Semiglobal Oblique Projection Exponential Dynamical Observers for Nonautonomous Semilinear Parabolic-Like Equations

Abstract

The estimation of the full state of a nonautonomous semilinear parabolic equation is achieved by a Luenberger-type dynamical observer. The estimation is derived from an output given by a finite number of average measurements of the state on small regions. The state estimate given by the observer converges exponentially to the real state, as time increases. The result is semiglobal in the sense that the error dynamics can be made stable for an arbitrary given initial condition, provided a large enough number of measurements, depending on the norm of the initial condition, are taken. The output injection operator is explicit and involves a suitable oblique projection. The results of numerical simulations are presented showing the exponential stability of the error dynamics.

Appendix

1.1 Proof of Proposition 3.7

With ${{\widehat{y}}}_1{:=}y+z_1$ and ${\widehat{y}}_2{:=}y+z_2$, we write

$$\begin{aligned} {{\mathfrak {N}}}_y(t,z_1)-{{\mathfrak {N}}}_y(t,z_2)&={{\mathcal {N}}}(t,{{\widehat{y}}}_1)-{{\mathcal {N}}}(t, y)-({{\mathcal {N}}}(t,{{\widehat{y}}}_2)-{{\mathcal {N}}}(t, y)) \\&={{\mathcal {N}}}(t,{{\widehat{y}}}_1)-{{\mathcal {N}}}(t,{{\widehat{y}}}_2) ={{\mathcal {N}}}(t,y+z_1)-{{\mathcal {N}}}(t,y+z_2), \end{aligned}$$

which leads us to ${{\widehat{y}}}_1-{{\widehat{y}}}_2=z_1-z_2{=:}d$ and, using Lemma 3.4,

$$\begin{aligned}&2\Bigl ( {{\mathfrak {N}}}_y(t,z_1)-{{\mathfrak {N}}}_y(t,z_2),A(z_1-z_2)\Bigr )_{H} =2\Bigl ( {{\mathcal {N}}}(t,{{\widehat{y}}}_1)-{{\mathcal {N}}}(t,{{\widehat{y}}}_2),A({{\widehat{y}}}_1-{{\widehat{y}}}_2))\Bigr )_{H} \\&\quad \le {{\widehat{\gamma }}}_0 \left| d\right| _{D(A)}^{2} +\!\left( \!1+{{\overline{\gamma }}}_0^{-\frac{1+\Vert \delta _2\Vert }{1-\Vert \delta _2\Vert } }\right) \!{{\overline{C}}}_{{{\mathcal {N}}}1}\sum \limits _{j=1}^n\left| d\right| _{V}^{{ 2}}\sum \limits _{k=1}^2 \left| {{\widehat{y}}}_k\right| _{V}^\frac{2\zeta _{1j}}{{ \delta _{1j}}} \left| {{\widehat{y}}}_k\right| _{D(A)}^\frac{2\zeta _{2j}}{{ \delta _{1j}}} . \end{aligned}$$

Therefore, (3.5) follows with $\widetilde{C}_{{{\mathfrak {N}}}1}={{\overline{C}}}_{{{\mathcal {N}}}1}= {{\overline{C}}}_{\left[ n,\frac{1}{1-\Vert \delta _{2}\Vert },C_{{\mathcal {N}}}\right] }$.

By setting $z_2=0$ in (3.5), we obtain for each ${{\widetilde{\gamma }}}_0>0$,

$$\begin{aligned}&2\Bigl ( {{\mathfrak {N}}}_y(t,z_1),Az_1\Bigr )_{H}\nonumber \\&\quad \le {{\widetilde{\gamma }}}_0 \left| z_1\right| _{D(A)}^{2} +\!\left( \!1+{{\widetilde{\gamma }}}_0^{-\frac{1+\Vert \delta _2\Vert }{1-\Vert \delta _2\Vert } }\right) \!{{\widetilde{C}}}_{{{\mathfrak {N}}}1} \sum \limits _{j=1}^n\left| z_1\right| _{V}^{{ 2}} \sum \limits _{l=0}^1\left| y+lz_1\right| _{V}^\frac{2\zeta _{1j}}{{ \delta _{1j}}} \left| y+lz_1\right| _{D(A)}^\frac{2\zeta _{2j}}{{ \delta _{1j}}}. \end{aligned}$$

(A.1)

Now, for simplicity we fix j, and set

$$\begin{aligned}&p =p_j{:=}\frac{2\zeta _{1j}}{ \delta _{1j}}\quad \text{ and }\quad q=q_j{:=}\frac{2\zeta _{2j}}{ \delta _{1j}}<2. \end{aligned}$$

Note that $p\ge 0$ and $q\in [0,2)$ due to the relations $\delta _{1j}+\delta _{2j} =1$ and $\zeta _{2j}+\delta _{2j}<1$, in Assumption 2.4.

We consider first the case $q\ne 0$. By the triangle inequality and Phan and Rodrigues (2017, Prop. 2.6), we obtain

$$\begin{aligned} \Upsilon _j&{:=}\left| z_1\right| _{V}^{{ 2}}\sum \limits _{l=0}^1\left| y+lz_1\right| _{V}^p \left| y+lz_1\right| _{D(A)}^q\nonumber \\&=\left| z_1\right| _{V}^{{ 2}}\left| y\right| _{V}^p \left| y\right| _{D(A)}^q +\left| z_1\right| _{V}^{{ 2}}\left| y+z_1\right| _{V}^p \left| y+z_1\right| _{D(A)}^q\nonumber \\&\le \left| z_1\right| _{V}^{{ 2}}\left| y\right| _{V}^p \left| y\right| _{D(A)}^q\nonumber \\&\quad +\left| z_1\right| _{V}^{{ 2}}(1+2^{p-1})(1+2^{q-1})\left( \left| y\right| _{V}^{p}+\left| z_1\right| _{V}^{p}\right) \left( \left| y\right| _{D(A)}^q+\left| z_1\right| _{D(A)}^q\right) . \end{aligned}$$

(A.2)

Setting $D_{p,q}{:=}(1+2^{p-1})(1+2^{q-1})$ and using the Young inequality, the last term satisfies for each $\gamma _2>0$,

$$\begin{aligned} D_{p,q}^{-1}{{\mathcal {T}}}_j&{:=}\left| z_1\right| _{V}^{{ 2}}\left( \left| y\right| _{V}^{p}+\left| z_1\right| _{V}^{p}\right) \left( \left| y\right| _{D(A)}^q+\left| z_1\right| _{D(A)}^q\right) \\&\le \left| z_1\right| _{V}^{{ 2}}\left( \left| y\right| _{V}^{p}+\left| z_1\right| _{V}^{p}\right) \left| z_1\right| _{D(A)}^q +\left| z_1\right| _{V}^{{ 2}}\left( \left| y\right| _{V}^{p}+\left| z_1\right| _{V}^{p}\right) \left| y\right| _{D(A)}^q \\&\le \gamma _2^\frac{2}{q}\left| z_1\right| _{D(A)}^2 + \gamma _2^{-(1-\frac{q}{2})^{-1}}\left( \left| z_1\right| _{V}^{{ 2}} \left( \left| y\right| _{V}^{p}+\left| z_1\right| _{V}^{p}\right) \right) ^{(1-\frac{q}{2})^{-1}}\\&\quad +\left| z_1\right| _{V}^{{ 2}}\left( \left| y\right| _{V}^{p}+\left| z_1\right| _{V}^{p}\right) \left( 1+\left| y\right| _{D(A)}^{\left\| \frac{2\zeta _{2}}{ \delta _{1}}\right\| _{}}\right) , \end{aligned}$$

which implies, since $1-\frac{q}{2}=\frac{2-q}{2}$,

$$\begin{aligned} D_{p,q}^{-1}{{\mathcal {T}}}_j&\le \gamma _2^\frac{2}{q}\left| z_1\right| _{D(A)}^2 + \gamma _2^{-\frac{2}{2-q}}\left| z_1\right| _{V}^{\frac{ 4}{2-q}}\left( \left| y\right| _{V}^{p} +\left| z_1\right| _{V}^{p}\right) ^{\frac{2}{2-q}}\\&\quad +\left| z_1\right| _{V}^{{ 2}}\left( \left| y\right| _{V}^{p}+\left| z_1\right| _{V}^{p}\right) \left( 1+\left| y\right| _{D(A)}^{\left\| \frac{2\zeta _{2}}{ \delta _{1}}\right\| _{}}\right) \\&\le \gamma _2^\frac{2}{q}\left| z_1\right| _{D(A)}^2 + \gamma _2^{-\frac{2}{2-q}}(1+2^{\frac{2}{2-q}-1})\left| z_1\right| _{V}^{\frac{ 4}{2-q}} \left( \left| y\right| _{V}^{\frac{2p}{2-q}}+\left| z_1\right| _{V}^{\frac{2p}{2-q}}\right) \\&\quad +\left| z_1\right| _{V}^{{ 2}}\left( \left| y\right| _{V}^{p}+\left| z_1\right| _{V}^{p}\right) \left( 1+\left| y\right| _{D(A)}^{\left\| \frac{2\zeta _{2}}{ \delta _{1}}\right\| _{}}\right) \\&\le \gamma _2^\frac{2}{q}\left| z_1\right| _{D(A)}^2\\&\quad + D_{q,\gamma _2}\left( \left| z_1\right| _{V}^{\frac{ 4}{2-q}} \left| y\right| _{V}^{\frac{2p}{2-q}}+\left| z_1\right| _{V}^{\frac{2( 2 +p)}{2-q}} +\left| z_1\right| _{V}^{{ 2}}\left| y\right| _{V}^{p}+\left| z_1\right| _{V}^{ 2 +p}\right) \left( 1+\left| y\right| _{D(A)}^{\left\| \frac{2\zeta _{2}}{ \delta _{1}}\right\| _{}}\right) \end{aligned}$$

with

$$\begin{aligned} D_{q,\gamma _2}{:=}1+\gamma _2^{-\frac{2}{2-q}}(1+2^{\frac{2}{2-q}-1}), \end{aligned}$$

and then

$$\begin{aligned}&D_{q,\gamma _2}^{-1}\left( D_{p,q}^{-1}{{\mathcal {T}}}_j-\gamma _2^\frac{2}{q}\left| z_1\right| _{D(A)}^2\right) \left( 1+\left| y\right| _{D(A)}^{\left\| \frac{2\zeta _{2}}{ \delta _{1}}\right\| _{}}\right) ^{-1}\nonumber \\&\quad \le \left( \left| z_1\right| _{V}^{\frac{ 4}{2-q}-2}\left| y\right| _{V}^{\frac{2p}{2-q}} +\left| z_1\right| _{V}^{\frac{2( 2 +p)}{2-q}-2} +\left| y\right| _{V}^{p}+\left| z_1\right| _{V}^{p}\right) \left| z_1\right| _{V}^2. \end{aligned}$$

