Controllability Properties for Equations with Memory of Fractional Type

Abstract

We study a general class of control systems with memory, which in particular includes systems with fractional derivatives and integrals as well as the standard heat equation. We prove that the approximate controllability property of the heat equation is inherited by every system with memory in this class while controllability to zero is a singular property, which holds solely in the special case that the system indeed reduces to the standard heat equation.

Appendices

Appendices

Here we collect ancillary results, as follows: Appendix A.1 shows that the assumptions of this paper are satisfied in significant cases and Appendix A.2 recall the properties of the Laplace transformation and Hardy spaces needed in the proofs.

Well posedness of the problem (1) with the boundary conditions (2) is proved in Appendix A.3. The proof uses the standing Assumptions in Sect. 1.1 and the properties of the Laplace transform recalled in Appendix A.2.

1.1 Discussion of the Assumptions

The powers of $\lambda$ denotes the principal value, so that $\arg \lambda ^\gamma =\gamma \arg \lambda$ with $-\pi \le \arg \lambda <\pi$.

We recall $A=\Delta$ with domain $H^2(\Omega )\cap H^1_0(\Omega )\subseteq L^2(\Omega )$ so that $\theta _A$ is any number in $(0,\pi /2)$ and we discuss the assumptions on the memory kernels in the following significant cases:

• 1. $K(t)=\delta (t)$, $N(t)=\frac{1}{\Gamma (\gamma )} t^{ \gamma -1 }$ so that ${{\hat{K}}}(\lambda )=1$, ${{\hat{N}}}(\lambda )=\frac{1}{ \lambda ^\gamma }$ where $\gamma \in (0,1)$. Then

  $$\begin{aligned} \frac{\lambda {{\hat{K}}}(\lambda )}{J(\lambda ) }=\frac{\lambda ^{1+\gamma }}{1+\lambda ^{\gamma }}. \end{aligned}$$

• 2. $K(t)=\frac{1}{\Gamma (1-\alpha )} t^{-\alpha }$, $N(t)=0$ with $\alpha \in (0,1)$. Then $\lambda {{\hat{K}}}(\lambda )=\lambda ^{\alpha }$ and

  $$\begin{aligned} \frac{\lambda {{\hat{K}}}(\lambda )}{J(\lambda ) }=\lambda ^{\alpha }. \end{aligned}$$

• 3. $K(t)=\frac{1}{\Gamma (1-\alpha )} t^{-\alpha }$, $N(t)=\frac{1}{\Gamma (\gamma )} t^{ \gamma -1}$ so that $\frac{\lambda {{\hat{K}}}(\lambda )}{ (1+{{\hat{N}}}(\lambda ) )}=\lambda ^{1+\alpha }/\left( 1+1/\lambda ^\gamma \right)$ where $\alpha$ and $\gamma$ belong to (0, 1). Then,

  $$\begin{aligned} \frac{\lambda {{\hat{K}}}(\lambda )}{J(\lambda ) }=\frac{\lambda ^{ \alpha +\gamma }}{1+\lambda ^{\gamma }}. \end{aligned}$$

The conditions in the items 1-3 and the asymptotic conditions of item 5 clearly hold in the three examples, and we must prove the existence of a sector $\Sigma _{\theta +\pi /2}$ over which the condition in item 4 hold, i.e. $\lambda {{\hat{K}}}(\lambda )/J(\lambda )$ transforms $\Sigma _{\theta +\pi /2}$ into a sector $\Sigma _{\pi -\epsilon }$ for some $\epsilon >0$. This assumption is clearly satisfied in case 2 since $\alpha \in (0,1)$. We prove that it is satisfied in case 1. We consider $\arg \lambda \ge 0$ (similar computations when $\arg \lambda \le 0$ lead to the same condition on $\epsilon$). If $\arg \lambda >0$ we have

$$\begin{aligned} \arg \frac{\lambda ^{1+\gamma }}{1+\lambda ^\gamma }=(1+\gamma )\arg \lambda - \arg (1+\lambda ^\gamma )<(1+\gamma )\arg \lambda . \end{aligned}$$

The required condition holds of we choose $\epsilon \in \left( 0,(1-\gamma )\pi /2\right)$.

