The stabilization problem of a class of fuzzy systems with particular uncertainty restrictions is addressed in this paper using an observer design. We construct a fuzzy controller that ensures the Takagi-Sugeno fuzzy systems' uncertain solutions will converge. The ability to examine the convergence of trajectories using an estimated state controller towards a certain region of the origin that defines the system's asymptotic behavior is one benefit of the methodology employed in this work. Additionally, we provide an example to demonstrate the primary result's validity.
Citation: Kahouli Omar, Maatoug Tarak, Delmotte François, Ali Hammami Mohamed, Ben Ali Naim, Alshammari Mohammad. Analysis and design for a class of fuzzy control nonlinear systems with state observer under disturbances[J]. AIMS Mathematics, 2025, 10(4): 9595-9613. doi: 10.3934/math.2025442
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The stabilization problem of a class of fuzzy systems with particular uncertainty restrictions is addressed in this paper using an observer design. We construct a fuzzy controller that ensures the Takagi-Sugeno fuzzy systems' uncertain solutions will converge. The ability to examine the convergence of trajectories using an estimated state controller towards a certain region of the origin that defines the system's asymptotic behavior is one benefit of the methodology employed in this work. Additionally, we provide an example to demonstrate the primary result's validity.
The kinetic equation has become a powerful mathematical tool to describe the dynamics of many interacting particle systems, such as electrons, ions, stars, and galaxy or galactic aggregations. Since the nineteenth century, when Boltzmann formalized the concepts of kinetic equations, they have been used to model a variety of phenomena in different fields, such as rarefied gas dynamics, plasma physics, astrophysics, and socioeconomics. Particularly, in the life and social sciences, kinetic theory is used to model the dynamics of a large number of individuals, for example biological cells, animal flocks, pedestrians, or traders in large economic markets [1,2,3,4,5]. Moreover, it has applications in aerospace engineering [6], semi-conductor technology [7], nuclear engineering [8], chemotaxis, and immunology [9].
In this article, we consider the kinetic equation
∂tu(x,p,t)+n∑j=1pj∂xju(x,p,t)=f(t)g(x,p), | (1.1) |
under the conditions:
u(x,p,t)|x1≤0=u1(x,p,t), | (1.2) |
u(x,p,0)=u0(x,p) | (1.3) |
in the domain Ω={(x,p,t): x1>0, x∈D, p∈Rn, t∈R}, where D⊂Rn, x=(x1,ˉx)∈Rn, ˉx=(x2,...,xn)∈Rn−1. Throughout the paper, we used the following notations:
∂tu=∂u∂t,∂xju=∂u∂xj,∂xj∂yju=∂2u∂xj∂yj,∂2ηju=∂2u∂η2j,Δηu=n∑j=1∂2ηju(1≤j≤n)andi=√−1. |
In applications, u represents the number (or the mass) of particles in the unit volume element of the phase space in the neighbourhood of the point (x,p,t), where x, p and t are the space, momentum and time variables, respectively.
Before stating our problem, we reduce Eq (1.1) to a second-order partial differential equation. Toward this aim, we apply the Fourier transform with respect to p, then from (1.1) we obtain
∂tˆu(x,y,t)+in∑j=1∂xj∂yjˆu(x,y,t)=f(t)ˆg(x,y), | (1.4) |
where y is the parameter of the Fourier transform.
By (1.2) and (1.3), we find
ˆu(x,y,t)|x1≤0=ˆu1(x,y,t), ˆu(x,y,0)=ˆu0(x,y). | (1.5) |
Making the change of variables
x−y=ξ, x+y=η |
and introducing the function z(ξ,η,t)=ˆu(ξ+η2,η−ξ2,t), h(ξ,η)=ˆg(ξ+η2,η−ξ2) in (1.4), we get
∂tz(ξ,η,t)+i(Δη−Δξ)z(ξ,η,t)=f(t)h(ξ,η). | (1.6) |
From (1.5), it follows that
z(ξ,η,t)|ξ1+η1≤0=z1, z(ξ,η,0)=z0(ξ,η). | (1.7) |
We introduce the set Z of functions z∈C2(R2n+1)∩H2(R2n+1) such that z=0 for x1−y1≥ξ0, and the Fourier transform of z with respect to t is finite. We assume that f∈C(R)∩H2(R).