(A.3a)

Observe that

$$\begin{aligned} \frac{2p}{2-q}&=\frac{4\zeta _{1j}}{ 2\delta _{1j} -2\zeta _{2j}} \le 2\left\| \frac{\zeta _{1}}{\delta _{1}-\zeta _{2}} \right\| _{}, \end{aligned}$$

(A.4a)

$$\begin{aligned} \frac{2( 2 +q+p)-4}{2-q}&=\frac{4(\zeta _{1j}+\zeta _{2j})}{ 2\delta _{1j} -2\zeta _{2j}} \le 2\left\| \frac{\zeta _{1}+\zeta _{2}}{\delta _{1}-\zeta _{2}}\right\| _{}, \end{aligned}$$

(A.4b)

and that $\frac{2c}{2-q}\ge c\Longleftrightarrow 0\ge -qc$. Thus, we obtain that $\frac{2c}{2-q}\ge c$ for all $c\ge 0$, and hence,

$$\begin{aligned} \frac{2( 2 +q+p)-4}{2-q}&\ge \frac{2p}{2-q}\ge p\quad \text{ and }\quad \frac{2( 2 +q+p)-4}{2-q}\\&\ge \frac{2( 2 +q)-4}{2-q} = \frac{2q}{2-q}\ge q, \end{aligned}$$

which together with (A.3) give us

$$\begin{aligned}&D_{q,\gamma _2}^{-1}\left( D_{p,q}^{-1}{{\mathcal {T}}}_j-\gamma _2^\frac{2}{q}\left| z_1\right| _{D(A)}^2\right) \left( 1+\left| y\right| _{D(A)}^{\left\| \frac{2\zeta _{2}}{ \delta _{1}}\right\| _{}}\right) ^{-1} \\&\quad \le \left( \left| z_1\right| _{V}^{\frac{2( 2 +q)-4}{2-q}}+\left| z_1\right| _{V}^{\frac{2( 2 +q+p)-4}{2-q}} +1+\left| z_1\right| _{V}^{p}\right) \left( 2+\left| y\right| _{V}^{\frac{2p}{2-q}} +\left| y\right| _{V}^{p}\right) \left| z_1\right| _{V}^2 \\&\quad \le \left( 4+3\left| z_1\right| _{V}^{{ 2\left\| \frac{\zeta _{1}+\zeta _{2}}{\delta _{1}-\zeta _{2}}\right\| _{}}}\right) \left( 4+2\left| y\right| _{V}^{{ 2\left\| \frac{\zeta _{1}}{\delta _{1}-\zeta _{2}}\right\| _{}}} \right) \left| z_1\right| _{V}^2, \end{aligned}$$

which implies

$$\begin{aligned}&{{\mathcal {T}}}_j \le D_{p,q}\gamma _2^\frac{2}{q}\left| z_1\right| _{D(A)}^2 + {{\widehat{D}}}\left( 1+\left| y\right| _{V}^{\chi _1} \right) \left( 1+\left| y\right| _{D(A)}^{\chi _2}\right) \left( 1+\left| z_1\right| _{V}^{\chi _3}\right) \left| z_1\right| _{V}^2, \end{aligned}$$

(A.5a)

$$\begin{aligned}&D_{p,q}{:=}(1+2^{p-1})(1+2^{q-1}),\quad {{\widehat{D}}}{:=}16D_{p,q}D_{q,\gamma _2}, \end{aligned}$$

(A.5b)

$$\begin{aligned}&\chi _1{:=}2\left\| \frac{\zeta _{1}}{\delta _{1}-\zeta _{2}}\right\| _{} \ge 0, \quad \chi _2{:=}{\left\| \frac{2\zeta _{2}}{ \delta _{1}}\right\| _{}}\ge 0 ,\quad \chi _3{:=}2\left\| \frac{\zeta _{1}+\zeta _{2}}{\delta _{1}-\zeta _{2}}\right\| _{} \ge 0. \end{aligned}$$

(A.5c)

For the first term on the right-hand side of (A.2), we also obtain

$$\begin{aligned} {{\mathcal {F}}}_j{:=}\left| z_1\right| _{V}^2\left| y\right| _{V}^p \left| y\right| _{D(A)}^q \le \left( 1+\left| y\right| _{V}^{\chi _1}\right) \left( 1+\left| y\right| _{D(A)}^{\chi _2}\right) \left| z_1\right| _{V}^2 \end{aligned}$$

(A.6)

because $p\le \frac{2p}{2-q}\le \chi _1$ and $0<q\le \chi _2$; see (A.4).

Therefore, by (A.2), (A.5), and (A.6), it follows that for all $\gamma _2\in (0,1]$,

$$\begin{aligned} \Upsilon _j&\le {{\mathcal {F}}}_j+{{\mathcal {T}}}_j\nonumber \\&\le D_{p,q}\gamma _2^\frac{2}{q}\left| z_1\right| _{D(A)}^2 + (1+{\widehat{D}})\left( 1+\left| y\right| _{V}^{\chi _1} \right) \left( 1+\left| y\right| _{D(A)}^{\chi _2}\right) \left( 1+\left| z_1\right| _{V}^{\chi _3}\right) \left| z_1\right| _{V}^2,\nonumber \\&\le D_{p,q}\gamma _2^{\frac{2}{\chi _2}}\left| z_1\right| _{D(A)}^2 + (1+{{\widehat{D}}})\left( 1+\left| y\right| _{V}^{\chi _1} \right) \left( 1+\left| y\right| _{D(A)}^{\chi _2}\right) \left( 1+\left| z_1\right| _{V}^{\chi _3}\right) \left| z_1\right| _{V}^2, \end{aligned}$$

(A.7a)

$$\begin{aligned} \text{ for }&q\ne 0,\quad \text{ with }\quad \gamma _2\le 1. \end{aligned}$$

(A.7b)

Note that $\gamma _2^\frac{2}{q}\le \gamma _2^\frac{2}{\chi _2}$ because $\gamma _2\le 1$ and $0<q\le \chi _2$.

Finally, we consider the case $q=0$. We find

$$\begin{aligned} \Upsilon _j&{:=}\left| z_1\right| _{V}^{{ 2}}\sum \limits _{l=0}^1\left| y+lz_1\right| _{V}^p =\left| z_1\right| _{V}^{{ 2}}\left| y\right| _{V}^p +\left| z_1\right| _{V}^{{ 2}}\left| y+z_1\right| _{V}^p\nonumber \\&\le \left| z_1\right| _{V}^{{ 2}}\left( (2+2^{p-1})\left| y\right| _{V}^{p}+(1+2^{p-1})\left| z_1\right| _{V}^p\right) \nonumber \\&\le (2+2^{p-1}) \left( 1+\left| y\right| _{V}^{p}\right) \left( 1 +\left| z_1\right| _{V}^{p}\right) \left| z_1\right| _{V}^2\nonumber \\&\le \left( 2+2^{p-1}\right) \left( 2+\left| y\right| _{V}^{\chi _1}\right) \left( 2+\left| z_1\right| _{V}^{\chi _3}\right) \left| z_1\right| _{V}^2 \nonumber \\&\le 8 \left( 1+2^{p-1}\right) \left( 1+\left| y\right| _{V}^{\chi _1}\right) \left( 1+\left| z_1\right| _{V}^{\chi _3}\right) \left| z_1\right| _{V}^2,\qquad q=0. \end{aligned}$$

(A.8a)

$$\begin{aligned} \text{ for }&q=0,\quad \text{ with }\quad \gamma _2\le 1, \end{aligned}$$

(A.8b)

with $\chi _1$ and $\chi _3$ as in (A.5), where we have used (A.4).

