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

Abstract

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 using a representation theorem for analytic functions of exponential type.

1 Introduction

This work analyzes the null controllability for 1D degenerate/singular parabolic equations when the control acts at the endpoint where the degeneracy/singularity arises. In this paper, we consider a weighted Neumann boundary control.

Let $T > 0$ and set $Q:= (0, 1) \times (0, T )$. For $\alpha ,\beta \in {\mathbb {R}}$ with $0\le \alpha <2$, $\alpha +\beta >1$, consider the system

$$\begin{aligned} \left\{ \begin{array}{ll} u_t-(x^\alpha u_x)_x-\beta x^{\alpha -1}u_x-\frac{\mu }{x^{2-\alpha }} u=0 &{} \quad \text{ in } Q, \\ \left( x^{-\gamma }u_x\right) (0, t) =f(t),\quad u(1, t)=0 &{} \quad \text{ on } (0, T), \\ u(x, 0) =u_{0}(x) &{} \quad \text{ in } (0, 1), \end{array}\right. \end{aligned}$$

(1)

provided that $\mu \in {\mathbb {R}}$ satisfies

$$\begin{aligned} -\infty<\mu <\mu (\alpha +\beta ), \end{aligned}$$

(2)

where

$$\begin{aligned} \mu (\alpha +\beta ):= & {} \frac{(\alpha +\beta -1)^2}{4},\quad \gamma =\gamma (\alpha , \beta ,\mu )\nonumber \\ {}:= & {} -(1+\alpha +\beta )/2-\sqrt{\mu (\alpha +\beta )-\mu }. \end{aligned}$$

(3)

We notice the constant $\mu (\alpha +\beta )$ appears in the Hardy inequality (7); thus, the condition (2) corresponds to a so-called sub-critical potential, see [1, p. 765]. We chose the constant $\gamma$ in such a way that the limit in (A10) is finite, see also (20).

1.1 Dirichlet boundary conditions

Most published articles about degenerate/singular parabolic equations, like system (1), consider a weighted Dirichlet boundary condition $\lim _{x\rightarrow 0^+}x^{\sigma }u(x,t)=f(t)$ at the point of controllability $x=0$, with a suitable constant $\sigma$. In this context, the first result was obtained in [2] in the case of a weak degeneracy $0 \le \alpha < 1$, without singularity and drift, i.e., $\beta =\mu =0$. In that work, the null controllability is achieved using $L^2$ controls. The last result was improved in [3] where the authors prove that the controllability cost behaves as $(1-\alpha )^{-1}\textrm{e}^{1/T}$, both upper and lower estimate when $\alpha <1$.

In [4], the author analyzes the controllability of the heat equation with a singular inverse-square potential, i.e., with $\alpha =\beta =0$ and $\mu <\mu (0)=1/4$. In [5], the authors also study the weakly degenerate case without drift ($\beta =0$) and $\mu$ satisfying (2). All these results give the null controllability using $H^1$ controls. Unfortunately, the proofs of [4, Theorem 2.1] and [5, Theorem 2.2] are not complete, see [6, Remark 4.2] for details.

Concerning the strongly degenerate case ($1<\alpha <2$) with no singularity ($\mu =0$), in [7], the authors analyze the null controllability of a degenerate parabolic equation with a degenerate one-order transport term, but they consider a unweighted Dirichlet boundary control at the left endpoint. In [8,9,10], the authors consider the null controllability of 1D degenerate parabolic equations with first-order terms, but they use interior controls.

Concerning the case $\alpha +\beta <1$, in [6], the authors prove the null controllability of the equation in the system (1) with initial conditions in $L^2_\beta (0,1)$ but considering a weighted Dirichlet boundary condition at the left endpoint. There and here, suitable versions of a Hardy inequality are proved to assure the well-posedness of the corresponding systems, see Proposition 3.3.

In the case $\alpha +\beta =1$, the Hardy inequality does not hold anymore, although there exist some generalized Hardy-type inequalities that can be used to obtain controllability results, see [11]. In this work, we choose to utilize certain results from singular Sturm–Liouville theory to establish the well-posedness of the system (6).

1.2 Neumann boundary conditions

In this context, in [12], the author considers the system (1) with a strong degeneracy $1\le \alpha <2$, without singularity and drift, i.e., $\beta =\mu =0$, and a homogeneous weighted Neumann condition at $x=0$. In that work, the author uses the flatness approach to prove the null controllability by means of controls (in the Gevrey class) acting in the right endpoint. The last result was improved in [13], where the authors use $L^2$ controls and prove that the controllability cost behaves as $\textrm{e}^{(2-\alpha )^{-2}/T}$.

Recently, in [14], the authors use the flatness approach to prove the null controllability of the following system,

$$\begin{aligned} \begin{array}{ll} \left( a(x) u_x\right) _x+q(x) u=\rho (x) u_t, \quad &{} x \in (0,1), \,t\in (0, T), \\ \left( a u_x\right) (0, t)=0, \quad &{} t \in (0, T), \\ \alpha u(1, t)+\beta \left( a u_x\right) (1, t)=h(t), \quad &{}t \in (0, T), \\ u(x, 0)=u_0(x), \quad &{} x \in (0,1). \end{array} \end{aligned}$$

where the nonnegative function a may vanish strongly at $x = 0$, the potential q may be singular at $x = 0$, and $\rho (x)>0$. In this case, the control h is in a Gevrey class and acts at the right endpoint. Only weakly degenerate parabolic equations were analyzed in [15].

As far as we know, this is the first time the null controllability of the system (1) has been proved with a weighted Neumann boundary control at the left endpoint. We also get lower and upper estimates of the cost of controllability.

2 Main results

The first goal of this work is to provide a notion of a weak solution for the system (1) and show the well-posedness 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 [16] 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.

Consider the weighted Lebesgue space $L^2_\beta (0,1):=L^2((0,1);x^\beta \textrm{d}x)$, $\beta \in {\mathbb {R}}$, endowed with the inner product

$$\begin{aligned} \langle f,g\rangle _\beta :=\int _0^1 f(x)g(x)x^\beta \textrm{d}x, \end{aligned}$$

and its corresponding norm is denoted by $\Vert \cdot \Vert _{\beta }$.

In Sect. 3, we show the well-posedness of the system (1) with initial data in $L^2_\beta (0,1)$, although the solution u(t) lives in an interpolation space ${\mathcal {H}}^{-s}$. Once we know the system (1) admits a unique solution for initial conditions in $L^2_\beta (0,1)$, it is said the system (1) is null controllable at time $T>0$ with controls in $L^2(0,T)$ if for any $u_0\in L^2_\beta (0,1)$, there exists $f\in L^2(0,T)$ such that the corresponding solution satisfies $u(\cdot ,T)\equiv 0$.

We are also interested in the behavior of the cost of the null controllability. Consider the set of admissible controls

$$\begin{aligned} U(T,\alpha ,\beta ,\mu ,u_0)=\{f\in L^2(0,T): u \text { is solution of the system }(1)\text { that satisfies } u(\cdot ,T)\equiv 0\}. \end{aligned}$$

Then the cost of the controllability is defined as

$$\begin{aligned} {\mathcal {K}}(T,\alpha ,\beta ,\mu ):=\sup _{\Vert u_0\Vert _\beta \le 1}\inf \{\Vert f\Vert _{L^2(0,T)}:f\in U(T,\alpha ,\beta ,\mu ,u_0)\}. \end{aligned}$$

The main result of this work solves the case $\alpha +\beta >1$. First, we introduce the constants

$$\begin{aligned} \kappa _\alpha :=\frac{2-\alpha }{2},\quad \nu =\nu (\alpha ,\beta ,\mu ):=\sqrt{\mu (\alpha +\beta )-\mu }/\kappa _\alpha . \end{aligned}$$

(4)

The last constants are chosen so that the equation in (19) yields an eigenfunction of the operator ${\mathcal {A}}$ defined in (16), see (18).

Theorem 2.1

Let $T>0$ and $\alpha ,\beta ,\mu ,\gamma \in {\mathbb {R}}$ with $0\le \alpha <2$, $\alpha +\beta >1$, $\mu$ and $\gamma$ satisfying (2) and (3) respectively. The next statements hold.

• 1. Existence of a control. For any $u_0\in L^2_\beta (0,1)$, there exists a control $f \in L^2(0, T )$ such that the solution u to (1) satisfies $u(\cdot ,T ) \equiv 0$.

• 2. Upper bound of the cost. There exists a constant $c>0$ such that for every $\delta \in (0,1)$, we have

  $$\begin{aligned} {\mathcal {K}}(T,\alpha ,\beta ,\mu )\le c M(T,\alpha ,\nu ,\delta )T^{1/2}\kappa _\alpha ^{-1/2} \exp \left( -\frac{T}{2}\kappa _\alpha ^2 j_{\nu ,1}^2\right) , \end{aligned}$$

  where $j_{\nu ,1}$ is the first positive zero of the Bessel function $J_\nu$ (defined in the Appendix), and

  $$\begin{aligned} \begin{array}{rcl} M(T,\alpha ,\nu ,\delta ) &{} = &{}\displaystyle \left( 1+\frac{1}{(1-\delta )\kappa _\alpha ^2 T}\right) \left[ \exp \left( \frac{1}{\sqrt{2}\kappa _\alpha }\right) +\frac{1}{\delta ^3}\exp \left( \frac{3}{(1-\delta )\kappa _\alpha ^2 T}\right) \right] \\ &{}&{}\\ &{} &{}\displaystyle \times \exp \left( -\frac{(1-\delta )^{3/2}T^{3/2}}{8(1+T)^{1/2}}\kappa _\alpha ^3 j_{\nu ,1}^2\right) . \end{array} \end{aligned}$$

• 3. Lower bound of the cost. There exists a constant $c>0$ such that

  $$\begin{aligned} \displaystyle \frac{c2^{\nu } \Gamma (\nu +1) \left| J_{\nu }^{\prime }\left( j_{\nu , 1}\right) \right| \exp {\left( \left( \frac{1}{2}-\frac{\log 2}{\pi }\right) j_{\nu ,2}\right) }}{\left( {2T \kappa _\alpha }\right) ^{1/2}\left( j_{\nu , 1}\right) ^{\nu }}&\exp&\left( -\left( j_{\nu ,1}^2+\frac{j_{\nu ,2}^2}{2}\right) \kappa _\alpha ^2 T\right) \\\le & {} {\mathcal {K}}(T,\alpha ,\beta ,\mu ), \end{aligned}$$

where $j_{\nu ,2}$ is the second positive zero of the Bessel function $J_\nu$.

To prove this result, we use the moment method by Fattorini and Russell, in particular, we use the biorthogonal family $(\psi _k)_k$ defined in (33) and constructed in [6].

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

$$\begin{aligned} \left\{ \begin{array}{ll} u_t-(x^\alpha u_x)_x-\beta x^{\alpha -1}u_x-\frac{\mu }{x^{2-\alpha }} u=0 \quad &{} \text{ in } Q, \\ \left( x^{-\gamma }u_x\right) (0, t) = 0, u(1, t) = f(t) \quad &{} \text{ on } (0, T), \\ u(x, 0) =u_{0}(x) \quad &{} \text{ in } (0, 1), \end{array}\right. \end{aligned}$$

(5)

the corresponding set of admissible controls

$$\begin{aligned} {\widetilde{U}}(T,\alpha ,\beta ,\mu ,u_0)=\{f\in L^2(0,T): u \text { is solution of the system }(5)\text { that satisfies } u(\cdot ,T)\equiv 0\}, \end{aligned}$$

and the cost of the controllability given by

$$\begin{aligned} \widetilde{{\mathcal {K}}}(T,\alpha ,\beta ,\mu ):=\sup _{\Vert u_0\Vert _\beta \le 1}\inf \{\Vert f\Vert _{L^2(0,T)}:f\in {\widetilde{U}}(T,\alpha ,\beta ,\mu ,u_0)\}. \end{aligned}$$

Theorem 2.2

Let $T>0$ and $\alpha ,\beta ,\mu ,\gamma \in {\mathbb {R}}$ with $0\le \alpha <2$, $\alpha +\beta >1$, $\mu$ and $\gamma$ satisfying (2) and (3), respectively. The next statements hold.

• 1. Existence of a control. For any $u_0\in L^2_\beta (0,1)$, there exists a control $f \in L^2(0, T )$ such that the solution u to (5) satisfies $u(\cdot ,T ) \equiv 0$.

• 2. Upper bound of the cost. There exists a constant $c>0$ such that for every $\delta \in (0,1)$, we have

  $$\begin{aligned} \widetilde{{\mathcal {K}}}(T,\alpha ,\beta ,\mu )\le \frac{c M(T,\alpha ,\nu ,\delta )T^{1/2}}{(2\kappa _\alpha )^{\nu }\Gamma (\nu +1)} \left( \dfrac{2\nu +1}{T\textrm{e}}\right) ^{(2\nu +1)/4} \exp \left( -\frac{T}{4}\kappa _\alpha ^2 j_{\nu ,1}^2\right) . \end{aligned}$$

• 3. Lower bound of the cost. There exists a constant $c>0$ such that

  $$\begin{aligned} \frac{c\exp {\left( \left( \frac{1}{2}-\frac{\log 2}{\pi }\right) j_{\nu ,2}\right) }}{T^{1/2} \kappa _\alpha ^{3/2}j_{\nu , 1}}\exp \left( -\left( j_{\nu ,1}^2+\frac{j_{\nu ,2}^2}{2}\right) \kappa _\alpha ^2 T\right) \le \widetilde{{\mathcal {K}}}(T,\alpha ,\beta ,\mu ). \end{aligned}$$

Remark 2.3

In [17], the authors prove the null controllability for a parabolic equation with inverse-square potential (i.e., $\alpha =\beta =0$) using Dirichlet $L^2$-controls acting at the right endpoint and considering a homogenous Dirichlet condition at the left endpoint. They also show that the cost of controllability behaves as $\textrm{e}^{1/T}$. In [5], the authors improve the last result, they also prove that the cost of controllability behaves as $\textrm{e}^{1/T}$ for $0\le \alpha <1$, and the lower estimate is uniform with respect to $\mu \le \mu (\alpha )$. Thus, with respect to the lower estimate, our technique is worst than the one used in [5, 17].

Finally, we also analyze the null controllability of the system when the parameters satisfy $0\le \alpha <2$, $\beta =1-\alpha$, and $\mu <0$. Thus, consider the next system

$$\begin{aligned} \left\{ \begin{array}{ll} u_t-(x^\alpha u_x)_x-(1-\alpha ) x^{\alpha -1}u_x-\frac{\mu }{x^{2-\alpha }} u=0 &{} \quad \text{ in } Q, \\ \left( x^{\sqrt{-\mu }}u\right) (0, t) = f(t), u(1, t) = 0 &{} \quad \text{ on } (0, T), \\ u(x, 0) =u_{0}(x) &{} \quad \text{ in } (0, 1). \end{array}\right. \end{aligned}$$

(6)

The corresponding set of admissible controls is given by

$$\begin{aligned} {\widehat{U}}(T,\alpha ,\mu ,u_0)=\{f\in L^2(0,T): u \text { is solution of the system }(6)\text { that satisfies } u(\cdot ,T)\equiv 0\}, \end{aligned}$$

and the cost of the controllability is given by

$$\begin{aligned} \widehat{{\mathcal {K}}}(T,\alpha ,\mu ):=\sup _{\Vert u_0\Vert _{1-\alpha }\le 1}\inf \{\Vert f\Vert _{L^2(0,T)}:f\in {\widehat{U}}(T,\alpha ,\mu ,u_0)\}. \end{aligned}$$

Theorem 2.4

Let $T>0$ and $\alpha ,\mu \in {\mathbb {R}}$ with $0\le \alpha <2$, $\mu <0$. The next statements hold.

