Tychonoff Solutions of the Time-Fractional Heat Equation

Abstract

In the literature, one can find several applications of the time-fractional heat equation, particularly in the context of time-changed stochastic processes. Stochastic representation results for such an equation can be used to provide a Monte Carlo simulation method, upon proving that the solution is actually unique. In the classical case, however, this is not true if we do not consider any additional assumption, showing, thus, that the Monte Carlo simulation method identifies only a particular solution. In this paper, we consider the problem of the uniqueness of the solutions of the time-fractional heat equation with initial data. Precisely, under suitable assumptions about the regularity of the initial datum, we prove that such an equation admits an infinity of classical solutions. The proof mimics the construction of the Tychonoff solutions of the classical heat equation. As a consequence, one has to add some addtional conditions to the time-fractional Cauchy problem to ensure the uniqueness of the solution.

1. Introduction

The heat equation is the prototype of a parabolic partial differential equation. It was introduced by Fourier in 1822 in his work on heat flows, called Théorie analytique de la chaleur (see the reprint [1]), and it is used to describe how heat diffuses in a certain region of space. Such an equation is not only important due to its physical interpretation, but also from the purely mathematical point of view: the heat equation can be generalized to much wider geometrical settings, such as manifolds [2] and metric spaces [3]. A microscopic interpretation of the heat equation has been given, for instance, by Einstein in 1905 [4] (see also [5,6]), but the germs of such a connection were already present in [7]. For an almost full overview of the history of the Brownian motion, see [8]. A full formalization of the connection between the heat equation and the Brownian motion is given by both the backward and forward Kolmogorov equations. Here, let us focus on the backward equation (a simple reference for it is [9]).

Theorem 1.

Let $(\Omega ,\mathcal{F},{\left({\mathcal{F}}_t\right)}_{t\ge 0},\mathbb{P})$ be a complete filtered probability space supporting a d-dimensional standard Brownian motion $W=\left\{W\right(t),t\ge 0\}$. Let $u_0\in C_c^{\infty }\left(\mathbb{R}\right)$, i.e., an infinitely continuously differentiable function with compact support. Then, the function

$$u(t,x)=\mathbb{E}\left[u_0\left(W\left(t\right)\right)\right|W\left(0\right)=x]$$

solves the Cauchy problem:

$$\left\{\begin{array}{cc}{\displaystyle \frac{\partial u}{\partial t}(t,x)=\frac{1}{2}\Delta u(t,x)}\hfill & t>0,x\in {\mathbb{R}}^d\hfill \\ u(0,x)=u_0\left(x\right)\hfill & x\in {\mathbb{R}}^d.\hfill \end{array}\right.$$

Let us stress that, if we remove the constant $\frac{1}{2}$, we have to use $\sqrt{2}W\left(t\right)$ (or, equivalently, $W\left(2t\right)$) in place of $W\left(t\right)$. Such a connection can be exploited to obtain several mathematical properties of the solution of both the heat equation and the Laplace/Poisson equation (see, for instance, [10,11,12]). Moreover, it can be also used to obtain some numerical algorithms to solve the heat equation via Monte Carlo simulation methods (see [13,14,15]). A natural question arising from these applications is the following: is the solution of the heat equation given in Theorem 1 unique? Unluckily, the answer is negative.

Indeed, despite that the heat equation on bounded domains with Dirichlet or Neumann conditions admits a unique solution, this is not true for unbounded domains, even in the one-dimensional case, as the following theorem states.

Theorem 2.

For $u_0\in C_c^{\infty }\left(\mathbb{R}\right)$, the Cauchy problem

$$\left\{\begin{array}{cc}{\displaystyle \frac{\partial u}{\partial t}(t,x)=\frac{{\partial }^{2}u}{\partial x^2}(t,x)}\hfill & t>0,x\in \mathbb{R}\hfill \\ u(0,x)=u_0\left(x\right)\hfill & x\in \mathbb{R}\hfill \end{array}\right.$$

(1)

admits an infinity of solutions.

Here, by solution we mean the function $u:{\mathbb{R}}_0^{+}\times \mathbb{R}$, where ${\mathbb{R}}_0^{+}:=[0,+\infty )$ belongs to $C^0({\mathbb{R}}_0^{+}\times \mathbb{R})$, i.e., the space of continuous functions on ${\mathbb{R}}_0^{+}\times \mathbb{R}$, such that ${\displaystyle \frac{\partial u}{\partial t},\frac{\partial u}{\partial x},\frac{{\partial }^{2}u}{\partial x^2}}$ are well defined in $C^0({\mathbb{R}}^{+}\times \mathbb{R})$, where ${\mathbb{R}}^{+}:=(0,+\infty )$, and u solves Equation (1) pointwise. Clearly, thanks to Theorem 1, to prove this result one only needs to construct a family of solutions, in the case $f\equiv 0$, that are called Tychonoff solutions of the heat equation. This is accomplished by using a particular class of functions called the Holmgren class (see [16]).

Adding some conditions to the behavior of the solution leads to its uniqueness, as stated in the following theorem (see [16]).

Theorem 3.

For $u_0\in C_c^{\infty }\left(\mathbb{R}\right)$, the Cauchy problem (Equation (1)) admits a unique solution, such that

$$\left|u(t,x)\right|\le C_1{e}^{C_2{\left|x\right|}^2},x\in \mathbb{R},t\ge 0,$$

for some $C_1,C_2>0$.

Such a condition is known as Tychonoff condition. Clearly, the solutions provided by Theorem 1 satisfy them. One usually refers to solutions satisfying the Tychonoff condition as physical solutions of the heat equation. It can be also shown, by considering the heat equation as a linear abstract Cauchy problem on $H^{-1}\left(\mathbb{R}\right)$ (the dual space of the Sobolev space $H^1\left(\mathbb{R}\right)$), that there exists a unique solution $u\in C^1({\mathbb{R}}_0^{+};H^{-1}\left(\mathbb{R}\right))$, i.e., a continuously differentiable function with $H^{-1}\left(\mathbb{R}\right)$ values. The latter is actually the solution one finds by means of semigroup theory, and then it is, again, the one provided by Theorem 1. One can also see [17] for further details.

In recent years, fractional calculus has received a lot of attention thanks to its numerous applications. To cite some of them, Caputo derivatives have been considered in [18] to study wave dissipation in geophysics, but after that, they have been adopted in several contexts such as image processing [19], signal processing [20], viscoelasticity [21,22], biology [23,24], and so on. As in the classical case, the link between fractional differential equations and a suitable class of stochastic processes has been investigated. This is the case, for instance, of the fractional Poisson process, whose state probabilities are shown to be solutions of a fractional-order system of infinite ODEs (see [25,26]). These results have been then extended to a wider family of continuous-time semi-Markov chains (see, for instance, [27,28,29]) as well as time-changed continuous processes (see, for instance, [30,31]). Such a link has been exploited to develop suitable models for applications in, for instance, queueing theory [32,33,34,35], epidemiology [36,37], or anomalous diffusion-aggregation and particle trapping [38]. For an almost full discussion on the link between anomalous diffusion and fractional calculus, we refer to [39].

Among this wide range of applications, let us focus on the time-fractional heat conduction (see [40]). This theory, introduced in [41] in the case of finite wave speed, and then extended, for instance, in [42] in the anomalous diffusive setting, describes the heat conduction in materials with memory by means of a time-nonlocal heat flux. Such a model leads to the formulation, with a suitable choice of the memory kernel, of the time-fractional heat equation

$$\frac{{\partial }^{\alpha }u}{\partial t^{\alpha }}(t,x)=\frac{{\partial }^{2}u}{\partial x^2}(t,x),t>0,x\in \mathbb{R},$$

where $\frac{{\partial }^{\alpha }·}{\partial t^{\alpha }}$ is the Caputo derivative of order $\alpha \in (0,1)$. Together with this equation, one has to consider different conditions, such as Dirichlet conditions or Neumann conditions or also Stefan conditions, that arise in models of latent heat accumulation [43]. To cite some other recent papers on fractional heat transfer, we refer to [44], in which the Boltzmann theory has been studied under the fractional heat flux assumption, and [45], in which anomalous heat diffusion is obtained from a fractional Fokker–Planck equation. We also refer to [46], which discusses heat transfer in low dimesions from the macroscopic to the micro/nanoscopic scale. In the applications, one usually considers a bounded domain to describe the heat transfer in a certain object. However, in probability theory, one is also interested in the solution of the heat equation in the whole space, as its fundamental solution provides the transition density of the Brownian motion. At the same time, if the object in which one has to study heat conduction is sufficiently big, with respect to the unit of measure, then one can approximate it as an unbounded domain. For instance, if one considers a sufficiently long rod, it can be approximated as an infinite rod; thus, the heat equation has to be solved in the whole space $\mathbb{R}$ (see, for instance, [47]). As for the classical heat equation, one can provide a stochastic representation result of a certain solution of the initial value problem.

Theorem 4.

Let $(\Omega ,\mathcal{F},{\left({\mathcal{F}}_t\right)}_{t\ge 0},\mathbb{P})$ be a complete filtered probability space supporting a 1-dimensional standard Brownian motion $W=\left\{W\right(t),t\ge 0\}$ and an independent inverse α-stable subordinator $L_{\alpha }=\{L_{\alpha }\left(t\right),t\ge 0\}$. Define $W_{\alpha }\left(t\right)=\sqrt{2}W\left(L_{\alpha }\left(t\right)\right)$ for any $t\ge 0$, and let $u_0\in C_c^{\infty }\left(\mathbb{R}\right)$, i.e., an infinitely continuously differentiable function with compact support. Then, the function

$$u(t,x)=\mathbb{E}\left[u_0\left(W_{\alpha }\left(t\right)\right)\right|W\left(0\right)=x]$$

is a mild solution of the Cauchy problem

$$\left\{\begin{array}{cc}{\displaystyle \frac{{\partial }^{\alpha }u}{\partial t^{\alpha }}(t,x)=\frac{{\partial }^{2}u}{\partial x^2}(t,x)}\hfill & t>0,x\in {\mathbb{R}}^d\hfill \\ u(0,x)=u_0\left(x\right)\hfill & x\in \mathbb{R},\hfill \end{array}\right.$$

(2)

i.e., the Laplace transform $\overline{u}$ of u, with respect to the variable t, solves

$$s^{\alpha }\overline{u}(s,x)-s^{\alpha -1}{u}_0\left(x\right)=\frac{{\partial }^2\overline{u}}{\partial x^2}(s,x),s>0,x\in \mathbb{R}.$$

This is a specific case of ([48], Theorem 3.1). The process $W_{\alpha }$ involved in the previous Theorem is usually called delayed Brownian motion (see [49]). Here, we are interested in more regular solutions. Precisely, we will consider pointwise solutions $u:{\mathbb{R}}_0^{+}\times \mathbb{R}\to \mathbb{R}$ of the equation belonging to $C^0({\mathbb{R}}_0^{+}\times \mathbb{R})$, such that ${\displaystyle \frac{{\partial }^{\alpha }u}{\partial t^{\alpha }},\frac{\partial u}{\partial x},\frac{{\partial }^{2}u}{\partial x^2}\in C^0({\mathbb{R}}^{+}\times \mathbb{R})}$, which we call classical solutions. One can show that the solutions provided by Theorem 4 are actually classical solutions of the time-fractional heat Equation (2)) by an argument such as that found in ([50], Theorem 3.11), as we will do in Theorem 7. Thus, if we prove that such a solution is unique, Theorem 4 can be used to provide a Monte Carlo method to obtain a numerical solution of the time-fractional heat Equation (2). A direct consequence of the weak maximum principle proved in [51] (see also [50] for the unbounded domain case) gives us the following result.