Now, we observe that

$$\begin{aligned} (1+2^{p-1})&\le D_{p,q}= (1+2^{p-1})(1+2^{q-1})\nonumber \\&\le \left( 1+2^{ \left\| \frac{2\zeta _1-\delta _1}{\delta _1}\right\| _{}}\right) \left( 1+2^{ \left\| \frac{2\zeta _2-\delta _1}{\delta _1}\right\| _{}}\right) {=:}\widetilde{D}_{1} \end{aligned}$$

(A.9a)

and, since $0<\frac{2}{2-q}\le \left\| \frac{\delta _1}{\delta _1-\zeta _2}\right\| _{}$,

$$\begin{aligned} {{\widehat{D}}}&= 16D_{p,q}D_{q,\gamma _2}\le 16\widetilde{D}_{1}(1+2^{\frac{2q}{2-q}}) \left( 1+\gamma _2^{-\frac{2}{2-q }}\right) \nonumber \\&\le 16{{\widetilde{D}}}_{1}\left( 2+2^{ \chi _3}\right) \left( 2+\gamma _2^{{- \left\| \frac{\delta _1}{\delta _1-\zeta _2}\right\| _{}}}\right) \le {{\widetilde{D}}}_{2}\left( 1+\gamma _2^{-\chi _4 }\right) , \end{aligned}$$

(A.9b)

$$\begin{aligned} {{\widetilde{D}}}_{2}&{:=}32{{\widetilde{D}}}_{1} \left( 2+2^{ \chi _3}\right) , \end{aligned}$$

(A.9c)

$$\begin{aligned} \chi _4&{:=}\left\| \frac{\delta _1}{\delta _1-\zeta _2}\right\| _{} \ge 1. \end{aligned}$$

(A.9d)

Further, from ${{\widetilde{D}}}_{2}>64{{\widetilde{D}}}_{1} \ge 8\left( 1+2^{p-1}\right)$ and from (A.7), (A.8), and (A.9), we conclude that for both cases, $q>0$ and $q=0$, we have

$$\begin{aligned}&\Upsilon _j\le \vartheta \left| z_1\right| _{D(A)}^2 + \widetilde{D}_2\left( 1+\gamma _2^{-\chi _4}\right) \left( 1+\left| y\right| _{V}^{\chi _1} \right) \left( 1+\left| y\right| _{D(A)}^{\chi _2}\right) \left( 1+\left| z_1\right| _{V}^{\chi _3}\right) \left| z_1\right| _{V}^2, \end{aligned}$$

(A.10a)

$$\begin{aligned}&\qquad \text{ for } \text{ all }\quad \gamma _2\in (0,1],\quad \text{ with }\quad \vartheta {:=}{\left\{ \begin{array}{ll} {{\widetilde{D}}}_{1}\gamma _2^{\frac{2}{\chi _2}},&{} \text{ for } \chi _2>0,\\ 0,&{} \text{ for } \chi _2=0, \end{array}\right. } \end{aligned}$$

(A.10b)

$$\begin{aligned}&\qquad \text{ and }\quad \widetilde{D}_{1}={{\overline{C}}}_{\left[ \left\| \zeta _{1}\right\| _{}, \left\| \zeta _{2}\right\| _{}, \left\| \frac{1}{\delta _1}\right\| _{}\right] } ,\qquad {{\widetilde{D}}}_{2}= {{\overline{C}}}_{\left[ \left\| \zeta _{1}\right\| _{},\left\| \zeta _{2}\right\| _{}, \left\| \frac{1}{\delta _1}\right\| _{}, \left\| \frac{\zeta _1+\zeta _2}{\delta _1-\zeta _2}\right\| _{}\right] }. \end{aligned}$$

(A.10c)

Now, from (A.1) and (A.10), for all ${{\widetilde{\gamma }}}_0>0$ and $\gamma _2\in (0,1]$, and with

$$\begin{aligned} \chi _5{:=}\frac{1+\Vert \delta _2\Vert }{1-\Vert \delta _2\Vert } \ge 1, \end{aligned}$$

we derive that

$$\begin{aligned}&2\Bigl ( {{\mathfrak {N}}}_y(t,z_1),Az_1\Bigr )_{H}\le {{\widetilde{\gamma }}}_0 \left| z_1\right| _{D(A)}^{2} +\left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) {{\widetilde{C}}}_{{{\mathfrak {N}}}1} \sum \limits _{j=1}^n\Upsilon _j,\nonumber \\&\quad \le \left( {{\widetilde{\gamma }}}_0+n\vartheta \left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) {{\widetilde{C}}}_{{{\mathfrak {N}}}1}\right) \left| z_1\right| _{D(A)}^{2}\nonumber \\&\quad \quad +n{{\widetilde{D}}}_{2}\left( 1+\gamma _2^{-\chi _4}\right) \left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) {{\widetilde{C}}}_{{{\mathfrak {N}}}1} \left( 1+\left| y\right| _{V}^{\chi _1} \right) \left( 1+\left| y\right| _{D(A)}^{\chi _2}\right) \left( 1+\left| z_1\right| _{V}^{\chi _3}\right) \left| z_1\right| _{V}^2. \end{aligned}$$

(A.11)

For an arbitrary ${{\widehat{\gamma }}}_0>0$, we can choose

$$\begin{aligned} \gamma _2 =\left( \frac{{{\widehat{\gamma }}}_0}{(n+1)\widetilde{D}_{1}{{\widetilde{C}}}_{{{\mathfrak {N}}}1}\left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) +{{\widehat{\gamma }}}_0}\right) ^{\frac{\chi _2}{2}}\le 1, \end{aligned}$$

and ${{\widetilde{\gamma }}}_0=\frac{{{\widehat{\gamma }}}_0}{n+1}$. Note that, in particular,

$$\begin{aligned} \widetilde{D}_{1}\gamma _2^{\frac{2}{\chi _2}}\left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) {{\widetilde{C}}}_{{{\mathfrak {N}}}1}< \gamma _2^{\frac{2}{\chi _2}}\left( {{\widetilde{D}}}_{1} \widetilde{C}_{{{\mathfrak {N}}}1} \left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) +\frac{{{\widehat{\gamma }}}_0}{n+1}\right) =\frac{{{\widehat{\gamma }}}_0}{n+1},\quad \text{ for }\quad \chi _2>0. \end{aligned}$$

and thus for the coefficient of $\left| z_1\right| _{D(A)}^{2}$ in (A.11), we find

$$\begin{aligned} {\left\{ \begin{array}{ll} {{\widetilde{\gamma }}}_0+n\vartheta \left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) {{\widetilde{C}}}_{{{\mathfrak {N}}}1}< \frac{{{\widehat{\gamma }}}_0}{n+1}+n\frac{{{\widehat{\gamma }}}_0}{n+1}={{\widehat{\gamma }}}_0, &{}\qquad \text{ if } \chi _2>0;\\ {{\widetilde{\gamma }}}_0+n\vartheta \left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) {{\widetilde{C}}}_{{{\mathfrak {N}}}1} =\frac{{{\widehat{\gamma }}}_0}{n+1}< {{\widehat{\gamma }}}_0, &{}\qquad \text{ if } \chi _2=0. \end{array}\right. } \end{aligned}$$

(A.12)

Observe, next, that

$$\begin{aligned}&1+{{\widetilde{\gamma }}}_0^{-\chi _5 }=1+(n+1)^{\chi _5 }{{\widehat{\gamma }}}_0^{-\chi _5 }, \end{aligned}$$

(A.13)

and

$$\begin{aligned} 1+\gamma _2^{-\chi _4}&=2,\quad \text{ if }\quad \chi _2=0,\\ 1+\gamma _2^{-\chi _4}&\le 1+\left( (n+1){{\widetilde{D}}}_{1}\widetilde{C}_{{{\mathfrak {N}}}1}\left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) +{{\widehat{\gamma }}}_0\right) ^{\frac{\chi _2\chi _4}{2}}{{\widehat{\gamma }}}_0^{-\frac{\chi _2\chi _4}{2}} \\&\le 1+(1+2^{\frac{\chi _2\chi _4}{2}-1})\left( \left( (n+1)\widetilde{D}_{1} {{\widetilde{C}}}_{{{\mathfrak {N}}}1}\left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) \right) ^{\frac{\chi _2\chi _4}{2}} +{{\widehat{\gamma }}}_0^{\frac{\chi _2\chi _4}{2}}\right) {{\widehat{\gamma }}}_0^{-\frac{\chi _2\chi _4}{2}}\\&\le {{\widehat{C}}}_1+{{\widehat{C}}}_2\left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) ^{\frac{\chi _2\chi _4}{2}} {{\widehat{\gamma }}}_0^{-\frac{\chi _2\chi _4}{2}},\quad \text{ if }\quad \chi _2>0; \end{aligned}$$

with

$$\begin{aligned}&{{\widehat{C}}}_1{:=}1+(1+2^{\frac{\chi _2\chi _4}{2}-1}) ={{\overline{C}}}_{\left[ \left\| \zeta _{2}\right\| _{}, \left\| \frac{1}{\delta _1}\right\| _{}, \left\| \frac{\delta _1}{\delta _1-\zeta _2}\right\| _{}\right] },\\&{{\widehat{C}}}_2{:=}(1+2^{\frac{\chi _2\chi _4}{2}-1})\left( (n+1) {{\widetilde{D}}}_{1}{{\widetilde{C}}}_{{{\mathfrak {N}}}1}\right) ^{\frac{\chi _2\chi _4}{2}} ={{\overline{C}}}_{\left[ n, C_{{{\mathcal {N}}}},\frac{1}{1-\left\| \delta _{2}\right\| _{}}, \left\| \zeta _{1}\right\| _{},\left\| \zeta _{2}\right\| _{}, \left\| \frac{1}{\delta _1}\right\| _{}, \left\| \frac{\delta _1}{\delta _1-\zeta _2}\right\| _{}\right] }. \end{aligned}$$

Since ${{\widehat{C}}}_1\ge 2$ holds for $\chi _2\ge 0$ we can write

$$\begin{aligned} 1+\gamma _2^{-\chi _4}\le {{\widehat{C}}}_1+{\widehat{C}}_2\left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) ^{\frac{\chi _2\chi _4}{2}} {{\widehat{\gamma }}}_0^{-\frac{\chi _2\chi _4}{2}},\quad \text{ for }\quad \chi _2\ge 0. \end{aligned}$$