• 1. Existence of a control. For any $u_0\in L^2_{1-\alpha }(0,1)$, there exists a control $f \in L^2(0, T)$ such that the solution u to (6) satisfies $u(\cdot ,T ) \equiv 0$.

• 2. Upper bound of the cost. There exists a constant $c>0$ such that for every $\delta \in (0,1)$, we have

  $$\begin{aligned} \widehat{{\mathcal {K}}}(T,\alpha ,\mu )\le \frac{c M(T,\alpha ,\nu ,\delta )T^{1/2}}{\kappa _\alpha ^{1/2}\sqrt{-\mu }} \exp \left( -\frac{T}{2}\kappa _\alpha ^2 j_{\nu ,1}^2\right) , \end{aligned}$$

  where $\nu =\nu (\alpha ,\mu ):=\sqrt{-\mu }/\kappa _\alpha$.

• 3. Lower bound of the cost. There exists a constant $c>0$ such that

  $$\begin{aligned} \displaystyle \frac{c2^{\nu } \Gamma (\nu +1) \left| J_{\nu }^{\prime }\left( j_{\nu , 1}\right) \right| \exp {\left( \left( \frac{1}{2}-\frac{\log 2}{\pi }\right) j_{\nu ,2}\right) }}{({2T \kappa _\alpha })^{1/2}\sqrt{-\mu }\left( j_{\nu , 1}\right) ^{\nu }}&\exp&\left( -\left( j_{\nu ,1}^2+\frac{j_{\nu ,2}^2}{2}\right) \kappa _\alpha ^2 T\right) \\{} & {} \le \widehat{{\mathcal {K}}}(T,\alpha ,\mu ). \end{aligned}$$

Remark 2.5

Notice that the lower bound blows up as $\mu \rightarrow 0^-=\mu (1)$. If $\alpha =1, \mu =0$, the system (6) becomes the degenerate parabolic equation with a pure Dirichlet boundary condition.

This paper is organized as follows. In Sect. 3, 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 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 Sect. 4, we prove Theorem 2.1 using the moment method, as a consequence, we get an upper estimate of ${\mathcal {K}}(T,\alpha ,\beta ,\mu )$. Then we use the representation theorem in Theorem A.5 to obtain a lower estimate of ${\mathcal {K}}(T,\alpha ,\beta ,\mu )$. In Sect. 5, we proceed as before to prove Theorem 2.2. In Sect. 6, we sketch the proof of Theorem 2.4. Finally in Sect. 7, we give some conclusions and open problems.

3 Functional setting and well-posedness

In the context of degenerate/singular parabolic equations, in [5, 6, 14, 15, 18], the authors noted that a suitable state space is a weighted Lebesgue space ($L^2_\beta (0,1)$ in this work) for this kind of problem, together with weighted Sobolev spaces. The idea of using $L^2_\beta (0,1)$ is that the expression $x^\beta {\mathcal {A}}u(x)$, where ${\mathcal {A}}$ is the stationary operator given in (16), can be written in divergence form with a singular potential. This issue motivates us to use the inner product $\langle \cdot , \cdot \rangle _{\alpha , \beta }$ provided below.

Thus, for $\alpha ,\beta \in {\mathbb {R}}$, here we consider the weighted Sobolev space

$$\begin{aligned} H_{\alpha , \beta }^{1}(0,1)=\left\{ u \in L_{\beta }^{2}(0,1)\cap H^1_{loc}(0,1): x^{\alpha / 2} u_{x} \in L_{\beta }^{2}(0,1)\right\} \end{aligned}$$

endowed with the inner product

$$\begin{aligned} \langle u, v\rangle _{\alpha , \beta }:=\int _{0}^{1} u v\, x^{\beta } \!dx+\int _{0}^{1} x^{\alpha +\beta } u_{x} v_{x} dx, \end{aligned}$$

and its corresponding norm denoted by $\Vert \cdot \Vert _{\alpha ,\beta }$.

3.1 Trace and integration by parts formula

The next result implies that we can talk about the trace at $x=1$ of functions in $H_{\alpha , \beta }^{1}(0,1)$, so in our systems, we can impose the usual homogeneous Dirichlet condition at the right endpoint.

Proposition 3.1

Let $\alpha ,\beta \in {\mathbb {R}}$. Then $H_{\alpha , \beta }^{1} \subset W^{1,1}(\varepsilon ,1)$ for all $\varepsilon \in (0,1)$. In particular, $H_{\alpha ,\beta }^{1}(0,1)\subset C((0,1])$, and $|u|^2\in W^{1,1}(\varepsilon ,1)$ for all $u\in H_{\alpha , \beta }^{1}(0,1),\varepsilon \in (0,1)$.

Proof

Let $u\in H_{\alpha , \beta }^{1}(0,1)$. For $\varepsilon \in (0,1),\delta \in {\mathbb {R}}$ fixed, there exists a constant $c(\varepsilon ,\delta )>0$ such that $x^\delta \le c(\varepsilon ,\delta )$, $x\in (\varepsilon ,1]$, thus

$$\begin{aligned}{} & {} \int _{\varepsilon }^{1}|u| \textrm{d}x \le (1-\varepsilon )^{1/2}\left( \int _{\varepsilon }^{1} |u|^{2} \textrm{d}x\right) ^{1 / 2}\\{} & {} \quad \le (1-\varepsilon )^{1/2}c(\varepsilon ,-\beta )^{1/2}\left( \int _{0}^{1} |u|^{2} x^{\beta }\textrm{d}x\right) ^{1 / 2},\,\,\,\text {and} \\{} & {} \quad \int _{\varepsilon }^{1}\left| u_{x}\right| \textrm{d}x\le (1-\varepsilon )^{1/2}c(\varepsilon ,-\alpha -\beta )^{1/2}\left( \int _{0}^{1} |u_{x}|^{2} x^{\alpha +\beta }\textrm{d}x\right) ^{1 / 2}. \end{aligned}$$

Hence, we get the existence of the limit $u(1):=\lim _{x\rightarrow 1^-}u(x)$, and $u\in C([\varepsilon ,1])$. $\square$

Definition 3.2

For $\alpha ,\beta \in {\mathbb {R}}$, consider the space

$$\begin{aligned} H_{\alpha , \beta ,N}^{1}=H_{\alpha , \beta ,N}^{1}(0,1):=\left\{ u \in H_{\alpha , \beta }^{1}(0,1): u(1)=0\right\} . \end{aligned}$$

The classic Hardy inequality is typically proven for test functions defined on a finite interval. In our context, the following result shows that the Hardy inequality can be extended to functions in the weighted Sobolev space $H_{\alpha , \beta ,N}^{1}$. As an application, this generalized Hardy inequality implies the weighted Poincaré inequality (10) for functions in $H_{\alpha , \beta ,N}^{1}$.

Proposition 3.3

For $\alpha ,\beta \in {\mathbb {R}}$ with $\alpha +\beta >1$, the Hardy inequality

$$\begin{aligned} \mu (\alpha +\beta )\int _0^1\frac{|u|^2}{x^{2-(\alpha +\beta )}}\textrm{d}x\le \int _0^1x^{\alpha +\beta } |u_x|^2\textrm{d}x \end{aligned}$$

(7)

holds for any $u\in H_{\alpha , \beta ,N}^{1}$. In particular, $H_{\alpha , \beta ,N}^{1}\hookrightarrow L^2_{\alpha +\beta -2}(0,1)$.

Proof

Let $u\in H_{\alpha , \beta ,N}^{1}$ and $\varepsilon \in (0,1)$. Set $\delta =\alpha +\beta$. Since $|u|^2\in W^{1,1}(\varepsilon ,1)$ we have

$$\begin{aligned} \begin{array}{l} \displaystyle \int _\varepsilon ^1\left( x^{\delta /2}u_x-\frac{1-\delta }{2}\frac{u}{x^{(2-\delta )/2}}\right) ^2\textrm{d}x \\ \\ = \displaystyle \int _\varepsilon ^1x^{\delta }|u_x|^2\textrm{d}x+\mu (\delta )\int _\varepsilon ^1\frac{u^2}{x^{2-\delta }}\textrm{d}x-\frac{1-\delta }{2}\int _\varepsilon ^1\frac{(u^2)_x}{x^{1-\delta }}\textrm{d}x\\ \\ = \displaystyle \int _\varepsilon ^1x^{\delta }|u_x|^2\textrm{d}x-\mu (\delta )\int _\varepsilon ^1\frac{u^2}{x^{2-\delta }}\textrm{d}x-\frac{1-\delta }{2}\left( \lim _{x\rightarrow 1^{-}}\frac{|u(x)|^2}{x^{1-\delta }}-\frac{|u(\varepsilon )|^2}{\varepsilon ^{1-\delta }}\right) \\ \\ = \displaystyle \int _\varepsilon ^1x^{\delta }|u_x|^2\textrm{d}x-\mu (\delta )\int _\varepsilon ^1\frac{u^2}{x^{2-\delta }}\textrm{d}x+\frac{1-\delta }{2}\frac{|u(\varepsilon )|^2}{\varepsilon ^{1-\delta }}, \end{array} \end{aligned}$$

since $\delta > 1$ we get

$$\begin{aligned} \mu (\delta )\int _\varepsilon ^1\frac{u^2}{x^{2-\delta }}\textrm{d}x\le \int _\varepsilon ^1x^\delta u_x^2\textrm{d}x \end{aligned}$$

for all $\varepsilon \in (0,1)$. The result follows by the dominated convergence theorem. $\square$

The next result will allow us analyze the behavior at $x=0$ of functions in $H_{\alpha , \beta , N}^{1}$, see (9).

Proposition 3.4

Let $\alpha ,\beta \in {\mathbb {R}}$ with $\alpha +\beta >1$. Then $x^{\delta }u\in W^{1,1}(0,1)$ for all $u\in H_{\alpha , \beta , N}^{1}$ provided that $\delta > (\alpha +\beta -1)/2$.

Proof

Let $u\in H_{\alpha , \beta , N}^{1}$ and assume $2\delta > \alpha +\beta -1$.

Since $(x^{\delta }u)_{x} = x^{\delta }u_{x}+\delta x^{\delta -1}u$, we compute

$$\begin{aligned} \int _{0}^{1}x^{\delta }|u_{x}| \textrm{d}x \le \frac{1}{(2\delta -(\alpha +\beta ) +1)^{1/2}}\left( \int _{0}^{1} x^{\alpha +\beta }|u_{x}|^{2} \textrm{d}x\right) ^{1 / 2}<\infty , \end{aligned}$$

(8)

and Proposition 3.3 implies

$$\begin{aligned} \int _{0}^{1}x^{\delta -1}|u| \textrm{d}x\le & {} \frac{1}{(2\delta -(\alpha +\beta ) +1)^{1/2}}\left( \int _{0}^{1}\frac{|u|^{2}}{x^{2-(\alpha +\beta )}}\textrm{d}x\right) ^{1 / 2}\\\le & {} \frac{1}{(2\delta -(\alpha +\beta ) +1)^{1/2}}\frac{1}{\mu (\alpha +\beta )}\left( \int _{0}^{1}x^{\alpha +\beta }u_{x}^{2}\textrm{d}x\right) ^{1 / 2}<\infty . \end{aligned}$$

Hence, $(x^{\delta }u)_{x}\in L^1(0,1)$. Notice that $x^\delta \le x^{\delta -1}$ on (0, 1), thus $x^{\delta }u\in L^1(0,1)$ and the result follows. $\square$

Remark 3.5

For $\alpha ,\beta \in {\mathbb {R}}$ with $\alpha +\beta >1$, the last result implies the existence of $L_{\delta }:=\lim _{x\rightarrow 0^{+}}x^{\delta }u(x)$ provided that $\delta > (\alpha + \beta - 1)/2$, in fact, $L_{\delta } = 0$. Now choose any $\delta > (\alpha + \beta - 1)/2$ so

$$\begin{aligned} x^{\delta }|u(x)|\le & {} \int _0^x \left| \frac{\textrm{d}}{\textrm{d}s} (s^\delta u(s))\right| \textrm{d}s\\\le & {} \frac{x^{\delta -(\alpha +\beta -1)/2}}{(2\delta -(\alpha +\beta )+1)^{1/2}}\left[ \left( \int _{0}^{x}s^{\alpha +\beta }|u_x|^2\textrm{d}s\right) ^{1/2} \right. \\ {}{} & {} \left. +\delta \left( \int _{0}^{x}\frac{|u|^2}{s^{2-(\alpha +\beta )}}\textrm{d}s\right) ^{1/2}\right] , \end{aligned}$$

therefore

$$\begin{aligned} \lim _{x\rightarrow 0^+}x^{(\alpha +\beta -1)/2}|u(x)|=0, \quad u\in H_{\alpha , \beta ,N}^{1}. \end{aligned}$$

(9)

The last equality will be useful to prove the integration by parts formula given in (15).

From now on, we assume $\alpha < 2$ and $\alpha + \beta >1$. For any $u\in H^1_{\alpha ,\beta ,N}$, we obtain the weighted Poincaré inequality from Proposition 3.3:

$$\begin{aligned} \int _{0}^{1}x^{\beta }|u|^{2}\textrm{d}x \le \int _{0}^{1}\dfrac{|u|^2}{x^{2-(\alpha +\beta )}} \textrm{d}x\le \frac{1}{\mu (\alpha +\beta )}\int _{0}^{1} x^{\alpha +\beta }\left| u_{x}\right| ^{2} \textrm{d}x, \end{aligned}$$

(10)

therefore

$$\begin{aligned} \Vert u\Vert _{\alpha ,\beta ,N}:=\left( \int _{0}^{1} x^{\alpha +\beta }\left| u_{x}\right| ^{2} \textrm{d}x\right) ^{1 / 2} \end{aligned}$$

is an equivalent norm to $\Vert u\Vert _{\alpha , \beta }$ in $H_{\alpha , \beta , N}^{1}$.

For $\mu <\mu (\alpha +\beta )$, Proposition 3.3 also implies that

$$\begin{aligned} \Vert u\Vert _{*}=\left( \int _{0}^{1} x^{\alpha +\beta }\left[ \left| u_{x}\right| ^{2}-\frac{\mu }{x^{2}} u^{2}\right] \textrm{d}x\right) ^{1 / 2} \end{aligned}$$

is an equivalent norm to $\Vert u\Vert _{\alpha ,\beta ,N}$ in $H_{\alpha , \beta , N}^{1}$. In fact, we have

$$\begin{aligned} \Vert u\Vert _{\alpha ,\beta ,N}\le & {} \Vert u\Vert _{*} \le \left( 1-\frac{\mu }{\mu (\alpha +\beta )}\right) ^{1/2}\Vert u \Vert _{\alpha ,\beta ,N},\quad \mu<0, \\ \left( 1-\frac{\mu }{\mu (\alpha +\beta )}\right) ^{1/2}\Vert u\Vert _{\alpha ,\beta ,N}\le & {} \Vert u\Vert _{*} \le \left\| u \right\| _{\alpha ,\beta ,N},\quad 0\le \mu <\mu (\alpha +\beta ). \end{aligned}$$

Since $C_c^\infty (0,1)\subset H_{\alpha , \beta , N}^{1}\subset L^2_\beta (0,1),$ and (10) implies that the inclusion $(H_{\alpha , \beta ,N}^1,\Vert \cdot \Vert _*)\hookrightarrow L^2_\beta (0,1)$ is continuous, the following definition makes sense.

