Eq (2.1) as an ODE in a Hilbert space. By writing Eq (2.1) as the first-order system
{∂tu(t,x)−v(t,x)=0,(t,x)∈(0,+∞)×Ω,∂tv(t,x)−Δu(t,x)+P[v(t,⋅)](x)=0,(t,x)∈(0,+∞)×Ω,u(t,x)=0,(t,x)∈(0,+∞)×∂Ω,u(0,x)=u0(x),v(0,x)=v0(x),x∈Ω, |
then Eq (2.1) can be interpreted as the Cauchy problem for curves U:[0,+∞)→H
{dUdt(t)+AU(t)=0,t∈(0,+∞),U(0)=U0, | (2.3) |
where
● U(t)=(u(t),v(t)) and U0=(u0,v0);
● H=H10(Ω)×L2(Ω) is the Hilbert space endowed with the scalar product (we are exploiting Poincaré 's inequality)
⟨(u,v),(w,z)⟩H=⟨∇u,∇w⟩L2(Ω)+⟨v,z⟩L2(Ω); |
● the linear operator A:Dom(A)→H is defined by
A(u,v)=(−v,−Δu+P[v]),(u,v)∈Dom(A); |
● the domain Dom(A) is given by
Dom(A)={(u,v)∈H:u∈H2(Ω)∩H10(Ω),v∈H10(Ω)}. |
We start with the key result that guarantees the well-posedness of the problem, relying on the theory of maximal monotone operators [17].
Lemma 2.4. Assume (Ω1) and (P1)–(P2). Then the operator A:Dom(A)→H is maximal monotone.
Proof. We have to show the following properties:
(i) A is monotone, i.e., for all (u,v),(w,z)∈Dom(A), we have
⟨A(u,v)−A(w,z),(u−w,v−z)⟩H≥0; |
(ii) A is maximal monotone, which is equivalent to saying that A+λIdH is surjective for some (and hence all) λ>0.
Let us prove (i). By linearity of A, it is enough to show that A satisfies
⟨A(u,v),(u,v)⟩H≥0,(u,v)∈Dom(A). |
Let (u,v)∈Dom(A). Integrating by parts and by (P2), we have that
⟨A(u,v),(u,v)⟩H=⟨(−v,−Δu+P[v]+u),(u,v)⟩H=−⟨∇v,∇u⟩L2(Ω)+⟨−Δu,v⟩L2(Ω)+⟨P[v],v⟩L2(Ω)=−⟨∇v,∇u⟩L2(Ω)+⟨∇u,∇v⟩L2(Ω)+⟨P[v],v⟩L2(Ω)=⟨P[v],v⟩L2(Ω)≥0. |
Let us prove (ii). We show that A+IdH is surjective. Let (f,g)∈H. We show that it is possible to find (u,v)∈Dom(A) such that
{−v+u=f,−Δu+P[v]+v=g. |
Indeed, the first equation gives v=−f+u. We substitute in the second equation to obtain that
−Δu+P[−f+u]+u−f=g⟹−Δu+P[u]+u=f+P[f]+g. |
The problem then reduces to finding u∈H2(Ω) such that
{−Δu+P[u]+u=h,in Ω,u=0,on ∂Ω, | (2.4) |
where h=f+P[f]+g∈L2(Ω). This problem has a unique solution u∈H10(Ω) by the Lax-Milgram theorem. Indeed, the bilinear form
b(u,u′)=⟨∇u,∇u′⟩L2(Ω)+⟨u,u′⟩L2(Ω)+⟨P[u],u′⟩L2(Ω),u,u′∈H10(Ω), |
satisfies
|b(u,u′)|≤C‖u‖H10(Ω)‖u′‖H10(Ω),u,u′∈H10(Ω), |
by the continuity of P and Poincaré's inequality. Moreover, it is coercive, since
b(u,u)=‖∇u‖2L2(Ω)+‖u‖2L2(Ω)+⟨P[u],u⟩L2(Ω)≥‖∇u‖2L2(Ω)+‖u‖2L2(Ω)≥c‖u‖2H10(Ω), |
by (P2) and Poincaré's inequality.
Then we read Eq (2.4) as
{−Δu=h−P[u]−u,in Ω,u=0,on ∂Ω, |
with the right-hand side h−P[u]−u∈L2(Ω). By the elliptic regularity theory, we conclude that u∈H2(Ω)∩H10(Ω). This concludes the proof.
Thanks to the maximal monotonicity of A, we can apply the theory of maximal monotone operators to guarantee the well-posedness of the Cauchy problem (2.3). We have the following classical result.
Theorem 2.5. Assume (Ω1) and (P1) and (P2). Then the operator A generates a continuous semigroup {S(t)}t≥0 on H, i.e., for all t≥0 the operator S(t):H→H is linear with ‖S(t)‖≤1, S(0)=IdH, S(t+s)=S(t)S(s) for all t,s≥0, and limt→0‖S(t)U0−U0‖H=0 for all U0∈H. Moreover, if U0∈Dom(A), then U(t)=S(t)U0 belongs to C1([0,+∞);H)∩C([0,+∞);Dom(A)) and is the unique solution to Eq (2.3).