Equation (1.6) is called an ultrahyperbolic Schrödinger equation. Here, for technical reasons, we consider the equation
∂tz(ξ,η,t)+i(Δη−a−1Δξ)z(ξ,η,t)=f(t)h(ξ,η), | (1.8) |
where a∈C1(¯D) and a>0.
Since z=0 for x1−y1≥ξ0, by (1.7) we have
z(ξ,ξ0,ˉη,t)=∂η1z(ξ,ξ0,ˉη,t)=0, z(ξ,η,0)=z0(ξ,η), | (1.9) |
where ˉη=(η2,...,ηn).
We deal with the following problem:
Problem 1. Determine the pair of functions z(ξ,η,t) and h(ξ,η) from relations (1.8) and (1.9).
We investigate uniqueness of solution of Problem 1. For the proof, we shall use the Fredholm alternative theorem, so we consider the related homogeneous problem.
Then, by the condition z(ξ,η,0)=z0(ξ,η)=0 from (1.8), we can write
h(ξ,η)=if(0)∫Rωˆz(ξ,η,ω)dω, |
so we have
∂tz(ξ,η,t)+n∑j=1pj(x)∂xjz(ξ,η,t)=if(t)f(0)∫Rωˆz(ξ,η,ω)dω. | (1.10) |
The solvability of various inverse problems for kinetic equations was studied by Amirov [10] and Anikonov [11] on a bounded domain, where the problem is reduced to a Dirichlet problem for a third-order partial differential equation. See also [12] for an inverse problem for the transport equation, where the equation is reduced to a second-order differential equation with respect to the time variable t. Moreover, in [10,11], the problem of determining the potential in a quantum kinetic equation was discussed in an unbounded domain. Numerical algorithms to obtain the approximate solutions of some inverse problems were developed in [13,14]. The main difference between the current work and the existing works is that here the problem of finding the source function is considered in an unbounded domain with the Cauchy data which is given on a planar part of the boundary.
The main result of this paper is given below:
Theorem 1.1. Let ∂η1a>0 and f(0)≠0. Then, Problem 1 has at most one solution (z,h) such that z∈Z and h∈L1(R2n).
We apply the Fourier transform to Eq (1.10) and for condition (1.9) with respect to (ξ,t), we get
−ωˆz−Δηˆz−a−1|s|2ˆz=−ˆf(ω)f(0)∫Rωˆz(s,η,ω)dω, | (1.11) |
ˆz(s,0,ˉη,ω)=∂η1ˆz(s,0,ˉη,ω)=0. | (1.12) |
We write ˆz=z1+iz2 and ˆf=f1+if2 in (1.11), and so we obtain the following system of equations:
Δηzk+a−1|s|2zk=lk (k=1,2), | (1.13) |
where
l1=1f(0)(f1∫Rωz1dω−f2∫Rωz2dω)−ωz1,l2=1f(0)(f1∫Rωz2dω+f2∫Rωz1dω)−ωz2. | (1.14) |
By (1.12), we have
z1(s,0,ˉη,ω)=0, ∂η1(z1)(s,0,ˉη,ω)=0, z2(s,0,ˉη,ω)=0, ∂η1(z2)(s,0,ˉη,ω)=0. | (1.15) |
Thus, we shall show that this homogeneous problem has only the trivial solution. In the proof of Theorem 1.1, the main tool is a pointwise Carleman estimate, which will be presented in the next section. The proof of Theorem 1.1 will be given in the last section.
The Carleman estimate is a key tool for proving uniqueness and stability results in determining a source or a coefficient for ill-posed Cauchy problems. Carleman [15] established the first Carleman estimate in 1939 for proving the unique continuation for a two-dimensional elliptic equation. Later, Müller [16], Calderón [17], and Hörmander [18] obtained more general results. In the theory of inverse problems, Carleman estimates were first introduced by Bukgeim and Klibanov in [19]. After that, there have been many works relying on that method with modified arguments. We refer to Puel and Yamamoto [20], Isakov and Yamamoto [21], Imanuvilov and Yamamoto [22,23], Bellassoued and Yamamoto [24], and Klibanov and Yamamoto [25] for hyperbolic equations; Yamamoto [26] for parabolic equations; Amirov [10], Lavrentiev et al. [27], Romanov [28], and Gölgeleyen and Yamamoto [29] for ultrahyperbolic equations; Gölgeleyen and Kaytmaz [30,31,32] for ultrahyperbolic Schrödinger equations; and Cannarsa et al. [33], Gölgeleyen and Yamamoto [34], and Klibanov and Pamyatnykh [12] for transport equations.