Further, we see that

$$\begin{aligned} 1+\gamma _2^{-\chi _4}&\le {{\widehat{C}}}_1+{\widehat{C}}_2(1+2^{\frac{\chi _2\chi _4}{2}-1}) \left( {{\widehat{\gamma }}}_0^{-\frac{\chi _2\chi _4}{2}}+{{\widetilde{\gamma }}}_0^{-\frac{\chi _5\chi _2\chi _4}{2} } {{\widehat{\gamma }}}_0^{-\frac{\chi _2\chi _4}{2}}\right) \end{aligned}$$

and

$$\begin{aligned}&{{\widetilde{\gamma }}}_0^{-\frac{\chi _5\chi _2\chi _4}{2} }{{\widehat{\gamma }}}_0^{-\frac{\chi _2\chi _4}{2}} =(n+1)^{\frac{\chi _5\chi _2\chi _4}{2}}{{\widehat{\gamma }}}_0^{-\frac{(\chi _5+1)\chi _2\chi _4}{2} }, \\&{{\widehat{\gamma }}}_0^{-\frac{\chi _2\chi _4}{2}}\le 1+{{\widehat{\gamma }}}_0^{-\frac{(\chi _5+1)\chi _2\chi _4}{2}}, \end{aligned}$$

from which we obtain

$$\begin{aligned} 1+\gamma _2^{-\chi _4}&\le {{\widehat{C}}}_1 +{\widehat{C}}_2(1+2^{\frac{\chi _2\chi _4}{2}-1})\left( 1+(1+(n+1)^{\frac{\chi _5\chi _2\chi _4}{2}}) {{\widehat{\gamma }}}_0^{-\frac{(\chi _5+1)\chi _2\chi _4}{2}}\right) \nonumber \\&\le {{\widehat{C}}}_3 +{{\widehat{C}}}_4{{\widehat{\gamma }}}_0^{-\frac{(\chi _5+1)\chi _2\chi _4}{2}}, \end{aligned}$$

(A.14a)

with

$$\begin{aligned}&{{\widehat{C}}}_3{:=}{{\widehat{C}}}_1+{\widehat{C}}_2(1+2^{\frac{\chi _2\chi _4}{2}-1}),\qquad {{\widehat{C}}}_4{:=}{\widehat{C}}_2(1+2^{\frac{\chi _2\chi _4}{2}-1})(1+(n+1)^{\frac{\chi _5\chi _2\chi _4}{2}}) . \end{aligned}$$

(A.14b)

Therefore, (A.13) and (A.14) lead us to

$$\begin{aligned}&\left( 1+\gamma _2^{-\chi _4}\right) \left( 1+{{\widetilde{\gamma }}}_0^{-\chi _5 }\right) \le {{\widehat{C}}}_5\left( 1+{{\widehat{\gamma }}}_0^{-\chi _5 }\right) \left( 1+{{\widehat{\gamma }}}_0^{-\frac{(\chi _5+1)\chi _2\chi _4}{2} }\right) , \end{aligned}$$

(A.15a)

with, recalling that $\chi _2$, $\chi _4$, and $\chi _5$ are nonnegative,

$$\begin{aligned}&{{\widehat{C}}}_5{:=}(n+1)^{\chi _5 } ({{\widehat{C}}}_3+{{\widehat{C}}}_4) ={{\overline{C}}}_{\left[ n,C_{{{\mathcal {N}}}},\left\| \zeta _{1}\right\| _{},\left\| \zeta _{2}\right\| _{}, \left\| \frac{1}{\delta _1}\right\| _{}, \left\| \frac{\delta _1}{\delta _1-\zeta _2}\right\| _{}, \frac{1}{1-\left\| \delta _{2}\right\| _{}}\right] }. \end{aligned}$$

(A.15b)

Hence, (A.11), (A.12), and (A.15) give us

$$\begin{aligned}&2\Bigl ( {{\mathfrak {N}}}_y(t,z_1),Az_1\Bigr )_{H} \\&\quad \le {{\widehat{\gamma }}}_0 \left| z_1\right| _{D(A)}^{2}\\&\quad \quad +{{\widetilde{C}}}_{{{\mathfrak {N}}}2}\left( 1+{{\widehat{\gamma }}}_0^{-\chi _5 }\right) \! \left( 1+{{\widehat{\gamma }}}_0^{-\frac{(\chi _5+1)\chi _2\chi _4}{2} }\right) \! \left( 1+\left| y\right| _{V}^{\chi _1} \right) \! \left( 1+\left| y\right| _{D(A)}^{\chi _2}\right) \! \left( 1+\left| z_1\right| _{V}^{\chi _3}\right) \! \left| z_1\right| _{V}^2, \end{aligned}$$

with ${{\widetilde{C}}}_{{{\mathfrak {N}}}2}{:=}n{{\widetilde{D}}}_{2}{\widehat{C}}_5{{\widetilde{C}}}_{{{\mathfrak {N}}}1} ={{\overline{C}}}_{\left[ n,C_{{{\mathcal {N}}}},\left\| \zeta _{1}\right\| _{}, \left\| \zeta _{2}\right\| _{}, \left\| \frac{1}{\delta _1}\right\| _{}, \left\| \frac{\delta _1}{\delta _1-\zeta _2}\right\| _{}, \left\| \frac{\zeta _1+\zeta _2}{\delta _1-\zeta _2}\right\| _{}, \frac{1}{1-\left\| \delta _{2}\right\| _{}}\right] }$. This ends the proof of Proposition 3.7. $\square$

1.2 Proof of Proposition 3.8

Recall that , for $\xi \ge 0$, and $H={{\widetilde{{{\mathcal {W}}}}}}_{S}\oplus {{\mathcal {W}}}_{S}^\perp$. We prove firstly that ${{\widetilde{{{\mathcal {W}}}}}}_{S}$ and ${{\mathcal {W}}}_{S}^\perp \cap D(A^\xi )$ are closed subspaces of $D(A^\xi )$. Clearly, ${{\widetilde{{{\mathcal {W}}}}}}_{S}$ is closed, because it is finite-dimensional. Let now  $(h_n)_{n\in {\mathbb {N}}_0}$ be an arbitrary sequence in ${{\mathcal {W}}}_{S}^\perp \cap D(A^\xi )$ and a vector ${{\overline{h}}}\in D(A^\xi )$, so that $\left| h_n-{{\overline{h}}}\right| _{D(A^\xi )}\rightarrow 0$, as $n\rightarrow +\infty$. Since $\left| h_n-{{\overline{h}}}\right| _{H}\le C\left| h_n-{{\overline{h}}}\right| _{D(A^\xi )}$, for a suitable constant $C>0$, it follows that $\left| h_n-{{\overline{h}}}\right| _{H}\rightarrow 0$, and since ${{\mathcal {W}}}_{S}^\perp$ is closed in H, it follows that ${{\overline{h}}}\in {{\mathcal {W}}}_{S}^\perp$. Thus, ${{\overline{h}}}\in {{\mathcal {W}}}_{S}^\perp \cap D(A^\xi )$, and we can conclude that ${{\mathcal {W}}}_{S}^\perp \cap D(A^\xi )$ is a closed subspace of $D(A^\xi )$. Next, we observe that $D(A^\xi )={{\widetilde{{{\mathcal {W}}}}}}_{S}\oplus ({{\mathcal {W}}}_{S}^\perp \cap D(A^\xi ))$, which is a straightforward consequence of $H={{\widetilde{{{\mathcal {W}}}}}}_{S}\oplus {{\mathcal {W}}}_{S}^\perp$. To show that the oblique projection $P_{{{\widetilde{{{\mathcal {W}}}}}}_{S}}^{{{\mathcal {W}}}_{S}^\perp \cap D(A^\xi )}$ in $D(A^\xi )$ coincides with the restriction $\left. P_{{{\widetilde{{{\mathcal {W}}}}}}_{S}}^{{{\mathcal {W}}}_{S}^\perp }\right| _{D(A^\xi )}$ of the oblique projection $P_{{{\widetilde{{{\mathcal {W}}}}}}_{S}}^{{{\mathcal {W}}}_{S}^\perp }$ in H, it is enough to observe that by definition of a projection we have that

$$\begin{aligned}&\left. P_{{{\widetilde{{{\mathcal {W}}}}}}_{S}}^{{{\mathcal {W}}}_{S}^\perp \cap D(A^\xi )}\right| _{D(A^\xi )}w_1=w_1 =P_{{{\widetilde{{{\mathcal {W}}}}}}_{S}}^{{{\mathcal {W}}}_{S}^\perp }w_1,\quad \text{ for } \text{ all }\quad w_1\in {{\widetilde{{{\mathcal {W}}}}}}_{S},\\&\left. P_{{{\widetilde{{{\mathcal {W}}}}}}_{S}}^{{{\mathcal {W}}}_{S}^\perp \cap D(A^\xi )}\right| _{D(A^\xi )}w_2=0 =P_{{{\widetilde{{{\mathcal {W}}}}}}_{S}}^{{{\mathcal {W}}}_{S}^\perp }w_2,\quad \text{ for } \text{ all } \quad w_2\in {{\mathcal {W}}}_{S}^\perp \cap D(A^\xi ). \end{aligned}$$

Finally, we have $P_{{{\widetilde{{{\mathcal {W}}}}}}_{S}}^{{{\mathcal {W}}}_{S}^\perp \cap D(A^\xi )}\in {{\mathcal {L}}}(D(A^\xi ))$ because (oblique) projections are continuous; see Brezis (2011, Sect. 2.4, Thm. 2.10). $\square$

1.3 Proof of Proposition 3.9

It is clear that $P_{{{\widetilde{{{\mathcal {W}}}}}}_{S}}^{{{\mathcal {W}}}_{S}^\perp }\Bigr |^{D(A^{-\xi })}$ is an extension of the oblique projection $P_{{{\widetilde{{{\mathcal {W}}}}}}_{S}}^{{{\mathcal {W}}}_{S}^\perp }\in {{\mathcal {L}}}(H)$ to $D(A^{-\xi })\supseteq H$, because for $z\in H$ we have that $\left\langle P_{{{\mathcal {W}}}_{S}}^{{{\widetilde{{{\mathcal {W}}}}}}_{S}^\perp }\Bigr |^{D(A^{-\xi })} z,w\right\rangle _{D(A^{-\xi }),D(A^{\xi })}= (z,{{\mathcal {P}}}_{{{\widetilde{{{\mathcal {W}}}}}}_{S}}^{{{\mathcal {W}}}_{S}^\perp }w)_{H} =(P_{{{\mathcal {W}}}_{S}}^{{{\widetilde{{{\mathcal {W}}}}}}_{S}^\perp }z,w)_{H}$, where for the last identity we have used Lemma 3.5. By relation (3.9) and Proposition 3.8, the inequalities $\left| P_{{{\mathcal {W}}}_{S}}^{{{\widetilde{{{\mathcal {W}}}}}}_{S}^\perp }\Bigr |^{D(A^{-\xi })}\right| _{{{\mathcal {L}}}(D(A^{-\xi }))} \le \left| P_{{{\widetilde{{{\mathcal {W}}}}}}_{S}}^{{{\mathcal {W}}}_{S}^\perp }\Bigr |_{D(A^{\xi })}\right| _{{{\mathcal {L}}}(D(A^{\xi }))}<+\infty$ follow. Afterward, the same relation (3.9) gives us the converse inequality. Hence, we obtain the stated norm identity. Finally, by definition of the adjoint operator we also have the stated adjoint identity. $\square$