Proof. The result follows from the theory of maximal monotone operators; see, e.g., [17, Theorem Ⅰ.2.2.1 & Remark I.2.2.3].
In this subsection we show that the energy of the solution to Eq (2.1) is non-increasing in time, and we provide an explicit expression for the energy dissipated by the damping term. Given (u,v)∈H10(Ω)×L2(Ω), we define the energy functional by
E(u,v)=12∫Ω|v(x)|2dx+12∫Ω|∇u(x)|2dx. |
Theorem 2.6. Assume (Ω1) and (P1) and (P2). Let U0=(u0,v0)∈H10(Ω)×L2(Ω). Let U(t)=(u(t),v(t)) be the unique solution to Eq (2.3) with initial datum U0 provided by Theorem 2.5. Then the following energy-dissipation balance holds:
E(u(t),v(t))+∫t0⟨P[v(s)],v(s)⟩L2(Ω)ds=E(u0,v0),t∈[0,+∞). | (2.5) |
In particular, the energy of the solution is non-increasing in time, i.e.,
E(u(s),v(s))≥E(u(t),v(t)),0≤s≤t. |
Proof. Let us start by proving Eq (2.5) in the case U0=(u0,v0)∈Dom(A). In this case, by Theorem 2.5, the solution U(t)=(u(t),v(t)) belongs to C1([0,+∞);H)∩C([0,+∞);Dom(A)). It follows that t↦E(u(t),v(t)) is differentiable. Differentiating the energy functional, substituting the equation in (2.1), and integrating by parts, we obtain
ddtE(u(t),v(t))=ddtE(u(t),∂tu(t))=⟨∂tu(t),∂ttu(t)⟩L2(Ω)+⟨∇u(t),∇∂tu(t)⟩L2(Ω)=⟨∂tu(t),Δu(t)−P[∂tu(t)]⟩L2(Ω)−⟨Δu(t),∂tu(t)⟩L2(Ω)=−⟨P[∂tu(t)],∂tu(t)⟩L2(Ω). |
Integrating in time, we obtain Eq (2.5).
Assuming that U0=(u0,v0)∈H10(Ω)×L2(Ω), we approximate it by a sequence Un0=(un0,vn0)∈Dom(A) such that Un0→U0 in H as n→+∞. By the continuity of the semigroup obtained in Theorem 2.5, we have that Un(t)=(un(t),vn(t))=S(t)Un0→U(t)=(u(t),v(t))=S(t)U0 in H as n→+∞, for all t≥0. The energy-dissipation balance Eq (2.5) holds for Un0 and Un(t), i.e.,
E(un(t),vn(t))+∫t0⟨P[vn(s)],vn(s)⟩L2(Ω)ds=E(un0,vn0),t∈[0,+∞). |
By the continuity of the energy functional with respect to the norm of H, we pass to the limit as n→+∞ in the above equation to obtain Eq (2.5) for U0 and U(t).
Finally, to show that the energy is non-increasing in time, we observe that
E(u(t),v(t))+∫ts⟨P[v(τ)],v(τ)⟩L2(Ω)dτ=E(u(s),v(s)),0≤s≤t, |
and, by the monotonicity of P, we have that ⟨P[v(τ)],v(τ)⟩L2(Ω)≥0 for all τ≥0. This concludes the proof.
Remark 2.7. If P=IdL2(Ω), Eq (2.1) reduces to the damped wave Eq (2.2); see Remark 2.1. A classical result states that the energy of the solution to the damped wave equation decays exponentially fast to zero. We recall the proof of this result, which relies on the perturbed energy functional
Eλ(u,v)=12∫Ω|v(x)|2dx+12∫Ω|∇u(x)|2dx+λ∫Ωu(x)v(x)dx, |
where λ>0 is a parameter to be chosen. Assuming that the initial datum U0=(u0,v0) belongs to Dom(A), we have that the solution U(t)=(u(t),v(t)) provided by Theorem 2.5 belongs to C1([0,+∞);H)∩C([0,+∞);Dom(A)). We can differentiate the perturbed energy functional and substitute the equation to obtain that
ddtEλ(u(t),v(t))=−‖v(t)‖2L2(Ω)+λ‖v(t)‖2L2(Ω)+λ⟨u(t),Δu(t)⟩L2(Ω)−λ⟨u(t),v(t)⟩L2(Ω)=−(1−λ)‖v(t)‖2L2(Ω)−λ‖∇u(t)‖2L2(Ω)−λ⟨u(t),v(t)⟩L2(Ω). |
Choosing 0<λ<12, we obtain that
ddtEλ(u(t),v(t))≤−CλEλ(u(t),v(t)), |
for a suitable Cλ>0, which implies that
Eλ(u(t),v(t))≤Eλ(u0,v0)e−γt, |
with γ>0 depending on λ. Then we estimate, by Poincaré's inequality, that
|⟨u(t),v(t)⟩L2(Ω)|≤12‖u(t)‖2L2(Ω)+12‖v(t)‖2L2(Ω)≤C‖∇u(t)‖L2(Ω)+12‖v(t)‖2L2(Ω), |
where C>0 depends on the Poincaré constant of Ω. By choosing λ suitably small, we obtain the estimate
1CEλ(u(t),v(t))≤E(u(t),v(t))≤CEλ(u(t),v(t)), |
which implies that
E(u(t),v(t))≤CE(u0,v0)e−γt. |
The estimate above is extended to the case U0=(u0,v0)∈H10(Ω)×L2(Ω) by approximation. This energy decay implies that the solution to the damped wave equation converges exponentially fast to zero in the energy space.