Theorem 5.

The time-fractional heat Equation (2) admits a unique classical solution satisfying ${lim}_{x\to \pm \infty }u(t,x)=0$ locally uniformly, with respect to $t\in {\mathbb{R}}_0^{+}$.

Moreover, if we consider the time-fractional heat equation as an abstract Cauchy problem on $H^{-1}\left(\mathbb{R}\right)$, we obtain the uniqueness of the mild solution $u\in C({\mathbb{R}}_0^{+};H^{-1}\left(\mathbb{R}\right))$ by ([52], Theorem $3.3$).

In this paper, we want to prove the following non-uniqueness result, which is analogous to Theorem 2.

Theorem 6.

For any $u_0\in C_c^{\infty }\left(\mathbb{R}\right)$ and $\alpha \in (0,1)$, the Cauchy problem (Equation (2)) admits an infinity of classical solutions.

Let us stress that the uniqueness theory for the heat equation, both classical and fractional, is quite different depending on whether the domain is bounded or unbounded. There are several strategies to prove that the solution of the heat equation with Dirichlet conditions on a bounded domain is unique (see [17]). Up to now, only one of these strategies has been extended to the fractional case. Indeed, in [51], a weak maximum principle for fractional diffusion–advection equations is proved: this clearly implies the uniqueness of classical solutions of the fractional heat equation with Dirichlet conditions on bounded domains. With this in mind, it is clear that Theorem 6 does not extend to the bounded domain case. However, with a similar strategy, one can prove the same result for the fractional heat equation on the half axis $(-\infty ,a)$ (or, equivalently, $(a,+\infty )$) with the Dirichlet condition on the boundary. Let us also remark that Theorem 4 (and Theorem 7, which we will state in the following) allows one to apply Monte Carlo methods to describe a solution of the time-fractional heat equation. In general, when one adopts a numerical method to solve a PDE, it is important to identify the solution that is being approximated. This is simple to understand if the solution is unique. Otherwise, the identification is almost mandatory, since we could apply a numerical scheme without knowing what solution we are capturing. Moreover, to use such equations in the theory of semi-Markov processes and anomalous diffusions, for instance, to identify conditional expectations of functions of semi-Markov processes by means of nonlocal equations, the identification of the solution among a multiplicity of them is indispensable. The main motivation of the paper is to underline, via Theorem 6, the existence of multiple (precisely, infinite) solutions for the time-fractional heat equation with initial data, thus remarking the need for the aforementioned identification. In particular, as a direct consequence of Theorem 7, despite the existence of infinite solutions, we are able to identify the conditional expected value of a $C^{\infty }$ function with compact support applied to a delayed Brownian motion as the unique classical solution of the time-fractional heat equation satisfying the conditions dictated by Theorem 5 (see Remark 3). Subsequently, it is clear that a Monte Carlo method based on Theorem 7 is able to capture such a solution among all of them.

To prove our main result, we will mimic the usual arguments adopted to prove Theorem 2, adapting them to the time-fractional case. The form of the Tychonoff-type solutions of the time-fractional heat equation suggests that the uniqueness of the solutions should be guaranteed under the same condition as in Theorem 3. We will investigate Tychonoff-type uniqueness results in a future work.

Precisely, the paper is structured as follows. In Section 2, we introduce the fractional integrals and the Caputo fractional derivatives and we provide some preliminary properties. In Section 3, we define a fractional analogue of the Holmgren class and we prove that a certain family of functions belongs to such a class. Finally, Section 4 is devoted to the proof of Theorem 6.

2. Fractional Integrals and Derivatives

Let us introduce some notation. We say that a function $f:{\mathbb{R}}_0^{+}\to \mathbb{R}$ belongs to $C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$ if and only if f is continuous on ${\mathbb{R}}_0^{+}$ with $f\left(0\right)=0$, infinitely continuously differentiable on ${\mathbb{R}}^{+}:=(0,+\infty )$, and, for any $n\in \mathbb{N}$, where $\mathbb{N}$ is the set of positive integers, it holds that ${lim}_{t\to 0^{+}}{f}^{\left(n\right)}\left(t\right)=0$. Thanks to the latter property, if $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$, we can set $f^{\left(n\right)}\left(0\right)=0$ for any $n\in \mathbb{N}$, so that each derivative $f^{\left(n\right)}$ is continuous on ${\mathbb{R}}_0^{+}$ and belongs to $C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$.

Following the lines of ([53], Chapter I), we now define the fractional integral operator.

Definition 1.

Let $f\in L_{\mathrm{loc}}^1\left({\mathbb{R}}_0^{+}\right)$ and $\alpha >0$. The fractional integral of order α of f is defined as

$${\mathcal{I}}^{\alpha }f\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_0^t{(t-\tau )}^{\alpha -1}f\left(\tau \right)d\tau ,t>0.$$

One can easily verify that if $f:{\mathbb{R}}_0^{+}\to \mathbb{R}$ is continuous, then ${\mathcal{I}}^{\alpha }f:{\mathbb{R}}^{+}\to \mathbb{R}$ is also continuous. Now, let us show two preliminary lemmas. Let us stress that both the results clearly hold for less regular functions, but we will apply them on $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$.

Lemma 1.

Let $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$ and $\alpha >0$. Then,

$$\underset{t\to 0^{+}}{lim}{\mathcal{I}}^{\alpha }f\left(t\right)=0$$

Proof. 

Let us consider any $\epsilon >0$. Then, there exists $\delta >0$, such that $\left|f\right(\tau \left)\right|<\Gamma (\alpha +1)\epsilon$ for any $\tau \in [0,\delta ]$. Without loss of generality, we can suppose $\delta <1$. Then, for any $t<\delta$, we have

$$|{\mathcal{I}}^{\alpha }f\left(t\right)|\le \frac{1}{\Gamma \left(\alpha \right)}{\int }_0^t{(t-\tau )}^{\alpha -1}\left|f\left(\tau \right)\right|d\tau \le \epsilon t^{\alpha }<\epsilon ,$$

concluding the proof. □

Lemma 2.

Let $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$ and $\alpha >0$. Then,

$$\frac{d}{dt}{\mathcal{I}}^{\alpha }f\left(t\right)={\mathcal{I}}^{\alpha }{f}^{\prime }\left(t\right),t>0$$

Proof. 

Fix $t>0$, and let us consider, for $h>0$,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \frac{{\mathcal{I}}^{\alpha }f(t+h)-{\mathcal{I}}^{\alpha }f\left(t\right)}{h}& =\frac{1}{\Gamma \left(\alpha \right)}\left({\int }_0^{t+h}{\tau }^{\alpha -1}f(t+h-\tau )d\tau -{\int }_0^t{\tau }^{\alpha -1}f(t-\tau )d\tau \right)\hfill \\ \hfill & =\frac{1}{\Gamma \left(\alpha \right)}\left(\frac{1}{h}{\int }_t^{t+h}{\tau }^{\alpha -1}f(t+h-\tau )d\tau +{\int }_0^t{\tau }^{\alpha -1}\frac{f(t+h-\tau )-f(t-\tau )}{h}d\tau \right)\hfill \\ \hfill & =\frac{1}{\Gamma \left(\alpha \right)}(I_1(t,h)+I_2(t,h)).\hfill \end{array}\end{array}$$

(3)

Let us first consdier $I_1(t,h)$. We have

$$I_1(t,h)\le \frac{\left({max}_{s\in [0,h]}\right|f\left(s\right)\left|\right)}{h}{\int }_t^{t+h}{\tau }^{\alpha -1}d\tau .$$

If $\alpha \le 1$, it holds that

$$I_1(t,h)\le (\underset{s\in [0,h]}{max}\left|f\left(s\right)\right|)t^{\alpha -1}.$$

On the other hand, if $\alpha >1$, we can suppose $h<1$ to conclude that

$$I_1(t,h)\le (\underset{s\in [0,h]}{max}{\left|f\left(s\right)\right|)(t+1)}^{\alpha -1}.$$

Thus, we conclude that

$$I_1(t,h)\le max\{t^{\alpha -1},{(t+1)}^{\alpha -1}\}(\underset{s\in [0,h]}{max}\left|f\left(s\right)\right|).$$

Being $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$, it is also uniformly continuous in $[0,1]$, and then ${max}_{s\in [0,h]}\left|f\left(s\right)\right|\to 0$ as $h\to 0$. Thus, we achieve

$$\underset{h\to 0^{+}}{lim}{I}_1(t,h)=0.$$

(4)

Concerning $I_2(t,h)$, let us observe that, as $\tau \in [0,t]$,

$${\tau }^{\alpha -1}\frac{\left|f\right(t+h-\tau )-f(t-\tau \left)\right|}{h}\le {\tau }^{\alpha -1}\underset{s\in [0,t+1]}{max}\left|f^{\prime }\left(s\right)\right|,$$

which belongs to $L^1(0,t)$. Thus, we can use the dominated convergence theorem to conclude that

$$\underset{h\to 0^{+}}{lim}{I}_2(t,h)={\int }_0^t{\tau }^{\alpha -1}{f}^{\prime }\left(\tau \right)d\tau .$$

(5)

Combining Equations (4) and (5) in (3), we obtain

$$\underset{h\to 0^{+}}{lim}\frac{{\mathcal{I}}^{\alpha }f(t+h)-{\mathcal{I}}^{\alpha }f\left(t\right)}{h}={\mathcal{I}}^{\alpha }{f}^{\prime }\left(t\right).$$

A similar argument also holds for $h<0$, concluding the proof. □

Remark 1.

