Boundary controllability for a 1D degenerate parabolic equation with drift and a singular potential and a Neumann boundary condition

We prove the null controllability of a one-dimensional degenerate parabolic equation with drift and a singular potential. Here, we consider a weighted Neumann boundary control at the left endpoint, where the potential arises. We use a spectral decomposition of a suitable operator, defined in a weighted Sobolev space, and the moment method by Fattorini and Russell to obtain an upper estimate of the cost of controllability. We also obtain a lower estimate of the cost of controllability by using a representation theorem for analytic functions of exponential type.


Introduction and main results
Let T > 0 and set Q := (0, 1) × (0, T ).For α, β ∈ R with 0 ≤ α < 2, α + β > 1, consider the system x −γ ux (0, t) = f (t), u(1, t) = 0 on (0, T ), u(x, 0) = u0(x) in (0, 1), (1) provided that µ ∈ R satisfies − ∞ < µ < µ(α + β), where µ(δ) := (δ − 1) 2 4 , δ ∈ R, and γ = γ(α, β, µ) := −(1 The first goal of this work is to provide a notion of a weak solution for the system (1) and show the wellposedness of this problem in suitable interpolation spaces.Here we consider a weighted Neumann boundary condition at the left endpoint to compensate for the singularity of the potential at this point.Then, we use the moment method introduced by Fattorini and Russell in [11] to prove the null controllability and show an upper bound estimate of the cost of controllability.Next, we use a representation theorem for analytic functions of exponential type to get a lower bound estimate of the cost of controllability.
In particular, when β = 0 this work solves the case of strong degeneracy with singularity.Concerning the strongly degenerate case (1 < α < 2) with no singularity (µ = 0), in [7] the authors study the null controllability of a degenerate parabolic equation with a degenerate one-order transport term.In [4][5][6] the authors prove the null controllability of 1D degenerate parabolic equations with first-order terms by means of Carleman inequalities, so they use interior controls.Now, assume the system (1) admits a unique solution for initial conditions in a certain Hilbert space H, which is described in the next section.We say that the system (1) in null controllable in H at time T > 0 with controls in L 2 (0, T ) if for any u0 ∈ H there exists f ∈ L 2 (0, T ) such that the corresponding solution satisfies u(•, T ) ≡ 0.
Once we know the system (1) is null controllable we study the behavior of the cost of the controllability.Consider the set of admissible controls U (T, α, β, µ, u0) = {f ∈ L 2 (0, T ) : u is solution of the system (1) that satisfies u(•, T ) ≡ 0}.

Lower bound of the cost
There exists a constant c > 0 such that where jν,2 is the second positive zero of the Bessel function Jν .
To prove this result we proceed as in [12], in particular, we use the biorthogonal family (ψ k ) k defined in (32) and constructed in [12].We also exploit this approach to show the null controllability of the system when the control is located at the right endpoint.Hence, consider the following system the corresponding set of admissible controls and the cost of the controllability given by Theorem 2. Let T > 0 and α, β, µ, γ ∈ R with 0 ≤ α < 2, α + β > 1, µ and γ satisfying (2) and ( 3) respectively.The next statements hold.

Upper bound of the cost
There exists a constant c > 0 such that for every δ ∈ (0, 1) we have

Lower bound of the cost
There exists a constant c > 0 such that Finally, we also analyze the null controllability of the system when the parameters satisfy 0 ≤ α < 2, β = 1 − α, and µ < 0. Thus, we consider the following system.
The corresponding set of admissible controls is given by and the cost of the controllability is given by We use some result from the singular Sturm-Liouville theory to show the well-posedness of system (6).