1.4 Proof of Proposition 3.10

Since ${{\mathfrak {s}}}\in (0,1)$, we have that $g(\tau ){:=}-\eta _1\tau +\eta _2\tau ^{{\mathfrak {s}}}$ satisfies $g(0)=0$, $\lim \limits _{\tau \rightarrow +\infty }g(\tau )=-\infty$, and $\frac{\mathrm d}{\mathrm d\tau }\left. \right| _{\tau =\tau _0}g(\tau )=-\eta _1+{{\mathfrak {s}}}\eta _2\tau _0^{{{\mathfrak {s}}}-1}$, for $\tau _0>0$. In particular, g is differentiable at each $\tau _0>0$. Furthermore,

$$\begin{aligned} \frac{\mathrm d}{\mathrm d\tau }\left. \right| _{\tau =\tau _0}g(\tau )>0&\Longleftrightarrow \tau _0^{{{\mathfrak {s}}}-1} >\eta _1({{\mathfrak {s}}}\eta _2)^{-1}\Longleftrightarrow \tau _0^{1-{{\mathfrak {s}}}}<({{\mathfrak {s}}}\eta _2)\eta _1^{-1}\\&\Longleftrightarrow \tau _0 <({{\mathfrak {s}}}\eta _2)^\frac{1}{1-{{\mathfrak {s}}}}\eta _1^{-\frac{1}{1-{{\mathfrak {s}}}}}. \end{aligned}$$

Thus, $g(\tau )$ strictly increases if $\tau \in (0,{{\overline{\tau }}})$ with ${{\overline{\tau }}}{:=}({{\mathfrak {s}}}\eta _2)^\frac{1}{1-{{\mathfrak {s}}}}\eta _1^{-\frac{1}{1-{{\mathfrak {s}}}}} =({{\mathfrak {s}}}\eta _2)^\frac{1}{1-{{\mathfrak {s}}}}\eta _1^{\frac{1}{{{\mathfrak {s}}}-1}}$. Analogously, we find that $\frac{\mathrm d}{\mathrm d\tau }\left. \right| _{\tau =\tau _0}g(\tau ) < 0\Longleftrightarrow \tau _0>{{\overline{\tau }}}$. Necessarily, the maximum is attained at ${{\overline{\tau }}}>0$ and can be computed as

$$\begin{aligned} -\eta _1{{\overline{\tau }}}+\eta _2{{\overline{\tau }}}^{{\mathfrak {s}}}&=-\eta _1({{\mathfrak {s}}}\eta _2)^\frac{1}{1-{{\mathfrak {s}}}}\eta _1^\frac{1}{{{\mathfrak {s}}}-1} + \eta _2\left( ({{\mathfrak {s}}}\eta _2)^\frac{1}{1-{{\mathfrak {s}}}}\eta _1^\frac{1}{{{\mathfrak {s}}}-1}\right) ^{{{\mathfrak {s}}}} \\&= \eta _2^\frac{1}{1-{{\mathfrak {s}}}}\eta _1^\frac{{{\mathfrak {s}}}}{{{\mathfrak {s}}}-1}\left( -{{\mathfrak {s}}}^\frac{1}{1-{{\mathfrak {s}}}}+{{\mathfrak {s}}}^\frac{{{\mathfrak {s}}}}{1-{{\mathfrak {s}}}}\right) . \end{aligned}$$

Thus, $-\eta _1{{\overline{\tau }}}+\eta _2{{\overline{\tau }}}^{{\mathfrak {s}}}=(1-{{\mathfrak {s}}}){{\mathfrak {s}}}^\frac{{{\mathfrak {s}}}}{1-{{\mathfrak {s}}}}\eta _2^\frac{1}{1-{{\mathfrak {s}}}}\eta _1^\frac{{{\mathfrak {s}}}}{{{\mathfrak {s}}}-1}$, which finishes the proof. $\square$

1.5 Proof of Proposition 3.11

For the sake of simplicity, we shall omit the subscript in the usual norm in ${{\mathbb {R}}}$, that is, . The solution of (3.12) is given by

$$\begin{aligned} v(t)=\mathrm {e}^{-{{\overline{\mu }}} (t-s)+\int _s^t\left| h(\tau )\right| _{}\,\mathrm d\tau }v(s),\quad t\ge s\ge 0,\quad v(0)=v_0. \end{aligned}$$

(A.16)

Observe that the exponent satisfies, using (3.10),

$$\begin{aligned} -{{\overline{\mu }}} (t-s)+\int _s^t\left| h(\tau )\right| _{}\,\mathrm d\tau&\le -{{\overline{\mu }}} (t-s)+(t-s)^\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}\left( \int _s^t \left| h(\tau )\right| _{}^{{\mathfrak {r}}}\,\mathrm d\tau \right) ^\frac{1}{{{\mathfrak {r}}}} \\&\le -{{\overline{\mu }}} (t-s)+(t-s)^\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}\left( \int _s^{s+T\lceil \frac{t-s}{T}\rceil } \left| h(\tau )\right| _{}^{{\mathfrak {r}}}\,\mathrm d\tau \right) ^\frac{1}{{{\mathfrak {r}}}} \\&\le -{{\overline{\mu }}} (t-s)+(t-s)^\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}\left( \lceil \frac{t-s}{T}\rceil C_h^{{\mathfrak {r}}}\right) ^\frac{1}{{{\mathfrak {r}}}}, \end{aligned}$$

where  $\lceil \tfrac{t-s}{T}\rceil \in {\mathbb {N}}$ is the nonnegative integer defined in (3.27). Hence,

$$\begin{aligned} -{{\overline{\mu }}} (t-s)+\int _s^t\left| h(\tau )\right| _{}\,\mathrm d\tau&\le -{{\overline{\mu }}} (t-s)+(t-s)^\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}\left( \frac{t-s}{T}+1\right) ^\frac{1}{{{\mathfrak {r}}}} C_h\nonumber \\&\le T^{-\frac{1}{{{\mathfrak {r}}}}}(-{{\overline{\mu }}} T^\frac{1}{{{\mathfrak {r}}}}+C_h )(t-s)+(t-s)^\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}} C_h, \end{aligned}$$

(A.17)

where we have used $\left( \tfrac{t-s}{T}+1\right) ^\frac{1}{{{\mathfrak {r}}}}\le (\tfrac{t-s}{T})^\frac{1}{{{\mathfrak {r}}}}+1$, since ${{\mathfrak {r}}}>1$; see Phan and Rodrigues (2017, Proposition 2.6).

By (3.11), we have that

$$\begin{aligned} {{\widehat{\mu }}}{:=}T^{-\frac{1}{{{\mathfrak {r}}}}}({{\overline{\mu }}} T^\frac{1}{{{\mathfrak {r}}}}-C_h ) \ge \max \left\{ 2\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}\left( \frac{C_h^{{\mathfrak {r}}}}{{{\mathfrak {r}}}\log (\varrho )}\right) ^\frac{1}{{{\mathfrak {r}}}-1}, 2\mu \right\} >0, \end{aligned}$$

(A.18)

from which, together with (A.17) and Proposition 3.10, we obtain

$$\begin{aligned} -{{\overline{\mu }}} (t-s)+\int _s^t\left| h(\tau )\right| _{}\,\mathrm d\tau&\le -\frac{1}{2}{{\widehat{\mu }}}(t-s)-\frac{1}{2}{{\widehat{\mu }}}(t-s)+(t-s)^\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}} C_h\nonumber \\&\le -\frac{{{\widehat{\mu }}}}{2}(t-s) +\frac{1}{{{\mathfrak {r}}}}(\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}})^{{{\mathfrak {r}}}-1}C_h^{{\mathfrak {r}}}(\frac{{{\widehat{\mu }}}}{2})^{1-{{\mathfrak {r}}}}, \end{aligned}$$

(A.19)

because by Proposition 3.10, with ${{\mathfrak {s}}}=\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}$ and $\eta _1 =\tfrac{{{\widehat{\mu }}}}{2}$, $\eta _2=C_h$,

$$\begin{aligned} \max _{t-s\ge 0}\{-\eta _1(t-s)+(t-s)^\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}} \eta _2\} =(1-{{\mathfrak {s}}}){{\mathfrak {s}}}^\frac{{{\mathfrak {s}}}}{1-{{\mathfrak {s}}}}\eta _2^\frac{1}{1-{{\mathfrak {s}}}}\eta _1^\frac{{{\mathfrak {s}}}}{{{\mathfrak {s}}}-1} =\frac{1}{{{\mathfrak {r}}}}(\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}})^{{{\mathfrak {r}}}-1}\eta _2^{{{\mathfrak {r}}}}\eta _1^{1-{{\mathfrak {r}}}}. \end{aligned}$$