Clearly, one can prove by induction that for $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$, $\alpha >0$, and $n\in \mathbb{N}$, it holds that

$$\frac{d^n}{d{t}^n}{\mathcal{I}}^{\alpha }f\left(t\right)={\mathcal{I}}^{\alpha }{f}^{\left(n\right)}\left(t\right),t>0.$$

(6)

Combining the two previous Lemmas, we obtain the following simple regularity result.

Lemma 3.

Let $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$ and $\alpha >0$. Then, if we set ${\mathcal{I}}^{\alpha }f\left(0\right)=0$, we have ${\mathcal{I}}^{\alpha }f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$.

Proof. 

Let us first observe that, by Lemma 1, ${\mathcal{I}}^{\alpha }f$ is continuous on ${\mathbb{R}}_0^{+}$ with ${\mathcal{I}}^{\alpha }f\left(0\right)=0$. Lemma 2 and Remark 1 tell us that ${\mathcal{I}}^{\alpha }f$ is infinitely differentiable. Finally, since if $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$ also $f^{\left(n\right)}\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$ for any $n\in \mathbb{N}$, Lemma 1, applied to Equation (6), guarantees that ${lim}_{t\to 0^{+}}\frac{d^n}{d{t}^n}{\mathcal{I}}^{\alpha }f\left(t\right)=0$, concluding the proof. □

Now, let us introduce the Caputo fractional derivatives. Since we will work with $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$, we define them directly by using ([54], Theorem 2.1).

Definition 2.

For any function $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$ and any $\alpha >0$, we define the Caputo fractional derivative of f of order α as, for $t>0$,

$$\frac{d^{\alpha }}{d{t}^{\alpha }}f\left(t\right)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{\Gamma (\lfloor \alpha \rfloor +1-\alpha )}{\int }_0^t{(t-\tau )}^{\lfloor \alpha \rfloor -\alpha }{f}^{(\lfloor \alpha \rfloor )+1}\left(\tau \right)d\tau ,}\hfill & \alpha \notin \mathbb{N}\hfill \\ {\displaystyle f^{\left(\alpha \right)}\left(t\right)}\hfill & \alpha \in \mathbb{N},\hfill \end{array}\right.$$

where $\lfloor \alpha \rfloor =max\{n\in {\mathbb{N}}_0:\alpha \ge n\}$ is the integer part of α and ${\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\}$.

It is clear by definition that if $\alpha \notin \mathbb{N}$, then

$$\frac{d^{\alpha }}{d{t}^{\alpha }}f={\mathcal{I}}^{\lfloor \alpha \rfloor +1-\alpha }{f}^{(\lfloor \alpha \rfloor +1)}.$$

Let us recall that the fractional integral admits a semigroup property with respect to the order of integration, i.e., for any $\alpha ,\beta >0$ (see ([53], Equation $\left(2.21\right)$)),

$${\mathcal{I}}^{\alpha }{\mathcal{I}}^{\beta }={\mathcal{I}}^{\alpha +\beta }.$$

This is not true in general for the Caputo derivative, as only a partial semigroup property holds (see [55]). Thus, in general, we have to define the iterated Caputo derivative by recursion.

Definition 3.

Given a function $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$, we define, for any $\alpha >0$, the sequence of iterated Caputo derivatives of f as

$$\left\{\begin{array}{c}{f}^{(\alpha ,0)}=f\hfill \\ f^{(\alpha ,n)}=\frac{d^{\alpha }}{d{t}^{\alpha }}{f}^{(\alpha ,n-1)}\hfill & n\in \mathbb{N}.\hfill \end{array}\right.$$

If $\alpha \in \mathbb{N}$, then we obtain $f^{(\alpha ,n)}=f^{\left(\alpha n\right)}$. Moreover, if $n=1$, we denote $f^{(\alpha ,1)}=f^{\left(\alpha \right)}$. This does not lead to any ambiguity with the previous notation, as, if $\alpha ,n\in \mathbb{N}$, then $f^{(\alpha ,n)}=f^{\left(\alpha n\right)}=f^{(\alpha n,1)}$ is a standard derivative.

The previous definition is necessary, in general, for functions that are infinitely continuously differentiable. However, the additional condition, that ${lim}_{t\to 0^{+}}{f}^{\left(n\right)}\left(t\right)=0$, tells us that we can always reduce to the case of a single Caputo derivative, thanks to an improved semigroup property—which will come handy in the proof of the main result.

Proposition 1.

Let $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$, $\alpha >0$, and $n\in \mathbb{N}$. Then, $f^{(\alpha ,n)}=f^{\left(\alpha n\right)}$.

Proof. 

Fix $\alpha >0$ with $\alpha \notin \mathbb{N}$, and let us prove that $f^{(\alpha ,n)}=f^{\left(\alpha n\right)}$ by induction on n. This is clear by the definition of $f^{\left(\alpha \right)}$ if $n=1$. Thus, let us suppose that $f^{(\alpha ,n)}=f^{\left(\alpha n\right)}$, and let us prove the statement for $n+1$. First, suppose that $\alpha n\in \mathbb{N}$. Then,

$$f^{(\alpha ,n+1)}=\frac{d^{\alpha }}{d{t}^{\alpha }}{f}^{\left(\alpha n\right)}={\mathcal{I}}^{\lfloor \alpha \rfloor +1-\alpha }{f}^{(\lfloor \alpha \rfloor +1+\alpha n)}.$$

(7)

Now, observe that $\lfloor \alpha \rfloor +\alpha n=\lfloor \alpha (n+1)\rfloor$, and then that

$$\lfloor \alpha \rfloor +1-\alpha =\lfloor \alpha (n+1)\rfloor +1-\alpha (n+1),$$

so that Equation (7) becomes

$$f^{(\alpha ,n+1)}=\lfloor \alpha (n+1)\rfloor +1-\alpha (n+1)={\mathcal{I}}^{\lfloor \alpha (n+1)\rfloor +1-\alpha (n+1)}{f}^{(\lfloor \alpha (n+1)\rfloor +1)}=f^{\left(\alpha \right(n+1\left)\right)}.$$

Now, let us suppose that $\alpha n\notin N$. We have

$$f^{(\alpha ,n+1)}=\frac{d^{\alpha }}{d{t}^{\alpha }}{f}^{\left(\alpha n\right)}={\mathcal{I}}^{\lfloor \alpha \rfloor +1-\alpha }\frac{d^{\lfloor \alpha \rfloor +1}}{d{t}^{\lfloor \alpha \rfloor +1}}{f}^{\left(\alpha n\right)}={\mathcal{I}}^{\lfloor \alpha \rfloor +1-\alpha }\frac{d^{\lfloor \alpha \rfloor +1}}{d{t}^{\lfloor \alpha \rfloor +1}}{\mathcal{I}}^{\lfloor \alpha n\rfloor +1-\alpha n}{f}^{(\lfloor \alpha n\rfloor +1)}.$$

Since $f^{(\lfloor \alpha n\rfloor +1)}\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$, we can use Lemma 2, Remark 1, and the semigroup property of the fractional integral to rewrite

$$f^{(\alpha ,n+1)}={\mathcal{I}}^{\lfloor \alpha \rfloor +\lfloor \alpha n\rfloor +2-\alpha (n+1)}{f}^{(\lfloor \alpha n\rfloor +\lfloor \alpha \rfloor +2)}.$$

(8)

First of all, since $\lfloor \alpha \rfloor \le \alpha$ and $\lfloor \alpha n\rfloor \le \alpha n$, it holds that $\lfloor \alpha \rfloor +\lfloor \alpha n\rfloor \le \alpha (n+1)$, and then, by definition,

$$\lfloor \alpha \rfloor +\lfloor \alpha n\rfloor \le \lfloor \alpha (n+1)\rfloor$$

(9)

Now, let us define, for any $\beta >0$, $d\left(\beta \right)=\beta -\lfloor \beta \rfloor$, and let us observe that for any ${\beta }_1,{\beta }_2>0$, it holds that

$$0\le d\left({\beta }_1\right)+d\left({\beta }_2\right)<2.$$

(10)

Recalling that

$$\lfloor \alpha (n+1)\rfloor +d\left(\alpha \right(n+1\left)\right)=\alpha (n+1)=\alpha n+\alpha =\lfloor \alpha \rfloor +d\left(\alpha \right)+\lfloor \alpha n\rfloor +d\left(\alpha n\right),$$

we obtain

$${\mathbb{N}}_0\ni \lfloor \alpha (n+1)\rfloor -\lfloor \alpha \rfloor -\lfloor \alpha n\rfloor =d\left(\alpha \right)+d\left(\alpha n\right)-d\left(\alpha (n+1)\right).$$

Hence, by Equation (9), we know that $d\left(\alpha \right)+d\left(\alpha n\right)-d\left(\alpha \right(n+1\left)\right)\ge 0$, while Equation (10) tells us that $d\left(\alpha \right)+d\left(\alpha n\right)-d\left(\alpha \right(n+1\left)\right)<2$. Thus, we know that either $d\left(\alpha \right)+d\left(\alpha n\right)-d\left(\alpha \right(n+1\left)\right)=0$ or $d\left(\alpha \right)+d\left(\alpha n\right)-d\left(\alpha \right(n+1\left)\right)=1$. Let us distinguish these two cases.