Note that the constant γ appearing in the exponential decay estimate depends on the domain Ω (through the Poincaré constant) and does not depend on the initial datum U0.
In this section we provide an explicit solution to the problem (2.1) in the one-dimensional case, i.e., Ω=(0,L), assuming a specific structure for the operator P. This will allow us to understand the asymptotic behavior of the solution in some specific cases.
We aim to solve the problem
{∂ttu−∂xxu+P[∂tu(t,⋅)](x)=0,(t,x)∈(0,+∞)×(0,L),u(t,0)=u(t,L)=0,t∈[0,+∞),u(0,x)=u0(x),ut(0,x)=v0(x)x∈(0,L), | (3.1) |
where u0∈H10(0,L) and v0∈L2(0,L).
To solve the problem explicitly, we will regard solutions as L-periodic functions and use Fourier series.† We write
†Because of the Dirichlet boundary conditions, one could use a different orthonormal system, e.g., writing u0(x)=∑n∈Nansin(nπxL). We prefer to use Fourier series to make the computations easier to follow.
u0(x)=∑k∈Zˆuk(0)e2πikx/L,v0(x)=∑k∈Zˆvk(0)e2πikx/L,u(t,x)=∑k∈Zˆuk(t)e2πikx/L, |
where the Fourier coefficients are given by
ˆuk(t)=1L∫L0u(t,x)e−2πikx/Ldx,k∈Z, |
and analogously for u0 and v0. Since u0 and v0 are real-valued, we have that
ˆu−k(0)=¯ˆuk(0),ˆv−k(0)=¯ˆvk(0). |
We analyze a specific form for the damping term P[∂tu(t,⋅)], assuming that P acts as a frequency filter. More precisely, we assume that
(P3) there exists a bounded sequence (ˆϕk)k∈Z with ˆϕk≥0 for all k∈Z such that
(^P[v])k=ˆϕkˆvk. |
Remark 3.1. If P satisfies (P3), by Parseval's identity, we have that
‖P[v]‖2L2(0,L)=∑k∈Z|ˆϕkˆvk|2≤supk∈Z|ˆϕk|2‖v‖2L2(0,L), |
whence (P1) holds. Moreover,
⟨P[v],v⟩L2(0,L)=∑k∈Zˆϕkˆv2k≥0, |
i.e., (P2) holds.
Example 3.2. If P is obtained via a convolution with an L1 function, i.e.,
P[v](x)=ϕ∗v(x)=1L∫L0ϕ(x−y)v(y)dy, |
with ϕ∈L1(0,L) extended periodically, then (P3) is satisfied. Indeed, we have that
(^P[v])k=(^ϕ∗v)k=ˆϕkˆvk. |
Taking the Fourier series in the equation in (3.1), we obtain that the Fourier coefficients ˆuk(t) satisfy the following system of ordinary differential equations:
∂ttˆuk(t)+(2πkL)2ˆuk(t)+ˆϕk∂tˆuk(t)=0,k∈Z. | (3.2) |
This can be solved explicitly. The characteristic equation is
λ2+ˆϕkλ+(2πkL)2=0, |
and its solutions are, possibly counted with multiplicity,
λ±k=−ˆϕk±√ˆϕ2k−(4πkL)22. | (3.3) |
Case 1: ˆϕ2k−(4πkL)2>0 (Overdamping). The roots obtained in Eq (3.3) are real and distinct. The general solution of Eq (3.2) is
ˆuk(t)=Aketλ+k+Bketλ−k, | (3.4) |
where Ak and Bk are complex constants that depend on the initial data. To find them, we impose
{Ak+Bk=ˆuk(0),Akλ+k+Bkλ−k=ˆvk(0). |
Solving this system, we obtain
Ak=−ˆuk(0)λ−k−ˆvk(0)λ+k−λ−k,Bk=ˆuk(0)λ+k−ˆvk(0)λ+k−λ−k. |
Substituting in Eq (3.4), we obtain
ˆuk(t)=−ˆuk(0)λ−k−ˆvk(0)λ+k−λ−ketλ+k+ˆuk(0)λ+k−ˆvk(0)λ+k−λ−ketλ−k=ˆuk(0)−λ−ketλ+k+λ+ketλ−kλ+k−λ−k+ˆvk(0)etλ+k−etλ−kλ+k−λ−k=ˆuk(0)(12ˆϕk+12√ˆϕ2k−(4πkL)2)etλ+k+(−12ˆϕk+12√ˆϕ2k−(4πkL)2)etλ−k√ˆϕ2k−(4πkL)2+ˆvk(0)etλ+k−etλ−k√ˆϕ2k−(4πkL)2=ˆuk(0)etλ+k+etλ−k2+(ˆϕkˆuk(0)+2ˆvk(0))1√ˆϕ2k−(4πkL)2etλ+k−etλ−k2. |
Writing the explicit expression of λ±k, we obtain
ˆuk(t)=e−tˆϕk2[cosh(t2√ˆϕ2k−(4πkL)2)ˆuk(0)+112√ˆϕ2k−(4πkL)2sinh(t2√ˆϕ2k−(4πkL)2)(ˆϕk2ˆuk(0)+ˆvk(0))]. |
Case 2: ˆϕ2k−(4πkL)2<0 (Damped oscillations). The roots obtained in Eq (3.3) are complex. The algebra to obtain the solution is the same as in the previous case. We obtain