Case 3 is more interesting since in this case the condition in item 4 imposes a link between the exponents $\alpha$ and $\gamma$. It is convenient to rename $\epsilon$ as $\pi \epsilon$ and $\theta$ as $\theta \pi /2$ so that now $\theta \in (0,1)$. Then, the condition in item 4 of the Assumptions 5 can be written as follows: there exists $\theta \in (0,1)$ and $\epsilon >0$ such that both the following conditions hold when $-(\pi /2)(\theta +1)\le \arg \lambda \le (\pi /2)(\theta +1)$:

$$\begin{aligned} -\pi (1-\epsilon )<\arg \frac{\lambda ^{\alpha +\gamma }}{1+\lambda ^\gamma }<\pi (1-\epsilon ). \end{aligned}$$

(28)

We examine the case $0\le \arg \lambda \le \frac{\pi }{2}\left( \theta +1\right)$. Computing the limit for $\lambda \rightarrow 0$ along the line $\arg \lambda =\left( \theta +1\right) (\pi /2)$ we get

$$\begin{aligned} (\alpha +\gamma )(\pi /2)(\theta +1)<\pi (1-\epsilon )\quad \text{ which } \text{ implies } \text{ the } \text{ necessary } \text{ condition } \alpha +\gamma <2. \end{aligned}$$

When $\mathfrak {R}{\textstyle e}\,\lambda \ge 0$ the left inequality in (28) is always satisfied since $\arg (1+\lambda ^\gamma )\le \gamma \arg \lambda$. We prove that the inequality from above is satisfied, for a suitable $\theta \in (0,1)$ and $\epsilon >0$, when $\alpha +\gamma <2$. In fact,

$$\begin{aligned} (\alpha +\gamma )\arg \lambda -\arg (1+\lambda ^\gamma )\le (\alpha +\gamma )\arg \lambda \le (\alpha +\gamma )\frac{\pi }{2}(\theta +1) \end{aligned}$$

and we want the existence of $\theta _0>0$ such that the following holds for $\theta <\theta _0$:

$$\begin{aligned} \frac{1}{2}(\alpha +\gamma )(\theta +1)\pi <\pi . \end{aligned}$$

This is achieved provided that

$$\begin{aligned} \theta <\theta _0\quad \mathrm{where}\quad \theta _0=\frac{2-(\alpha +\gamma )}{\alpha +\gamma }>0. \end{aligned}$$

A similar analysis shows that when $\alpha +\gamma <2$ the required condition (28) can be satisfied also in a sector $-(\pi /2)(\theta +1)\le \arg \lambda \le 0$.

1.2 Information on the Laplace Transform and Hardy Spaces

References on this subject are [1, Ch. 2] and [16, Part One Ch. 6]. Let X be any separable Hilbert space. We use $\Pi _+$ to denote the right half plane, $\Pi _+=\{\mathfrak {R}{\textstyle e}\,\lambda >0\}$. The Hardy space $H^2(\Pi _+;X)$ is the space of the functions $F(\lambda )$ which are analytic in $\mathfrak {R}{\textstyle e}\,\lambda >0$ and such that

$$\begin{aligned} \Vert F\Vert ^2_{H^2(\Pi _+;X)}=\sup _{x>0}\int _{-\infty }^{+\infty } \Vert F(x+iy)\Vert _{X}^2\;\text{ d }y<+\infty . \end{aligned}$$

The linear space $H^2(\Pi _+;X)$ endowed with this norm is a Hilbert space and the Laplace transformation $f\mapsto {{\hat{f}}}=F$ is bounded and boundedly invertible between $L^2(0,+\infty ;X)$ and $H^2(\Pi _+;X)$.

We need the following additional pieces of information:

• if $F\in H^2(\Pi _+;X)$ then $\lim _{x\rightarrow 0^+}F(x+iy)=F(iy)$ exists a.e. and exists in the sense of $L^2(i{\mathbb {R}})$, i.e. $\lim _{x\rightarrow 0^+}\Vert F(x+iy)-F(iy)\Vert _{L^2(-\infty ,+\infty ;X)}= 0$.