If $d\left(\alpha \right)+d\left(\alpha n\right)-d\left(\alpha \right(n+1\left)\right)=0$, it holds that $\lfloor \alpha (n+1)\rfloor =\lfloor \alpha \rfloor +\lfloor \alpha n\rfloor$; then, from Equation (8), we obtain

$$\begin{array}{cc}\hfill f^{(\alpha ,n+1)}& ={\mathcal{I}}^{\lfloor \alpha (n+1)\rfloor +2-\alpha (n+1)}{f}^{(\lfloor \alpha (n+1)\rfloor +2)}\hfill \\ \hfill & ={\mathcal{I}}^{\lfloor \alpha (n+1)\rfloor +1-\alpha (n+1)}{\mathcal{I}}^1\frac{d}{dt}\left(f^{(\lfloor \alpha (n+1)\rfloor +1)}\right).\hfill \end{array}$$

Since $f^{(\lfloor \alpha (n+1)\rfloor +1)}\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$, by the fundamental theorem of calculus, we have ${\mathcal{I}}^1\frac{d}{dt}\left(f^{(\lfloor \alpha (n+1)\rfloor +1)}\right)=f^{(\lfloor \alpha (n+1)\rfloor +1)}$, and then

$$\begin{array}{c}\hfill f^{(\alpha ,n+1)}={\mathcal{I}}^{\lfloor \alpha (n+1)\rfloor +1-\alpha (n+1)}{f}^{(\lfloor \alpha (n+1)\rfloor +1)}=f^{\left(\alpha \right(n+1\left)\right)}.\end{array}$$

If $d\left(\alpha \right)+d\left(\alpha n\right)-d\left(\alpha \right(n+1\left)\right)=1$, then it holds that $\lfloor \alpha (n+1)\rfloor =\lfloor \alpha \rfloor +\lfloor \alpha n\rfloor +1$; then, from Equation (8), we obtain

$$\begin{array}{cc}\hfill f^{(\alpha ,n+1)}& ={\mathcal{I}}^{\lfloor \alpha (n+1)\rfloor +3-\alpha (n+1)}{f}^{(\lfloor \alpha (n+1)\rfloor +3)}\hfill \\ \hfill & ={\mathcal{I}}^{\lfloor \alpha (n+1)\rfloor +2-\alpha (n+1)}{\mathcal{I}}^1\frac{d}{dt}\left(f^{(\lfloor \alpha (n+1)\rfloor +2)}\right).\hfill \end{array}$$

Again, since $f^{(\lfloor \alpha (n+1)\rfloor +2)}\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$, by the fundamental theorem of calculus, we have ${\mathcal{I}}^1\frac{d}{dt}\left(f^{(\lfloor \alpha (n+1)\rfloor +2)}\right)=f^{(\lfloor \alpha (n+1)\rfloor +2)}$, and so

$$\begin{array}{cc}\hfill f^{(\alpha ,n+1)}& ={\mathcal{I}}^{\lfloor \alpha (n+1)\rfloor +2-\alpha (n+1)}{f}^{(\lfloor \alpha (n+1)\rfloor +2)}\hfill \\ \hfill & ={\mathcal{I}}^{\lfloor \alpha (n+1)\rfloor +1-\alpha (n+1)}{\mathcal{I}}^1\frac{d}{dt}\left(f^{(\lfloor \alpha (n+1)\rfloor +1)}\right)\hfill \\ \hfill & ={\mathcal{I}}^{\lfloor \alpha (n+1)\rfloor +1-\alpha (n+1)}{f}^{(\lfloor \alpha (n+1)\rfloor +1)}=f^{\left(\alpha \right(n+1\left)\right)}.\hfill \end{array}$$

This concludes the proof. □

Remark 2.

Let $f\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$. If we set $f^{\left(\alpha n\right)}\left(0\right)=0$, then the previous Lemma also guarantees that the functions $f^{\left(\alpha n\right)}:{\mathbb{R}}_0^{+}\to \mathbb{R}$ are continuous.

3. The Fractional Holmgren Class

Let us recall here the definition of Holmgren class in the classical setting (see [16], Definition 2.2.1)), which is needed to provide Tychonoff solutions for the heat equation.

Definition 4.

Let ${\gamma }_1,{\gamma }_2,C_1,t_0>0$. We say that a function f belongs to the Holmgren class $\mathcal{H}({\gamma }_1,{\gamma }_2,C_1,t_0)$ if f is infinitely continuously differentiable on the set $B_{{\gamma }_2}\left(t_0\right)=\{t\in \mathbb{R}:|t-t_0|<{\gamma }_2\}$, and if it holds that

$$|f^{\left(n\right)}\left(t\right)|\le C_1{\gamma }_1^{-2n}\left(2n\right)!,\forall n\in \mathbb{N},\forall t\in B_{{\gamma }_2}\left(t_0\right).$$

We want to extend this definition to the fractional case. To do this, we will make use of the iterated Caputo derivatives.

Definition 5.

Let $\alpha \in (0,1)$, ${\gamma }_1,C_1,t_0>0$, and $0<{\gamma }_2<t_0$. We say that a function f belongs to the α-fractional Holmgren class ${\mathcal{H}}_{\alpha }({\gamma }_1,{\gamma }_2,C_1,t_0)$ if $f^{(\alpha ,n)}$ is well defined on $(0,t_0+{\gamma }_2]$ for any $n\in \mathbb{N}$, and, setting $B_{{\gamma }_2}\left(t_0\right)=\{t\in \mathbb{R}:|t-t_0|<{\gamma }_2\}$, it holds that

$$|f^{(\alpha ,n)}\left(t\right)|\le C_1{\gamma }_1^{-2n}\left(2n\right)!,\forall n\in \mathbb{N},\forall t\in B_{{\gamma }_2}\left(t_0\right).$$

Let $\alpha \in (0,1)$, ${\gamma }_1,C_1,{\gamma }_2>0$. We say that $f\in {\mathcal{H}}_{\alpha }^0({\gamma }_1,{\gamma }_2,C_1)$ if $f^{(\alpha ,n)}$ is well defined on $(0,{\gamma }_2]$ for any $n\in \mathbb{N}$, and if it holds that

$$|f^{(\alpha ,n)}\left(t\right)|\le C_1{\gamma }_1^{-2n}\left(2n\right)!,\forall n\in \mathbb{N},\forall t\in (0,{\gamma }_2).$$

(11)

Clearly, for any $t_0<{\gamma }_2$ and $0<{\tilde{\gamma }}_2<min\{t_0,{\gamma }_2-t_0\}$, it holds that ${\mathcal{H}}_{\alpha }^0({\gamma }_1,{\gamma }_2,C_1)\subset {\mathcal{H}}_{\alpha }({\gamma }_1,{\tilde{\gamma }}_2,C_1,t_0)$.

Here, we have given two possible generalizations of the Holmgren class. However, we will only practically use the second one, which takes into consideration the nonlocal nature of the fractional Caputo derivative.

One could ask whether such a class is actually empty or not. Let us show an example of a parametric family of functions belonging to a suitable $\alpha$-fractional Holmgren class (in analogy to what is conducted in ([16], Section 2.4)). To do this, we first need the following two technical Lemmas.

Lemma 4.

Let $\beta >1$, and let us denote by $z^{-\beta }$, for any complex number $z\in \mathbb{C}$, the principal value of the power, i.e., if $z=\rho e^{i\theta }$ for some $\rho >0$ and $\theta \in (-\pi ,\pi ]$, then

$$z^{-\beta }={\rho }^{-\beta }{e}^{-i\beta \theta }.$$

Then, there exists a constant $r_{\beta }\in (0,1)$ such that

$$\Re \left({\left(1+r{e}^{i\theta }\right)}^{-\beta }\right)\ge 1/2,\forall (r,\theta )\in (0,r_{\beta })\times (-\pi ,\pi ]$$

Proof. 

Let us denote $z:(r,\theta )\in (0,1)\times (-\pi ,\pi ]\mapsto 1+r{e}^{i\theta }\in \mathbb{C}$ and $\Theta (r,\theta )=\mathrm{Arg}\left(z\right(r,\theta \left)\right)$ as the principal argument of z, and let $\rho (r,\theta )=\left|z\right(r,\theta \left)\right|$ be its modulus. First of all, let us observe that, being $r<1$, $\Re (1+r{e}^{i\theta })=1+rcos\left(\theta \right)>0$. Hence, we have $\Theta (r,\theta )\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$. Moreover, being

$$\rho (r,\theta )=\sqrt{1+r^2+2rcos\left(\theta \right)},\Im (1+r{e}^{i\theta })=rsin\left(\theta \right)$$

we conclude that

$$\Theta (r,\theta )=arcsin\left(\frac{rsin\left(\theta \right)}{\sqrt{1+r^2+2rcos\left(\theta \right)}}\right).$$

Let us first show that ${lim}_{r\to 0^{+}}\Theta (r,\theta )=0$ uniformly for $\theta \in (-\pi ,\pi ]$. To do this, let us observe that $\Theta (r,\theta )=-\Theta (r,-\theta )$; thus, we only need to focus on $\theta \in [0,\pi ]$. In such a case, $sin\left(\theta \right)\ge 0$, and then we have

$$\Theta (r,\theta )\le arcsin\left(\frac{r}{\sqrt{1+r^2-2r}}\right)\to 0.$$

Once this is accomplished, let us observe that

$$F(r,\theta ):=\Re \left({\left(1+r{e}^{i\theta }\right)}^{-\beta }\right)=\frac{cos(\beta \Theta (r,\theta \left)\right)}{{(1+r^2+2rcos\left(\theta \right))}^{\frac{\beta }{2}}}.$$

We want to show that ${lim}_{r\to 0}F(r,\theta )=1$ uniformly in $\theta \in (-\pi ,\pi ]$. Again, if we notice that $F(r,\theta )=F(r,-\theta )$, we can focus on the case $\theta \in [0,\pi ]$.

We have

$$|1-F(r,\theta )|\le \frac{|{(1+r^2+2rcos\left(\theta \right))}^{\frac{\beta }{2}}-1|+|1-cos(\beta \Theta (r,\theta ))|}{{(1+r^2-2r)}^{\frac{\beta }{2}}}.$$

(12)

Let us consider, for fixed $r>0$, the function $G\left(\theta \right)={(1+r^2+2rcos\left(\theta \right))}^{\frac{\beta }{2}}-1$, and let us observe that

$$G^{\prime }\left(\theta \right)=-\beta rsin\left(\theta \right){(1+r^2+2rcos\left(\theta \right))}^{\frac{\beta }{2}-1}$$

which implies that G is decreasing as $\theta \in [0,\pi ]$. Thus, we conclude that

$$\begin{array}{c}\hfill \left|G\left(\theta \right)\right|\le max\left\{\right|G\left(0\right)|,|G\left(\pi \right)\left|\right\}\le max\left\{\right|{(1+r^2-2r)}^{\frac{\beta }{2}}-1|,|{(1+r^2+2r)}^{\frac{\beta }{2}}-1\left|\right\}=:G_M\left(r\right),\forall \theta \in [0,\pi ].\end{array}$$

(13)

On the other hand, let us recall that

$$1-cos(\beta \Theta (r,\theta ))\le \frac{{\Theta }^2(r,\theta )}{2}\le \frac{1}{2}{(\underset{\theta \in [0,\pi ]}{max}\Theta (r,\theta ))}^2=:{\Theta }_M\left(r\right).$$

(14)

Hence, using Inequalities (13) and (14) in Equation (12), we obtain

$$\underset{\theta \in [0,\pi ]}{sup}|1-F(r,\theta )|\le \frac{G_M\left(r\right)+{\Theta }_M\left(r\right)}{{(1+r^2-2r)}^{\frac{\beta }{2}}}.$$

(15)

Sending $r\to 0^{+}$ in Equation (15), we conclude the proof. □

Before stating the second technical lemma, let us define the following family of functions. For $\beta >1$, we set

$$f_{\beta }\left(t\right)=\left\{\begin{array}{cc}{e}^{-t^{-\beta }}\hfill & t>0\hfill \\ 0\hfill & t=0.\hfill \end{array}\right.$$

(16)

Such functions clearly belong to $C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$. We would like to show that they belong to a fractional Holmgren class for a suitable choice of constants. To do this, we first need to provide an estimate on the integer order derivatives of $f_{\beta }$. Such an estimate is well known (see, for instance, ([16], Equation (2.4.6)) in the case $\beta =2$), but we give here a proof for completeness.