ˆuk(t)=e−tˆϕk2[cos(t2√(4πkL)2−ˆϕ2k)ˆuk(0)+112√(4πkL)2−ˆϕ2ksin(t2√(4πkL)2−ˆϕ2k)(ˆϕk2ˆuk(0)+ˆvk(0))]. |
Case 3: ˆϕ2k−(4πkL)2=0. The roots obtained in Eq (3.3) are real and coincide. This means that the general solution of Eq (3.2) is
ˆuk(t)=Ake−tˆϕk2+Bkte−tˆϕk2. | (3.5) |
To find the constants Ak and Bk, we impose
{Ak=ˆuk(0),Ak(−ˆϕk2)+Bk=ˆvk(0). |
Solving this system, we obtain
Ak=ˆuk(0),Bk=ˆϕk2ˆuk(0)+ˆvk(0). |
Substituting in Eq (3.5), we obtain
ˆuk(t)=e−tˆϕk2[ˆuk(0)+(ˆϕk2ˆuk(0)+ˆvk(0))t]. |
In the previous subsection we have obtained the explicit solution to the problem (3.1) in the one-dimensional case. The result is summarized in the following theorem.
Theorem 3.3. Assume that Ω=(0,L) and that the operator P satisfies (P3). Let u0∈H10(0,L) and v0∈L2(0,L). Let u(t,x) be the unique solution to Eq (3.1) with initial data u0 and v0. Then the solution u(t,x) is given by
u(t,x)=∑k∈Zˆuk(t)e2πikx/L, |
where the Fourier coefficients ˆuk(t) are given by the following expressions:
● If ˆϕ2k−(4πkL)2>0, then
ˆuk(t)=e−tˆϕk2[cosh(t2√ˆϕ2k−(4πkL)2)ˆuk(0)+112√ˆϕ2k−(4πkL)2sinh(t2√ˆϕ2k−(4πkL)2)(ˆϕk2ˆuk(0)+ˆvk(0))]. |
● If ˆϕ2k−(4πkL)2<0, then
ˆuk(t)=e−tˆϕk2[cos(t2√(4πkL)2−ˆϕ2k)ˆuk(0)+112√(4πkL)2−ˆϕ2ksin(t2√(4πkL)2−ˆϕ2k)(ˆϕk2ˆuk(0)+ˆvk(0))]. | (3.6) |
● If ˆϕ2k−(4πkL)2=0, then
ˆuk(t)=e−tˆϕk2[ˆuk(0)+(ˆϕk2ˆuk(0)+ˆvk(0))t]. | (3.7) |
We provide some examples of the explicit solution in the one-dimensional case. In the next two examples, we show that the solution converges exponentially fast to the solution projected on the null space of the frequency filter P.
Example 3.4. Let (ˆϕk)k∈Z be such that
ˆϕk={1,if |k|≥k0,0,if |k|<k0, |
for some k0∈N. This means that the operator P:L2(0,L)→L2(0,L) is an orthogonal projection on Fourier modes with high frequencies. Indeed, it is idempotent, i.e., P2=P, and self-adjoint.
Assume that 1−4(2πkL)2<0 for |k|≥k0. In this case, the solution is given by Eq (3.6). In particular,
ˆuk(t)=cos(t2π|k|L)ˆuk(0)+12π|k|Lsin(t2π|k|L)ˆvk(0), |
for |k|<k0, and
ˆuk(t)=e−t2[cos(t2√(4πkL)2−1)ˆuk(0)+2√(4πkL)2−1sin(t2√(4πkL)2−1)(12ˆuk(0)+ˆvk(0))], | (3.8) |
for |k|≥k0.
From the explicit solutions we can deduce the asymptotic behavior of the solution. First of all, we observe that the null space of the linear operator P:L2(0,L)→L2(0,L) is given by
N(P)={v=∑|k|<k0ˆvke2πikx/L∈L2(0,L)}. |
We define Q=Id−P:L2(0,L)→L2(0,L), which is the orthogonal projection on the null space of P, i.e., the space of Fourier modes with low frequencies. We observe that
^Q[u(t,⋅)]k={cos(t2π|k|L)ˆuk(0)+12π|k|Lsin(t2π|k|L)ˆvk(0),if |k|<k0,0,if |k|≥k0. |
This implies that ‖u(t,⋅)−Q[u(t,⋅)]‖2H1(0,L) can be estimated, using Parseval's identity, simply in terms of the Fourier coefficients in Eq (3.8), giving
‖u(t,⋅)−Q[u(t,⋅)]‖H1(0,L)≤Ce−t2(‖u0‖H1(0,L)+‖v0‖L2(0,L)). |
Moreover, by linearity, Q[u(t,⋅)] is the solution to the problem (3.1) (which becomes an undamped wave equation, since PQ=QP=0) with initial data Q[u0] and Q[v0], i.e., the initial data projected on the null space of P. We have shown that the solution u(t,x) to the problem (3.1) converges exponentially fast to the solution to the problem with initial data projected on the null space of P. See also Figure 1 for a numerical simulation.