Lower bound of the cost
There exists a constant c > 0 such that This paper is organized as follows.In Section 2, we introduce suitable weighted Sobolev spaces and prove some results about the trace (at the endpoints) of functions in these spaces, as well as on the behavior of these functions at the endpoints, we also show an integration by parts formula.In that section, we prove that the autonomous operator given in ( 16) is diagonalizable, which allows the introduction of interpolation spaces for the initial data.Then, we prove the system (1) is well-posed in this setting.
In Section 3 we prove Theorem 1 by using the moment method, as a consequence, we get an upper estimate of K(T, α, β, µ).Then we use the representation theorem in Theorem A.1 to obtain a lower estimate of K(T, α, β, µ).In Section 4 we proceed as before to prove Theorem 2. Finally, in Section 5 we sketch the proof of Theorem 3.
The next result implies that we can talk about the trace at x = 1 of functions in H 1 α,β (0, 1).
The next result will allow us analyze the behavior at x = 0 of functions in H 1 α,β,N , see (9).
α,β,N stands for the dual space of (H 1 α,β,N , • * ) with respect to the pivot space L 2 β (0, 1): The inner product •, • * induces an isomorphism A : The next result gives a handy characterization of D(A).It shows the behavior of the derivative of functions in D(A) at the endpoints, see (11) and (12), and also provides an integration by parts formula, see (15).
Proof.Let H be the set on the right-hand side, we will show that D(A) = H.
In particular, Now let u ∈ H.We claim that x δ ux ∈ W 1,1 (0, 1) for all δ > (α + β + 1) /2.Just apply (8) with δ − 1 instead of δ to get that x δ−1 ux ∈ L 1 (0, 1), in particular x δ ux ∈ L 1 (0, 1).On the other hand, we have Notice the last quantity is finite by Proposition 6.Thus, we get the existence of the limit ux(1) := lim and we also have that lim x→0 + x δ ux(x) = 0 provided that δ > (α + β + 1)/2, see Remark 8.As in the proof of (9), we can see that lim Now consider any v ∈ H 1 α,β,N .We claim that x α+β uxv ∈ W 1,1 (0, 1): On the other hand, ( 9), ( 11) and ( 12) imply that lim x→0 + x α+β ux(x)v(x) = 0 and lim Thus, from (13) we get 1 0 From Proposition 9 in [3, p. 370] we have that A is a closed operator with D(A) dense in L 2 β (0, 1).We also have that The next result shows that A is a diagonalizable self-adjoint operator whose Hilbert basis of eigenfunctions can be written in terms of a Bessel function of the first kind Jν and its corresponding zeros j ν,k , k ≥ 1, located in the positive half line.In the appendix, we give some properties of Bessel functions and their zeros.Proposition 11. −A is a negative self-adjoint operator.Furthermore, the family is an orthonormal basis for L 2 β (0, 1) such that where ν is defined in (4).
Proof.From (15) we get that A is a symmetric operator.Letting u = v ∈ D(A) in ( 15) and using Proposition 6 we obtain that −A ≤ 0.
We claim that Ran( ) ′ , the Riesz representation theorem implies that there exists a unique u ∈ H 1 α,β,N such that ) and Corollary 2.4.10 in [2, p. 24] implies that −A is self-adjoint.
In [8] was proved that the family is an orthonormal basis for L 2 (0, 1).
Now we set w(x) = y(z) with z = cx a , a, c > 0. Assume that y = Jν .Therefore y satisfies the differential equation (61), i.e Then (A, D(A)) is the infinitesimal generator of a diagonalizable analytic semigroup of contractions in L 2 β (0, 1).Thus, we consider interpolation spaces for the initial data.For any s ≥ 0, we define and we also consider the corresponding dual spaces It is well known that H −s is the dual space of H s with respect to the pivot space L 2 β (0, 1), i.e Equivalently, H −s is the completion of L 2 β (0, 1) with respect to the norm It is well known that the linear mapping given by defines a self-adjoint semigroup S(t), t ≥ 0, in H s for all s ∈ R.