Therefore, from (A.16), (A.18), and (A.19), we derive that

$$\begin{aligned} \left| v(t)\right| _{}\le \mathrm {e}^{\frac{C_h^{{\mathfrak {r}}}}{{{\mathfrak {r}}}}\left( \frac{2({{\mathfrak {r}}}-1)}{{{\mathfrak {r}}}}\right) ^{{{\mathfrak {r}}}-1} {{\widehat{\mu }}}^{1-{{\mathfrak {r}}}}}\mathrm {e}^{-\frac{{{\widehat{\mu }}}}{2}(t-s)}\left| v(s)\right| _{} \le \varrho \mathrm {e}^{-\mu (t-s)}\left| v(s)\right| _{}. \end{aligned}$$

Indeed, observe that

$$\begin{aligned}&\mathrm {e}^{\frac{C_h^{{\mathfrak {r}}}}{{{\mathfrak {r}}}}\left( \frac{2({{\mathfrak {r}}}-1)}{{{\mathfrak {r}}}}\right) ^{{{\mathfrak {r}}}-1}{{\widehat{\mu }}}^{1-{{\mathfrak {r}}}}}\le \varrho \quad \Longleftrightarrow \quad \frac{C_h^{{\mathfrak {r}}}}{{{\mathfrak {r}}}}\left( \frac{2({{\mathfrak {r}}}-1)}{{{\mathfrak {r}}}}\right) ^{{{\mathfrak {r}}}-1}{{\widehat{\mu }}}^{1-{{\mathfrak {r}}}}\le \log (\varrho ) \\ \Longleftrightarrow \quad&\frac{C_h^{{\mathfrak {r}}}}{{{\mathfrak {r}}}\log (\varrho )}\left( \frac{2({{\mathfrak {r}}}-1)}{{{\mathfrak {r}}}}\right) ^{{{\mathfrak {r}}}-1}\le {{\widehat{\mu }}}^{{{\mathfrak {r}}}-1} \quad \Longleftrightarrow \quad {{\widehat{\mu }}}\ge \left( \frac{C_h^{{\mathfrak {r}}}}{{{\mathfrak {r}}}\log (\varrho )}\right) ^\frac{1}{{{\mathfrak {r}}}-1}\frac{2({{\mathfrak {r}}}-1)}{{{\mathfrak {r}}}}, \end{aligned}$$

and the last inequality follows from (A.18), which also gives us $\frac{{{\widehat{\mu }}}}{2}>\mu$. $\square$

1.6 Proof of Proposition 3.12

We shall use a fixed point argument, through the contraction principle, in the closed subset

$$\begin{aligned} {{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}{:=}\left\{ g\in L^\infty ({{\mathbb {R}}}_0,{{\mathbb {R}}})\left| \;\; \left| \mathrm {e}^{\mu _0 t}g(t)\right| _{}\le \varrho \left| \varpi _0\right| _{}\right. \!\right\} \end{aligned}$$

of the Banach space

We show now that since (3.15) holds true, the mapping

$$\begin{aligned} \varPsi :{{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}\rightarrow {{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0},\qquad \breve{\varpi }\mapsto \varpi , \end{aligned}$$

where $\varpi$ solves

$$\begin{aligned} {\dot{\varpi }}=-({{\overline{\mu }}}-\left| h\right| _{})\varpi +\left| h\right| _{}\left| \breve{\varpi }\right| _{}^p\breve{\varpi },\quad \varpi (0)=\varpi _0, \end{aligned}$$

(A.20)

is well defined and is a contraction in ${{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{{\mu _0}}$.

We look at (A.20) as a perturbation of the nominal linear system

$$\begin{aligned} {\dot{v}}=-({{\overline{\mu }}}-\left| h\right| _{})v,\quad v(0)=v_0=\varpi _0\in {{\mathbb {R}}}. \end{aligned}$$

(A.21)

Note that (3.15) implies that

$$\begin{aligned} {{\overline{\mu }}} \ge \max \left\{ 2\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}} \left( \frac{C_h^{{\mathfrak {r}}}}{{{\mathfrak {r}}}\log \left( \varrho ^\frac{1}{2}\right) }\right) ^\frac{1}{{{\mathfrak {r}}}-1}, 4{\mu _0}\right\} + T^{-\frac{1}{{{\mathfrak {r}}}}}C_h \end{aligned}$$

which we use together with Proposition 3.11 to conclude that the solution

$$\begin{aligned} v(t)\,{=:}\,{{\mathcal {S}}}\,(t,s)v(s) \end{aligned}$$

of (A.21) satisfies

$$\begin{aligned} \left| v(t)\right| _{}=\left| {{\mathcal {S}}}(t,s)v(s)\right| _{}\le \varrho ^\frac{1}{2}\mathrm {e}^{-2{\mu _0}(t-s)}\left| v(s)\right| _{}, \quad t\ge s\ge 0,\quad v(0)=v_0. \end{aligned}$$

(A.22)

By the Duhamel formula, we have that the solution w of (A.20) is given as

$$\begin{aligned} \varpi (t)={{\mathcal {S}}}(t,s)\varpi (s)+\int _s^t{{\mathcal {S}}}(t,\tau ) \left| h(\tau )\right| _{}\left| \breve{\varpi }(\tau )\right| _{}^p\breve{\varpi }(\tau )\,\mathrm d\tau , \quad \varpi =\varPsi (\breve{\varpi }). \end{aligned}$$

(A.23)

•  Step 1: $\varPsi$ maps ${{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}$ into itself, if $\left| \varpi _0\right| _{}<\varrho R$. We observe that (A.22) and (A.23) give us the estimate

  $$\begin{aligned} \left| \varpi (t)\right| _{}\le \varrho ^\frac{1}{2}\mathrm {e}^{-2{\mu _0} t}\left| \varpi _0\right| _{} +\int _0^t\varrho ^\frac{1}{2}\mathrm {e}^{-2{\mu _0}(t-\tau )}\left| h(\tau )\right| _{} \left| \breve{\varpi }(\tau )\right| _{}^{p+1}\,\mathrm d\tau . \end{aligned}$$

  (A.24)

Next, we also find, since $\breve{\varpi }\in {{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}$,

$$\begin{aligned}&\int _0^t\mathrm {e}^{-2{\mu _0}(t-\tau )}\left| h(\tau )\right| _{}\left| \breve{\varpi }(\tau )\right| _{}^{p+1}\,\mathrm d\tau \le \varrho ^{p+1}\left| \varpi _0\right| _{}^{p+1}\int _0^t\mathrm {e}^{-2{\mu _0}(t-\tau )} \mathrm {e}^{-{\mu _0} (p+1)\tau }\left| h(\tau )\right| _{}\,\mathrm d\tau \nonumber \\&\quad \le \varrho ^{p+1}\left| \varpi _0\right| _{}^{p+1} \mathrm {e}^{-{\mu _0} t} \int _0^t\mathrm {e}^{-{\mu _0} (t-\tau )}\mathrm {e}^{-{\mu _0} p\tau }\left| h(\tau )\right| _{}\,\mathrm d\tau \nonumber \\&\quad \le \varrho ^{p+1}\left| \varpi _0\right| _{}^{p+1} \mathrm {e}^{-{\mu _0} t} \left( \int _0^t\mathrm {e}^{-\frac{{{\mathfrak {r}}}}{{{\mathfrak {r}}}-1}{\mu _0} (t-\tau )}\,\mathrm d\tau \right) ^{\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}} \left( \int _0^t\mathrm {e}^{-{{\mathfrak {r}}}{\mu _0} p\tau }\left| h(\tau )\right| _{}^{{\mathfrak {r}}}\,\mathrm d\tau \right) ^{\frac{1}{{{\mathfrak {r}}}}} \nonumber \\&\quad \le \varrho ^{p+1}\left| \varpi _0\right| _{}^{p+1} \mathrm {e}^{-{\mu _0} t} \left( \frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}{\mu _0}}\right) ^{\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}} \left( \sum \limits _{i=1}^{\lceil \frac{t}{T}\rceil }\mathrm {e}^{-{{\mathfrak {r}}}{\mu _0} p(i-1)T } \int _{(i-1)T}^{iT} \left| h(\tau )\right| _{}^{{\mathfrak {r}}}\,\mathrm d\tau \right) ^{\frac{1}{{{\mathfrak {r}}}}} \nonumber \\&\quad \le \varrho ^{p+1}\left| \varpi _0\right| _{}^{p+1} \mathrm {e}^{-{\mu _0} t} \left( \frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}{\mu _0}}\right) ^{\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}}C_h \left( \sum \limits _{i=1}^{\lceil \frac{t}{T}\rceil }\mathrm {e}^{-{{\mathfrak {r}}}{\mu _0} p(i-1)T }\right) ^{\frac{1}{{{\mathfrak {r}}}}} \nonumber \\&\quad \le \varrho ^{p+1}\left| \varpi _0\right| _{}^{p+1}C_h \left( \frac{1}{1-\mathrm {e}^{-{{\mathfrak {r}}}{\mu _0} pT }} \right) ^{\frac{1}{{{\mathfrak {r}}}}} \left( \frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}{\mu _0}}\right) ^{\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}} \mathrm {e}^{-{\mu _0} t}. \end{aligned}$$

(A.25)

By combining (A.24) with (A.25), we arrive at

$$\begin{aligned} \mathrm {e}^{{\mu _0} t}\left| \varpi (t)\right| _{}&\le \varrho ^\frac{1}{2}\mathrm {e}^{-{\mu _0} t}\left| \varpi _0\right| _{} +\varrho ^{p+\frac{3}{2}}\left| \varpi _0\right| _{}^{p+1}C_h\left( \frac{1}{1-\mathrm {e}^{-{{\mathfrak {r}}}{\mu _0} pT }} \right) ^{\frac{1}{{{\mathfrak {r}}}}} \left( \frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}\right) ^{\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}} {\mu _0}^{\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}}} \nonumber \\&\le \varrho ^\frac{1}{2}\left( 1 +\varrho ^{p+1}\left| \varpi _0\right| _{}^{p}C_h\left( \frac{1}{1-\mathrm {e}^{-{{\mathfrak {r}}}{\mu _0} pT }} \right) ^{\frac{1}{{{\mathfrak {r}}}}} \left( \frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}\right) ^{\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}} {\mu _0}^{\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}}}\right) \left| \varpi _0\right| _{}. \end{aligned}$$