• The space $\{ F(iy),\ F(x+iy)\in H^2(\Pi _+;X)\}$ is a closed subspace of $L^2(-\infty ,+\infty ;X)$ and it turns out that

  $$\begin{aligned} \langle F,G\rangle _{H^2(\Pi _+;X)}= & {} \sup _{x>0}\left\{ \int _{-\infty }^{+\infty } {{\bar{G}}}(x+iy)F(x+iy)\;\text{ d }y\right\} \\= & {} \int _{-\infty }^{+\infty } {{\bar{G}}}( iy)F( iy)\;\text{ d }y \end{aligned}$$

  (i.e. $H^2(\Pi _+;X)$ and such closed subspace of $L^2(-\infty ,+\infty ;X)$ are isometric).

• In the special case of a real valued function f(t), $\overline{{{\hat{f}}}}(x+iy)={{\hat{f}}}(x-iy)$ and so, if f and g are real valued and $F={{\hat{f}}}$, $G={{\hat{g}}}$, we have

  $$\begin{aligned} \langle F,G\rangle _{H^2(\Pi _+;X)}= & {} \sup _{x>0}\left\{ \int _{-\infty }^{+\infty } {{\bar{G}}}(x+iy)F(x+iy)\;\text{ d }y\right\} \\= & {} \int _{-\infty }^{+\infty } G(- iy)F( iy)\;\text{ d }y. \end{aligned}$$

• Let f and g belong to $L^2(0,+\infty ;X)$. Then,

  $$\begin{aligned} h(t)=\int _0^t\langle f(t-s),g(s)\rangle \;\text{ d }s\in L^\infty (0,+\infty )\cap C([0,+\infty )) \end{aligned}$$

  and we have:

  • The Laplace transform of h(t) is

    $$\begin{aligned} {{\hat{h}}}(\lambda )=\langle {{\hat{f}}}(\lambda ),\overline{{{\hat{g}}}}(\lambda )\rangle _X. \end{aligned}$$

  • In the special case that f(t) and g(t) take values in a real Hilbert space X then h(t) is real valued and

    $$\begin{aligned} {{\hat{h}}}(i\omega )= \langle {{\hat{f}}}(i\omega ),\overline{{{\hat{g}}}}(i\omega )\rangle _X=\langle {{\hat{f}}}(i\omega ), {{\hat{g}}}(-i\omega )\rangle _X. \end{aligned}$$

• when $f\in L^2(0,+\infty ;X)$ i.e. ${{\hat{f}}}\in H^2(\Pi _+;X)$, the usual formula of the inverse Laplace transform holds

  $$\begin{aligned} f(t)=\frac{1}{2\pi } \int _{-\infty }^{+\infty }e^{i\omega t}{{\hat{f}}}(i\omega )\;\text{ d }\omega \end{aligned}$$

  (29)

  in the sense that the integral computed on $[-T,T]$ converges to f in $L^2(0,+\infty ;X)$ (and converges to zero for $t<0$) when $T\rightarrow +\infty$. If it happens that ${{\hat{f}}}(i\omega )\in L^1(-\infty ,+\infty ;X)$ then f(t) is continuous and the convergence (for $T\rightarrow +\infty$) is uniform on compact subsets of $[0,+\infty )$.

• If there exists $\alpha \in {\mathbb {R}}$ such that $F(\lambda +\alpha )\in H^2(\Pi _+;X)$ then $F(\lambda )={{\hat{f}}}(\lambda )$ where the function f(t) is given by (c is any real number larger then $\alpha$)

  $$\begin{aligned} f(t)=\frac{1}{2\pi }\int _{c-i\infty }^{c+i\infty } e^{(c+i\omega ) t} F(c+i\omega )\;\text{ d }\omega = \frac{1}{2\pi i}\int _{R_c}e^{\lambda t} F(\lambda )\;\text{ d }\lambda \end{aligned}$$

  (30)

  where $R_c=c+i\omega$, $-\infty<\omega <+\infty$.