Example 3.5. Let L=1 for simplicity. Assume that
ˆϕk={0,if |k|≠1,4π,if |k|=1. |
Consider the initial data
u0=0,v0=2cos(2πx)=e2πix+e−2πix. |
Then
ˆvk(0)={0,if |k|≠1,1,if |k|=1. |
By Eq (3.7), the solution u(t,x) satisfies
ˆuk(t)={0,if |k|≠1,te−t2,if |k|=1. |
We observe that the null space of the operator P is given by
N(P)={v=∑|k|≠1ˆvke2πikx∈L2(0,1)}. |
The orthogonal projection Q=Id−P on the null space of P is simply the projection on the modes with |k|≠1. This implies that u(t,⋅)−Q[u(t,⋅)] is given by the modes with |k|=1. It follows that
‖u(t,⋅)−Q[u(t,⋅)]‖2H1(0,1)=(1+(2π)2)|ˆu1(t)|2+(1+(−2π)2)|ˆu−1(t)|2=2(1+(2π)2)t2e−t. |
Taking the square root, we obtain that the rate of convergence is not e−t/2 as in Example 3.4, but it is still exponential e−γt for γ∈(0,12).
The exponential rate of convergence is not the general behavior of the solution. It is strongly related to the structure of the frequency filter P, as we show in the next example, where the rate of convergence is subexponential.
Example 3.6. In this example we show that the universal exponential decay is not always present. We provide an example of P such that the following statement is not true: There exist γ>0 and M>0 such that for all initial data u0∈H10(0,1) and v0∈L2(0,1), the solution u(t,⋅) with initial data u0 and v0 satisfies
lim supt→+∞‖u(t,⋅)‖H1(0,1)e−γt(‖u0‖H1(0,1)+‖v0‖L2(0,1))≤M. | (3.9) |
Let L=1 for simplicity. Let us fix (ˆϕk)k∈Z that satisfies the following properties:
● ˆϕk>0 for all k∈Z;
● ˆϕ2k−(4πk)2<0 for all k∈Z, k≠0;
● ˆϕk=ˆϕ−k for all k∈Z;
● lim inf|k|→+∞ˆϕk=0.
The first condition is required just to simplify the example. Indeed, it implies that the null space of the operator P is given by N(P)={0}. Hence, in this example we do not need to consider the orthogonal projection Q, and we simply have to analyze the convergence of the solutions to the zero function.
Our claim is the following.
Claim_: Let γ>0. Let M>0. There exist initial data u0∈H10(0,1) and v0∈L2(0,1) and a sequence of times tn→+∞ such that the solution u(tn,⋅) with initial data u0 and v0 satisfies‡
‡This is precisely the negation of Eq (3.9).
‖u(tn,⋅)‖H1(0,1)>Me−γtn(‖u0‖H1(0,1)+‖v0‖L2(0,1))for all n∈N. |
We divide the proof of the claim into several steps.
Step 1: Fixing a suitable frequency k0. Let us fix γ>0 and M>0. We let k0∈N∖{0} (depending on γ) be such that
ˆϕk0<γ2and2(1+(2πk0)2)>γ22. | (3.10) |
Step 2: Constructing the initial datum. In this step we define the initial data u0 and v0. We let
u0(x)=2sin(2πk0x). |
Note that u0∈H10(0,1) and
ˆuk(0)={−i,if k=k0,i,if k=−k0,0,otherwise. |
Moreover, we let v0=−12P[u0]∈L2(0,1), so that
ˆvk(0)=−ˆϕk2ˆuk(0)⟹ˆϕk2ˆuk(0)+ˆvk(0)=0,for all k∈Z. |
Note that
‖u0‖2H1(0,1)=∑k∈Z(1+(2πk)2)|ˆuk(0)|2=2(1+(2πk0)2), |
and, by Eq (3.10),
‖v0‖2L2(0,1)=∑k∈Z|ˆvk(0)|2=|ˆvk0(0)|2+|ˆv−k0(0)|2=ˆϕ2k02<γ22<2(1+(2πk0)2)=‖u0‖2H1(0,1). | (3.11) |
Step 4: Computing the solution. By Eq (3.6), the solution u(t,x) to the problem (3.1) with initial data u0 and v0 has Fourier coefficients given by
ˆuk(t)=e−tˆϕk2cos(t2√(4πk)2−ˆϕ2k)ˆuk(0)={e−tˆϕk2cos(t2√(4πk)2−ˆϕ2k),if k=k0,−e−tˆϕk2cos(t2√(4πk)2−ˆϕ2k),if k=−k0,0,otherwise. |
Step 4: Estimating the H1 norm of the solution. By Eqs (3.10) and (3.11), we get
‖u(t,⋅)‖2H1=∑k∈Z(1+(2πk)2)|ˆuk(t)|2=e−tˆϕk0cos2(t2√(4πk0)2−ˆϕ2k0)2(1+(2πk0)2)≥e−γt/2cos2(t2√(4πk0)2−ˆϕ2k0)2(1+(2πk0)2)≥eγt/2e−γtcos2(t2√(4πk0)2−ˆϕ2k0)‖u0‖2H1(0,1)≥eγt/2e−γtcos2(t2√(4πk0)2−ˆϕ2k0)12(‖u0‖2H1(0,1)+‖v0‖2L2(0,1)). | (3.12) |
Step 5: Choice of sequence of times. First of all, we construct a sequence of times tn→+∞ such that
cos2(tn2√(4πk0)2−ˆϕ2k0)≥12, for all n∈N. | (3.13) |
For, it is enough to choose tn=2πn√(4πk0)2−ˆϕ2k0. By choosing n≥n0 with n0 large enough, we can additionally ensure that
14eγtn/2>M,for all n≥n0. | (3.14) |
Putting Eqs (3.13) and (3.14) in Eq (3.12), we obtain that
‖u(tn,⋅)‖2H1≥12eγtn/2e−γtncos2(tn2√(4πk0)2−ˆϕ2k0)(‖u0‖2H1(0,1)+‖v0‖2L2(0,1))≥Me−γtn(‖u0‖2H1(0,1)+‖v0‖2L2(0,1)). |
This concludes the proof of the claim.