Definition 3.6

For $\alpha ,\beta \in {\mathbb {R}}$ with $\alpha <2$ and $\alpha +\beta >1$, consider the Gelfand triple $\left( (H_{\alpha , \beta ,N}^1,\Vert \cdot \Vert _*), L^2_\beta (0,1), H_{\alpha , \beta ,N}^{-1}\right)$, i.e $H_{\alpha , \beta ,N}^{-1}$ stands for the dual space of $(H_{\alpha , \beta ,N}^1,\Vert \cdot \Vert _*)$ with respect to the pivot space $L^2_\beta (0,1)$:

$$\begin{aligned} \Big (H_{\alpha , \beta ,N}^1,\Vert \cdot \Vert _*\Big )\hookrightarrow L^2_\beta (0,1)=\left( L^2_\beta (0,1)\right) '\hookrightarrow H_{\alpha , \beta ,N}^{-1}:=\Big (H_{\alpha , \beta ,N}^1,\Vert \cdot \Vert _*\Big )'. \end{aligned}$$

The inner product $\langle \cdot ,\cdot \rangle _*$ induces an isomorphism ${\mathcal {A}}:H_{\alpha , \beta ,N}^1\rightarrow H_{\alpha , \beta ,N}^{-1}$ given by

$$\begin{aligned} \langle u,v \rangle _*=\langle {\mathcal {A}} u,v\rangle _{H_{\alpha , \beta ,N}^{-1},H_{\alpha , \beta ,N}^{1}},\quad u,v\in H_{\alpha , \beta ,N}^{1}. \end{aligned}$$

Let $D({\mathcal {A}}):={\mathcal {A}}^{-1}(L^2_\beta (0,1))=\{u\in H_{\alpha , \beta ,N}^{1}: {\mathcal {A}} u\in L^2_\beta (0,1)\}=\{u\in H_{\alpha , \beta ,N}^{1}: \exists f\in L^2_\beta (0,1) \text { such that }\langle u,v \rangle _*=\langle f,v\rangle _\beta , \text { for all } v\in H_{\alpha , \beta ,N}^{1} \}$.

The next result gives a handy characterization of $D({\mathcal {A}})$. It shows the behavior of the derivative of functions in $D({\mathcal {A}})$ at the endpoints, see (11) and (12), and also provides an integration by parts formula, see (15).

Proposition 3.7

For $\alpha ,\beta ,\mu \in {\mathbb {R}}$ with $0\le \alpha <2$, $\alpha +\beta >1$ and $\mu <\mu (\alpha +\beta )$, we have

$$\begin{aligned} D({\mathcal {A}})=\left\{ u\in H^1_{\alpha ,\beta ,N}\cap H^2_{loc}(0,1): (x^\alpha u_x)_x+\beta x^{\alpha -1}u_x+\frac{\mu }{x^{2-\alpha }} u\in L^2_\beta (0,1)\right\} . \end{aligned}$$

Proof

Let H be the set on the right-hand side, we will show that $D({\mathcal {A}})=H$.

Pick $u \in D({\mathcal {A}})$, then there exists $f \in L^{2}_\beta (0,1)$ such that

$$\begin{aligned} \int _{0}^{1} \left( x^{\alpha +\beta } u_{x} v_{x}-\frac{\mu }{x^{2-\alpha -\beta }} u v\right) \textrm{d}x=\int _0^1 f v x^{\beta } \textrm{d}x \quad \text {for all }v \in H_{\alpha , \beta , N}^{1}. \end{aligned}$$

In particular,

$$\begin{aligned} \int _{0}^{1} x^{\alpha +\beta } u_{x} v_{x}\textrm{d}x=\int _{0}^{1}\left( f+\frac{\mu }{x^{2-\alpha }} u\right) v x^{\beta } \textrm{d}x\quad \text{ for } \text{ all } v \in C_c^\infty (0,1), \end{aligned}$$

hence

$$\begin{aligned} -\left( x^{\alpha +\beta } u_{x}\right) _{x}=\left( f+\frac{\mu }{x^{2-\alpha }}u\right) x^{\beta } \text{ in } C_c^\infty (0,1)^{\prime }, \end{aligned}$$

which implies

$$\begin{aligned} \left( x^{\alpha } u_{x}\right) _{x}+\beta x^{\alpha -1} u_{x}+\frac{\mu }{x^{2-\alpha }} u=-f \text{ in } C_c^\infty (0,1)^{\prime }, \end{aligned}$$

therefore $u\in H$.

Now let $u \in H$. We claim that $x^{\delta } u_{x} \in W^{1,1}(0,1)$ for all $\delta > (\alpha +\beta +1)/2$. Just apply (8) with $\delta -1$ instead of $\delta$ to get that $x^{\delta -1}u_x\in L^1(0,1)$, in particular $x^{\delta }u_x\in L^1(0,1)$. On the other hand, we have

$$\begin{aligned}{} & {} \int _{0}^{1}x^{\delta }|u_{xx}|\textrm{d}x\\{} & {} \quad = \int _{0}^{1}x^{\delta -\alpha }\Bigg |x^{\alpha }u_{xx}+\alpha x^{\alpha -1}u_{x}+\beta x^{\alpha -1}u_{x}+\frac{\mu }{x^{2-\alpha }}u\\{} & {} \qquad -(\alpha +\beta )x^{\alpha -1}u_{x}-\frac{\mu }{x^{2-\alpha }}u\Bigg |\textrm{d}x\\{} & {} \quad \le \int _{0}^{1}x^{\delta -(\alpha +\beta /2)}|(x^{\alpha }u_{x})_{x}+\beta x^{\alpha -1}u_{x}+\dfrac{\mu }{x^{2-\alpha }}u|x^{\beta /2}\textrm{d}x\\{} & {} \qquad + (\alpha +\beta )\int _{0}^{1}x^{\delta -1}|u_{x}|\textrm{d}x +\quad |\mu |\int _{0}^{1}x^{\delta -(\alpha +\beta )/2-1}\dfrac{|u|}{x^{(2-\alpha -\beta )/2}}\textrm{d}x\\{} & {} \quad \le \dfrac{1}{(2\delta -(2\alpha +\beta )+1)^{1/2}}\left( \int _{0}^{1}|(x^{\alpha }u_{x})_{x}+\beta x^{\alpha -1}u_{x}+\dfrac{\mu }{x^{2-\alpha }}u|^2 x^{\beta }\textrm{d}x\right) ^{1/2}\\{} & {} \qquad + \frac{\alpha + \beta }{(2\delta -(\alpha +\beta +1))^{1/2}}\left( \int _{0}^{1}x^{\alpha +\beta } |u_{x}|^{2} \textrm{d}x\right) ^{1 / 2}\\{} & {} \qquad +\frac{|\mu |}{(2\delta -(\alpha +\beta +1))^{1/2}}\left( \int _{0}^{1}\frac{ |u|^{2}}{x^{2-(\alpha +\beta )}} \textrm{d}x\right) ^{1 / 2}. \end{aligned}$$

Notice the last quantity is finite by Proposition 3.3.

Thus, we get the existence of the limit

$$\begin{aligned} u_{x}(1):=\lim _{x \rightarrow 1^{-}}x^{\delta }u_{x}(x), \end{aligned}$$

(11)

and we also have that $\lim _{x\rightarrow 0^+}x^{\delta }u_{x}(x)=0$ provided that $\delta > (\alpha +\beta +1)/2$, see Remark 3.5. As in the proof of (9), we can see that

$$\begin{aligned} \lim _{x\rightarrow 0^+}x^{(\alpha +\beta +1)/2}u_{x}(x)=0. \end{aligned}$$

(12)

Now consider any $v\in H^{1}_{\alpha ,\beta ,N}$. We claim that $x^{\alpha +\beta } u_{x} v \in W^{1,1}(0,1)$:

$$\begin{aligned} \int _0^1x^{\alpha +\beta }|u_x v|\textrm{d}x\le & {} \left( \int _0^1x^{\alpha +\beta }|u_x|^2\textrm{d}x\right) ^{1/2}\left( \int _0^1x^{\alpha +\beta }|v|^2\textrm{d}x\right) ^{1/2}\\ {}\le & {} \Vert u\Vert _{\alpha ,\beta ,N}\Vert v\Vert _\beta <\infty , \end{aligned}$$

and

$$\begin{aligned} \left( x^{\alpha +\beta } u_{x} v\right) _{x}= & {} x^{\alpha +\beta } u_{x} v_{x}+x^{\beta }\left( \left( x^{\alpha } u_{x}\right) _x+\beta x^{\alpha -1} u_{x}+\frac{\mu }{x^{2-\alpha }} u\right) v \nonumber \\{} & {} -\frac{\mu \,uv}{x^{2-\alpha -\beta }} \in L^{1}(0,1). \end{aligned}$$

(13)

On the other hand, (9), (11), and (12) imply that

$$\begin{aligned} \lim _{x\rightarrow 0^+}x^{\alpha +\beta }u_{x}(x)v(x)= 0\quad \text {and}\quad \lim _{x\rightarrow 1^-}x^{\alpha +\beta }u_{x}(x)v(x) = 0. \end{aligned}$$

(14)

Thus, from (13), we get

$$\begin{aligned} \int _{0}^{1} \left( x^{\alpha +\beta } u_{x} v_{x}-\frac{\mu }{x^{2-\alpha -\beta }} u v\right) \textrm{d}x= & {} -\int _{0}^{1} x^{\beta }\Bigg (\left( x^{\alpha } u_{x}\right) _{x}\nonumber \\ {}{} & {} +\beta x^{\alpha -1} u_{x}+\frac{\mu }{{x^{2-\alpha }}} u\Bigg ) v \textrm{d}x \end{aligned}$$

(15)

for all $u \in H, v \in H_{\alpha , \beta , N}^{1}$. Therefore, $u \in D({\mathcal {A}})$. $\square$

3.2 Semigroup theory

For $\alpha ,\beta ,\mu \in {\mathbb {R}}$ with $0\le \alpha <2$, $\alpha +\beta >1$, $\mu <\mu (\alpha +\beta )$, we consider the unbounded operator ${\mathcal {A}}:D({\mathcal {A}})\subset L^2_\beta (0,1)\rightarrow L^2_\beta (0,1)$ given by

$$\begin{aligned} {\mathcal {A}}u:=-(x^\alpha u_x)_x-\beta x^{\alpha -1}u_x-\frac{\mu }{x^{2-\alpha }} u. \end{aligned}$$

(16)

From Proposition 9 in [19, p. 370], we have that ${\mathcal {A}}$ is a closed operator with $D({\mathcal {A}})$ dense in $L^2_\beta (0,1)$. We also have that ${\mathcal {A}}:(D({\mathcal {A}}),\Vert \cdot \Vert _{D({\mathcal {A}})})\rightarrow L^2_\beta (0,1)$ is an isomorphism, where

$$\begin{aligned} \Vert u\Vert _{D({\mathcal {A}})}=\Vert u\Vert _\beta +\Vert {\mathcal {A}} u\Vert _\beta ,\quad u\in D({\mathcal {A}}). \end{aligned}$$

The next result shows that ${\mathcal {A}}$ is a diagonalizable self-adjoint operator whose Hilbert basis of eigenfunctions can be written in terms of a Bessel function of the first kind $J_{\nu }$ and its corresponding zeros $j_{\nu ,k}$, $k\ge 1$, located in the positive half line. In the appendix, we give some properties of Bessel functions and their zeros.

Proposition 3.8

$-{\mathcal {A}}$ is a negative self-adjoint operator. Furthermore, the family

$$\begin{aligned} \Phi _k(x):=\frac{(2\kappa _\alpha )^{1/2}}{|J'_\nu (j_{\nu ,k})|}x^{(1-\alpha -\beta )/2}J_\nu (j_{\nu ,k}x^{\kappa _\alpha }),\quad k\ge 1, \end{aligned}$$

(17)

is an orthonormal basis for $L^2_\beta (0,1)$ such that

$$\begin{aligned} {\mathcal {A}}\Phi _k=\lambda _k \Phi _k, \quad \lambda _k=\kappa _\alpha ^2 (j_{\nu ,k})^2,\quad k\ge 1, \end{aligned}$$

(18)

where $\nu$ is defined in (4).

Proof

From (15), we get that ${\mathcal {A}}$ is a symmetric operator. Letting $u=v\in D({\mathcal {A}})$ in (15) and using Proposition 3.3, we obtain that $-{\mathcal {A}}\le 0$.

We claim that $\text {Ran}(I+{\mathcal {A}})=L^2_\beta (0,1)$: Let $f \in L_{\beta }^{2}(0,1)$ be given. Since the inner product $\langle \cdot ,\cdot \rangle _\beta +\langle \cdot ,\cdot \rangle _*$ is equivalent to $\langle \cdot ,\cdot \rangle _{\alpha ,\beta }$ in $H_{\alpha , \beta , N}^{1}$ and $f\in (H_{\alpha , \beta , N}^{1},\Vert \cdot \Vert _{\alpha ,\beta })'$, the Riesz representation theorem implies that there exists a unique $u\in H_{\alpha , \beta , N}^{1}$ such that

$$\begin{aligned} \int _{0}^{1} u v x^{\beta } \textrm{d}x+\int _{0}^{1} x^{\alpha +\beta }\left( u_{x} v_{x}-\frac{\mu }{x^{2}} u v\right) \textrm{d}x=\int _{0}^{1} f v x^{\beta } \textrm{d}x \end{aligned}$$

for all $v\in H_{\alpha , \beta , N}^{1}$. Therefore,

$$\begin{aligned} u-\left( x^{\alpha } u_{x}\right) _{x}-\beta x^{\alpha -1} u_{x}-\frac{\mu }{x^{2-\alpha }} u=f \text{ in } C_c^\infty (0,1)^{\prime }, \end{aligned}$$

thus $u\in D({\mathcal {A}})$ and $u+{\mathcal {A}} u=f$.

It follows that $-{\mathcal {A}}$ is m-dissipative in $L^2_\beta (0,1)$ and Corollary 2.4.10 in [20, p. 24] implies that $-{\mathcal {A}}$ is self-adjoint.

In [21], it was proved that the family

$$\begin{aligned} \Theta _k(x):=\frac{2^{1/2}}{|J'_\nu (j_{\nu ,k})|}x^{1/2}J_\nu (j_{\nu ,k}x),\quad k\ge 1, \end{aligned}$$

is an orthonormal basis for $L^2(0,1)$.

Let ${\mathcal {U}}$ be the unitary operator ${\mathcal {U}}:L^2(0,1)\rightarrow L^2_\beta (0,1)$ given by

$$\begin{aligned} {\mathcal {U}}u(x):=\kappa _\alpha ^{1/2}x^{-\alpha /4-\beta /2}u(x^{\kappa _\alpha }), \quad u\in L^2(0,1). \end{aligned}$$

Notice that ${\mathcal {U}}\Theta _k=\Phi _k$, $k\ge 1$, therefore $\Phi _k$, $k\ge 1$, is an orthonormal basis for $L^2_\beta (0,1)$. We also can see that $\Phi _k\in H^1_{\alpha ,\beta ,N}$ using that $\nu >0,$ (A1) and (A2).