(A.26)

Next, we use (3.14) and $\left| \varpi _0\right| _{}\le \varrho R$ to obtain

$$\begin{aligned} \frac{1}{1-\mathrm {e}^{-{{\mathfrak {r}}}{\mu _0} pT }}\le \frac{1}{1-\mathrm {e}^{-{\mu _0} pT }}\le 2, \end{aligned}$$

(A.27a)

and

$$\begin{aligned}&1 +\varrho ^{p+1}\left| \varpi _0\right| _{}^{p}C_h\left( \frac{1}{1-\mathrm {e}^{-{{\mathfrak {r}}}{\mu _0} pT }} \right) ^{\frac{1}{{{\mathfrak {r}}}}} \left( \frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}\right) ^{\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}} {\mu _0}^{\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}}}\nonumber \\&\quad \le 1 +\varrho ^{2p+1}R^{p}C_h\left( \frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}\right) ^{\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}} {\mu _0}^{\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}}}2^\frac{1}{{{\mathfrak {r}}}}\nonumber \\&\quad \le 1 +\varrho ^{2p+1}R^{p}C_h\left( \frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}\right) ^{\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}}2^\frac{1}{{{\mathfrak {r}}}} \left( \frac{\varrho ^{2p+1}R^{p}C_h}{\varrho ^\frac{1}{2}-1}\right) ^{-1}2^{-\frac{1}{{{\mathfrak {r}}}}} \left( \frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}\right) ^{\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}}}\nonumber \\&\quad =1 +\left( \frac{1}{\varrho ^\frac{1}{2}-1}\right) ^{-1}\nonumber \\&\quad =\varrho ^\frac{1}{2}. \end{aligned}$$

(A.27b)

From (A.26) and (A.27), we find $\mathrm {e}^{{\mu _0} t}\left| \varpi (t)\right| _{}\le \varrho \left| \varpi _0\right| _{}$, and hence, $\varpi =\varPsi (\breve{\varpi })\in {{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}$.

•  Step 2: $\varPsi$ is a contraction in ${{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}$, if $\left| \varpi _0\right| _{}<\varrho R$. For an arbitrary given $(\breve{\varpi }_1,\breve{\varpi }_2)\in {{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0} \times {{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}$, we have that the difference

  $$\begin{aligned} D{:=}\varPsi (\breve{\varpi }_1)-\varPsi (\breve{\varpi }_2) \end{aligned}$$

solves

$$\begin{aligned} \dot{D}=-({{\overline{\mu }}}-\left| h\right| _{})D +\left| h\right| _{}\left( \left| \breve{\varpi }_1\right| _{}^p\breve{\varpi }_1-\left| \breve{\varpi }_2\right| _{}^p\breve{\varpi }_2\right) ,\quad D(0)=0, \end{aligned}$$

By the Duhamel formula and the mean value theorem, we obtain

$$\begin{aligned} \left| D(t)\right| _{}&=\left| {{\mathcal {S}}}(t,0)D(0)\right| _{} +\left| \int _0^t{{\mathcal {S}}}(t,\tau )\left| h(\tau )\right| _{} \left| \left| \breve{\varpi }_1\right| _{}^p\breve{\varpi }_1-\left| \breve{\varpi }_2\right| _{}^p\breve{\varpi }_2\right| _{} \,\mathrm d\tau \right| _{}\nonumber \\&\le \varrho ^\frac{1}{2} (p+1)\int _0^t\mathrm {e}^{-{\mu _0}(t-\tau )}\left| h(\tau )\right| _{} \left( \left| \breve{\varpi }_1(\tau )\right| _{}^p+\left| \breve{\varpi }_2(\tau )\right| _{}^p\right) \left| \breve{\varpi }_1(\tau )-\breve{\varpi }_2(\tau )\right| _{}\,\mathrm d\tau \nonumber \\&\le \varrho ^\frac{1}{2} (p+1) \left| \breve{\varpi }_1-\breve{\varpi }_2\right| _{{{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}}\mathrm {e}^{-{\mu _0} t} \int _0^t\left| h(\tau )\right| _{}\left( \left| \breve{\varpi }_1(\tau )\right| _{}^p+\left| \breve{\varpi }_2(\tau )\right| _{}^p\right) \,\mathrm d\tau \nonumber \\&\le 2\varrho ^{p+\frac{1}{2}} (p+1)\left| \varpi _0\right| _{}^p \left| \breve{\varpi }_1-\breve{\varpi }_2\right| _{{{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}} \mathrm {e}^{-{\mu _0} t}\int _0^t\mathrm {e}^{-{\mu _0}\tau p}\left| h(\tau )\right| _{}\,\mathrm d\tau . \end{aligned}$$

(A.28)

Note that

$$\begin{aligned}&\int _0^t\mathrm {e}^{-{\mu _0}\tau p}\left| h(\tau )\right| _{}\,\mathrm d\tau =\int _0^t\mathrm {e}^{-\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}}{\mu _0}\tau p}\mathrm {e}^{-\frac{1}{{{\mathfrak {r}}}}{\mu _0}\tau p} \left| h(\tau )\right| _{}\,\mathrm d\tau \nonumber \\&\quad \le \left( \int _0^t\mathrm {e}^{-{\mu _0}\tau p}\,\mathrm d\tau \right) ^\frac{{{\mathfrak {r}}}-1}{{{\mathfrak {r}}}} \left( \int _0^t\mathrm {e}^{-{\mu _0}\tau p}\left| h(\tau )\right| _{}^{{\mathfrak {r}}}\,\mathrm d\tau \right) ^\frac{1}{{{\mathfrak {r}}}}\nonumber \\&\quad \le \left( {\mu _0} p\right) ^\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}} \left( \sum \limits _{i=1}^{\lceil \frac{t}{T}\rceil } \mathrm {e}^{-{\mu _0} p(i-1)T}\int _{(i-1)T}^{iT} \left| h(\tau )\right| _{}^{{\mathfrak {r}}}\,\mathrm d\tau \right) ^\frac{1}{{{\mathfrak {r}}}}\nonumber \\&\quad \le \left( {\mu _0} p\right) ^\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}}C_h \left( \sum \limits _{i=1}^{\lceil \frac{t}{T}\rceil } \mathrm {e}^{-{\mu _0} p(i-1)T}\right) ^\frac{1}{{{\mathfrak {r}}}} \le \left( {\mu _0} p\right) ^\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}}C_h \left( \frac{1}{1- \mathrm {e}^{-{\mu _0} pT}}\right) ^\frac{1}{{{\mathfrak {r}}}}. \end{aligned}$$

(A.29)

From (A.28) and (A.29),

$$\begin{aligned} \mathrm {e}^{{\mu _0} t}\left| D(t)\right| _{}&\le 2\varrho ^{p+\frac{1}{2}} (p+1) p^\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}} C_h \left( \frac{1}{1- \mathrm {e}^{-{\mu _0} pT}}\right) ^\frac{1}{{{\mathfrak {r}}}}\left| \varpi _0\right| _{}^p{\mu _0}^\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}} \left| \breve{\varpi }_1-\breve{\varpi }_2\right| _{{{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}}, \end{aligned}$$

which together $\left| \varpi _0\right| _{}\le \varrho R$ and ${\mu _0}\ge \frac{\log (2)}{pT}$, see (3.14), give us $\tfrac{1}{1- \mathrm {e}^{-{\mu _0} pT}}\le 2$ and

$$\begin{aligned}&\mathrm {e}^{{\mu _0} t}\left| D(t)\right| _{}\nonumber \\&\quad \le 2^\frac{{{\mathfrak {r}}}+1}{{{\mathfrak {r}}}}\varrho ^{2p+\frac{1}{2}} (p+1) p^\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}} C_h R^p{\mu _0}^\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}} \left| \breve{\varpi }_1-\breve{\varpi }_2\right| _{{{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}}\nonumber \\&\quad \le 2^\frac{{{\mathfrak {r}}}+1}{{{\mathfrak {r}}}}\varrho ^{2p+\frac{1}{2}} (p+1) p^\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}} C_h R^p\left( 2^\frac{{{\mathfrak {r}}}+1}{{{\mathfrak {r}}}-1} \left( \varrho ^{2p+\frac{1}{2}} C_h\frac{p+1}{p}R^pc\right) ^\frac{{{\mathfrak {r}}}}{{{\mathfrak {r}}}-1} p^\frac{1}{{{\mathfrak {r}}}-1}\right) ^\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}} \left| \breve{\varpi }_1-\breve{\varpi }_2\right| _{{{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}}\nonumber \\&\quad \le c^{-1} p^\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}} \left( \left( \frac{1}{p}\right) ^\frac{{{\mathfrak {r}}}}{{{\mathfrak {r}}}-1} p^\frac{1}{{{\mathfrak {r}}}-1}\right) ^\frac{1-{{\mathfrak {r}}}}{{{\mathfrak {r}}}} \left| \breve{\varpi }_1-\breve{\varpi }_2\right| _{{{\mathcal {Z}}}_{\varrho ,\left| w_0\right| _{}}^{\mu _0}} =c^{-1}\left| \breve{\varpi }_1-\breve{\varpi }_2\right| _{{{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}} \end{aligned}$$

(A.30)

with $c>1$ as in (3.14). Therefore, (A.30) implies that

$$\begin{aligned} \left| \varPsi (\breve{\varpi }_1)-\varPsi (\breve{\varpi }_2)\right| _{{{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}}&=\left| D\right| _{{{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}}\le c^{-1}\left| d\right| _{{{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}} =c^{-1}\left| \breve{\varpi }_1-\breve{\varpi }_2\right| _{{{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}}, \end{aligned}$$

which shows that $\varPsi$ is a contraction.