Lemma 5.

For any $\beta >1$, there exists a constant $K\left(\beta \right)\in (0,1)$ such that

$$|f_{\beta }^{\left(m\right)}\left(t\right)|\le \frac{m!}{{\left(K\left(\beta \right)t\right)}^m}{e}^{-\frac{1}{2t^{\beta }}},\forall m\in \mathbb{N},\forall t\ge 0$$

Proof. 

Let us observe that, being a composition of analytic functions, $f_{\beta }$ is analytic as soon as $t>0$. Let $r_{\beta }\in (0,1)$ be the constant defined in Lemma 4, and consider any $r\in (0,r_{\beta })$. Let ${\Gamma }_{r,t}=\{z\in \mathbb{C}:z=t(1+r{e}^{i\theta }),\theta \in (-\pi ,\pi ]\}$ be the circle centred in t with radius $rt$. Clearly, ${\Gamma }_{r,t}$ is separated from 0, and we can use it as a path in the Cauchy formula to obtain

$$f_{\beta }^{\left(m\right)}\left(t\right)=\frac{m!}{2\pi i}{\int }_{{\Gamma }_{r,t}}\frac{e^{-z^{-\beta }}}{{(z-t)}^{m+1}}dz.$$

Using the parametrization $z=t(1+r{e}^{i\theta })$, we obtain

$$f_{\beta }^{\left(m\right)}\left(t\right)=\frac{m!}{2\pi }{\int }_{-\pi }^{\pi }\frac{e^{-t^{-\beta }{(1+r{e}^{i\theta })}^{-\beta }}}{{\left(rt{e}^{i\theta }\right)}^m}d\theta .$$

Taking the modulus, we conclude that, thanks to Lemma 4,

$$|f_{\beta }^{\left(m\right)}\left(t\right)|\le \frac{m!}{2\pi {\left(rt\right)}^m}{\int }_{-\pi }^{\pi }{e}^{-t^{-\beta }\Re \left({(1+r{e}^{i\theta })}^{-\beta }\right)}d\theta \le \frac{m!}{{\left(rt\right)}^m}{e}^{-\frac{1}{2t^{\beta }}}.$$

□

Now, we can prove the main result of this section.

Proposition 2.

For any ${\gamma }_1,{\gamma }_2>0$, $\alpha \in (0,1)$, and $\beta >1$, there exists a constant $C_1({\gamma }_1,{\gamma }_2,\alpha ,\beta )>0$ such that $f_{\beta }\in {\mathcal{H}}_{\alpha }^0({\gamma }_1,{\gamma }_2,C_1({\gamma }_1,{\gamma }_2,\alpha ,\beta ))$.

Proof. 

Let us fix $\alpha \in (0,1)$, ${\gamma }_1,{\gamma }_2>0$, and $\beta >1$. First of all, let us observe that $f_{\beta }\in C_{0+}^{\infty }\left({\mathbb{R}}_0^{+}\right)$; thus, by Proposition 1, we know that $f_{\beta }^{(\alpha ,n)}=f_{\beta }^{\left(\alpha n\right)}$. By Lemma 5, we have

$$|f_{\beta }^{(\lfloor \alpha n\rfloor +1)}\left(t\right)|\le \frac{(\lfloor \alpha n\rfloor +1)!}{{\left(K\left(\beta \right)\right)}^{\lfloor \alpha n\rfloor +1}}{t}^{-\lfloor \alpha n\rfloor -1}{e}^{-\frac{1}{2t^{\beta }}},t\ge 0;$$

hence, we obtain

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill |f_{\beta }^{\left(\alpha n\right)}\left(t\right)|& \le \frac{1}{\Gamma (\lfloor \alpha n\rfloor -\alpha n+1)}{\int }_0^t{(t-\tau )}^{\lfloor \alpha n\rfloor -\alpha n}\left|f_{\beta }^{(\lfloor \alpha n\rfloor +1)}\left(\tau \right)\right|d\tau \hfill \\ \hfill & \le \frac{1}{\Gamma (\lfloor \alpha n\rfloor -\alpha n+1)}{\int }_0^t{(t-\tau )}^{\lfloor \alpha n\rfloor -\alpha n}\frac{(\lfloor \alpha n\rfloor +1)!}{{\left(K\left(\beta \right)\right)}^{\lfloor \alpha n\rfloor +1}}{\tau }^{-\lfloor \alpha n\rfloor -1}{e}^{-\frac{1}{2{\tau }^{\beta }}}d\tau \hfill \\ \hfill & =\frac{(\lfloor \alpha n\rfloor +1)!}{{\left(K\left(\beta \right)\right)}^{\lfloor \alpha n\rfloor +1}\Gamma (\lfloor \alpha n\rfloor -\alpha n+1)}{\int }_0^t\tau {(t-\tau )}^{\lfloor \alpha n\rfloor -\alpha n}{\left({\tau }^{-\beta }\right)}^{\frac{\lfloor \alpha n\rfloor +2}{\beta }}{e}^{-\frac{{\tau }^{-\beta }}{2}}d\tau .\hfill \end{array}\end{array}$$

(17)

Now, let us recall that, for any $c_1,c_2>0$, it holds that

$$x^{c_1}{e}^{-c_{2}x}\le {\left(\frac{c_1}{e{c}_2}\right)}^{c_1},x\ge 0.$$

Setting $x={\tau }^{-\beta }$, $c_1=\frac{\lfloor \alpha n\rfloor +2}{\beta }$, and $c_2=\frac{1}{2}$, we have

$${\left({\tau }^{-\beta }\right)}^{\frac{\lfloor \alpha n\rfloor +2}{\beta }}{e}^{-\frac{{\tau }^{-\beta }}{2}}\le {\left(\frac{2(\lfloor \alpha n\rfloor +2)}{\beta e}\right)}^{\frac{\lfloor \alpha n\rfloor +2}{\beta }},$$

and plugging the previous estimate into Inequality (17), we achieve

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill |f_{\beta }^{\left(\alpha n\right)}\left(t\right)|& \le {\left(\frac{2(\lfloor \alpha n\rfloor +2)}{e\beta }\right)}^{\frac{\lfloor \alpha n\rfloor +2}{\beta }}\frac{(\lfloor \alpha n\rfloor +1)!}{{\left(K\left(\beta \right)\right)}^{\langle \alpha n\rangle +1}\Gamma (\lfloor \alpha n\rfloor -\alpha n+1)}{\int }_0^t\tau {(t-\tau )}^{\lfloor \alpha n\rfloor -\alpha n}d\tau .\hfill \end{array}\end{array}$$

(18)

Now, let us recall, by ([53], Section 2.5), that

$$\frac{1}{\Gamma (\lfloor \alpha n\rfloor -\alpha n+1)}{\int }_0^t\tau {(t-\tau )}^{\lfloor \alpha n\rfloor -\alpha n}d\tau =\frac{1}{\Gamma (\lfloor \alpha n\rfloor -\alpha n+3)}{t}^{\lfloor \alpha n\rfloor -\alpha n+2};$$

then, Inequality (18) becomes

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill |f_{\beta }^{\left(\alpha n\right)}\left(t\right)|& \le {\left(\frac{2(\lfloor \alpha n\rfloor +2)}{e\beta }\right)}^{\frac{\lfloor \alpha n\rfloor +2}{\beta }}\frac{(\lfloor \alpha n\rfloor +1)!}{{\left(K\left(\beta \right)\right)}^{\langle \alpha n\rangle +1}\Gamma (\lfloor \alpha n\rfloor -\alpha n+3)}{t}^{\lfloor \alpha n\rfloor -\alpha n+2}.\hfill \end{array}\end{array}$$

(19)

Now, let us observe that $\frac{2(\lfloor \alpha n\rfloor +2)}{e\beta }>1$ if and only if

$$\lfloor \alpha n\rfloor >\frac{e\beta }{2}-2.$$

It is clear that there exists $n_{\alpha }\in \mathbb{N}$ such that the inequality holds for any $n>n_{\alpha }$. However, let us stress that $f_{\beta }^{\left(\alpha n\right)}:{\mathbb{R}}_0^{+}\to \mathbb{R}$ is continuous; thus, for any $n\le n_{\alpha }$, it holds that

$$|f_{\beta }^{\left(\alpha n\right)}\left(t\right)|\le C_2({\gamma }_1,{\gamma }_2,\alpha ,\beta ){\gamma }_1^{-2n}\left(2n\right)!,t\in [0,{\gamma }_2]$$

where

$$C_2({\gamma }_1,{\gamma }_2,\alpha ,\beta )=\underset{n\le n_{\alpha }}{max}\left(\frac{{\gamma }_1^{2n}}{\left(2n\right)!}\underset{t\in [0,{\gamma }_2]}{max}\left|f_{\beta }^{\left(\alpha n\right)}\left(t\right)\right|\right).$$

Let us now consider the case $n>n_{\alpha }$. Since $\alpha \in (0,1)$, $\alpha n<n$, and then $\lfloor \alpha n\rfloor \le n-1$, for any $n\in \mathbb{N}$. This implies that

$${\left(\frac{2(\lfloor \alpha n\rfloor +2)}{e\beta }\right)}^{\frac{\lfloor \alpha n\rfloor +2}{\beta }}\le {\left(\frac{2(n+1)}{e\beta }\right)}^{\frac{n+1}{\beta }}.$$

(20)

Next, observe that $0<K\left(\beta \right)<1$; thus, the function $m\in \mathbb{N}\mapsto {\left(K\left(\beta \right)\right)}^m$ is decreasing, while $m\in \mathbb{N}\mapsto (m+1)!$ is increasing. This leads to

$$\frac{(\lfloor \alpha n\rfloor +1)!}{{\left(K\left(\beta \right)\right)}^{\lfloor \alpha n\rfloor +1}}\le \frac{n!}{{\left(K\left(\beta \right)\right)}^n}.$$

(21)