In order to obtain a Carleman estimate for Eq (1.13), we write
P0zk=Δηzk+a−1|s|2zk (k=1,2). | (2.1) |
Next, we define
Ω0={(s,η):s∈ Rn, η∈Rn, η1>0, 0<δη1<γ−n∑j=2(ηj−η0j)2}, |
where 0<γ<1, δ>1, η0=(η01,...,η0n), γ+α0=ρ<1. Moreover, we introduce a Carleman weight function
φ=eλψ−ν, | (2.2) |
where α0>0, the parameters δ, λ and v are positive numbers, and
ψ(η)=δη1+12n∑j=2(ηj−η0j)2+α0. | (2.3) |
Obviously, η0∈Ω0 and Ω0⊂Ω for sufficiently small γ>0.
To establish a Carleman estimate we first present an auxiliary lemma.
Lemma 2.1. The following equality holds for any function zk∈Z:
zk(P0zk)φ2=a−1|s|2z2kφ2−|∇ηzk|2φ2+z2kn∑j=1(2λ2ν2(∂ηjψ)2ψ−2ν−2+λν(v+1)ψ−ν−2(∂ηjψ)−λνψ−ν−1(∂2ηjψ))φ2+d1(zk), | (2.4) |
where
d1(zk)=n∑j=1∂ηj((zk(∂ηjzk)+ψ−ν−1λv(∂ηjψ)z2k)φ2). |
Proof of Lemma 2.1. Since P0zk=(Δη+a−1|s|2)zk, we can write
zk(P0zk)φ2=zk(Δηzk)φ2+a−1|s|2z2kφ2, | (2.5) |
and using the equalities
zk(Δηzk)φ2=∂ηj(zkφ2(∂ηjzk))−∂ηj(zkφ2)(∂ηjzk), | (2.6) |
−∂ηj(zkφ2)(∂ηjzk)=−(∂ηjzk)2φ2−zk∂ηj(φ2)(∂ηjzk), | (2.7) |
−zk∂ηj(φ2)(∂ηjzk)=−∂ηj(z2k(−λv(∂ηjψ)ψ−ν−1)φ2)+12z2k∂2ηj(φ2), | (2.8) |
12z2k∂2ηj(φ2)=(2λ2ν2(∂ηjψ)2ψ−2ν−2+λv(v+1)(∂ηjψ)2ψ−ν−2−λv(∂2ηjψ)ψ−ν−1)z2kφ2, | (2.9) |
we obtain
zk(P0zk)φ2=−|∇ηzk|2φ2+z2kn∑j=1(2λ2ν2(∂ηjψ)2ψ−2ν−2+λν(v+1)ψ−ν−2(∂ηjψ)−λνψ−ν−1(∂2ηjψ))φ2+n∑j=1∂ηj((zk(∂ηjzk)+ψ−ν−1λv(∂ηjψ)z2k)φ2)+a−1|s|2z2kφ2. |
Proposition 2.1. Under the assumptions of Theorem 1.1, the following inequality is valid for all zk∈Z:
−2nλvzk(P0zk)φ2+ψν+1(P0zk)2φ2≥2λ3v3ψ−2ν−2z2kφ2+2λva−1|s|2z2kφ2+2λv|∇ηzk|2φ2−2nλvd1(zk)+d2(zk), | (2.10) |
where λ and ν are large parameters to be specified in the proof below. Moreover, d2(zk) denotes the sum of divergence terms, which will be given explicitly later.