Now we set $w(x)=y(z)$ with $z=cx^a$, $a,c>0$. Assume that $y=J_\nu$. Therefore, y satisfies the differential Eq. (A3), i.e.,

$$\begin{aligned} z\frac{d}{dz}\left( z\frac{dy}{dz}\right) +(z^2-\nu ^2)y=0, \end{aligned}$$

which implies that

$$\begin{aligned} x\frac{d}{dx}\left( x\frac{dw}{dx}\right) +a^2(c^2x^{2a}-\nu ^2)w=0. \end{aligned}$$

Then we set $v(x)=x^bw(x)$, $b\in {\mathbb {R}}$. Hence,

$$\begin{aligned} x^{2-2a}\frac{d^2v}{dx^2}+(1-2b)x^{1-2a}\frac{dv}{dx}+(b^2-a^2\nu ^2 )x^{-2a}v=-a^2c^2v. \end{aligned}$$

(19)

Finally, we take $a=\kappa _\alpha , b=(1-\alpha -\beta )/2$, and $c=j_{\nu ,k}$, $k\ge 1$, to get $\Phi _k(1)=0$ and ${\mathcal {A}}\Phi _k=\lambda _k\Phi _k$ for all $k \ge 1$. $\square$

Then $({\mathcal {A}},D({\mathcal {A}}))$ is the infinitesimal generator of a diagonalizable self-adjoint analytic semigroup of contractions in $L^2_{\beta }(0,1)$. Thus, we consider interpolation spaces for the initial data. For any $s\ge 0$, we define

$$\begin{aligned} {\mathcal {H}}^{s}={\mathcal {H}}^{s}(0,1):=D({\mathcal {A}}^{s/2})=\left\{ u=\sum _{k=1}^\infty a_{k} \Phi _{k}:\Vert u\Vert _{{\mathcal {H}}^{s}}^{2}=\sum _{k=1}^\infty |a_{k}|^{2} \lambda _{k}^{s}<\infty \right\} , \end{aligned}$$

and we also consider the corresponding dual spaces

$$\begin{aligned} {\mathcal {H}}^{-s}:=\left[ {\mathcal {H}}^{s}(0,1)\right] ^{\prime }. \end{aligned}$$

It is well known that ${\mathcal {H}}^{-s}$ is the dual space of ${\mathcal {H}}^{s}$ with respect to the pivot space $L^2_\beta (0,1)$, i.e

$$\begin{aligned} {\mathcal {H}}^s\hookrightarrow {\mathcal {H}}^0=L^2_{\beta }(0,1)=\left( L^2_{\beta }(0,1)\right) '\hookrightarrow {\mathcal {H}}^{-s},\quad s>0. \end{aligned}$$

Equivalently, ${\mathcal {H}}^{-s}$ is the completion of $L^2_\beta (0,1)$ with respect to the norm

$$\begin{aligned} \Vert u\Vert ^2_{-s}:=\sum _{k=1}^{\infty }\lambda _k^{-s}|\langle u,\Phi _k\rangle _\beta |^2. \end{aligned}$$

It is well known that the linear mapping given by

$$\begin{aligned} S(t)u_0=\sum _{k=1}^\infty \text {e}^{-\lambda _k t}a_k\Phi _k\quad \text {if}\quad u_0=\sum _{k=1}^\infty a_{k} \Phi _{k}\in {\mathcal {H}}^s \end{aligned}$$

defines a self-adjoint semigroup S(t), $t\ge 0$, in ${\mathcal {H}}^s$ for all $s\in {\mathbb {R}}$.

3.3 Existence and uniqueness of solutions to system (1)

For $\delta \in {\mathbb {R}}$ and a function $z:(0,1)\rightarrow {\mathbb {R}}$, we introduce the notion of $\delta$-generalized limit of z at $x=0$ as follows

$$\begin{aligned} {\mathcal {O}}_\delta (z):=\lim _{x\rightarrow 0^+} x^\delta z(x). \end{aligned}$$

Now we consider a convenient definition of a weak solution for system (1), we multiply the equation in (1) by $x^\beta \varphi (\tau )=x^\beta S(\tau -t)z^{\tau }$, integrate by parts (formally), and take the expression obtained.

Definition 3.9

Let $T>0$ and $\alpha ,\beta ,\mu \in {\mathbb {R}}$ with $0\le \alpha <2$, $\alpha + \beta >1$, $\mu <\mu (\alpha +\beta )$. Let $f \in L^2(0,T)$ and $u_0\in {\mathcal {H}}^{-s}$ for some $s > 0$. A weak solution of (1) is a function $u \in C^0([0,T];{\mathcal {H}}^{-s})$ such that for every $\tau \in (0,T]$ and for every $z^\tau \in {\mathcal {H}}^s$ we have

$$\begin{aligned} \left\langle u(\tau ), z^{\tau }\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}= & {} -\int _{0}^{\tau } f(t) {\mathcal {O}}_{\alpha +\beta +\gamma }\left( S(\tau -t) z^{\tau }\right) \textrm{d} t\nonumber \\ {}{} & {} +\left\langle u_{0}, S(\tau )z^\tau \right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}, \end{aligned}$$

(20)

where $\gamma =\gamma (\alpha ,\beta ,\mu )$ is given in (3).

The next result shows the existence of weak solutions for the system (1) under suitable conditions on the parameters $\alpha ,\beta ,\mu ,\gamma$ and s. The proof is similar to the proof of Proposition 2.9 in [6].

Proposition 3.10

Let $T>0$ and $\alpha ,\beta \in {\mathbb {R}}$ with $0\le \alpha <2$, $\alpha + \beta >1$. Let $f \in L^2(0,T)$ and $u_0\in {\mathcal {H}}^{-s}$ such that $s>\nu$, where $\nu$ is given in (4). Then formula (20) defines for each $\tau \in [0, T ]$ a unique element $u(\tau ) \in {\mathcal {H}}^{-s}$ that can be written as

$$\begin{aligned} u(\tau )=S(\tau ) u_{0}-B(\tau ) f, \quad \tau \in (0, T], \end{aligned}$$

where $B(\tau )$ is the strongly continuous family of bounded operators $B(\tau ): L^{2}(0,T) \rightarrow {\mathcal {H}}^{-s}$ given by

$$\begin{aligned} \left\langle B(\tau ) f, z^{\tau }\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}=\int _{0}^{\tau } f(t) {\mathcal {O}}_{\alpha +\beta +\gamma }\left( S(\tau -t) z^{\tau }\right) \textrm{d} t, \quad \text {for all }z^{\tau } \in {\mathcal {H}}^{s}. \end{aligned}$$

Furthermore, the unique weak solution u on [0, T] to (1) (in the sense of (20)) belongs to ${C}^{0}\left( [0, T]; {\mathcal {H}}^{-s}\right)$ and fulfills

$$\begin{aligned} \Vert u\Vert _{L^{\infty }\left( [0, T]; {\mathcal {H}}^{-s}\right) } \le C\left( \left\| u_{0}\right\| _{{\mathcal {H}}^{-s}}+\Vert f\Vert _{L^{2}(0, T)}\right) . \end{aligned}$$

Proof

Fix $\tau >0$. Let $u(\tau )\in {\mathcal {H}}^{-s}$ be determined by the condition (20), hence

$$\begin{aligned} -u(\tau )+S(\tau ) u_{0}=\zeta (\tau )f, \end{aligned}$$

where

$$\begin{aligned} \left\langle \zeta (\tau )f, z^{\tau }\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}=\int _{0}^{\tau } f(t) {\mathcal {O}}_{\alpha +\beta +\gamma }\left( S(\tau -t) z^{\tau }\right) \textrm{d} t, \quad \text {for all } z^{\tau } \in {\mathcal {H}}^{s}. \end{aligned}$$

We claim that $\zeta (\tau )$ is a bounded operator from $L^{2}(0, T)$ into ${\mathcal {H}}^{-s}$: consider $z^{\tau } \in {\mathcal {H}}^{s}$ given by

$$\begin{aligned} z^{\tau }=\sum _{k=1}^\infty a_{k} \Phi _{k}, \end{aligned}$$

(21)

therefore

$$\begin{aligned} S(\tau -t) z^{\tau }=\sum _{k=1}^\infty \textrm{e}^{\lambda _{k}(t-\tau )} a_{k} \Phi _{k}, \quad \text {for all } t \in [0, \tau ]. \end{aligned}$$

Using Lemma A.3 and (A10), we obtain that there exists a constant $C=C(\alpha ,\beta ,\mu )>0$ such that

$$\begin{aligned} |{\mathcal {O}}_{\alpha +\beta +\gamma }\left( \Phi _k\right) |\le C |j_{\nu ,k}|^{\nu +1/2},\quad k\ge 1; \end{aligned}$$

hence, (A9) implies that there exists a constant $C=C(\alpha ,\beta ,\mu )>0$ such that

$$\begin{aligned} \left( \int _{0}^{\tau }\left| {\mathcal {O}}_{\alpha +\beta +\gamma }\left( S(\tau -t) z^{\tau }\right) \right| ^{2} \mathrm {~d} t\right) ^{1/2}\le & {} \sum _{k=1}^\infty |a_k| |{\mathcal {O}}_{\alpha +\beta +\gamma }(\Phi _k)| \left( \int _0^\tau \textrm{e}^{2\lambda _{k}(t-\tau )} \mathrm {~d} t\right) ^{1/2} \\\le & {} C\left\| z^{\tau }\right\| _{{\mathcal {H}}^{s}}\left( \sum _{k=1}^\infty |\lambda _k|^{\nu -1/2-s}(1-\textrm{e}^{-2\lambda _k \tau })\right) ^{1/2}\\\le & {} C\left\| z^{\tau }\right\| _{{\mathcal {H}}^{s}}\left( \sum _{k=1}^\infty \frac{1}{k^{2(s-\nu +1/2)}}\right) ^{1/2}=C\left\| z^{\tau }\right\| _{{\mathcal {H}}^{s}}. \end{aligned}$$

Therefore, $\Vert \zeta (\tau ) f\Vert _{{\mathcal {H}}^{-s}}\le C\Vert f\Vert _{L^2(0,T)}$ for all $f\in L^2(0,T)$, $\tau \in (0,T]$.

Finally, we fix $f\in L^2(0,T)$ and show that the mapping $\tau \mapsto \zeta (\tau ) f$ is right-continuous on [0, T). Let $h>0$ be small enough and $z\in {\mathcal {H}}^s$ given as in (21). Thus, proceeding as in the last inequalities, we have

$$\begin{aligned}{} & {} \displaystyle |\left\langle \zeta (\tau +h)f-\zeta (\tau )f, z\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}|\\{} & {} \quad =\left| \int _{0}^{\tau +h}f(t){\mathcal {O}}_{\alpha +\beta +\gamma }\left( S(\tau +h-t) z\right) \textrm{d}t -\int _{0}^{\tau }f(t){\mathcal {O}}_{\alpha +\beta +\gamma }\left( S(\tau -t) z\right) \textrm{d}t\right| \\ \\{} & {} \quad \displaystyle \le \int _{0}^{\tau }|f(t)||{\mathcal {O}}_{\alpha +\beta +\gamma }\left( \left( S(\tau +h-t)-S(\tau -t)\right) z\right) |\textrm{d}t \\{} & {} \qquad + \int _{\tau }^{\tau +h}|f(t)||{\mathcal {O}}_{\alpha +\beta +\gamma }\left( S(\tau +h-t) z\right) |\textrm{d}t\\ \\{} & {} \quad \displaystyle \le \left\| f\right\| _{L^2(0,T)}\left[ \sum _{k=1}^{\infty }|a_{k}||{\mathcal {O}}_{\alpha +\beta +\gamma }(\Phi _{k})|\left( \int _{0}^{\tau }\left( \textrm{e}^{\lambda _k(t-\tau -h)}-\textrm{e}^{\lambda _k(t-\tau )}\right) ^2\textrm{d}t\right) ^{1/2}\right. \\{} & {} \qquad \displaystyle \left. +\sum _{k=1}^{\infty }|a_{k}||{\mathcal {O}}_{\alpha +\beta +\gamma }(\Phi _{k})|\left( \int _{\tau }^{\tau +h}\textrm{e}^{2\lambda _k(t-\tau -h)}\textrm{d}t\right) ^{1/2}\right] \\{} & {} \quad \displaystyle \le C\left\| f\right\| _{L^2(0,T)}\left[ \sum _{k=1}^{\infty }|a_{k}||j_{\nu ,k}|^{v+1/2}\left( \int _{0}^{\tau }\left( \textrm{e}^{\lambda _k(t-\tau -h)}-\textrm{e}^{\lambda _k(t-\tau )}\right) ^2\textrm{d}t\right) ^{1/2}\right. \\{} & {} \qquad \displaystyle \left. +\sum _{k=1}^{\infty }|a_{k}||j_{\nu ,k}|^{v+1/2}\left( \int _{\tau }^{\tau +h}\textrm{e}^{2\lambda _k(t-\tau -h)}\textrm{d}t\right) ^{1/2}\right] \\{} & {} \quad \displaystyle \le C\left\| f\right\| _{L^2(0,T)}\left[ \sum _{k=1}^{\infty }|a_{k}|\lambda _{k}^{s/2}|\lambda _{k}|^{v/2+1/4-s/2}\left( \int _{0}^{\tau }\left( \textrm{e}^{\lambda _k(t-\tau -h)}-\textrm{e}^{\lambda _k(t-\tau )}\right) ^2\textrm{d}t\right) ^{1/2}\right. \\{} & {} \qquad \displaystyle \left. +\sum _{k=1}^{\infty }|a_{k}|\lambda _{k}^{s/2}|\lambda _{k}|^{v/2+1/4-s/2}\left( \int _{\tau }^{\tau +h}\textrm{e}^{2\lambda _k(t-\tau -h)}\textrm{d}t\right) ^{1/2}\right] \\{} & {} \quad \displaystyle \le C\left\| z\right\| _{{\mathcal {H}}^{s}}\Vert f\Vert _{L^2(0,T)}\left[ \left( \sum _{k=1}^{\infty }\lambda _{k}^{\nu -1/2-s}I(\tau ,k,h)\right) ^{1/2}\right. \\{} & {} \qquad \left. +\left( \sum _{k=1}^{\infty }\lambda _{k}^{\nu -1/2-s}\left( 1-\textrm{e}^{-2\lambda _{k}h}\right) \right) ^{1/2}\right] \\{} & {} \quad \displaystyle \le C\left\| z\right\| _{{\mathcal {H}}^{s}}\Vert f\Vert _{L^2(0,T)}\left[ \left( \sum _{k=1}^\infty \frac{I(\tau ,k,h)}{k^{2(s-\nu +1/2)}}\right) ^{1/2}+\left( \sum _{k=1}^\infty \frac{1-\textrm{e}^{-2\lambda _k h}}{k^{2(s-\nu +1/2)}}\right) ^{1/2}\right] , \end{aligned}$$

where

$$\begin{aligned} I(\tau ,k,h)= & {} \lambda _k\int _0^\tau \left( \textrm{e}^{\lambda _k(t-\tau -h)}-\textrm{e}^{\lambda _k(t-\tau )}\right) ^2\mathrm {~d}t\nonumber \\= & {} \frac{1}{2}(1-\textrm{e}^{-\lambda _k h})^2(1-\textrm{e}^{-2\lambda _k \tau })\rightarrow 0 \end{aligned}$$

(22)

as $h\rightarrow 0^+$.