It is also clear that, since $1\le \lfloor \alpha n\rfloor -\alpha n+2\le 2$, if $t\le 1$, then it holds that $t^{\lfloor \alpha n\rfloor -\alpha n+2}\le 1$. If we suppose ${\gamma }_2>1$ and $1<t<{\gamma }_2$, it clearly holds that $t^{\lfloor \alpha n\rfloor -\alpha n+2}\le t^2\le {\gamma }_2^2$. Thus, we conclude that

$$t^{\lfloor \alpha n\rfloor -\alpha n+2}\le max\left\{1,{\gamma }_2^2\right\}.$$

(22)

Finally, recall that the function $x\in {\mathbb{R}}^{+}\mapsto \Gamma \left(x\right)$ is increasing as $x\ge 2$ and $\lfloor \alpha n\rfloor -\alpha n+3\ge 2$; thus, we obtain

$$\frac{1}{\Gamma (\lfloor \alpha n\rfloor -\alpha n+3)}\le \frac{1}{\Gamma \left(2\right)}=1.$$

(23)

Combining Inequalities (20), (21)‖(23) into Equation (19), we have, for any $n>n_{\alpha }$,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill |f_{\beta }^{\left(\alpha n\right)}\left(t\right)|& \le {\left(\frac{2(n+1)}{e\beta }\right)}^{\frac{n+1}{\beta }}\frac{n!}{{\left(K\left(\beta \right)\right)}^n}max\{1,{\gamma }_2^2\}\hfill \\ \hfill & ={\left(\frac{2(n+1)}{e\beta }\right)}^{\frac{n+1}{\beta }}\frac{n!}{{\left(K\left(\beta \right)\right)}^n{\left({\gamma }_1\right)}^{-2n}\left(2n\right)!}max\{1,{\gamma }_2^2\}{\gamma }_1^{-2n}\left(2n\right)!\hfill \\ \hfill & =:A_{n}max\{1,{\gamma }_2^2\}{\gamma }_1^{-2n}\left(2n\right)!.\hfill \end{array}\end{array}$$

(24)

Now, we want to show that $A_n$ is bounded. To do this, we will actually prove that ${lim}_{n\to \infty }{A}_n=0$. Indeed, let us observe that

$$\begin{array}{cc}\hfill \frac{A_{n+1}}{A_n}& ={\left(1+\frac{1}{n+1}\right)}^{\frac{n+1}{\beta }}{\left(\frac{2}{e\beta }\right)}^{\frac{1}{\beta }}\frac{{(n+2)}^{\frac{1}{\beta }}(n+1){\gamma }_1^2}{(2n+2)(2n+1)}\to 0,\hfill \end{array}$$

since $\beta >1$. The ratio test guarantees that $A_n\to 0$, and thus, there exists $C_3({\gamma }_1,{\gamma }_2,\alpha ,\beta )$ such that $A_{n}max\{1,{\gamma }_2^2\}\le C_3({\gamma }_1,{\gamma }_2,\alpha ,\beta )$ for any $n>n_{\alpha }$. Hence, Inequality (24) guarantees that, as $n>n_{\alpha }$ and $t<{\gamma }_2$,

$$|f_{\beta }^{\left(\alpha n\right)}\left(t\right)|\le C_3({\gamma }_1,{\gamma }_2,\alpha ,\beta ){\gamma }_1^{-2n}\left(2n\right)!.$$

Setting $C_1({\gamma }_1,{\gamma }_2,\alpha ,\beta )=max\{C_2({\gamma }_1,{\gamma }_2,\alpha ,\beta ),C_3({\gamma }_1,{\gamma }_2,\alpha ,\beta )\}$, we conclude the proof.

□

In the next section, we will use the functions $f_{\beta }$ to construct a suitable family of solutions for the time-fractional heat equation.

4. Tychonoff Solutions of the Time-Fractional Heat Equation

Let us consider the time-fractional heat equation with the initial condition $u_0\in C_c^{\infty }\left(\mathbb{R}\right)$:

$$\left\{\begin{array}{cc}{\displaystyle \frac{{\partial }^{\alpha }u}{\partial t^{\alpha }}(t,x)=\frac{{\partial }^{2}u}{\partial x^2}(t,x)}\hfill & t>0,x\in \mathbb{R}\hfill \\ u(0,x)=u_0\left(x\right)\hfill & x\in \mathbb{R}.\hfill \end{array}\right.$$

(25)

Definition 6.

We say that $u:{\mathbb{R}}_0^{+}\times \mathbb{R}\mapsto \mathbb{R}$ is a classical solution of Equation (25) if:

(i)

$u\in C({\mathbb{R}}_0^{+}\times \mathbb{R})$;

(ii)

${\displaystyle \frac{{\partial }^{\alpha }u}{\partial t^{\alpha }}}$ is well defined and belongs to $C({\mathbb{R}}^{+}\times \mathbb{R})$;

(iii)

${\displaystyle \frac{\partial u}{\partial x}\frac{{\partial }^{2}u}{\partial x^2}}$ are well defined and belong to $C({\mathbb{R}}^{+}\times \mathbb{R})$;

(iv)

Equation (25) holds pointwise.

First of all, we need to exhibit at least a classical solution of Equation (2). To do this, let us first fix a complete filtered probability space $(\Omega ,\mathcal{F},{\left\{{\mathcal{F}}_t\right\}}_{t\ge 0},\mathbb{P})$ and a standard Brownian motion W on it. We recall the following definition (see [56]).

Definition 7.

An α-stable subordinator ${\sigma }_{\alpha }=\{{\sigma }_{\alpha }\left(t\right),t\ge 0\}$ is an increasing Lévy process on $(\Omega ,\mathcal{F},{\left\{{\mathcal{F}}_t\right\}}_{t\ge 0},\mathbb{P})$ such that $\mathbb{E}\left[e^{-\lambda {\sigma }_{\alpha }\left(t\right)}\right]=e^{-t{\lambda }^{\alpha }}$. The inverse α-stable subordinator $L_{\alpha }=\{L_{\alpha }\left(t\right),t\ge 0\}$ is defined as

$$L_{\alpha }\left(t\right)=inf\{s>0:{\sigma }_{\alpha }\left(s\right)>t\}.$$

For a fixed $t>0$, we denote, by $h_{\alpha }(s;t)$, the probability density function of $L_{\alpha }\left(t\right)$.

Let us also denote, for $t>0$ and $(x,y)\in {\mathbb{R}}^2$,

$$p_{\alpha }(t,x;y)={\int }_0^{+\infty }\frac{1}{\sqrt{4\pi s}}{e}^{-\frac{{|x-y|}^2}{4s}}{h}_{\alpha }(s;t)ds.$$

Now, we are ready to prove that Equation (25) admits at least a classical solution, as shown, for instance, in ([50], Remark 3.10).

Theorem 7.

The function

$$u(t,x)={\int }_{\mathbb{R}}{p}_{\alpha }(t,x;y)u_0\left(y\right)dy=\mathbb{E}\left[u_0\left(\sqrt{2}W\left(L_{\alpha }\left(t\right)\right)\right)\right|W\left(0\right)=x]$$

(26)

is a classical solution of Equation (25), where $L_{\alpha }$ is an inverse α-stable subordinator that is independent of the Brownian motion W.

Proof. 

By an argument such as that in ([50], Proposition 3.8), we have, as $x\ne y$,

$$\begin{array}{cc}\hfill \frac{\partial p_{\alpha }}{\partial x}(t,x;y)& =-\frac{1}{4\sqrt{\pi }}{\int }_0^{\infty }\frac{x-y}{s^{\frac{3}{2}}}{e}^{-\frac{{|x-y|}^2}{4s}}{h}_{\alpha }(s;t)ds\hfill \\ \hfill \frac{{\partial }^2{p}_{\alpha }}{\partial x^2}(t,x;y)& =\frac{1}{4\sqrt{\pi }}{\int }_0^{\infty }\frac{1}{s^{\frac{3}{2}}}\left(\frac{{(x-y)}^2}{2s}-1\right)e^{-\frac{{|x-y|}^2}{4s}}{h}_{\alpha }(s;t)ds.\hfill \end{array}$$

Moreover, by an argument such as that in ([50], Proposition 3.4), and by using ([50], Lemma 1.1), we know that, as $x\ne y$,

$$\frac{{\partial }^{\alpha }{p}_{\alpha }}{\partial t^{\alpha }}(t,x;y)={\int }_0^{\infty }\frac{1}{2\sqrt{\pi s}}{e}^{-\frac{{|x-y|}^2}{4s}}\frac{{\partial }^{\alpha }{h}_{\alpha }}{\partial t^{\alpha }}(s;t)ds.$$

(27)

The same exact proof of ([50], Theorem 3.9) guarantees that, as $x\ne y$,

$$\frac{{\partial }^{\alpha }{p}_{\alpha }}{\partial t^{\alpha }}(t,x;y)=\frac{{\partial }^2{p}_{\alpha }}{\partial x^2}(t,x;y).$$

(28)

Since the aforementioned functions depend only on $x-y$, we can suppose, without loss of generality, that $y=0$. We need to prove that $\frac{\partial p_{\alpha }}{\partial x}(·,·;0)$ is bounded in a neighborhood of $(t,0)$ for any $t>0$. To do this, for $x>0$, set $z=\frac{x}{2\sqrt{s}}$ to obtain

$$\frac{\partial p_{\alpha }}{\partial x}(t,x;0)=-\frac{1}{\sqrt{\pi }}{\int }_0^{\infty }{e}^{-z^2}{h}_{\alpha }\left(\frac{x^2}{4z^2};t\right)dz.$$

Now, let us remark (see, for instance, ([56], Equation (2.9))) that $h_{\alpha }(s;t)$ is a continuous function for $(s,t)\in {\mathbb{R}}_0^{+}\times {\mathbb{R}}^{+}$, and, by ([50], Proposition 3.6), it is also bounded for $t\in [a,b]\subset {\mathbb{R}}^{+}$ and $s\to \infty$. Moreover, recall that $h_{\alpha }(0;t)=\frac{t^{-\alpha }}{\Gamma (1-\alpha )}$ (see ([56], Equation (4.3))). Thus, for any $[a,b]\subset {\mathbb{R}}^{+}$, we obtain

$$\begin{array}{c}\underset{t\in [a,b]}{max}\left|\frac{\partial p_{\alpha }}{\partial x}(t,x;0)+\frac{1}{2}\frac{t^{-\alpha }}{\Gamma (1-\alpha )}\right|\le \frac{1}{\sqrt{\pi }}{\int }_0^{\infty }{e}^{-z^2}\underset{t\in [a,b]}{max}\left|h_{\alpha }\left(\frac{x^2}{4z^2};t\right)-\frac{t^{-\alpha }}{\Gamma (1-\alpha )}\right|dz\to 0\hfill \\ \hfill asx\to 0^{+}.\end{array}$$