Proof of Proposition 2.1. We first define a new function w=φzk, and then we can write
ψν+1(P0zk)2φ2=ψν+1((Δηzk)φ+a−1|s|2φzk)2≥4λν(n∑j=1(∂ηjψ)(∂ηjw))(Δηw+n∑r=1λvψ−ν−1(λvψ−ν−1(∂ηrψ)2−(v+1)ψ−1(∂ηrψ)2+(∂2ηrψ))w+a−1|s|2w):=5∑j=1Kj. | (2.11) |
Now, we calculate the terms Kj, 1≤j≤5 as follows:
K1=4λνn∑j,r=1(∂ηjψ)(∂ηjw)∂2ηrw=4λνn∑j,r=1∂ηj((∂ηjψ)(∂ηjw)(∂ηrw))−2λνn∑j,r=1∂ηj((∂ηjψ)(∂ηrψ)2)+2λν(n−1)2n∑j=1(∂ηjψ)2−4λν(n−1)2n∑j=2(∂ηjψ)2. | (2.12) |
Similarly, we can calculate
K2=4λ3v3ψ−2ν−2n∑j,r=1(∂ηjψ)(∂ηjw)(∂ηrψ)2w=2λ3v3ψ−2ν−2n∑j,r=1∂ηj((∂ηjψ)(∂ηrψ)2w2)+4λ3v3(v+1)ψ−2ν−3|∇ηψ|4w2−2(n−1)λ3v3ψ−2ν−2|∇ηψ|2w2−4λ3v3ψ−2ν−2|∇ηψ|2w2≥d22(w)+4λ3v4δ4ψ−2ν−2w2. | (2.13) |
Next, we have
K3=−4λ2ν2(v+1)ψ−ν−2n∑j,r=1(∂ηjψ)(∂ηjw)(∂ηrψ)2w=−2λ2ν2(v+1)ψ−ν−2n∑j,r=1∂ηj((∂ηjψ)(∂ηrψ)2w2)+2λ2ν2(v+1)(n−1)ψ−ν−2|∇ηψ|2w2+4λ2ν2(v+1)|∇ηψ|2ψ−ν−2w2−2λ2ν2(v+1)(v+2)ψ−ν−3|∇ηψ|4w2≥d23(w)−2λ2ν2(v+1)(v+2)ψ−ν−3|∇ηψ|4w2. | (2.14) |
As for the fourth term,
K4=4λ2ν2ψ−ν−1n∑j,r=1(∂ηjψ)(∂ηjw)(∂2ηrψ)w=2λ2ν2ψ−ν−1(n−1)n∑j=1∂ηj((∂ηjψ)w2)−2λ2ν2ψ−ν−1(n−1)2w2+2λ2ν2(n−1)(v+1)ψ−ν−2|∇ηψ|2w2≥d24(w)−2λ2ν2ψ−ν−1(n−1)2w2. | (2.15) |
Moreover, we have
K5=4λνa−1|s|2n∑j=1(∂ηjψ)(∂ηjw)w=2λν|s|2n∑j=1∂ηj((∂ηjψ)a−1w2)+4λν|s|2a−2w2(δ(∂η1a)+n∑j=2(∂ηja)(ηj−η0j))−2λν(n−1)|s|2a−1w2≥d25(w)+2λν(n+1)|s|2a−1w2, | (2.16) |
where we choose
δ>maxη∈¯D{n∑j=2|(∂ηja)(ηj−η0j)|+na}/minη∈¯D(∂η1a). |
If we replace the functions w with zk, then from (2.12)–(2.16), we can write
ψν+1φ2((Δηzk)+a−1|s|2zk)2≥d2(zk)−2λv(n−1)|∇ηzk|2φ2+2λv(n+1)|s|2a−1z2kφ2+3λ3v4δ4ψ−2ν−2z2kφ2+λ2v2z2kφ2(λv2δ4ψ−ν−1−2(v+1)(v+2)|∇ηψ|4ψ−1−2(n−1)2)ψ−ν−1, | (2.17) |
where
d2(zk)=5∑j=1d2j(zk),d21(zk)=4λνn∑j,r=1∂ηr((∂ηjψ)((∂ηjzk)−λνψ−ν−1(∂ηjψ)zk)((∂ηrzk)−λνψ−ν−1(∂ηrψ)zk)φ2−2λνn∑j,r=1∂ηj((∂ηjψ)((∂ηrzk)−λνψ−ν−1(∂ηrψ)zk)2φ2)+2λ2ν2ψ−ν−1(n−1)n∑j=1∂ηj((∂ηjψ)z2kφ2),d22(zk)=2λ3v3ψ−2ν−2∂ηj(|∇ηψ|2z2kn∑j=1∂ηjψ),d23(zk)=−2λ2ν2(v+1)ψ−ν−2n∑j=1∂ηj((∂ηjψ)|∇ηψ|2z2k),d24(zk)=2λ2ν2(n−1)ψ−ν−1n∑j=1∂ηj((∂ηjψ)z2k),d25(zk)=2λν|s|2n∑j=1∂ηj((∂ηjψ)a−1z2k). |
Here, we can choose λ≥λ0 such that
λv2δ4ψ−ν−1−2(v+1)(v+2)|∇ηψ|4ψ−1−2(n−1)2≥0, |
and so from (2.17) we obtain
ψν+1(P0zk)2φ2≥3λ3v4δ4ψ−2ν−2z2kφ2+2λv(n+1)|s|2a−1z2kφ2−2λv(n−1)|∇ηzk|2φ2+d2(zk). | (2.18) |
Finally, multiplying inequality (2.4) by −2nλv and summing with (2.18), we get
−2nλvzk(P0zk)φ2+ψν+1(P0zk)2φ2≥2λv|s|2a−1z2kφ2+2λv|∇ηzk|2φ2+(3λ3v4δ4ψ−2ν−2−4nλ3v3ψ−2ν−2|∇ηψ|2−2nλ2v2(v+1)ψ−ν−2n∑j=1(∂ηjψ)2+2nλ2v2ψ−ν−1(n−1))z2kφ2−2nλvd1(zk)+d2(zk). | (2.19) |
By choosing v≥v1, we obtain (2.10). Thus, the proof of Proposition 2.1 is complete.