Since $0\le I(\tau ,k,h)\le 1/2$ uniformly for $\tau , h>0$, $k\ge 1$, the result follows by the dominated convergence theorem. $\square$

Remark 3.11

In the following section, we will consider initial conditions in $L^2_\beta (0,1)$. Notice that $L^2_\beta (0,1)\subset H^{-\nu -\delta }$ for all $\delta >0$, and we can apply Proposition 3.10 with $s=\nu +\delta$, $\delta >0$, then the corresponding solutions will be in $C^0([0, T ], H^{-\nu -\delta })$.

4 Proof of Theorem 2.1: control at the left endpoint

4.1 Upper estimate of the cost of the null controllability

In this section, we use the method moment, introduced by Fattorini and Russell in [16], to prove the null controllability of the system (1). In [6, Sect. 3], the authors construct a biorthogonal family $\displaystyle \{\psi _k\}_{k\ge 1}\subset L^2(0,T)$ to the family of exponential functions $\{\textrm{e}^{-\lambda _{k}(T-t)}\}_{k\ge 1}$ on [0, T], i.e., that satisfies

$$\begin{aligned} \int _{0}^{T}\psi _k(t) \textrm{e}^{-\lambda _{l}(T-t)} dt = \delta _{kl},\quad \text {for all}\quad k,l\ge 1. \end{aligned}$$

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 $\displaystyle \{\psi _k\}_{k\ge 1}$, see [6, Sect. 3] for details.

4.1.1 Construction of the biorthogonal family

Consider the Weierstrass infinite product

$$\begin{aligned} \Lambda (z):=\prod _{k=1}^{\infty }\left( 1+\dfrac{iz}{(\kappa _\alpha j_{\nu , k})^2}\right) . \end{aligned}$$

(23)

From (A8), we have that $j_{\nu , k}=O(k)$ for k large; thus, the infinite product is well-defined and converges absolutely in ${\mathbb {C}}$. Hence, $\Lambda (z)$ is an entire function with simple zeros at $i(\kappa _\alpha j_{\nu , k})^2=i\lambda _k$, $k\ge 1$. It follows that

$$\begin{aligned} \Psi _{k}(z):=\dfrac{\Lambda (z)}{\Lambda '(i\lambda _{k})(z-i\lambda _{k})},\quad k\ge 1, \end{aligned}$$

(24)

is an entire function that satisfies $\Psi _{k}(i\lambda _{l})=\delta _{kl}$ for all $k,l\ge 1$. Since $\Psi _{k}(x)$ is not in $L^2({\mathbb {R}})$, we need to fix this by using a suitable “complex multiplier", to do this we follow the approach introduced in [22].

For $\theta ,a>0$, we define

$$\begin{aligned} \sigma _{\theta }(t):=\exp \left( -\frac{\theta }{1-t^2}\right) ,\quad t\in (-1,1), \end{aligned}$$

and extended by 0 outside of $(-1, 1)$. Clearly $\sigma _{\theta }$ is analytic on $(-1,1)$. Set $C_{\theta }^{-1}:=\int _{-1}^{1}\sigma _{\theta }(t)\textrm{d}t$ and define

$$\begin{aligned} H_{a,\theta }(z)=C_{\theta }\int _{-1}^{1}\sigma _{\theta }(t)\exp (-iatz)\textrm{d}t. \end{aligned}$$

(25)

Clearly $H_{a,\theta }(z)$ is an entire function. The following result gives additional information about $H_{a,\theta }(z)$.

Lemma 4.1

The function $H_{a,\theta }$ fulfills the following inequalities

$$\begin{aligned} H_{a,\theta }(ix)&\ge \frac{\exp \left( a|x|/(2\sqrt{\theta +1})\right) }{11\sqrt{\theta +1}},\quad x\in {\mathbb {R}}, \end{aligned}$$

(26)

$$\begin{aligned} |H_{a,\theta }(z)|&\le \exp (a|\Im (z)|),\quad z\in {\mathbb {C}}, \end{aligned}$$

(27)

$$\begin{aligned} |H_{a,\theta }(x)|&\le \chi _{\{|x|\le 1\}}(x)+c\sqrt{\theta +1}\sqrt{a\theta |x|}\exp \Big (3\theta /4\nonumber \\ {}&\quad -\sqrt{a\theta |x|}\Big )\chi _{\{|x|> 1\}}(x),\quad x\in {\mathbb {R}}, \end{aligned}$$

(28)

where $c>0$ does not depend on a and $\theta$.

We refer to [22, pp. 85–86] for the details.

For $k\ge 1$, consider the entire function $F_{k}$ given as

$$\begin{aligned} F_{k}(z):=\Psi _{k}(z)\dfrac{H_{a,\theta }(z)}{H_{a,\theta }(i\lambda _{k})},\quad z\in {\mathbb {C}}. \end{aligned}$$

(29)

For $\delta \in (0,1)$, we set

$$\begin{aligned} a:=\frac{T(1-\delta )}{2}>0,\quad \text {and}\quad \theta :=\dfrac{(1+\delta )^2}{\kappa _\alpha ^2 T(1-\delta )}>0. \end{aligned}$$

(30)

Lemma 4.2

For each $k\ge 1$, the function $F_{k}(z)$ satisfies the following properties:

• i) $F_{k}$ is of exponential type T/2.

• ii) $F_{k}\in L^1({\mathbb {R}})\cap L^2({\mathbb {R}})$.

• iii) $F_k$ satisfies $F_{k}(i\lambda _{l})=\delta _{kl}$ for all $k,l\ge 1$.

• iv) Furthermore, there exists a constant $c>0$, independent of $T,\alpha$ and $\delta$, such that

  $$\begin{aligned} \left\| F_{k}\right\| _{L^{1}({\mathbb {R}})} \le \frac{C(T, \alpha ,\delta )}{\lambda _{k}\left| \Lambda ^{\prime }\left( i \lambda _{k}\right) \right| } \exp \left( -\frac{a\lambda _k}{2\sqrt{\theta +1}}\right) , \end{aligned}$$

  (31)

  where

  $$\begin{aligned} C(T, \alpha ,\delta )=c\sqrt{\theta +1}\left( \exp \left( {\frac{1}{\sqrt{2}\kappa _\alpha }}\right) +\sqrt{\theta +1}\frac{\kappa _\alpha ^2}{\delta ^3}\exp \left( \frac{3 \theta }{4}\right) \right) . \end{aligned}$$

  (32)

The $L^2$-version of the Paley–Wiener theorem implies that there exists $\eta _k\in L^2({\mathbb {R}})$ with support in $[-T/2,T/2]$ such that $F_k(z)$ is the analytic extension of the Fourier transform of $\eta _k$. Hence,

$$\begin{aligned} \psi _k(t):=\textrm{e}^{\lambda _k T/2}\eta _k(t-T/2),\quad t\in [0,T],\,k\ge 1, \end{aligned}$$

(33)

is the family we are looking for.

Since $\eta _k, F_k\in L^1({\mathbb {R}})$, the inverse Fourier theorem yields

$$\begin{aligned} \eta _k(t)=\frac{1}{2\pi }\int _{{\mathbb {R}}}\textrm{e}^{it\tau }F_k(\tau )\textrm{d}\tau ,\quad t\in {\mathbb {R}}, k\ge 1, \end{aligned}$$

hence (33) implies that $\psi _k\in C([0,T])$, and using (31), we have

$$\begin{aligned} \Vert \psi _k\Vert _{\infty }\le \frac{C(T, \alpha ,\delta )}{\lambda _{k}\left| \Lambda ^{\prime }\left( i \lambda _{k}\right) \right| } \exp \left( \frac{T\lambda _k}{2}-\frac{a\lambda _k}{2\sqrt{\theta +1}}\right) ,\quad k\ge 1. \end{aligned}$$

(34)

4.1.2 Proof of Theorem 2.1

Now, we are ready to prove the null controllability of the system (1). Let $u_{0}\in L^{2}_\beta (0,1)$. Then consider its Fourier series with respect to the orthonormal basis $\{\Phi _{k}\}_{k\ge 1}$,

$$\begin{aligned} u_{0}(x)=\sum _{k=1}^{\infty } a_{k} \Phi _{k}(x). \end{aligned}$$

(35)

We set

$$\begin{aligned} f(t):=\sum _{k=1}^{\infty }\frac{a_{k} \textrm{e}^{-\lambda _{k} T}}{{\mathcal {O}}_{\alpha +\beta +\gamma }\left( \Phi _{k}\right) } \psi _{k}(t). \end{aligned}$$

(36)

Since $\{\psi _k\}$ is biorthogonal to $\{\textrm{e}^{-\lambda _k(T-t)}\}_{k\ge 1}$, we have

$$\begin{aligned} \int _{0}^{T} f(t) {\mathcal {O}}_{\alpha +\beta +\gamma }\left( \Phi _{k}\right) \textrm{e}^{-\lambda _{k}(T-t)} \textrm{d} t= & {} a_{k} \textrm{e}^{-\lambda _{k} T} = \left\langle u_{0}, \textrm{e}^{-\lambda _{k} T}\Phi _{k}\right\rangle _\beta \\ {}= & {} \left\langle u_{0}, \textrm{e}^{-\lambda _{k} T}\Phi _{k}\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}. \end{aligned}$$

Let $u\in C([0,T];H^{-s})$ that satisfies (20) for all $\tau \in (0,T]$, $z^\tau \in H^s$. In particular, for $\tau =T$, we take $z^T=\Phi _k$, $k\ge 1$, then the last equality implies that

$$\begin{aligned} \left\langle u(\cdot , T), \Phi _{k}\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}=0\quad \text {for all}\quad k \ge 1, \end{aligned}$$

hence $u(\cdot , T)\equiv 0$.

It just remains to estimate the norm of the control f. From (34) and (36), we get

$$\begin{aligned} \Vert f\Vert _{\infty }\le & {} C(T, \alpha ,\delta )\sum _{k=1}^{\infty } \frac{\left| a_{k}\right| }{\left| {\mathcal {O}}_{\alpha +\beta +\gamma }\left( \Phi _{k}\right) \right| } \frac{1}{\lambda _{k}\left| \Lambda ^{\prime }\left( i \lambda _{k}\right) \right| }\nonumber \\{} & {} \exp \left( -\frac{T\lambda _k}{2}-\frac{a\lambda _k}{2\sqrt{\theta +1}}\right) . \end{aligned}$$

(37)

Using [23, Chap. XV, p. 498, eq. (3)], we can write

$$\begin{aligned} \Lambda (z)=\Gamma (\nu +1)\left( \dfrac{2\kappa _\alpha }{\sqrt{-iz}}\right) ^\nu J_{\nu }\left( \dfrac{\sqrt{-iz}}{\kappa _\alpha }\right) , \end{aligned}$$

therefore

$$\begin{aligned} \left| \Lambda ^{\prime }\left( i \lambda _{k}\right) \right| =\Gamma (\nu +1)\frac{2^{\nu }}{|j_{\nu , k}|^{\nu }} \frac{1}{2\kappa _\alpha ^{2} j_{\nu , k}} |J_{\nu }^{\prime }\left( j_{\nu , k}\right) |, \quad k\ge 1, \end{aligned}$$

(38)

and by using (18) and (A10) we get

$$\begin{aligned} \displaystyle \left| {\mathcal {O}}_{\alpha +\beta +\gamma }\left( \Phi _{k}\right) \lambda _{k}\Lambda ^{\prime }\left( i \lambda _{k}\right) \right| = 2^{-1/2}\sqrt{\kappa _\alpha } j_{\nu , k}. \end{aligned}$$

From (37), the last two equalities and using that $\lambda _k\ge \lambda _1$, it follows that

$$\begin{aligned} \Vert f\Vert _{\infty } \le \frac{C(T, \alpha ,\delta )}{\sqrt{\kappa _\alpha }} \exp \left( -\frac{T\lambda _1}{2}-\frac{a\lambda _1}{2\sqrt{\theta +1}}\right) \sum _{k=1}^{\infty } \frac{|a_{k}| }{j_{\nu , k}}. \end{aligned}$$

Using the Cauchy–Schwarz inequality, the fact that $j_{\nu ,k}\ge (k-1/4)\pi$ (by (A9)) and (35), we obtain that

$$\begin{aligned} \Vert f\Vert _{\infty }\le & {} \frac{C(T, \alpha ,\delta )}{\sqrt{\kappa _\alpha }} \exp \left( -\frac{T\lambda _1}{2}-\frac{a\lambda _1}{2\sqrt{\theta +1}}\right) \left\| u_{0}\right\| _{\beta }. \end{aligned}$$

Using the definition of $a,\theta$ in (30) and the facts $\theta >0$, $\delta \in (0,1)$, $0<\kappa _\alpha \le 1$, we get that

$$\begin{aligned} \theta \le \frac{4}{(1-\delta )\kappa _\alpha ^2 T },\quad \sqrt{\theta +1}\le \frac{2(1+T)^{1/2}}{(1-\delta )^{1/2}\kappa _\alpha T^{1/2}},\quad \sqrt{\theta +1}\le \theta +1, \end{aligned}$$

therefore

$$\begin{aligned} \displaystyle \frac{a}{\sqrt{\theta +1}}\ge & {} \frac{\kappa _\alpha (1-\delta )^{3/2}T^{3/2}}{4(1+T)^{1/2}},\nonumber \\ \displaystyle C(T, \alpha ,\delta )\le & {} c\left( 1+\frac{1}{(1-\delta )\kappa _\alpha ^2 T}\right) \Bigg [\exp \left( \frac{1}{\sqrt{2}\kappa _\alpha }\right) \nonumber \\ {}{} & {} +\frac{1}{\delta ^3}\exp \left( \frac{3}{(1-\delta )\kappa _\alpha ^2 T}\right) \Bigg ], \end{aligned}$$

(39)

and using the definition of $\lambda _{1}$ the result follows.