In this section we show that, under suitable assumptions on P, the solution is split into two components: One that decays exponentially fast (the projected solution) and one that solves the undamped wave equation (the orthogonal component).
The precise assumptions on the bounded linear operator P:L2(Ω)→L2(Ω) are the following:
(A1) P commutes pointwise§ with the Dirichlet Laplacian, i.e., for every φ∈H2(Ω)∩H10(Ω) we have that P[φ]∈H2(Ω)∩H10(Ω) and P[Δφ]=ΔP[φ].
§We use this nomenclature to distinguish the assumption from strong commutation. See, e.g., [18, Ⅷ.5]
(A2) P:L2(Ω)→L2(Ω) is an orthogonal projection, i.e., P2=P and P is self-adjoint;
Example 4.1. An operator P of the form of Example 3.4 satisfies (A1) and (A2).
Remark 4.2. Assume that P satisfies (A1) and (A2). Then it also satisfies (P1) and (P2); see Example 2.2.
Remark 4.3. Assume that P satisfies (A1) and (A2). Let us show that P preserves H10(Ω), i.e., P(H10(Ω))⊂H10(Ω). Let u∈H10(Ω) and let us show that P[u]∈H10(Ω). Let us fix an approximating sequence uj∈H2(Ω)∩H10(Ω) such that uj→u in H10(Ω). By (A1), we have that P[uj]∈H2(Ω)∩H10(Ω). Moreover, P[uj]→P[u] in L2(Ω), as P is bounded. Let us estimate supj‖∇P[uj]‖2L2(Ω). We integrate by parts and we exploit (A1) and (A2) to obtain that
‖∇P[uj]‖2L2(Ω)=−⟨P[uj](x),ΔP[uj]⟩L2(Ω)=−⟨P[uj],P[Δuj]⟩L2(Ω)=−⟨P2[uj],Δuj⟩L2(Ω)=−⟨P[uj],Δuj⟩L2(Ω)=⟨∇P[uj],∇uj⟩L2(Ω)≤‖∇P[uj]‖L2(Ω)‖∇uj‖L2(Ω), |
from which we deduce that
supj‖∇P[uj]‖L2(Ω)≤supj‖∇uj‖L2(Ω)<+∞. |
It follows that P[uj]⇀P[u] weakly in H10(Ω), proving the claim.
To the aim of the splitting result, we start with a preliminary result. The proof is classical, as it relies on the linearity of the equation. We provide a proof for the sake of completeness.
Theorem 4.4. Assume that P satisfies (A1) and (A2). Let u0∈H10(Ω) and v0∈L2(Ω). Let (u(t),v(t)) be the unique solution to Eq (2.1) with initial datum (u0,v0) provided by Theorem 2.5. Then (P[u(t)],P[v(t)]) is the unique solution (in the sense of Theorem 2.5) to the problem:
{∂ttw−Δw+∂tw=0,(t,x)∈(0,+∞)×Ω,w=0,(t,x)∈(0,+∞)×∂Ω,w(0,x)=P[u0](x),∂tw(0,x)=P[v0](x),x∈Ω. | (4.1) |
Moreover, there exist constants M>0 and γ>0 such that
‖P[u(t)]‖2H1(Ω)+‖P[v(t)]‖2L2(Ω)≤Me−γt,for all t≥0, | (4.2) |
where M>0 depends on the energy of the initial data (u0,v0) and γ>0 depends on the domain Ω.
Proof. We split the proof into several steps.
Step 1: First of all, we show that t↦u(t,⋅)∈H10(Ω) is also a distributional solution to Eq (2.1), i.e., we have that
∫+∞0∫Ωu(t,x)(∂ttφ(t,x)−Δφ(t,x)−∂tP[φ(t,⋅)](x))dxdt=−∫Ωu0(x)∂tφ(0,x)dx+∫Ω(u0(x)P[φ(0,⋅)](x)+v0(x)φ(0,x))dx, | (4.3) |
for all φ∈C∞c(R×Ω).