Control at the left endpoint 3.1 Upper estimate of the cost of the null controllability
In this section we use the method moment, introduced by Fattorini & Russell in [11], to prove the null controllability of the system (1).In [12, Section 3] the authors construct a biorthogonal family {ψ k } k≥1 ⊂ L 2 (0, T ) to the family of exponential functions {e −λ k (T −t) } k≥1 on [0, T ], i.e that satisfies That construction will help us to get an upper bound for the cost of the null controllability of the system (1).Here, we sketch the process to get the biorthogonal family {ψ k } k≥1 , see [12,Section 3] for details.
Consider the Weierstrass infinite product From (63) we have that j ν,k = O(k) for k large, thus the infinite product is well-defined and converges absolutely in C. Hence Λ(z) is an entire function with simple zeros at i(καj ν,k Clearly H a,θ (z) is an entire function.The following result gives additional information about H a,θ (z).
Lemma 15.The function H a,θ fulfills the following inequalities where c > 0 does not depend on a and θ.
For k ≥ 1 consider the entire function F k given as For δ ∈ (0, 1) we set iv) Furthermore, there exists a constant c > 0, independent of T, α and δ, such that where The L 2 -version of the Paley-Wiener theorem implies that there exists η k ∈ L 2 (R) with support in [−T /2, T /2] such that F k (z) is the analytic extension of the Fourier transform of η k .Hence is the family we are looking for.
Since η k , F k ∈ L 1 (R), the inverse Fourier theorem yields ), and by using (30) we have Now, we are ready to prove the null controllability of the system (1).Let u0 ∈ L 2 β (0, 1).Then consider its Fourier series with respect to the orthonormal basis We set Since Let u ∈ C([0, T ]; H −s ) that satisfies (19) for all τ ∈ (0, T ], z τ ∈ H s .In particular, for τ = T we take z T = Φ k , k ≥ 1, then the last equality implies that It just remains to estimate the norm of the control f .From (33) and (35) we get Using [15, Chap.XV, p. 438, eq. ( 3)], we can write and by using ( 18) and (65) we get From (36), the last two equalities and using that λ k ≥ λ1, it follows that By using the Cauchy-Schwarz inequality, the fact that j ν,k ≥ (k − 1/4)π (by (64)) and (34), we obtain that Notice that 0 < κα ≤ 1, and θ > 0. Thus, by using (29) with δ ∈ (0, 1), we have that and by using the definition of λ1 the result follows.

Lower estimate of the cost of the null controllability
In this section, we get a lower estimate of the cost K = K(T, α, β, µ).We set For ε > 0 small enough, there exists f ∈ U (α, β, µ, T, u0) such that u(•, T ) ≡ 0, and Then, in (19) we set τ = T and take and (65) it follows that Now consider the function v : C → C given by v(s) := Fubini and Morera's theorems imply that v(s) is an entire function.Moreover, (41) implies that We also have that Consider the entire function F (z) given by for some δ > 0 that will be chosen later on.Clearly, From (39), ( 43) and (44) we obtain We apply Theorem A.1 to the function F (z) given in (44).In this case, (43 By using the definition of the constants a k 's we have where we have used Lemma A.2 and made the change of variables From (46) we get the estimate From ( 45), ( 47), ( 48) and (49) we have where The result follows by taking , and then letting ε → 0 + .

Control at the right endpoint
Here, we analyze the null controllability of the system (5) where α + β > 1, 0 ≤ α < 2, µ and γ satisfy ( 2) and ( 3) respectively.As in Section 3 we give a suitable definition of a weak solution for the system (5).
The next result shows the existence of weak solutions for the system (5) under certain conditions on the parameters α, β, µ, γ and s.

Upper estimate of the cost of the null controllability
We are ready to prove the null controllability of the system (5).Let u0 ∈ L 2 β (0, 1) given as follows We set Since the sequence {ψ k } is biorthogonal to {e −λ k (T −t) } we have Let u ∈ C([0, T ]; H −s ) be the weak solution of system (5).In particular, for τ = T we take z T = Φ k , k ≥ 1, then (51) and (56) imply that It just remains to estimate the norm of the control f .From (33), (37), ( 53) and (55) we get By using that e −x ≤ e −r r r x −r for all x, r > 0, the Cauchy-Schwarz inequality, Lemma A.3 and the fact that j ν,k ≥ (k − 1/4)π (by (64)) and (34), we obtain that and the result follows by (38).

Lower estimate of the cost of the null controllability at x = 1
Here, we just give a sketch of the proof of a lower estimate for the cost K = K(T, α, β, µ).Consider u0 ∈ L 2 β (0, 1) given in (39).
5 The case α + β = 1 Concerning the case α + β < 1, in [12] we showed the system (1) is well-posed when considering suitable weighted Dirichlet condition at the left endpoint and proved the null-controllability of the corresponding system.In both cases (α + β < 1 and α + β > 1) our approach is based on the validity of the Hardy inequality, see Proposition 6 and [12, Proposition 4].If α + β = 1 then µ(α + β) = 0, and the corresponding Hardy inequality does not provide any information.Thus, to solve the case α + β = 1 we use the singular Sturm-Liouville theory, see [16] for the definitions used here.
By using the notation in [16,Chapter 10] we can see that SF = S * min | D(S F ) = Smax| D(S F ) , and Smaxu = w −1 M u = Au, u ∈ Dmax.
The second part follows by using the computations in the proof of Proposition 11.
Thus, we are in the same position as in [12], so we can follow the same steps to get the proof of Theorem 3.