• When $f\in {\mathcal {D}} ( (0,+\infty ); X )$ its Laplace transform ${{\hat{f}}}(\lambda )$ decays faster then $1/|\lambda |^k$ when $|\lambda |\rightarrow +\infty$ in $\mathfrak {R}{\textstyle e}\,\lambda \ge 0$, for every $k>0$.

1.3 The Proofs of Theorems 7 and 9

The proof follows established routes, see for examples [31]. In fact, similar ideas are used in the study of analytic semigroups as in [30].

We sketch the proofs in order to see the role of the assumptions, in particular of the assumptions on ${{\hat{K}}}(\lambda )$.

The idea is to recover the candidate solution $w(t;w_0)=E(t)w_0$ as the inverse Laplace transform of $(\lambda {{\hat{K}}}(\lambda )I-J(\lambda ) A)^{-1}{{\hat{K}}}(\lambda )$. The restriction of this function to vertical lines is not integrable so that we cannot use formula (30) to compute its inverse Laplace transformation. Instead, we integrate on the path $G_\epsilon$ in (10) and represented in Fig. 1.

We recall that the angle $\alpha \in (\pi /2,\theta )$ is fixed.

Fig. 1

The path of integration $G_\epsilon$. The dotted angle is $\Sigma _{\theta }$

Then we define

$$\begin{aligned} E(z)=\frac{1}{ 2\pi i}\int _{G_\epsilon } e^{z\lambda } {{\hat{K}}}(\lambda )\left( \lambda {{\hat{K}}}(\lambda )I-J(\lambda )A\right) ^{-1}\;\text{ d }\lambda . \end{aligned}$$

Of course the improper integral is computed as the limit of integrals on $G_{R,\epsilon }=G_{\epsilon }\cap \{\lambda \,\ |\lambda |<R\}$. It is easy to see that the limit exists uniformly on compact subsets of the sector

$$\begin{aligned} \Sigma _{\theta }=\{z \,:\ |\arg z|<\theta ,\ z\ne 0\} \end{aligned}$$

and so E(z) is an analytic operator valued function on $\Sigma _\theta$.

The operator AE(z) has similar properties. This is simply seen as follows: the operator valued function AE(z) is bounded on every compact subset of $\Sigma _\theta$ as seen from

$$\begin{aligned} {{\hat{K}}}(\lambda )A\left( \lambda {{\hat{K}}}(\lambda )I-J(\lambda )A\right) ^{-1} =-I+\frac{\lambda {{\hat{K}}}(\lambda )}{J(\lambda )}\left( \frac{\lambda {{\hat{K}}}(\lambda )}{J(\lambda )}I-A\right) ^{-1}. \end{aligned}$$

The second addendum is bounded from the assumptions on the memory kernels and the first set of the inequalities (9) while $e^{z\lambda }$ is dominated by a decaying exponential if z belongs to a compact subset of $\Sigma _\theta$ (in particular it is far from 0).

Let now $w\in L^2(\Omega )$ and ${{\tilde{w}}}\in \mathrm{Dom}A$. Then we have (the crochet denotes the inner product in $L^2(\Omega )$)

$$\begin{aligned} \langle AE(z)w,{{\tilde{w}}}\rangle =\langle E(z)w,A{{\tilde{w}}}\rangle \end{aligned}$$

so that $z\mapsto \langle AE(z)w,{{\tilde{w}}}\rangle$ is an analytic valued function for every ${{\tilde{w}}}\in \mathrm{Dom}A$ (which is dense in $L^2(\Omega )$). We use [18, Remark 1.38 p. 139] and we deduce that $z\mapsto AE(z)$ is n analytic operator valued function on $\Sigma _\theta$.