First, for the right-hand side of (1.13), we can write
(l1)2+(l2)2≤3(f21+f22)f2(0)M1∫R(1+ω2)2(z21+z22)dω+3ω2(z21+z22). | (3.1) |
In (3.1), we used the following expressions:
(∫Rωzkdω)2=(∫R(1+ω2)−1/2(1+ω2)1/2ωzkdω)2≤∫R(1+ω2)−1dω∫Rω2(1+ω2)z2kdω≤M1∫R(1+ω2)2z2kdω |
and
M1=∫R(1+ω2)−1dω. |
Using (2.10), for k=1,2 we have
((lk)2+λ2ν2n2z2k)φ2+ψν+1(P0zk)2φ2≥−2λνnzk(P0zk)φ2+ψν+1(P0zk)2φ2≥2λ3ν3ψ−2ν−2z2kφ2+2λνa−1|s|2z2kφ2+2λν|∇ηzk|2φ2−2λνnd1(zk)+d2(zk). | (3.2) |
By (3.1), (3.2) and the equalities z21+z22=|ˆz|2, f21+f22=|ˆf|2, we obtain
(3|ˆf|2f2(0)M1∫R(1+ω2)2|ˆz|2dω+3ω2|ˆz|2)φ2≥2λ3ν3ψ−2ν−2|ˆz|2φ2+2λνa−1|s|2|ˆz|2φ2+2λν|∇ηˆz|2φ2+2∑k=1(d2(zk)−2λνnd1(zk)). | (3.3) |
If we multiply (3.3) by (1+ω2)2 and integrate with respect to the parameters ω over R, we have
∫Rφ2(|ˆz|2(2λ3ν3ψ−2ν−2+2λνa−1|s|2)+2λν|∇ηˆz|2)(1+ω2)2dω≤(3¯f0M2φ2∫R((1+ω2)2|ˆz|2)dω+3φ2∫R((1+ω2)2ω2|ˆz|2)dω)+2∑k=1∫R(1+ω2)2(2λνnd1(zk)−d2(zk))dω, | (3.4) |
where ¯f0=maxη∈¯D{M1f2(0)} and M2=∫R(1+ω2)2|ˆf|2dω.
In inequality (3.4), we can choose the big parameter λ, such that all terms on the right-hand side can be absorbed into the left-hand side. Then, we have
λ3ν3∫R|ˆz|2φ2dω≤div(ˆz), | (3.5) |
where
div(ˆz)=2∑k=1∫R(1+ω2)2(2λνnd1(zk)−d2(zk))dω. |
Since φ2>1 on Ω0, we have
∫R|ˆz|2dω≤∫R|ˆz|2φ2dω≤1λ3ν3div(ˆz). | (3.6) |
Integrating inequality (3.6) over Ω0 and passing to the limit as λ→∞, we have
∫Ω0∫R|ˆz|2dωdsdη≤0, | (3.7) |
which means that ˆz=0. Therefore, we conclude that z=0. Finally, by (1.8) we obtain h=0, which completes the proof of Theorem 1.1.
In this study, we considered an inverse problem for the kinetic equation in an unbounded domain. We reduced the equation to a second-order partial differential equation and proved the uniqueness of the solution of the problem by using the Carleman estimate. The method used in this paper can be applied to a variety of equations, including some first and second-order partial differential equations in mathematical physics, such as transport, ultrahyperbolic, and ultrahyperbolic Schrödinger equations. By similar arguments, a stability estimate can be obtained in an unbounded domain.
The author declares that she did not use Artificial Intelligence (AI) tools in the creation of this article.
The author declares no conflict of interest.
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