4.2 Lower estimate of the cost of the null controllability

In this section, we get a lower estimate of the cost ${\mathcal {K}}={\mathcal {K}}(T,\alpha ,\beta ,\mu )$. We set

$$\begin{aligned} u_0(x):= \frac{|J'_\nu (j_{\nu ,1})|}{(2\kappa _\alpha )^{1/2}}\Phi _1(x),\,x\in (0,1), \quad \text {hence}\quad \Vert u_0\Vert ^2_\beta =\frac{|J'_\nu (j_{\nu ,1})|^2}{2\kappa _\alpha }. \end{aligned}$$

(40)

For $\varepsilon >0$ small enough, there exists $f\in U(\alpha ,\beta ,\mu ,T,u_0)$ such that

$$\begin{aligned} u(\cdot ,T)\equiv 0,\quad \text {and}\quad \Vert f\Vert _{L^2(0,T)}\le ({\mathcal {K}}+\varepsilon )\Vert u_0\Vert _\beta . \end{aligned}$$

(41)

Then in (20), we set $\tau =T$ and take $z^\tau =\Phi _k$, $k\ge 1$, to obtain

$$\begin{aligned} \textrm{e}^{-\lambda _k T}\left\langle u_{0},\Phi _k\right\rangle _\beta= & {} \left\langle u_{0}, S(T)\Phi _k\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}= \int _{0}^{T} f(t) {\mathcal {O}}_{\alpha +\beta +\gamma }\left( S(T-t) \Phi _k\right) \textrm{d} t\\= & {} \textrm{e}^{-\lambda _k T}{\mathcal {O}}_{\alpha +\beta +\gamma }\left( \Phi _k\right) \int _{0}^{T} f(t) \textrm{e}^{\lambda _k t} \textrm{d} t, \end{aligned}$$

from (40) and (A10), it follows that

$$\begin{aligned} \int _{0}^{T} f(t) \textrm{e}^{\lambda _k t} \textrm{d} t= \frac{2^{\nu }\Gamma (\nu +1)|J'_{\nu }(j_{\nu , 1})|^2}{2\kappa _\alpha (j_{\nu ,1})^{\nu }}\delta _{1,k},\quad k\ge 1. \end{aligned}$$

(42)

Now consider the function $v: {\mathbb {C}} \rightarrow {\mathbb {C}}$ given by

$$\begin{aligned} v(s):=\int _{-T / 2}^{T / 2} f\left( t+\frac{T}{2}\right) \textrm{e}^{-i s t} \mathrm {~d} t, \quad s \in {\mathbb {C}}. \end{aligned}$$

Fubini and Morera’s theorems imply that v(s) is an entire function. Moreover, (42) implies that

$$\begin{aligned} v(i\lambda _k)=0\quad \text {for all }k\ge 2,\quad \text {and}\quad v(i\lambda _1)=\frac{2^{\nu }\Gamma (\nu +1)|J'_{\nu }(j_{\nu , 1})|^2}{2\kappa _\alpha (j_{\nu ,1})^{\nu }}\textrm{e}^{-\lambda _1 T/2}. \end{aligned}$$

We also have that

$$\begin{aligned} |v(s)| \le \textrm{e}^{T|\Im (s)|/2}\int _{0}^{T}|f(t)| \textrm{d} t \le ({\mathcal {K}}+\varepsilon ) T^{1/2}\textrm{e}^{T|\Im (s)|/2} \left\| u_{0}\right\| _{\beta }. \end{aligned}$$

(43)

Consider the entire function F(z) given by

$$\begin{aligned} F(s):=v\left( s-i \delta \right) , \quad s \in {\mathbb {C}}, \end{aligned}$$

(44)

for some $\delta >0$ that will be chosen later on. Clearly,

$$\begin{aligned} F\left( a_{k}\right)= & {} 0, \quad k\ge 2, \quad \text{ where } \quad a_{k}:=i\left( \lambda _k+\delta \right) ,\quad k\ge 1,\quad \text {and} \nonumber \\ F\left( a_{1}\right)= & {} \frac{2^{\nu }\Gamma (\nu +1)|J'_{\nu }(j_{\nu , 1})|^2}{2\kappa _\alpha (j_{\nu ,1})^{\nu }}\textrm{e}^{-\lambda _1 T/2}. \end{aligned}$$

(45)

From (40), (43), and (44), we obtain

$$\begin{aligned} \log |F(s)|\le \frac{T}{2}|\Im (s)-\delta |+\log \left( ({\mathcal {K}}+\varepsilon ) T^{1 / 2}\frac{|J'_\nu (j_{\nu ,1})|}{(2\kappa _\alpha )^{1/2}}\right) ,\quad s\in {\mathbb {C}}. \end{aligned}$$

(46)

We apply Theorem A.5 to the function F(z) given in (44). In this case, (43) implies that $A\le T/2$. Also notice that $\Im \left( a_{k}\right) >0$, $k\ge 1$, to get

$$\begin{aligned} \log \left| F\left( a_{1}\right) \right| \le \left( \lambda _1+\delta \right) \frac{T}{2}+\sum _{k=2}^{\infty } \log \left| \frac{a_{1}-a_{k}}{a_{1}-{\bar{a}}_{k}}\right| +\frac{\Im \left( a_{1}\right) }{\pi } \int _{-\infty }^{\infty } \frac{\log |F(s)|}{\left| s-a_{1}\right| ^{2}} \mathrm {~d}s. \end{aligned}$$

(47)

Using the definition of the constants $a_k$’s, we have

$$\begin{aligned} \sum _{k=2}^{\infty } \log \left| \frac{a_{1}-a_{k}}{a_{1}-{\bar{a}}_{k}}\right|= & {} \sum _{k=2}^{\infty } \log \left( \frac{\left( j_{\nu , k}\right) ^{2}-\left( j_{\nu , 1}\right) ^{2}}{2 \delta / \kappa _\alpha ^{2}+\left( j_{\nu , 1}\right) ^{2}+\left( j_{\nu , k}\right) ^{2}}\right) \nonumber \\\le & {} \sum _{k=2}^{\infty } \frac{1}{j_{\nu , k+1}-j_{\nu , k}} \int _{j_{\nu , k}}^{j_{\nu , k+1}} \log \left( \frac{ x^{2}}{2 \delta / \kappa _\alpha ^{2}+ x^{2}}\right) \textrm{d} x \nonumber \\\le & {} \frac{1}{\pi } \int _{j_{\nu , 2}}^{\infty } \log \left( \frac{ x^{2}}{2 \delta / \kappa _\alpha ^{2}+x^{2}}\right) \textrm{d} x,\nonumber \\= & {} -\frac{j_{\nu , 2}}{\pi } \log \left( \frac{1}{1+2 \delta /\left( \kappa _\alpha j_{\nu , 2}\right) ^{2}}\right) \nonumber \\ {}{} & {} - \frac{2\sqrt{2 \delta }}{\pi \kappa _\alpha }\left( \frac{\pi }{2}-\tan ^{-1}\left( \kappa _\alpha j_{\nu , 2} / \sqrt{2 \delta }\right) \right) , \end{aligned}$$

(48)

where we have used Lemma A.2 and made the change of variables

$$\begin{aligned} \tau =\frac{ \kappa _\alpha }{\sqrt{2 \delta }} x. \end{aligned}$$

From (46), we get the estimate

$$\begin{aligned} \frac{\Im \left( a_{1}\right) }{\pi } \int _{-\infty }^{\infty } \frac{\log |F(s)|}{\left| s-a_{1}\right| ^{2}} \mathrm {~d} s \le \frac{T \delta }{2}+\log \left( ({\mathcal {K}}+\varepsilon ) T^{1 / 2} \frac{\left| J_{\nu }^{\prime }\left( j_{\nu , 1}\right) \right| }{(2 \kappa _\alpha )^{1/2}}\right) . \end{aligned}$$

(49)

From (45), (47), (48), and (49), we have

$$\begin{aligned}{} & {} \frac{2\sqrt{2 \delta }}{\pi \kappa _\alpha }\tan ^{-1}\left( \frac{\sqrt{2 \delta }}{\kappa _\alpha j_{\nu , 2}}\right) -\frac{j_{\nu , 2}}{\pi } \log \left( 1+\frac{2 \delta }{ \left( \kappa _\alpha j_{\nu , 2}\right) ^{2}}\right) -(\lambda _1+\delta )T \nonumber \\ {}{} & {} \quad \le \log ({\mathcal {K}}+\varepsilon )+\log h(\alpha ,\beta ,\mu , T), \end{aligned}$$

(50)

where

$$\begin{aligned} h(\alpha ,\beta ,\mu , T)=\frac{({2T \kappa _\alpha })^{1/2}\left( j_{\nu , 1}\right) ^{\nu }}{2^{\nu } \Gamma (\nu +1) \left| J_{\nu }^{\prime }\left( j_{\nu , 1}\right) \right| }. \end{aligned}$$

The result follows by taking

$$\begin{aligned} \delta =\frac{\kappa _\alpha ^2 (j_{\nu ,2})^2}{2}, \quad \text {and then letting}\quad \varepsilon \rightarrow 0^+. \end{aligned}$$

5 Proof of Theorem 2.2: control at the right endpoint

5.1 Existence and uniqueness of solutions to system (5)

Here, we analyze the null controllability of the system (5) where $\alpha +\beta >1$, $0\le \alpha < 2$, $\mu$ and $\gamma$ satisfy (2) and (3) respectively. As in Sect. 4, we give a suitable definition of a weak solution for the system (5).

Definition 5.1

Let $T>0$ and $\alpha ,\beta ,\mu \in {\mathbb {R}}$ with $0\le \alpha <2$, $\alpha + \beta >1$, $\mu <\mu (\alpha +\beta )$. Let $f \in L^2(0,T)$ and $u_0\in {\mathcal {H}}^{-s}$ for some $s > 0$. A weak solution of (5) is a function $u \in C^0([0,T];{\mathcal {H}}^{-s})$ such that for every $\tau \in (0,T]$ and for every $z^\tau \in {\mathcal {H}}^s$, we have

$$\begin{aligned} \left\langle u(\tau ), z^{\tau }\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}=\left\langle u_{0}, S(\tau )z^\tau \right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}} - \int _{0}^{\tau } f(t)\lim _{x\rightarrow 1^- }S(\tau -t) z^{\tau }_{x}(x) \textrm{d} t, \end{aligned}$$

(51)

where $\gamma =\gamma (\alpha ,\beta ,\mu )$ is given by (3).

The next result shows the existence of weak solutions for the system (5) under certain conditions on the parameters $\alpha ,\beta ,\mu ,\gamma$ and s.

Proposition 5.2

Let $T>0$ and $\alpha ,\beta \in {\mathbb {R}}$ with $0\le \alpha <2$, $\alpha + \beta >1$. Let $f \in L^2(0,T)$ and $u_0\in {\mathcal {H}}^{-s}$ such that $s>1/2$. Then formula (51) defines for each $\tau \in [0, T ]$ a unique element $u(\tau ) \in {\mathcal {H}}^{-s}$ that can be written as

$$\begin{aligned} u(\tau )=S(\tau ) u_{0}-{\widetilde{B}}(\tau ) f, \quad \tau \in (0, T], \end{aligned}$$

where ${\widetilde{B}}(\tau )$ is the strongly continuous family of bounded operators ${\widetilde{B}}(\tau ): L^{2}(0,T) \rightarrow {\mathcal {H}}^{-s}$ given by

$$\begin{aligned} \left\langle {\widetilde{B}}(\tau ) f, z^{\tau }\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}=\int _{0}^{\tau } f(t)\lim _{x\rightarrow 1^- }S(\tau -t) z^{\tau }_{x}(x) \textrm{d} t, \quad \text {for all }z^{\tau } \in {\mathcal {H}}^{s}. \end{aligned}$$

Furthermore, the unique weak solution u on [0, T] to (5) (in the sense of (51)) belongs to ${C}^{0}\left( [0, T]; {\mathcal {H}}^{-s}\right)$ and fulfills

$$\begin{aligned} \Vert u\Vert _{L^{\infty }\left( [0, T]; {\mathcal {H}}^{-s}\right) } \le C\left( \left\| u_{0}\right\| _{{\mathcal {H}}^{-s}}+\Vert f\Vert _{L^{2}(0, T)}\right) . \end{aligned}$$

Proof

Fix $\tau >0$. Let $u(\tau )\in H^{-s}$ be determined by the condition (51); hence,

$$\begin{aligned} -u(\tau )+S(\tau ) u_{0}=\zeta (\tau )f, \end{aligned}$$

where

$$\begin{aligned} \left\langle \zeta (\tau )f, z^{\tau }\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}=\int _{0}^{\tau } f(t) \lim _{x\rightarrow 1^- }S(\tau -t) z^{\tau }_{x}(x)\textrm{d} t \quad \text {for all } z^{\tau } \in {\mathcal {H}}^{s}. \end{aligned}$$

Let $z^{\tau } \in {\mathcal {H}}^{s}$ given by

$$\begin{aligned} z^{\tau }=\sum _{k=1}^\infty a_{k} \Phi _{k}, \end{aligned}$$

(52)

therefore

$$\begin{aligned} \lim _{x\rightarrow 1^- }S(\tau -t) z^{\tau }_{x}(x)=\sum _{k=1}^\infty \textrm{e}^{\lambda _{k}(t-\tau )} a_{k} \Phi '_{k}(1) \quad \text {for all } t \in [0, \tau ]. \end{aligned}$$

By (17), we get

$$\begin{aligned} \left| \Phi '_{k}(1)\right| = 2^{1/2}\kappa _{\alpha }^{3/2}j_{\nu , k}, \quad k\ge 1, \end{aligned}$$

(53)

hence (A9) implies that there exists $C=C(\alpha ,\beta ,\mu )>0$ such that

$$\begin{aligned} \left( \int _{0}^{\tau }\left| \lim _{x\rightarrow 1^- }S(\tau -t) z^{\tau }_{x}(x)\right| ^{2} \mathrm {~d} t\right) ^{1/2}\le & {} \sum _{k=1}^\infty |a_k| |\Phi '_{k}(1)| \left( \int _0^\tau \textrm{e}^{2\lambda _{k}(t-\tau )} \mathrm {~d} t\right) ^{1/2} \\\le & {} C\left\| z^{\tau }\right\| _{{\mathcal {H}}^{s}}\left( \sum _{k=1}^\infty |\lambda _k|^{1-s}\int _0^\tau \textrm{e}^{2\lambda _{k}(t-\tau )} \mathrm {~d} t\right) ^{1/2}\\\le & {} C\left\| z^{\tau }\right\| _{{\mathcal {H}}^{s}}\left( \sum _{k=1}^\infty \frac{1}{k^{2s}}\right) ^{1/2}=C\left\| z^{\tau }\right\| _{{\mathcal {H}}^{s}}. \end{aligned}$$

Therefore, $\Vert \zeta (\tau ) f\Vert _{{\mathcal {H}}^{-s}}\le C\Vert f\Vert _{L^2(0,T)}$ for all $f\in L^2(0,T)$, $\tau \in (0,T]$.

Finally, we fix $f\in L^2(0,T)$ and show that the mapping $\tau \mapsto \zeta (\tau ) f$ is right-continuous on [0, T). Let $h>0$ be small enough and $z\in {\mathcal {H}}^s$ given as in (52). Thus, proceeding as in the last inequalities, we have

$$\begin{aligned}{} & {} \displaystyle |\left\langle \zeta (\tau +h)f-\zeta (\tau )f, z\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}|\\{} & {} \quad \displaystyle \le \int _{0}^{\tau } |f(t)|\left| \lim _{x\rightarrow 1^- }(S(\tau +h-t)-S(\tau -t)) z^{\tau }_{x}(x)\right| \textrm{d} t\\{} & {} \qquad \displaystyle + \int _{\tau }^{\tau +h} |f(t)|\left| \lim _{x\rightarrow 1^- }S(\tau +h-t) z^{\tau }_{x}(x)\right| \textrm{d} t\\{} & {} \quad \displaystyle \le C\left\| z\right\| _{{\mathcal {H}}^{s}}\Vert f\Vert _{L^2(0,T)}\left[ \left( \sum _{k=1}^\infty \frac{I(\tau ,k,h)}{k^{2s}}\right) ^{1/2}+\left( \sum _{k=1}^\infty \frac{1-\textrm{e}^{-2\lambda _k h}}{k^{2s}}\right] ^{1/2}\right) , \end{aligned}$$

where $I(\tau ,k,h)\rightarrow 0$ as $h\rightarrow 0^+$, see (22). $\square$

Remark 5.3

In the following subsections, we will consider initial conditions in $L^2_\beta (0,1)$. We can apply Proposition 5.2 with $s=1/2+\delta$, $\delta >0$, then the corresponding solutions will be in $C^0([0, T ], H^{-1/2-\delta })$.