•  Step 3: Existence of a solution in ${{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}$, if $\left| \varpi _0\right| _{}<\varrho R$. By the contraction mapping principle, there exists a fixed point for $\Psi$ in ${{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}$. Such fixed point is a solution for (3.13).

•  Step 4: Uniqueness of the solution in $L^\infty ({{\mathbb {R}}}_0,{{\mathbb {R}}})$. The uniqueness follows from the fact that the right-hand side of (3.13) is locally Lipschitz.

•  Step 5: Estimate (3.16) holds true. Fix $s\ge 0$ and note that ${{\widetilde{h}}}(\tau ){:=}h(\tau +s)$ also satisfies (3.10), with $C_{{{\widetilde{h}}}}\le C_h$.

Let $\varpi _{{\underline{s}}}{:=}\varpi \left. \right| _{{{\mathbb {R}}}_s}$ be the restriction to ${{\mathbb {R}}}_s=[s,+\infty )$ of the solution $\varpi \in {{\mathcal {Z}}}_{\varrho ,\left| \varpi _0\right| _{}}^{\mu _0}$ of (3.13), and observe that $z(\tau ){:=}\varpi _{{\underline{s}}}(\tau +s)$ solves

$$\begin{aligned} \frac{\mathrm d}{\mathrm d\tau } z=-({{\overline{\mu }}}-\left| \widetilde{h}\right| _{})z+\left| {{\widetilde{h}}}\right| _{}\left| z\right| _{}^pz, \quad z(0)=z_0,\qquad \tau \ge 0. \end{aligned}$$

If $\left| \varpi _0\right| _{}<R$, it follows that $\left| z_0\right| _{}=\left| \varpi (s)\right| _{} \le \varrho \mathrm {e}^{-{\mu _0} s}\left| \varpi _0\right| _{}\le \varrho R$. Then, by Step 3 we have that $z\in {{\mathcal {Z}}}_{\varrho ,\left| z_0\right| _{}}^{\mu _0}$, which implies that for $t\ge s$,

$$\begin{aligned} \left| \varpi (t)\right| _{}=\left| \varpi _{\underline{s}}(s+t-s)\right| _{}=\left| z(t-s)\right| _{} \le \varrho \mathrm {e}^{-{\mu _0} (t-s)}\left| z(0)\right| _{} =\varrho \mathrm {e}^{-{\mu _0} (t-s)}\left| \varpi (s)\right| _{}, \end{aligned}$$

which gives us (3.16).

The proof is finished. $\square$

1.7 Proof of Proposition 4.3

Let us denote by $\tau ^i=(\tau ^i_1,\tau ^i_2,\dots ,\tau ^i_d))\in {{\mathbb {R}}}^d$ the unit vector whose coordinates are $\tau _i^i=1$ and $\tau _j^i=0$ for $j\ne i$. Observe that ${{\mathbf {J}}}_{d,2}$ has exactly $d+1$ vectors. The only element in ${{\mathbf {J}}}_{d,2}$ with $\textstyle \sum _{j=1}^d{{\mathbf {j}}}_j= d$ is $\mathbf{1}^d{:=}(1,1,\dots ,1)$. All the other elements in ${{\mathbf {J}}}_{d,2}$ are of the form $\mathbf{1}^d+\tau ^i$, $i=1,2,\dots ,d$.

Let now $p\in {\mathbb {P}}_{\times ,1}$ such that ${{\mathfrak {S}}}(p)=0$, which implies that

$$\begin{aligned} \left| (p,1_{\omega _{\mathbf{1}^d,1}^\times })\right| _{{{\mathbb {R}}}}^{2}=0,\quad \text{ and }\quad \left| (p,1_{\omega _{\mathbf{1}^d+\tau ^i,1}^\times })\right| _{{{\mathbb {R}}}}^{2}=0, \text{ for } \text{ all } i=\{1,2,\dots ,d\}, \end{aligned}$$

that is, with $\omega _{*}{:=}\omega _{\mathbf{1}^d,1}$

$$\begin{aligned} \int _{\omega _{*}}p(x)\,\mathrm dx=0,\quad \text{ and }\quad \int _{\omega _{*}}p(x-\tau ^i)\,\mathrm dx=0. \quad 1\le i\le d, \end{aligned}$$

Denoting ${{\mathfrak {L}}}_ax{:=}\sum _{i=1}^da_ix_i$, and $p(x){=:}a_0+{{\mathfrak {L}}}_ax$, we obtain

$$\begin{aligned} \int _{\omega _{*}}c_0+{{\mathfrak {L}}}_ax\,\mathrm dx=0,\quad \text{ and }\quad \int _{\omega _{*}}c_0+{{\mathfrak {L}}}_a(x-\tau ^i)\,\mathrm dx=0, \quad 1\le i\le d, \end{aligned}$$

which implies

$$\begin{aligned} \int _{\omega _{*}}c_0+{{\mathfrak {L}}}_ax\,\mathrm dx=0,\quad \text{ and }\quad \int _{\omega _{*}}{{\mathfrak {L}}}_a\tau ^i\,\mathrm dx=0, \quad 1\le i\le d. \end{aligned}$$

(A.31)

Note that for fixed i, we have

$$\begin{aligned} \int _{\omega _{*}}{{\mathfrak {L}}}_a\tau ^i\,\mathrm dx=0\quad \Longleftrightarrow \quad \int _{\omega _{*}}a_i\,\mathrm dx=0 \quad \Longleftrightarrow \quad a_i=0, \end{aligned}$$

which together with (A.31) leads us to $a_i=0$, $1\le i\le d$, and $c_0=0$.

We have just shown that $p\in {\mathbb {P}}_{\times ,1}$ and ${{\mathfrak {S}}}(p)=0$ imply that $p=0$. Therefore, we can conclude that  is a norm on ${\mathbb {P}}_{\times ,1}$. $\square$

1.8 Proof of Proposition 4.7

Let $\theta =\sum \limits _{k=1}^{S_\sigma }\theta _k\Phi _k\in {{\widetilde{{{\mathcal {W}}}}}}_{S}$, with the auxiliary functions $\Phi _i$ as in (4.2b). Then, after a translation and denoting ${{\widehat{L}}}_i{:=}\frac{rL_i}{2S}$, for the H-norm we find that

and, with , since the $\Phi _i$s are pairwise orthogonal, we arrive at

$$\begin{aligned} \left| \theta \right| _{H}^2={\sum \limits _{k=1}^{S_\sigma }}\theta _k^2\left| \Phi _k\right| _{H}^2 =\left( \frac{3}{8}\right) ^d{\widehat{L}}^\times {\sum \limits _{k=1}^{S_\sigma }}\theta _k^2. \end{aligned}$$

Next, for the V-norm we find

$$\begin{aligned} \left| \theta \right| _{V}^2={\sum \limits _{k=1}^{S_\sigma }}\theta _k^2\left| \Phi _k\right| _{V}^2 =\nu {\sum \limits _{k=1}^{S_\sigma }}\theta _k^2\left| \nabla \Phi _k\right| _{L^2(\Omega )^d}^2+\left| \theta \right| _{H}^2 \end{aligned}$$

and, due to

we obtain

$$\begin{aligned} \left| \theta \right| _{V}^2&=\left( \nu \frac{4\pi ^2}{3}{\sum \limits _{i=1}^{d}} \frac{1}{{\widehat{L}}_i^2}+1\right) \left| \theta \right| _{H}^2=\left( \nu \frac{4\pi ^2}{3}(\frac{2S}{r})^2{\sum \limits _{i=1}^{d}} \frac{1}{ L_i^2}+1\right) \left| \theta \right| _{H}^2. \end{aligned}$$

That is,

$$\begin{aligned} \left| \theta \right| _{V}^2 =\left( C_1S^2 +1\right) \left| \theta \right| _{H}^2,\quad \text{ with }\quad C_1{:=}\frac{16}{3}\nu \pi ^2 r^{-2}{\sum \limits _{i=1}^{d}} L_i^{-2}. \end{aligned}$$

Finally, for the D(A)-norm we find

$$\begin{aligned} \left| \theta \right| _{D(A)}^2&=\left| -\nu \Delta \theta +\theta \right| _{H}^2 =\nu ^2\left| \Delta \theta \right| _{H}^2+2\nu \left| \nabla \Phi _k\right| _{L^2(\Omega )^d}^2+\left| \theta \right| _{H}^2\\&=\nu ^2\left| \Delta \theta \right| _{H}^2+2\left| \theta \right| _{V}^2-\left| \theta \right| _{H}^2 =\nu ^2\left| \Delta \theta \right| _{H}^2+\left( 2C_1S^2+1\right) \left| \theta \right| _{H}^2 \end{aligned}$$

and from

we obtain

$$\begin{aligned} \left| \Delta \theta \right| _{H}^2&={\sum \limits _{k=1}^{S_\sigma }}\theta _k^2\left| \Delta \Phi _k\right| _{H}^2=\frac{16\pi ^4}{3}{\sum \limits _{i=1}^{d}} \frac{1}{{{\widehat{L}}}_i^4}\left| \theta \right| _{H}^2 =\left( \frac{2S}{r}\right) ^4\frac{16\pi ^4}{3}{\sum \limits _{i=1}^{d}} \frac{1}{L_i^4}\left| \theta \right| _{H}^2, \end{aligned}$$

and hence,

$$\begin{aligned} \left| \theta \right| _{D(A)}^2 =\left( C_2S^4+ 2C_1S^2 +1\right) \left| \theta \right| _{H}^2,\quad \text{ with }\quad C_2{:=}\frac{\nu ^22^8\pi ^4}{3}r^{-4}{\sum \limits _{i=1}^{d}} L_i^{-4}, \end{aligned}$$

which finishes the proof. $\square$