Now we prove that E(t) is bounded for $t>0$. We proceed as in [7]: we change the variable $\lambda =\zeta /t$ and we see that E(t) is given by

$$\begin{aligned} E(t)=\frac{1}{2\pi i} \int _{G_{\epsilon t}} e^{\zeta }\left( 1/t\right) {{\hat{K}}} (\zeta /t )\left( \frac{\zeta }{t}{{\hat{K}}}(\zeta /t)I-J(\zeta /t)A\right) ^{-1}\;\text{ d }\zeta . \end{aligned}$$

The straight lines of the paths $G_{\epsilon t}$ and $G_\epsilon$ coincide but the paths are closed by the circular arcs of radius respectively $\epsilon t$ and $\epsilon$ (with the same center zero). The difference of the integration paths does not enclose singularities of the integrand so that we have also

$$\begin{aligned} E(t)=\frac{1}{2\pi i} \int _{G_{\epsilon }} e^{\zeta }\left( 1/t\right) {{\hat{K}}} (\zeta /t )\left( \frac{\zeta }{t}{{\hat{K}}}(\zeta /t)I-J(\zeta /t)A\right) ^{-1}\;\text{ d }\zeta . \end{aligned}$$

Boundedness uniformly in $t >0$ follows since

$$\begin{aligned} \left| \frac{1}{t}{{\hat{K}}} (\zeta /t ) \right| \,\left\| \left( \frac{\zeta }{t}{{\hat{K}}}(\zeta /t)I-J(\zeta /t)A\right) ^{-1} \right\| \le \frac{M}{|\zeta |}. \end{aligned}$$

Finally, we prove that

$$\begin{aligned} \lim _{t \rightarrow 0+} E(t)w_0=w_0\quad \forall w_0\in L^2(\Omega ). \end{aligned}$$

As we already noted that E(t) is a bounded function of t, we can confine ourselves to prove this property for $w_0=A^{-1} y_0$.

The first formula of the resolvent (see [16, p. 126]) gives the equality

$$\begin{aligned}&\frac{1}{2\pi i}\int _{G_\epsilon } e^{\lambda t}{{\hat{K}}}(\lambda )\left( \lambda {{\hat{K}}}(\lambda ) I-J(\lambda )A\right) ^{-1} w_0\;\text{ d }\lambda \\&\quad =\frac{1}{2\pi i}\int _{G_\epsilon } e^{\lambda t} \frac{1}{\lambda } A^{-1}y_0\;\text{ d }\lambda +\frac{1}{2\pi i}\int _{G_\epsilon }e^{\lambda t} \frac{1}{\lambda } \left( \frac{\lambda {{\hat{K}}}(\lambda )}{J(\lambda )}I-A\right) ^{-1}y_0\;\text{ d }\lambda . \end{aligned}$$

The first integral is $A^{-1}y_0$ from Cauchy integral formula and so it is sufficient to note that the limit of the second integral for $t\rightarrow 0+$ is equal zero. To prove this fact, we again use the transformation $\lambda =\zeta /t$ and write the second integral as

$$\begin{aligned} \int _{G_\epsilon } e^{\zeta }\left( 1/\zeta \right) \left( \frac{\zeta }{t} \frac{{{\hat{K}}}(\zeta /t)}{J(\zeta /t)}I-A\right) ^{-1} y_0\;\text{ d }\zeta . \end{aligned}$$

The condition in item 5a of the set of the Assumptions 5 shows that

$$\begin{aligned} \left\| \left( \frac{\zeta }{t} \frac{{{\hat{K}}}(\zeta /t)}{J(\zeta /t)}I-A\right) ^{-1}\right| \le M\frac{|J(\zeta /t)|}{|(\zeta /t){{\hat{K}}}(\zeta /t)|}\le \frac{M}{|\zeta /t|^{\gamma _0}} \end{aligned}$$

tends to zero when $t\rightarrow 0^+$. The integral tends to zero since $e^{\zeta }/\zeta$ tends to zero exponentially fast when $|\zeta |\rightarrow +\infty$, $\zeta \in G_\epsilon$.