Arguing in the same way for $x<0$, we obtain

$$\underset{x\to 0^{\pm }}{lim}\frac{\partial p_{\alpha }}{\partial x}(t,x;0)=\mp \frac{t^{-\alpha }}{2\Gamma (1-\alpha )}locallyuniformlyint>0.$$

This is enough to guarantee that $\frac{\partial p_{\alpha }}{\partial x}(·,·;0)$ is bounded in any sufficiently small neighborhood of $(t,0)$ for any $t>0$. To obtain an analogous result for $\frac{{\partial }^2{p}_{\alpha }}{\partial x^2}(·,·;0)$, it is clear, by Equation (28), that we only need to prove a similar bound on $\frac{{\partial }^{\alpha }{p}_{\alpha }}{\partial t^{\alpha }}(·,·;0)$. By ([56], Equation (5.7)), we obtain, from (27),

$$\frac{{\partial }^{\alpha }{p}_{\alpha }}{\partial t^{\alpha }}(t,x;0)=-{\int }_0^{\infty }\frac{1}{2\sqrt{\pi s}}{e}^{-\frac{x^2}{4s}}\frac{\partial h_{\alpha }}{\partial s}(s;t)ds.$$

(29)

Consider any $[a,b]\subset {\mathbb{R}}^{+}$, and observe that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \underset{t\in [a,b]}{max}\left|\frac{{\partial }^{\alpha }{p}_{\alpha }}{\partial t^{\alpha }}(t,x;0)+{\int }_0^{\infty }\frac{1}{2\sqrt{\pi s}}\frac{\partial h_{\alpha }}{\partial s}(s;t)ds\right|& \le {\int }_0^{\infty }\frac{1}{2\sqrt{\pi s}}\left(1-e^{-\frac{x^2}{4s}}\right)\underset{t\in [a,b]}{max}\left|\frac{\partial h_{\alpha }}{\partial s}(s;t)\right|ds\hfill \\ \hfill & ={\int }_0^1\frac{1}{2\sqrt{\pi s}}\left(1-e^{-\frac{x^2}{4s}}\right)\underset{t\in [a,b]}{max}\left|\frac{\partial h_{\alpha }}{\partial s}(s;t)\right|ds\hfill \\ & +{\int }_1^{\infty }\frac{1}{2\sqrt{\pi s}}\left(1-e^{-\frac{x^2}{4s}}\right)\underset{t\in [a,b]}{max}\left|\frac{\partial h_{\alpha }}{\partial s}(s;t)\right|ds\hfill \\ & =I_1\left(x\right)+I_2\left(x\right).\hfill \end{array}\end{array}$$

(30)

Concerning $I_1\left(x\right)$, it has been shown in ([50], Example 5.1) that ${lim}_{s\to 0^{+}}\frac{\partial h_{\alpha }}{\partial s}(s;t)=-\frac{2}{\Gamma (1-2\alpha )}{t}^{-2\alpha }$ (where we set $\frac{1}{\infty }=0$) locally uniformly in $t>0$; thus, there exists a constant $C_1$, such that, for $s\in [0,1]$,

$$\underset{t\in [a,b]}{max}\left|\frac{\partial h_{\alpha }}{\partial s}(s;t)\right|\le C_1.$$

This is enough to guarantee that ${lim}_{x\to 0}{I}_1\left(x\right)=0$. To handle $I_2\left(x\right)$, let us rewrite it as

$$I_2\left(x\right)=\frac{1}{2\sqrt{\pi }}{\int }_1^{\infty }{s}^{-\frac{3}{2}}\left(1-e^{-\frac{x^2}{4s}}\right)\underset{t\in [a,b]}{max}\left|s^2\frac{\partial h_{\alpha }}{\partial s}(s;t)\right|ds.$$

Now, we resort to ([50], Proposition 3.6) to guarantee that, for $s\ge 1$, there exists a constant $C_2$, such that

$$\underset{t\in [a,b]}{max}\left|s^2\frac{\partial h_{\alpha }}{\partial s}(s;t)\right|\le C^2,$$

and that is enough to imply that ${lim}_{x\to 0}{I}_2\left(x\right)=0$. Hence, taking the limit as $x\to 0$ in Equation (30),

$$\underset{x\to 0}{lim}\frac{{\partial }^2{p}_{\alpha }}{\partial x^2}(t,x;0)=\underset{x\to 0}{lim}\frac{{\partial }^{\alpha }{p}_{\alpha }}{\partial t^{\alpha }}(t,x;0)=-\frac{1}{2\sqrt{\pi }}{\int }_0^{+\infty }{s}^{-\frac{1}{2}}\frac{\partial h_{\alpha }}{\partial s}(s;t)ds$$

locally uniformly in $t>0$, where the integral at the right-hand side is finite.

Once we have proven that $\frac{\partial p_{\alpha }}{\partial x}(t,x;y)$, $\frac{{\partial }^2{p}_{\alpha }}{\partial x^2}(t,x;y)$, and $\frac{{\partial }^{\alpha }{p}_{\alpha }}{\partial t^{\alpha }}(t,x;y)$ are continuous in any $(t,x,y)$, such that $t>0$ and $x\ne y$ are bounded in any sufficiently small neighborhood of any point $(t,x;y)$ with $t>0$, it is clear that, considering Equation (26),

$$\begin{array}{cc}\hfill \frac{{\partial }^{2}u}{\partial x^2}(t,x)& =\frac{{\partial }^2}{\partial x^2}{\int }_{\mathbb{R}}{p}_{\alpha }(t,x;y)u_0\left(y\right)dy\hfill \\ \hfill & ={\int }_{\mathbb{R}}\frac{{\partial }^2{p}_{\alpha }}{\partial x^2}(t,x;y)u_0\left(y\right)dy\hfill \\ \hfill & ={\int }_{\mathbb{R}}\frac{{\partial }^{\alpha }{p}_{\alpha }}{\partial t^{\alpha }}(t,x;y)u_0\left(y\right)dy\hfill \\ \hfill & =\frac{{\partial }^{\alpha }}{\partial t^{\alpha }}{\int }_{\mathbb{R}}{p}_{\alpha }(t,x;y)u_0\left(y\right)dy=\frac{{\partial }^{\alpha }u}{\partial t^{\alpha }}(t,x),\hfill \end{array}$$

where we also used Equation (28). Moreover, by the dominated convergence theorem, we know that ${\displaystyle \frac{{\partial }^{2}u}{\partial x^2},\frac{{\partial }^{\alpha }u}{\partial t^{\alpha }}\in C({\mathbb{R}}^{+}\times \mathbb{R})}$. Finally, due to the fact that $u(0,x)=u_0\left(x\right)$ for any $x\in \mathbb{R}$, the continuity of u up to $\left\{0\right\}\times \mathbb{R}$ can be proven exactly, as in ([50], Theorem 3.11). □

Remark 3.

Let us stress that the stochastic representation in Equation (26) follows from the fact that, for fixed $t>0$, $p_{\alpha }(t,x;y)$ is the probability density function of $W_{\alpha }\left(t\right):=\sqrt{2}W\left(L_{\alpha }\left(t\right)\right)$, conditioned to $W_{\alpha }\left(0\right)=y$, and $p_{\alpha }(t,x;y)=p_{\alpha }(t,y;x)$. Moreover, one can prove that ${lim}_{\left|x\right|\to \infty }u(t,x)=0$ locally uniformly in $t\ge 0$; hence, it is the unique solution with such a property, as a consequence of Theorem 5.

Now that we proved that the time-fractional heat Equation (25) admits at least one solution, we are ready to show that it actually admits an infinity of solutions.

Proof 

(Proof of Theorem 6). Let $u(t,x)$ be the classical solution we found in Theorem 7. For any real number $\beta >1$, we want to construct a function $g_{\beta }\ne 0$, such that

$$u_{\beta }(t,x):=u(t,x)+g_{\beta }(t,x)$$

is still a classical solution of Equation (25). To do this, we only need to construct a family (parametrized by a real number $\beta >1$) of non-trivial classical solutions $g_{\beta }(t,x)$ of the Cauchy problem:

$$\left\{\begin{array}{cc}{\displaystyle \frac{{\partial }^{\alpha }{g}_{\beta }}{\partial t^{\alpha }}(t,x)=\frac{{\partial }^2{g}_{\beta }}{\partial x^2}(t,x)}\hfill & t>0,x\in \mathbb{R}\hfill \\ g_{\beta }(0,x)=0\hfill & x\in \mathbb{R}.\hfill \end{array}\right.$$

(31)

Let us define, for $t\ge 0$ and $x\in \mathbb{R}$,

$$g_{\beta }(t,x)=\sum _{n=0}^{\infty }{f}_{\beta }^{\left(\alpha n\right)}\left(t\right)\frac{x^{2n}}{\left(2n\right)!},$$

where $f_{\beta }$ is given in Equation (16). First of all, let us show that $g_{\beta }$ is well defined. Indeed, let us consider $a,{\gamma }_2>0$, $K_1=[0,{\gamma }_2]\subset {\mathbb{R}}_0^{+}$ and $K_2=[-a,a]\subset \mathbb{R}$. Let also ${\gamma }_1=2a$. For $(t,x)\in K_1\times K_2$, we have, by Proposition 2,

$$|f_{\beta }^{\left(\alpha n\right)}\left(t\right)|\frac{{\left|x\right|}^{2n}}{\left(2n\right)!}\le C_1({\gamma }_1,{\gamma }_2,\alpha ,\beta ){\left(\frac{a}{{\gamma }_1}\right)}^{2n}=C_1({\gamma }_1,{\gamma }_2,\alpha ,\beta )2^{-2n}.$$