To see this, let us first work in the case U0∈Dom(A), using the notation of Section 2. Then U(t)=(u(t),v(t))=S(t)U0 belongs to C1([0,+∞);H)∩C([0,+∞);Dom(A)) and is the unique solution to Eq (2.2). We fix a curve t↦Φ(t)∈H given by Φ(t)=(0,φ(t,⋅)) with φ∈C∞c(R×Ω). We have that
−⟨U0,Φ(0)⟩H=∫+∞0ddt⟨U(t),Φ(t)⟩Hdt=−∫+∞0⟨AU(t),Φ(t)⟩Hdt+∫+∞0⟨U(t),ddtΦ(t)⟩Hdt, |
which reads, using the self-adjointness of P,
−⟨v0,φ(0)⟩L2(Ω)=∫+∞0⟨Δu(t)−P[v(t)],φ(t)⟩L2(Ω)dt−∫+∞0⟨v(t),∂tφ(t)⟩L2(Ω)dt=∫+∞0(⟨u(t),Δφ(t)⟩L2(Ω)−⟨v(t),P[φ(t)]⟩L2(Ω))dt+∫+∞0⟨v(t),∂tφ(t)⟩L2(Ω)dt=∫+∞0(⟨u(t),Δφ(t)⟩L2(Ω)−⟨ddtu(t),P[φ(t)]⟩L2(Ω))dt+∫+∞0⟨ddtu(t),∂tφ(t)⟩L2(Ω)dt. |
Then we substitute in the previous equation the following two identities:
−⟨u0,∂tφ(0)⟩L2(Ω)=∫+∞0ddt⟨u(t),∂tφ(t)⟩L2(Ω)dt=∫+∞0⟨ddtu(t),∂tφ(t)⟩L2(Ω)dt+∫+∞0⟨u(t),∂ttφ(t)⟩L2(Ω)dt, |
and, using Remark 2.3,
−⟨u0,P[φ(0)]⟩L2(Ω)=∫+∞0ddt⟨u(t),P[φ(t)]⟩L2(Ω)dt=∫+∞0⟨ddtu(t),P[φ(t)]⟩L2(Ω)dt+∫+∞0⟨u(t),ddtP[φ(t)]⟩L2(Ω)dt=∫+∞0⟨ddtu(t),P[φ(t)]⟩L2(Ω)dt+∫+∞0⟨u(t),P[∂tφ(t)]⟩L2(Ω)dt, |
to obtain that
⟨u0,∂tφ(0)⟩L2(Ω)−⟨u0,P[φ(0)]⟩L2(Ω)−⟨v0,φ(0)⟩L2(Ω)=∫+∞0(⟨u(t),Δφ(t)⟩L2(Ω)+⟨u(t),P[∂tφ(t)]⟩L2(Ω))dt−∫+∞0⟨u(t),∂ttφ(t)⟩L2(Ω)dt, |
which is precisely Eq (4.3).
If U0∈H (not necessarily in Dom(A)), then Eq (4.3) is obtained by approximating U0 in the H-norm with a sequence in Dom(A) and then passing to the limit.
Step 2: In the condition (4.3), it is enough to test the equation with φ(t,x)=ζ(t)ψ(x), where ζ∈C∞c(R) and ψ∈C∞c(Ω). Hence, it reads
∫+∞0∫Ωu(t,x)(∂ttζ(t)ψ(x)−ζ(t)Δψ(x)−∂tζ(t)P[ψ](x))dxdt=−∫Ωu0(x)∂tζ(0)ψ(x)dx+∫Ω(u0(x)ζ(0)P[ψ](x)+v0(x)ζ(0)ψ(x))dx, | (4.4) |
for all ζ∈C∞c(R) and ψ∈C∞c(Ω).
Step 3: Note that, by an approximation argument, Eq (4.4) can be tested with ψ∈H2(Ω)∩H10(Ω).
Step 4: Given ψ∈H2(Ω)∩H10(Ω), by (A1) we have that P[ψ]∈H2(Ω)∩H10(Ω). Hence, we can use P[ψ] in Eq (4.4) instead of ψ to obtain that
∫+∞0∫Ωu(t,x)(∂ttζ(t)P[ψ](x)−ζ(t)ΔP[ψ](x)−∂tζ(t)P[P[ψ]](x))dxdt=−∫Ωu0(x)∂tζ(0)P[ψ](x)dx+∫Ω(u0(x)ζ(0)P[P[ψ]](x)+v0(x)ζ(0)P[ψ](x))dx. |
Using the properties P[Δψ]=ΔP[ψ] from (A1) and P2=P from (A2), we obtain that
∫+∞0∫Ωu(t,x)(∂ttζ(t)P[ψ](x)−ζ(t)P[Δψ](x)−∂tζ(t)P[ψ](x))dxdt=−∫Ωu0(x)∂tζ(0)P[ψ](x)dx+∫Ω(u0(x)+v0(x))ζ(0)P[ψ](x)dx. |
Finally, since P is self-adjoint, we have that
∫+∞0∫ΩP[u(t,⋅)](x)(∂ttζ(t)ψ(x)−ζ(t)Δψ(x)−∂tζ(t)ψ(x))dxdt=−∫ΩP[u0](x)∂tζ(0)ψ(x)dx+∫Ω(P[u0](x)+P[v0](x))ζ(0)ψ(x)dx. |
Step 6: Reasoning as for Eq (4.4), we conclude that
∫+∞0∫ΩP[u(t,⋅)](x)(∂ttφ(t,x)−Δφ(t,x)−∂tφ(t,x))dxdt=−∫ΩP[u0](x)∂tφ(0,x)dx+∫Ω(P[u0](x)+P[v0](x))φ(0,x)dx. |
for all φ∈C∞c(R×Ω), which we recognize as the definition of distributional solution to Eq (4.1).