Now we compute the Laplace transform ${{\hat{E}}}(s)w_0$ in $\mathfrak {R}{\textstyle e}\,s>0$ of $t\mapsto E(t)w_0$ and we prove that it is precisely ${{\hat{K}}}(\lambda ) (\lambda {{\hat{K}}}( \lambda )I-J(\lambda )A )^{-1}w_0$, as it must be if we want to choose $E(t) w_0$ as the definition of the solution $w(t;w_0)$:

$$\begin{aligned} {{\hat{E}}}(s)w_0= & {} \frac{1 }{2\pi i}\int _0^{+\infty } e^{-st}\left[ \int _{G_\epsilon } e^{t\lambda } {{\hat{K}}}(\lambda )\left( \lambda {{\hat{K}}}(\lambda )I-J(\lambda )A\right) ^{-1}w_0\;\text{ d }\lambda \right] \;\text{ d }t\\= & {} \frac{1}{2\pi i} \int _{G_\epsilon } \frac{1}{s-\lambda } {{\hat{K}}}(\lambda )\left( \lambda {{\hat{K}}}(\lambda )I-J(\lambda )A\right) ^{-1} w_0 \;\text{ d }\lambda \\= & {} {{\hat{K}}}(s)\left( s {{\hat{K}}}(s)I-J(s)A\right) ^{-1}w_0. \end{aligned}$$

The last integral is computed as the limit of the integrals on $G_{R,\epsilon }=G_{\epsilon }\cap \{\lambda \,\ |\lambda |<R\}$ completed with the circular arch of radius R and which intersect the right half plane $\mathfrak {R}{\textstyle e}\,\lambda >0$. The equality follows since the integrand has the sole singularity $\lambda =s$ in the interior region while the contribution of the circular arch tends to zero because the integrand decays as $1/\lambda ^2$ (we take into account the fact that the path is described in the negative sense).

This ends the proof of Theorem 7 and we proved also the representation formula (11). $\square$

The previous arguments show that it makes sense to use $w(t;w_0)=E(t)w_0$ as the mild solution of Eq. (1) (when $F=0$, $f=0$) and that the evolution operator E(t) admits an analytic extension to a sector enclosing the axis $t>0$.

In order to prove Theorem 9 we first examine $(\lambda {{\hat{K}}}(\lambda ) I-J(\lambda ) A )^{-1}{{\hat{F}}}(\lambda )$ and we prove that when $F\in L^2 (0,+\infty ;L^2(\Omega ) )$ this function is the Laplace transformation of a square integrable $L^2(\Omega )$-valued function.

By assumption, ${{\hat{F}}}(\lambda )\in H^2 (\Pi _+;L^2(\Omega ) )$ and

$$\begin{aligned} \left( \lambda {{\hat{K}}}(\lambda )-J(\lambda ) A\right) ^{-1}{{\hat{F}}}(\lambda )\in H^2 (\Pi _+;L^2(\Omega ) ) \end{aligned}$$

(31)

because (from (9))

$$\begin{aligned} \left\| \left( \lambda {{\hat{K}}}(\lambda ) I-J(\lambda ) A\right) ^{-1}\right\| \le \frac{M}{\left| \lambda {{\hat{K}}}(\lambda )+\omega J(\lambda )\right| }\;. \end{aligned}$$

In fact we noted in Remark 6 that the denominator is not zero, and it does not approach zero for $\lambda \rightarrow 0$ and $|\lambda |\rightarrow +\infty$ in $\mathfrak {R}{\textstyle e}\,\lambda >0$ as it follows from the Assumption 5 item 5a.

The fact that the multiplier $(\lambda {{\hat{K}}}(\lambda ) I-J(\lambda ) A )^{-1}$ is analytic and bounded in the right half plane implies also that the transformation from ${{\hat{F}}}\in H^2 (\Pi _+;L^2(\Omega ) )$ to the function in (31), as an element of $H^2 (\Pi _+;L^2(\Omega ) )$, is bounded and so, passing to the time domain, there exists a transformation ${\mathcal {E}}_d\in {\mathcal {L}} ( L^2 (0,T;L^2(\Omega ) ) )$ such that the Laplace transform of $({\mathcal {E}}F )(t)$ is $(\lambda {{\hat{K}}}(\lambda ) I-J(\lambda ) A )^{-1}{{\hat{F}}}(\lambda )$, as we wanted to prove.

For every $T>0$ the transformation $\left( {\mathcal {E}}_{d,T}F\right)$ is just the restriction of ${\mathcal {E}}F$ to (0, T) (in case that F is defined only on (0, T) we intend that it is extended with 0 for $t>T$).