Hence, by the Weierstrass M-test, the series defining $g_{\beta }$ is uniformly convergent in $K_1\times K_2$. With $a,{\gamma }_2>0$ being arbitrary, we know that $g_{\beta }$ is locally uniformly convergent in ${\mathbb{R}}_0^{+}\times \mathbb{R}$. This also guarantees that $g_{\beta }\in C({\mathbb{R}}_0^{+}\times \mathbb{R})$. In particular, it holds that $g_{\beta }(0,x)=0$ for any $x\in \mathbb{R}$; hence, $g_{\beta }$ satisfies the initial condition in the Cauchy problem (31). Now, let us prove that ${\displaystyle \frac{\partial g_{\beta }}{\partial x},\frac{{\partial }^2{g}_{\beta }}{\partial x^2}\in C({\mathbb{R}}^{+}\times \mathbb{R})}$. To do this, let us first prove that we can exchange the derivative with the summation sign. This is guaranteed as soon as we prove that the series

$$\sum _{n=1}^{\infty }{f}_{\beta }^{\left(\alpha n\right)}\left(t\right)\frac{x^{2n-1}}{(2n-1)!}$$

is locally uniformly convergent. To do this, let us consider again ${\gamma }_2,a>0$, ${\gamma }_1=2a$, $K_1=(0,{\gamma }_2]$, and $K_2=[-a,a]$. As before, we have, for $(t,x)\in K_1\times K_2$,

$$|f_{\beta }^{\left(\alpha n\right)}\left(t\right)|\frac{{\left|x\right|}^{2n-1}}{(2n-1)!}\le C_1({\gamma }_1,{\gamma }_2,\alpha ,\beta )2^{-2n+1}an;$$

thus, we obtain the uniform convergence thanks to the Weierstrass M-test. This proves that

$$\frac{\partial g_{\beta }}{\partial x}(t,x)=\sum _{n=1}^{\infty }{f}_{\beta }^{\left(\alpha n\right)}\left(t\right)\frac{x^{2n-1}}{(2n-1)!},$$

and, by local uniform convergence, all the summand being continuous, $\frac{\partial g_{\beta }}{\partial x}\in C_{0+}^{\infty }({\mathbb{R}}^{+}\times \mathbb{R})$. To apply the derivative a second time, we need to prove that

$$\sum _{n=1}^{\infty }{f}_{\beta }^{\left(\alpha n\right)}\left(t\right)\frac{x^{2(n-1)}}{\left(2\right(n-1\left)\right)!}$$

is locally uniformly convergent. Again, for $(t,x)\in K_1\times K_2$, we obtain

$$|f_{\beta }^{\left(\alpha n\right)}\left(t\right)|\frac{{\left|x\right|}^{2(n-1)}}{\left(2\right(n-1\left)\right)!}\le C_1({\gamma }_1,{\gamma }_2,\alpha ,\beta )2^{-2n+1}{a}^2(2n-1)n,$$

thereby obtaining the desired convergence, thanks to the Weierstrass M-test. Hence, we obtain

$$\frac{{\partial }^2{g}_{\beta }}{\partial x^2}(t,x)=\sum _{n=1}^{\infty }{f}_{\beta }^{\left(\alpha n\right)}\left(t\right)\frac{x^{2(n-1)}}{\left(2\right(n-1\left)\right)!}$$

and ${\displaystyle \frac{{\partial }^2{g}_{\beta }}{\partial x^2}\in C({\mathbb{R}}^{+}\times \mathbb{R})}$. Next, we prove that ${\displaystyle \frac{{\partial }^{\alpha }{g}_{\beta }}{\partial t^{\alpha }}\in C({\mathbb{R}}^{+}\times \mathbb{R})}$. To do this, let us first observe that we can exchange the fractional integral with the summation sign. Indeed, let us write

$$\begin{array}{cc}\hfill {\mathcal{I}}^{1-\alpha }{g}_{\beta }(t,x)& ={\int }_0^t{g}_{\beta }(\tau ,x)d\frac{{(t-\tau )}^{1-\alpha }}{\Gamma (2-\alpha )}\hfill \\ \hfill & =\sum _{n=0}^{\infty }{\int }_0^t{f}_{\beta }^{\left(\alpha n\right)}\left(\tau \right)d\frac{{(t-\tau )}^{1-\alpha }}{\Gamma (2-\alpha )}\frac{x^{2n}}{\left(2n\right)!}=\sum _{n=0}^{\infty }{\mathcal{I}}^{1-\alpha }{f}_{\beta }^{\left(\alpha n\right)}\left(t\right)\frac{x^{2n}}{\left(2n\right)!},\hfill \end{array}$$

where we used ([57], Theorem 7.16). Now, we need to differentiate the previous relation with respect to t. To do this, we would like to show that we can exchange the derivative with the summation sign: we can do this once we prove that

$$\sum _{n=1}^{\infty }{f}_{\beta }^{\left(\alpha \right(n+1\left)\right)}\left(t\right)\frac{x^{2n}}{\left(2n\right)!}$$

converges locally uniformly. This follows from the fact that, as $(t,x)\in K_1\times K_2$,

$$|f_{\beta }^{\left(\alpha \right(n+1\left)\right)}\left(t\right)|\frac{a^{2n}}{\left(2n\right)!}\le C_1({\gamma }_1,{\gamma }_2,\alpha ,\beta ){\gamma }_1^{-2}{2}^{-2n}(2n+1)(2n+2),$$

obtaining the desired convergence from the Weierstrass M-test. This proves that

$$\frac{{\partial }^{\alpha }{g}_{\beta }}{\partial t^{\alpha }}(t,x)=\sum _{n=0}^{\infty }{f}_{\beta }^{\left(\alpha \right(n+1\left)\right)}\left(t\right)\frac{x^{2n}}{\left(2n\right)!}$$

and that ${\displaystyle \frac{{\partial }^{\alpha }{g}_{\beta }}{\partial t^{\alpha }}\in C({\mathbb{R}}^{+}\times \mathbb{R})}$. Finally, let us observe that

$$\frac{{\partial }^2{g}_{\beta }}{\partial x^2}(t,x)=\sum _{n=1}^{\infty }{f}_{\beta }^{\left(\alpha n\right)}\left(t\right)\frac{x^{2(n-1)}}{\left(2\right(n-1\left)\right)!}=\sum _{n=0}^{\infty }{f}_{\beta }^{\left(\alpha \right(n+1\left)\right)}\left(t\right)\frac{x^{2n}}{\left(2n\right)!}=\frac{{\partial }^{\alpha }{g}_{\beta }}{\partial t^{\alpha }}(t,x),$$

concluding the proof. □

Remark 4.

Let us stress that we can construct other solutions of the time-fractional heat equation in addition to $u_{\beta }$. Indeed, if we define $u_f=u+g_f$, where

$$g_f(t,x)=\sum _{n=0}^{\infty }{f}^{(\alpha ,n)}\left(t\right)\frac{x^{2n}}{\left(2n\right)!},$$

$f^{(\alpha ,n)}\left(0^{+}\right)=0$ for any $n\in \mathbb{N}$, and $f\in {\mathcal{H}}_{\alpha }^0({\gamma }_1,{\gamma }_2,C)$ for any ${\gamma }_1,{\gamma }_2>0$, one can show that $u_f$ is a classical solution of the Cauchy problem (25) in the exact same way we proved Theorem 6.

Actually, if we fix $T>0$, and if we consider that

$$\left\{\begin{array}{cc}{\displaystyle \frac{{\partial }^{\alpha }u}{\partial t^{\alpha }}(t,x)=\frac{{\partial }^{2}u}{\partial x^2}(t,x)}\hfill & t\in (0,T),x\in \mathbb{R}\hfill \\ u(0,x)=u_0\left(x\right)\hfill & x\in \mathbb{R},\hfill \end{array}\right.$$

then $u_f$ is a classical solution for any f, such that $f\in {\mathcal{H}}_{\alpha }^0({\gamma }_1,{\gamma }_2,C)$ with ${\gamma }_2\ge T$ and any ${\gamma }_1>0$, and $f^{(\alpha ,n)}\left(0^{+}\right)=0$ for any $n\in \mathbb{N}$. Finally, if we consider $a>0$ and

$$\left\{\begin{array}{cc}{\displaystyle \frac{{\partial }^{\alpha }u}{\partial t^{\alpha }}(t,x)=\frac{{\partial }^{2}u}{\partial x^2}(t,x)}\hfill & t\in (0,T),x\in (-a,a)\hfill \\ u(0,x)=u_0\left(x\right)\hfill & x\in (-a,a),\hfill \end{array}\right.$$

then $u_f$ is a classical solution for any $f\in {\mathcal{H}}_{\alpha }^0({\gamma }_1,{\gamma }_2,C)$ with ${\gamma }_1\ge a$ and ${\gamma }_2\ge T$ such that $f^{(\alpha ,n)}\left(0^{+}\right)=0$ for any $n\in \mathbb{N}$.

5. Conclusions

In this paper, we considered the problem of the uniqueness of the classical solutions of the time-fractional heat equation with initial data on the whole real line. Precisely, once we proved that there exists at least a classical solution in Theorem 7, we showed that there are actually infinitely many solutions. Precisely, in Theorem 7 we proved the existence of a classical solution for such an equation with a $C_c^{\infty }\left(\mathbb{R}\right)$ initial data $u_0$ by (more or less) explicitly exhibiting one of them. Such a solution can be represented as the conditional expectation of $u_0$ applied to a delayed Brownian motion, which was already known to be a mild solution of the equation. To prove the existence of infinitely many solutions of the time-fractional heat equation with $C_c^{\infty }\left(\mathbb{R}\right)$ initial data, it is then sufficient to prove the statement only for the null initial data. This required some technical preliminaries. First, we proved an improved semigroup property for iterated Caputo derivatives of suitably regular functions (see Proposition 1). Then, we introduced a fractional analogue of the Holmgren class, which is the main tool to prove the existence of nontrivial null solutions also in the classical case. Once we recognized a parametrized family of functions belonging to such an Holmgren class, we used such a class of functions to construct the required solutions and to prove Theorem 6.

As we already stated in the introduction, there are some direct consequences of such a result. Indeed, we can identify, thanks to Theorem 7 and Remark 3, the aforementioned conditional expectation among all the other solutions as the unique one, satisfying the condition dictated by Theorem 5. This also implies that Monte Carlo methods based on the delayed Brownian motion could be developed to obtain a numerical estimate of the solution that satisfies the aforementioned condition, but it fails to approximate any other solution.

Moreover, the form of the nontrivial solutions of the time-fractional heat equation with null initial data suggests the possibility of a Tychonoff-like uniqueness condition (as in Theorem 3) for such an equation. Another open problem concerns more general operators than the classical Caputo derivative. Indeed, in [50], a generalized time-fractional heat equation is studied by means of more general delayed Brownian motions. Such an equation involves nonlocal integro-differential operators, which are defined starting from a suitable Bernstein function. This means that eventual composition properties are unclear and the definition of a suitable Holmgren-like class (with respect to iterated applications of such operators) is not straightforward. Again, once the existence of infinitely many solutions for the generalized time-fractional equation has been established, one should search for a Tychonoff-like uniqueness condition. We plan to carry on such studies in future works.