Step 7: The distributional solution w∈L2((0,+∞)×Ω) to Eq (4.1) is unique hence, a fortiori, it must coincide with the solution provided by Theorem 2.5. This follows from a duality argument. To be precise, let us assume that w∈L2((0,+∞)×Ω) is a distributional solution to Eq (2.2) with initial data P[u0]=0 and P[v0]=0, and let us show that
∫+∞0∫Ωw(t,x)φ(t,x)dxdt=0, |
for all φ∈C∞c(R×Ω). Given φ∈C∞c(R×Ω), let T>0 be such that φ(t,x)=0 for t≥T. We consider a (strong) solution ˜φ to the final-time dual problem:
{∂tt˜φ−Δ˜φ−∂t˜φ=φ,(t,x)∈(0,+∞)×Ω,˜φ=0,(t,x)∈(0,+∞)×∂Ω,˜φ(T,x)=0,∂t˜φ(T,x)=0,x∈Ω. |
We approximate ˜φ with a sequence ˜φj∈C∞c(R×Ω) such that ˜φj→˜φ in H2(R×Ω). Since w is a distributional solution to Eq (4.1) with zero initial data, we have that
∫+∞0∫Ωw(t,x)(∂tt˜φj(t,x)−Δ˜φj(t,x)−∂t˜φj(t,x))dxdt=0. |
Passing to the limit as j→+∞, we obtain that
0=∫+∞0∫Ωw(t,x)(∂tt˜φ(t,x)−Δ˜φ(t,x)−∂t˜φ(t,x))dxdt=∫+∞0∫Ωw(t,x)φ(t,x)dxdt, |
concluding the proof of the claim.
Step 8: The exponential decay follows from Remark 2.7.
Theorem 4.5. Assume that P satisfies (A1)–(A2). Let Q=Id−P. Let u0∈H10(Ω) and v0∈L2(Ω). Let (u(t),v(t)) be the unique solution to Eq (2.1) with initial datum (u0,v0) provided by Theorem 2.5. Then
(u(t),v(t))=(Q[u(t)],Q[v(t)])+(P[u(t)],P[v(t)]) | (4.5) |
where (Q[u(t)],Q[v(t)]) is the unique solution to the undamped wave equation with initial datum (Q[u0],Q[v0]):
{∂ttz−Δz=0,(t,x)∈(0,+∞)×Ω,z=0,(t,x)∈(0,+∞)×∂Ω,z(0,x)=Q[u0](x),∂tz(0,x)=Q[v0](x),x∈Ω, | (4.6) |
and there exist constants γ,M>0 such that
‖u(t)−Q[u(t)]‖H10(Ω)+‖v(t)−Q[v(t)]‖L2(Ω)≤Me−γt,for allt≥0, |
where M and γ are as in Theorem 4.4.
Proof. Note that Q is the orthogonal projection on the null space of P. The decomposition (4.5) follows from the fact that Q is the orthogonal projection on the orthogonal complement of the range of P. Moreover, Q satisfies (A1) and (A2) as well.
As in the proof of Proposition 4.4, one proves that (Q[u(t)],Q[v(t)]) is a distributional solution to the undamped wave equation (4.6). Finally, the exponential decay follows from Eq (4.2).
In this paper we have investigated a specific dissipation mechanism occurring in the damped linear wave equation, when a frequency selection operator affects the damping term. Well-posedness of the initial boundary value problem and qualitative decay properties of the energy are investigated. Owing to the linear structure of the problem, we have explicitly analyzed different cases pertaining the frequency selection. Eventually we have proved that in special cases solutions of the evolution problem split into a dissipative and a conservative part.
All authors contributed equally to the study and the writing of the manuscript.
The authors declare that they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors declare that they have no conflict of interest.
The authors are members of Gruppo Nazionale per l'Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM).
They have been partially supported by the Research Project of National Relevance "Evolution problems involving interacting scales" granted by the Italian Ministry of Education, University and Research (MUR Prin 2022, project code 2022M9BKBC, Grant No. CUP D53D23005880006).
They acknowledge financial support under the National Recovery and Resilience Plan (NRRP) funded by the European Union–NextGenerationEU–Project Title "Mathematical Modeling of Biodiversity in the Mediterranean sea: From bacteria to predators, from meadows to currents"–project code P202254HT8–CUP B53D23027760001–Grant Assignment Decree No. 1379 adopted on 01/09/2023 by the Italian Ministry of University and Research (MUR).
They were partially supported by the Italian Ministry of University and Research under the Programme "Department of Excellence" Legge 232/2016 (Grant No. CUP–D93C23000100001).
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