This justify taking the inverse Laplace transform of $(\lambda {{\hat{K}}}(\lambda ) I-J(\lambda ) A )^{-1} {{\hat{F}}}(\lambda )$ as the solution w(t; F) .

Now we prove the representation formula (12). Using the formula for the Laplace transform of a convolution, it is sufficient to prove that the integral (13) converges in $L^1 (0,T;{\mathcal {L}}(X) )$ (convergence in $C ( [a,T];{\mathcal {L}}(X) )$ for $a>0$ is clear) and then to compute its Laplace transform.

In order to prove convergence in $L^1 (0,T;{\mathcal {L}}(X) )$ we prove convergence of the integral of the norm. We consider the integrals on $G_+$ (the integral on $G_-$ is treated analogously). Using the parametrization of $G_+$ in (10) we see that

$$\begin{aligned} \left\| e^{\lambda t}\left( \lambda {{\hat{K}}}(\lambda )-J(\lambda ) A\right) ^{-1}\right\| \le e^{-st|\cos \alpha |} \frac{M}{s^{\gamma _0}}\;. \end{aligned}$$

We use this inequality to prove that

$$\begin{aligned} \lim _{R\rightarrow +\infty }\int _0^{+\infty } \int _R^{+\infty } e^{-stc}\frac{1}{s^{\gamma _0}} \;\text{ d }s\,\;\text{ d }t=0,\qquad c=|\cos \alpha |. \end{aligned}$$

We replace $stc=\nu$ in the inner integral and then $Rtc=\xi$ and we see that the integral is equal to

$$\begin{aligned}&\frac{1}{cR^{\gamma _0}} \int _0^{+\infty } \frac{1}{\xi ^{1-\gamma _0}} \int _\xi ^{+\infty } e^{-\nu }\frac{1}{\nu ^{\gamma _0}} \;\text{ d }\nu \,\;\text{ d }\xi \\&\quad =\frac{1}{cR^{\gamma _0}} \int _0^{+\infty }e^{-\nu }\frac{1}{\nu ^{\gamma _0}}\int _0^{\nu }\frac{1}{\xi ^{1-\gamma _0}}\;\text{ d }\xi \,\;\text{ d }\nu =\frac{1}{\gamma _0cR^{\gamma _0}}\rightarrow 0. \end{aligned}$$

Formula (12) is seen by proving that $(\lambda {{\hat{K}}}(\lambda )I-J(\lambda )A )^{-1}$ is the Laplace transform of (13), similar to what we did above. In fact, we can exchange the order of integration in the computation of the Laplace transformation:

$$\begin{aligned}&\int _0^{+\infty } e^{-\zeta t}\left[ \frac{1}{2\pi i}\int _{G_\epsilon } e^{\lambda t} \left( \lambda {{\hat{K}}}(\lambda )-J(\lambda ) A\right) ^{-1}\;\text{ d }\lambda \right] \;\text{ d }t\\&\quad =\frac{1}{2\pi i} \int _{G_\epsilon } \left[ \int _0^{+\infty } e^{t( \zeta -\lambda )}\;\text{ d }t\right] \left( \lambda {{\hat{K}}}(\lambda )-J(\lambda ) A\right) ^{-1}\;\text{ d }\lambda \\&\quad =\frac{1}{2\pi i} \int _{G_\epsilon } \frac{1}{\lambda -\zeta }\left( \lambda {{\hat{K}}}(\lambda )-J(\lambda ) A\right) ^{-1}\;\text{ d }\lambda = \left( \zeta {{\hat{K}}}(\zeta )-J(\zeta ) A\right) ^{-1} \end{aligned}$$

from Cauchy integral formula.

An analogous argument (based on the second inequality in (9)) shows that $f\mapsto w_f(t)$ ($t\in [0,T]$) is linear and continuous from $L^2 (0,T;L^2(\Gamma _a) )$ to $L^2 (0,T;L^2(\Omega ) )$ and the representation formula (12).