5.2 Upper estimate of the cost of the null controllability

5.2.1 Proof of Theorem 2.2

We are ready to prove the null controllability of the system (5). Let $u_{0}\in L^{2}_\beta (0,1)$ given as follows

$$\begin{aligned} u_{0}(x)=\sum _{k=1}^{\infty } a_{k} \Phi _{k}(x). \end{aligned}$$

(54)

We set

$$\begin{aligned} f(t):=\sum _{k=1}^{\infty }\frac{a_{k} \textrm{e}^{-\lambda _{k} T}}{\Phi '_{k}(1)} \psi _{k}(t). \end{aligned}$$

(55)

Since the sequence $\{\psi _k\}$ is biorthogonal to $\{\textrm{e}^{-\lambda _k(T-t)}\}_{k\ge 1}$, we have

$$\begin{aligned} \Phi '_{k}(1)\int _{0}^{T} f(t)\textrm{e}^{-\lambda _{k}(T-t)} \textrm{d} t= & {} a_{k} \textrm{e}^{-\lambda _{k} T} = \left\langle u_{0}, \textrm{e}^{-\lambda _{k} T}\Phi _{k}\right\rangle _\beta \nonumber \\ {}= & {} \left\langle u_{0}, \textrm{e}^{-\lambda _{k} T}\Phi _{k}\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}. \end{aligned}$$

(56)

Let $u\in C([0,T];H^{-s})$ be the weak solution of system (5). In particular, for $\tau =T$, we take $z^T=\Phi _k$, $k\ge 1$, then (51) and (56) imply that

$$\begin{aligned} \left\langle u(\cdot , T), \Phi ^{k}\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}=0\quad \text {for all}\quad k \ge 1, \end{aligned}$$

therefore $u(\cdot , T) = 0$.

It just remains to estimate the norm of the control f. From (34), (38), (53), and (55), we get

$$\begin{aligned} \Vert f\Vert _{\infty } \le \frac{C(T, \alpha ,\delta )\kappa _\alpha ^{1/2}}{2^{\nu }\Gamma (\nu +1)}\sum _{k=1}^{\infty } \frac{|j_{\nu ,k}|^{\nu }}{|J'_{\nu }(j_{\nu ,k})|}\dfrac{|a_{k}|}{\lambda _{k}}\exp \left( -\frac{T\lambda _k}{2}-\frac{a\lambda _k}{2\sqrt{\theta +1}}\right) . \end{aligned}$$

Using that $\textrm{e}^{-x}\le \textrm{e}^{-r}r^rx^{-r}$ for all $x,r>0,$ the Cauchy–Schwarz inequality, Lemma A.3 and the fact that $j_{\nu ,k}\ge (k-1/4)\pi$ (by (A9)) and (35), we obtain that

$$\begin{aligned} \displaystyle \Vert f\Vert _{\infty }\le & {} \displaystyle \frac{C(T, \alpha ,\delta )}{(2\kappa _\alpha )^{\nu }\Gamma (\nu +1)}\left( \dfrac{2\nu +1}{T}\right) ^{(2\nu +1)/4} \exp \left( -\frac{2\nu +1}{4}\right) \\{} & {} \exp \left( -\frac{a\lambda _1}{2\sqrt{\theta +1}}-\frac{T\lambda _1}{4}\right) \sum _{k=0}^{\infty }\frac{|a_k|}{\lambda _k}\\\le & {} \displaystyle \frac{C(T, \alpha ,\delta )}{(2\kappa _\alpha )^{\nu }\Gamma (\nu +1)}\left( \dfrac{2\nu +1}{T\textrm{e}}\right) ^{(2\nu +1)/4} \exp \left( -\frac{a\lambda _1}{2\sqrt{\theta +1}}-\frac{T\lambda _1}{4}\right) \left\| u_{0}\right\| _{\beta }, \end{aligned}$$

and the result follows by (39).

5.3 Lower estimate of the cost of the null controllability at the right endpoint

Here, we just give a sketch of the proof of a lower estimate for the cost $\widetilde{{\mathcal {K}}}=\widetilde{{\mathcal {K}}}(T,\alpha ,\beta ,\mu )$. Consider $u_0\in L^2_\beta (0,1)$ given in (40).

For $\varepsilon >0$ small enough, there exists $f\in {\widetilde{U}}(\alpha ,\beta ,\mu ,T,u_0)$ such that

$$\begin{aligned} u(\cdot ,T)\equiv 0,\quad \text {and}\quad \Vert f\Vert _{L^2(0,T)}\le (\widetilde{{\mathcal {K}}}+\varepsilon )\Vert u_0\Vert _\beta . \end{aligned}$$

(57)

Then in (51), we set $\tau =T$ and take $z^T=\Phi _k$, $k\ge 1$, to obtain

$$\begin{aligned} \textrm{e}^{-\lambda _k T}\left\langle u_{0},\Phi _k\right\rangle _\beta =\left\langle u_{0}, S(T)\Phi _k\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}} = \textrm{e}^{-\lambda _k T}\Phi '_k(1)\int _{0}^{T} f(t) \textrm{e}^{\lambda _k t} \textrm{d} t, \end{aligned}$$

from (40) and (53), it follows that

$$\begin{aligned} \int _{0}^{T} f(t) \textrm{e}^{\lambda _k t} \textrm{d} t = \frac{|J'_{\nu }(j_{\nu , 1})|}{2\kappa ^{2}_{\alpha }j_{\nu ,1}}\delta _{1,k},\quad k\ge 1. \end{aligned}$$

(58)

Next, we proceed as in (4.2)–(50). But in this case, the corresponding functions v and F satisfy

$$\begin{aligned}{} & {} v(i\lambda _k)=0\quad \text {for all }k\ge 2,\quad v(i\lambda _1)=\frac{|J'_{\nu }(j_{\nu , 1})|}{2\kappa ^{2}_{\alpha }j_{\nu ,1}}\textrm{e}^{-\lambda _1 T/2},\quad \text {and}\\{} & {} F(a_k)=0\quad \text {for all }k\ge 2,\quad F(a_1)=\frac{|J'_{\nu }(j_{\nu , 1})|}{2\kappa ^{2}_{\alpha }j_{\nu ,1}}\textrm{e}^{-\lambda _1 T/2}. \end{aligned}$$

Hence, we can see that

$$\begin{aligned}{} & {} \frac{2\sqrt{2 \delta }}{\pi \kappa _\alpha }\tan ^{-1}\left( \frac{\sqrt{2 \delta }}{\kappa _\alpha j_{\nu , 2}}\right) -\frac{j_{\nu , 2}}{\pi } \log \left( 1+\frac{2 \delta }{ \left( \kappa _\alpha j_{\nu , 2}\right) ^{2}}\right) -(\lambda _1+\delta )T \\{} & {} \quad \le \log (\widetilde{{\mathcal {K}}}+\varepsilon )+\log {\widetilde{h}}(\alpha ,\beta ,\mu , T), \end{aligned}$$

where ${\widetilde{h}}(\alpha ,\beta ,\mu , T)=T^{1/2}\kappa ^{3/2}_{\alpha }j_{\nu ,1}/\sqrt{2}$. The result follows by taking $\delta =\kappa _\alpha ^2 (j_{\nu ,2})^2/2$ and then letting $\varepsilon \rightarrow 0^+$.

6 Proof of Theorem 2.4: the critical case $\alpha +\beta =1$

Concerning the case $\alpha +\beta <1$, in [6], we show the system (1) is well-posed when considering suitable weighted Dirichlet condition at the left endpoint and prove the null controllability of the corresponding system. In both cases ($\alpha +\beta <1$ and $\alpha +\beta >1$), our approach is based on the validity of the Hardy inequality, see Proposition 3.3 and [6, Proposition 2.3]. If $\alpha +\beta =1$, then $\mu (\alpha +\beta )=0$, and the corresponding Hardy inequality does not provide any information. Thus, to solve the case $\alpha +\beta =1$, we use the singular Sturm–Liouville theory, see [24] for the definitions used here.

6.1 Singular Sturm–Liouville theory

Assume that $0\le \alpha <2$, $\beta \in {\mathbb {R}}$ such that $\beta = 1-\alpha$, and $\mu < 0$. Consider the differential expression M defined by

$$\begin{aligned} Mu=-(pu')'+qu, \end{aligned}$$

where $\displaystyle p(x) = x, q(x) = -\mu x^{-1}, w(x) = x^{1-\alpha }$, $'=\frac{\textrm{d}}{\textrm{d}x}$.

Clearly,

$$\begin{aligned} 1/p, q, w \in L_{\text {loc}}(0,1),\quad \text {and}\quad p,w >0\text { on } (0,1), \end{aligned}$$

thus, Mu is defined a.e. for functions u such that $u, pu'\in AC_{\text {loc}}(0,1)$, where $AC_{\text {loc}}(0,1)$ is the space of all locally absolutely continuous functions in (0, 1).

Notice that the operator ${\mathcal {A}}$ given in (16), with $\beta = 1-\alpha$, can be written as ${\mathcal {A}} =w^{-1}M$. Now, consider

$$\begin{aligned} D_{\max }:= & {} \left\{ u\in AC_{\text {loc}}(0,1)\, |\,pu'\in AC_{\text {loc}}(0,1),\, u, {\mathcal {A}} u\in L^2_{1-\alpha }(0,1)\right\} ,\quad \text {and} \\ D({\mathcal {A}}):= & {} \left\{ \begin{array}{ll} \{u\in D_{\max }\,| \lim _{x\rightarrow 0^+}x^{\sqrt{-\mu }}u(x)=0,u(1)=0\} &{} \quad \text {if } \sqrt{-\mu }<\kappa _\alpha , \\ \{u\in D_{\max }\,| u(1)=0\}&{} \quad \text {if } \sqrt{-\mu }\ge \kappa _\alpha . \end{array}\right. \end{aligned}$$

Recall that the Lagrange form associated with M is given as follows

$$\begin{aligned}{}[u,v]:=upv'-vpu',\quad u,v\in D_{\max }. \end{aligned}$$

According to [24, Definition 10.4.3], a pair of real-valued functions $y_{0}, z_{0}\in D_{\max }$ is called a (BC) basis at $x=0$ if they satisfy $[z_{0},y_{0}](0)=1$. Similarly, a pair of real-valued functions $y_{1}, z_{1}\in D_{\max }$ is called a (BC) basis at $x=1$ if they satisfy $[z_{1}, y_{1}](1)=1$. Moreover, a pair of real-valued functions $y,z\in D_{\max }$ is called a (BC) basis on (0, 1) if they form a (BC) basis at $x=0$ and at $x=1$.

Proposition 6.1

Let $0\le \alpha <2$, $\mu < 0$, and $\nu =\sqrt{-\mu }/\kappa _\alpha$. Then ${\mathcal {A}}:D({\mathcal {A}})\subset L^2_{1-\alpha }(0,1)\rightarrow L^2_{1-\alpha }(0,1)$ is a self-adjoint operator. Furthermore, when $\beta = 1-\alpha$ the family given in (17) is an orthonormal basis for $L^2_{1-\alpha }(0,1)$ such that

$$\begin{aligned} {\mathcal {A}} \Phi _k=\lambda _k \Phi _k, \quad k\ge 1. \end{aligned}$$

Proof

First, we refer to [24, Definition 7.3.1] to see that $x=1$ is a regular point because $1/p,q,w\in L^1(1/2,1)$.

Case i) Suppose $\sqrt{-\mu }<\kappa _\alpha$.

First, we will build a (BC) basis $\{y_{0},z_{0}\}$ at $x=0$ and a (BC) basis $\{y_{1}, z_{1}\}$ at $x=1$.

Consider the functions given by

$$\begin{aligned} y_{0}(x):=x^{\sqrt{-\mu }},\quad z_{0}(x):=\dfrac{x^{-\sqrt{-\mu }}}{2\sqrt{-\mu }},\quad x\in (0,1). \end{aligned}$$

(59)

Notice the assumption implies that $y_{0}, z_{0}\in D_{\max }$. Clearly, $[z_{0}, y_{0}](0)=1$, thus $\{y_{0}, z_{0}\}$ is a (BC) basis at $x=0$.

Since $y_{0}, z_{0}\in L^2_{1-\alpha }(0,1)$ are linearly independent solutions of $Mu=0u$, it follows that $x=0$ is limit-circle (LC), see [24, Definition 7.3.1, Theorem 7.2.2].

In the same way, consider also the functions

$$\begin{aligned} y_{1}(x):= & {} \frac{1}{2\sqrt{-\mu }}\left( x^{\sqrt{-\mu }}-x^{-\sqrt{-\mu }}\right) ,\nonumber \\ z_{1}(x):= & {} \frac{1}{2\sqrt{-\mu }}\left( x^{\sqrt{-\mu }}+x^{-\sqrt{-\mu }}\right) ,\quad x\in (0,1). \end{aligned}$$

(60)

Since $y_{1}, z_{1}\in D_{\max }$ and $[z_{1}, y_{1}](1)=1$, it follows that $\{y_{1}, z_{1}\}$ is a (BC) basis at $x=1$.

Now, we fix $c,d\in (0,1)$ with $c<d$. From the Patching Lemma, Lemma 10.4.1 in [24], there exist functions $g_1, g_2\in D_{\max }$ such that

$$\begin{aligned}{} & {} {\left\{ \begin{array}{ll}g_1(c)=y_0(c), &{} \!\! g_1(d)=y_1(d), \\ (p g_1^{\prime })(c)=(p y_0^{\prime })(c), &{}\!\! (p g_1^{\prime })(d)=(p y_1^{\prime })(d) \end{array}\right. },\\{} & {} {\left\{ \begin{array}{ll}g_2(c)=z_0(c), &{}\!\! g_2(d)=z_1(d), \\ (p g_2^{\prime })(c)=(p z_0^{\prime })(c), &{}\!\! (p g_2^{\prime })(d)=(p z_1^{\prime })(d). \end{array}\right. } \end{aligned}$$

Thus, the pair $\{y_+,y_-\}$ is a (BC) basis on (0, 1), see [24, Definition 10.4.3], where

$$\begin{aligned} y_{+}(x):=\left\{ \begin{array}{lll} y_0(x) &{} \text{ if } &{} x \in (0, c), \\ g_1(x) &{} \text{ if } &{} x \in [c, d], \\ y_1(x) &{} \text{ if } &{} x \in (d, 1), \end{array}\right. \,\,\,\,\,\, y_{-}(x):=\left\{ \begin{array}{lll} z_0(x) &{} \text{ if } &{} x \in (0, c), \\ g_2(x) &{} \text{ if } &{} x \in [c, d], \\ z_1(x) &{} \text{ if } &{} x \in (d, 1). \end{array}\right. \end{aligned}$$

The matrices

$$\begin{aligned} \textbf{A}=\begin{pmatrix}1 &{} 0\\ 0 &{} 0\end{pmatrix}\quad \text {and}\quad \textbf{B}=\begin{pmatrix}0 &{} 0\\ 1 &{} 0\end{pmatrix} \end{aligned}$$

satisfy the hypothesis in [24, Proposition 10.4.2], then

$$\begin{aligned} \begin{aligned} D({\mathcal {A}}):&=\left\{ u \in D_{\max }: \textbf{A} \left( \begin{array}{l} \left[ u, y_{+}\right] (0) \\ {\left[ u, y_{-}\right] (0)} \end{array}\right) +\textbf{B}\left( \begin{array}{l} {\left[ u, y_{+}\right] (1)} \\ {\left[ u, y_{-}\right] (1)} \end{array}\right) =\left( \begin{array}{l} 0 \\ 0 \end{array}\right) \right\} \\&=\left\{ u \in D_{\max }:\left[ u, y_{+}\right] (0)=\left[ u, y_{+}\right] (1)=0\right\} \\&=\left\{ u \in D_{\max }:\left[ u, y_{+}\right] (0)=u(1)=0\right\} \end{aligned} \end{aligned}$$

is a self-adjoint domain, therefore the operator ${\mathcal {A}}:D({\mathcal {A}})\subset L^2_\beta (0,1)\rightarrow L^2_\beta (0,1)$ is self-adjoint.