Now we observe that if F or f are $C^\infty$ with compact support then w(t; F) and $w_f(t)$ are continuous $L^2(\Omega )$ valued functions. In fact, the assumed regularity of F or f implies that for every k there exists $M_k>0$ such that, respectively,

$$\begin{aligned} \left\| {{\hat{F}}}(\lambda )\right\| _{L^2(\Omega )}\le \frac{M_k}{|\lambda |^k},\quad \left\| {{\hat{f}}}(\lambda )\right\| _{L^2(\Gamma )}\le \frac{M_k}{|\lambda |^k}. \end{aligned}$$

Hence:

$$\begin{aligned} \left( i\omega {{\hat{K}}}(i\omega )-J(i\omega )A\right) ^{-1}{{\hat{F}}}(i\omega ),\qquad \left( i\omega {{\hat{K}}}(i\omega )-J(i\omega )A\right) ^{-1}AG{{\hat{f}}}(i\omega ) \end{aligned}$$

are integrable on the imaginary axis and the inverse Laplace transformations are continuous $L^2(\Omega )$-valued functions.

Finally, if $F\in {\mathcal {D}} (\Omega \times (0,+\infty ) )$ then we have also $F\in C^{\infty } ([0,T],\mathrm{Dom}A^k )$ (and also $F^{(k)}(0)=0$) for every k so that the representation formula (12) shows that $w(t;F)\in C ([0,T];\mathrm{Dom}\, A^k )$ for every k. The proof of Theorem 9 is finished. $\square$

Finally, we give an example which shows that $t\mapsto w(t;F)$ needs not be continuous. We use the semigroup property of the Riemann-Liouville fractional integral:

$$\begin{aligned} J^\gamma \circ J^\sigma =J^{\gamma +\sigma }. \end{aligned}$$

Example 21

We fix $\epsilon \in (0,1/4)$ and $\gamma = \epsilon +1/2<1$. We consider the equation

$$\begin{aligned} J^\gamma w'=\Delta w+F\qquad w(0)=w_0=0,\quad f=0. \end{aligned}$$

We apply $J^{1-\gamma }$ to both the sides and we get the equation

$$\begin{aligned} J^1w'=w(t)=J^{1-\gamma }\Delta w+J^{1-\gamma } F. \end{aligned}$$

We choose $F=F(x,t)=F_0(t)$ and we put $c_n=\int _\Omega \phi _n(x)\;\text{ d }x$ where $\phi _n$ is an eigenfunction of the operator A such that $c_n\ne 0$ ($-\mu _n^2$ be the eigenvalue).

We project on the eigenfunction $\phi _n$. Let $w_n(t)=\int _{\Omega }w(x,t)\phi _n(x)\;\text{ d }x$. We get

$$\begin{aligned} w_n(t)+\mu _n^2 J^{1-\gamma } w_n = c_n J^{1-\gamma } F. \end{aligned}$$

If $t\mapsto w(\cdot ,t)$ is a continuous $L^2(\Omega )$ valued function, then the left hand side is a continuous function of t. In contrast with this, the right hand side is not continuous at $t=T$ when, for example,

$$\begin{aligned} F(t)=\frac{1}{\left( T-s\right) ^{(1/2)-\epsilon }}\in L^2(0,T). \end{aligned}$$

In fact in this case we have for $t<T$

$$\begin{aligned} (J^{1-\gamma } F)(t)=\frac{1}{\Gamma (\gamma )}\int _0^t\frac{1}{(t-s)^{(1/2)+\epsilon }} \frac{1}{(T-s)^{(1/2)-\epsilon }}\;\text{ d }s. \end{aligned}$$

This is a continuous function of t for $t<T$ but the limit for $t\rightarrow T^-$ is not finite. In fact, $t-s\le T-s$ so that

$$\begin{aligned} \lim _{t\rightarrow T^-}\int _0^t\frac{1}{(t-s)^{(1/2)+\epsilon }} \frac{1}{(T-s)^{(1/2)-\epsilon }}\;\text{ d }s\ge \lim _{t\rightarrow T^-} \int _0^t\frac{1}{T-s}\;\text{ d }s\\ = \lim _{t\rightarrow T^-}\left[ \log T-\log (T-t)\right] = +\infty . \end{aligned}$$

$\square$