Finally, we have that

$$\begin{aligned} \left[ u, y_{+}\right] (0)= & {} \lim _{x\rightarrow 0^+}[u, y_{0}](x)=\lim _{x\rightarrow 0^+}\left\{ \frac{u}{z_0}(x)[z_0,y_0](x)+[u,z_0](x)\frac{y_0}{z_0}(x)\right\} \\ {}= & {} \lim _{x\rightarrow 0^+}\frac{u}{z_0}(x), \end{aligned}$$

because $[z_0,y_0](0)=1$, $[u,z_0](0)$ is finite (see [24, Lemma 10.2.3]), and $\lim _{x\rightarrow 0^+}y_0/z_0(x)=0$. Hence, the result follows.

Case ii) $\sqrt{-\mu }\ge \kappa _\alpha$. Since $z_0\notin L^2_{1-\alpha }(0,1)$, then $x=0$ is limit point (LP). The result follows using Theorem 10.4.4 in [24] with $A_1=1, A_2=0$.

When $\alpha +\beta =1$, notice that $\Phi _k\in D({\mathcal {A}})$, $k\ge 1$, thus the second part of the proof follows using the computations in the proof of Proposition 3.8. $\square$

6.2 Existence and uniqueness of solutions to system (6)

As in Sect. 3, consider the corresponding interpolation spaces ${\mathcal {H}}^s$, $s\in {\mathbb {R}}$.

Definition 6.2

Let $T>0$ and $\alpha ,\mu \in {\mathbb {R}}$ with $0\le \alpha <2$, $\mu <0$. Let $f \in L^2(0,T)$ and $u_0\in {\mathcal {H}}^{-s}$ for some $s > 0$. A weak solution of (6) is a function $u \in C^0([0,T];{\mathcal {H}}^{-s})$ such that for every $\tau \in (0,T]$ and for every $z^\tau \in {\mathcal {H}}^s$ we have

$$\begin{aligned} \left\langle u(\tau ), z^{\tau }\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}=\int _{0}^{\tau } f(t) {\mathcal {O}}_{1-\sqrt{-\mu }}\left( S(\tau -t) z_x^{\tau }\right) \textrm{d} t+\left\langle u_{0}, S(\tau )z^\tau \right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}. \end{aligned}$$

(61)

The last definition coincides with the one given in [6, Definition 2.8], provided that $\beta =1-\alpha$. Thus, the next result about the existence of weak solutions for the system (6) can be proved in the same way, under suitable conditions on the parameters $\alpha ,\beta ,\mu ,$ and s.

Proposition 6.3

Let $T>0$ and $\alpha ,\mu \in {\mathbb {R}}$ with $0\le \alpha <2$, $\mu <0$. Let $f \in L^2(0,T)$ and $u_0\in {\mathcal {H}}^{-s}$ such that $s>\sqrt{-\mu }/\kappa _\alpha$. Then formula (61) defines for each $\tau \in [0, T ]$ a unique element $u(\tau ) \in {\mathcal {H}}^{-s}$ that can be written as

$$\begin{aligned} u(\tau )=S(\tau ) u_{0}+{\widehat{B}}(\tau ) f, \quad \tau \in (0, T], \end{aligned}$$

where ${\widehat{B}}(\tau )$ is the strongly continuous family of bounded operators ${\widehat{B}}(\tau ): L^{2}(0,T) \rightarrow {\mathcal {H}}^{-s}$ given by

$$\begin{aligned} \left\langle {\widehat{B}}(\tau ) f, z^{\tau }\right\rangle _{{\mathcal {H}}^{-s}, {\mathcal {H}}^{s}}=\int _{0}^{\tau } f(t) {\mathcal {O}}_{1-\sqrt{-\mu }}\left( S(\tau -t) z_x^{\tau }\right) \textrm{d} t, \quad \text {for all }z^{\tau } \in {\mathcal {H}}^{s}. \end{aligned}$$

Furthermore, the unique weak solution u on [0, T] to (6) (in the sense of (20)) belongs to ${C}^{0}\left( [0, T]; {\mathcal {H}}^{-s}\right)$ and fulfills

$$\begin{aligned} \Vert u\Vert _{L^{\infty }\left( [0, T]; {\mathcal {H}}^{-s}\right) } \le C\left( \left\| u_{0}\right\| _{{\mathcal {H}}^{-s}}+\Vert f\Vert _{L^{2}(0, T)}\right) . \end{aligned}$$

Thus, we can follow exactly the same steps as in [6, Sects. 2,3], just take $\beta =1-\alpha$, to get Theorem 2.4.

7 Conclusion, applications, and open problems

As a consequence of Theorems 2.1, 2.2, and 2.4, we obtain the following qualitative result.

Remark 7.1

Using (A4), (A5), (A6), and (A7), we obtain that

$$\begin{aligned} \begin{array}{rcl} \displaystyle \frac{2^{\nu }\Gamma (\nu +1)\left| J_{\nu }'(j_{\nu ,1})\right| }{\left( j_{\nu ,1}\right) ^\nu } &{}\sim &{} \displaystyle \textrm{e}\sqrt{\frac{\nu }{\nu +1}}\left( \frac{\nu }{\nu +1}\right) ^{ \nu }\dfrac{j_{\nu ,1}}{\nu +1}\\[0.5cm] &{}\ge &{}\displaystyle \displaystyle \textrm{e}\sqrt{\frac{\nu }{\nu +1}}\left( \frac{\nu }{\nu +1}\right) ^{ \nu }\dfrac{\nu }{\nu +1} \rightarrow 1,\quad \text {as } \nu \rightarrow \infty , \end{array} \end{aligned}$$

thus, the control costs ${\mathcal {K}}(T,\alpha , \beta ,\mu )$ and $\widehat{{\mathcal {K}}}(T,\alpha ,\mu )$ blow up when $\alpha \rightarrow 2^{-}$ and/or $T\rightarrow 0^{+}$. Since $j_{v,1}<j_{v,2}$, clearly $\widetilde{{\mathcal {K}}}(T,\alpha , \beta ,\mu )\rightarrow +\infty$ when $\alpha \rightarrow 2^{-}$ and/or $T\rightarrow 0^{+}$.

When $T\rightarrow 0^{+}$, notice that the upper bounds of the control costs for the systems (1) and (5) explode much faster than the lower bounds.

In the critical case $\alpha + \beta = 1$, we did not have Hardy’s inequality at our disposal, so we could not proceed as in Sect. 3. But we found out that the theory of singular Sturm–Liouville operators is a powerful tool to solve degenerate/singular evolution equations, for instance, in [25], we show the null controllability of the system (1) with a weighted Robin control acting at the left endpoint.

Next, we would like to highlight that the issues addressed in this work are closely connected to various applications. These include the Crocco-type equation, derived from the analysis of the velocity field in laminar flow over a flat plate [13], the Black–Scholes model stemming from option pricing concerns [26], and the Budyko–Sellers model originating from climate-related problems [27], among others.

In [28], the authors analyzed the rapid stabilization of the following system

$$\begin{aligned} \left\{ \begin{array}{ll} u_{t} = \left( x^{\alpha }u_{x}\right) _{x},&{} \quad (x,t)\in (0,1)\times {\mathbb {R}}^{+}\\ u(0,t) = f(t), u(1,t) = 0,&{} \quad t\in {\mathbb {R}}^{+},\\ u(x,0) = u_{0}(x), &{} \quad x\in (0,1), \end{array} \right. \end{aligned}$$

(62)

where f denotes the control, the initial condition $u_{0}$ is chosen in $L^2(0,1)$ and $0\le \alpha <1$.

Using spectral tools associated to the operator ${\mathcal {A}} u:= \left( x^{\alpha }u_{x}\right) _{x}$, they obtain

Theorem 7.2

For any $\lambda >0$, there exists $C(\lambda ) > 0$ and a feedback law $f (t) = K(u(t))$, where $K\in L^{2}(0,1)'$, such that for any $u_{0}\in L^2(0,1)$, there exists a unique solution u of (62) that verifies, for any $t \ge 0$:

$$\begin{aligned} \left\| u(t,\cdot )\right\| _{L^2(0,1)}\le C(\lambda )\left\| u_{0}\right\| _{L^2(0,1)}\textrm{e}^{-\lambda t}, \end{aligned}$$

An open problem is to analyze the rapid stabilization of the system (1), corresponding to the strongly degenerate case $1\le \alpha < 2$.

Another open problem would be to improve the given lower bounds on the control cost for each system we have considered.

Regarding the Sturm–Liouville theory, we believe that Theorem 10.6.1 in [24] would be useful for solving and controlling more general degenerate/singular evolution equations.

Bessel functions

We introduce the Bessel function of the first kind $J_{\nu }$ as follows

$$\begin{aligned} J_{\nu }(x)=\sum _{m \ge 0} \frac{(-1)^{m}}{m ! \Gamma (m+\nu +1)}\left( \frac{x}{2}\right) ^{2 m+\nu }, \quad x \ge 0, \end{aligned}$$

(A1)

where $\Gamma (\cdot )$ is the Gamma function. In particular, for $\nu >-1$ and $0<x \le \sqrt{\nu +1}$, from (A1) we have (see [29, 9.1.7, p. 360])

$$\begin{aligned} J_{\nu }(x) \sim \frac{1}{\Gamma (\nu +1)}\left( \frac{x}{2}\right) ^{\nu } \quad \text{ as } \quad x \rightarrow 0^{+}. \end{aligned}$$

(A2)

A Bessel function $J_\nu$ of the first kind solves the differential equation

$$\begin{aligned} x^2y''+xy'+(x^2-\nu ^2)y=0. \end{aligned}$$

(A3)

Bessel functions of the first kind satisfy the recurrence formula ([29], 9.1.27):

$$\begin{aligned} x J_{\nu }^{\prime }(x)-\nu J_{\nu }(x)=-x J_{\nu +1}(x). \end{aligned}$$

(A4)

Recall the asymptotic behavior of the Bessel function $J_{\nu }$ for large x, see [30, Lem. 7.2, p. 129].

Lemma A.1

For any $\nu \in {\mathbb {R}}$

$$\begin{aligned} J_{\nu }(x)=\sqrt{\frac{2}{\pi x}}\left\{ \cos \left( x-\frac{\nu \pi }{2}-\frac{\pi }{4}\right) +{\mathcal {O}}\left( \frac{1}{x}\right) \right\} \quad \text{ as } \quad x \rightarrow \infty . \end{aligned}$$

For $\nu >0$ the Bessel function $J_{\nu }$ has an infinite number of real zeros $(j_{\nu ,k})$ all of which are simple, with the possible exception of $x=0$, see [29, 9.5.2, p. 370], satisfying

$$\begin{aligned} \nu<j_{\nu , 1}<j_{\nu , 2}<\ldots . \end{aligned}$$

(A5)

We recall Stirling’s formula, see [29, 6.1.39, p. 257],

$$\begin{aligned} \Gamma (\nu +1)\sim \sqrt{2\pi \nu }\left( \frac{\nu }{\textrm{e}}\right) ^{\nu }\quad \text {as } \nu \rightarrow \infty . \end{aligned}$$

(A6)

In [29, 9.3.1, p. 365], we have

$$\begin{aligned} J_{\nu }(x)\sim \frac{1}{\sqrt{2\pi \nu }}\left( \frac{\textrm{e}x}{2\nu }\right) ^{\nu }\quad \text {as } \nu \rightarrow \infty . \end{aligned}$$

(A7)

In [30, Proposition 7.8] we can find the next information about the location of the zeros of the Bessel functions $J_{\nu }$:

Lemma A.2

Let $\nu \ge 0$.

• 1. The difference sequence $\left( j_{\nu , k+1}-j_{\nu , k}\right) _{k}$ converges to $\pi$ as $k \rightarrow \infty$.

• 2. The sequence $\left( j_{\nu , k+1}-j_{\nu , k}\right) _{k}$ is strictly decreasing if $|\nu |>\frac{1}{2}$, strictly increasing if $|\nu |<\frac{1}{2}$, and constant if $|\nu |=\frac{1}{2}$.

For $\nu \ge 0$ fixed, we consider the next asymptotic expansion of the zeros of the Bessel function $J_{\nu }$, see [23, Sect. 15.53],

$$\begin{aligned} j_{\nu , k}=\left( k+\frac{\nu }{2}-\frac{1}{4}\right) \pi -\frac{4 \nu ^{2}-1}{8\left( k+\frac{\nu }{2}-\frac{1}{4}\right) \pi }+O\left( \frac{1}{k^{3}}\right) , \quad \text{ as } k \rightarrow \infty . \end{aligned}$$

(A8)

In particular, we have

$$\begin{aligned} \begin{aligned}&j_{\nu , k} \ge \left( k-\frac{1}{4}\right) \pi \quad \text{ for } \nu \in \left[ 0, 1/2\right] , \\&j_{\nu , k} \ge \left( k-\frac{1}{8}\right) \pi \quad \text{ for } \nu \in \left[ 1/2,\infty \right) . \end{aligned} \end{aligned}$$

(A9)

Lemma A.3

For any $\nu \ge 0$ and any $k\ge 1$ we have

$$\begin{aligned} \sqrt{j_{\nu , k}}\left| J_{\nu }^{\prime }\left( j_{\nu , k}\right) \right| =\sqrt{\frac{2}{\pi }}+O\left( \frac{1}{j_{\nu , k}}\right) \quad \text {as}\quad k \rightarrow \infty . \end{aligned}$$

The proof of this result follows by using (A2) and the recurrence formula (A4).

Lemma A.4

Let $\gamma =\gamma (\alpha ,\beta ,\mu )$ and $\nu =\nu (\alpha ,\beta ,\mu )$ given in (3) and (4) respectively, then the following limit is finite for all $k\ge 1$, and

$$\begin{aligned} {\mathcal {O}}_{\alpha +\beta +\gamma }(\Phi _k)= \frac{(2\kappa _\alpha )^{1/2}(j_{\nu ,k})^{\nu }}{2^{\nu }\Gamma (\nu +1)|J'_{\nu }(j_{\nu , k})|}, \quad k\ge 1. \end{aligned}$$

(A10)

Proof

This result follows from (A2). $\square$

We recall the following representation theorem, see [31, p. 56].

Theorem A.5

Let g(z) be an entire function of exponential type and assume that

$$\begin{aligned} \int _{-\infty }^{\infty } \frac{\log ^{+}|g(x)|}{1+x^{2}} \textrm{d}x<\infty . \end{aligned}$$

Let $\left\{ b_{\ell }\right\} _{\ell \ge 1}$ be the set of zeros of g(z) in the upper half plane $\Im (z)>0$ (each zero being repeated as many times as its multiplicity). Then,

$$\begin{aligned} \log |g(z)|=A \Im (z)+\sum _{\ell =1}^{\infty } \log \left| \frac{z-b_{\ell }}{z-{\bar{b}}_{\ell }}\right| +\frac{\Im (z)}{\pi } \int _{-\infty }^{\infty } \frac{\log |g(s)|}{|s-z|^{2}} \textrm{d}s,\quad \Im (z)>0, \end{aligned}$$

where

$$\begin{aligned} A=\limsup _{y \rightarrow \infty } \frac{\log |g(i y)|}{y}. \end{aligned}$$