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Section 9.12 Supplemental Material

Consider the manipulation
\begin{align*} (A-\lambda I) R_\lambda(A) \amp=\amp I\\ A R_\lambda(A) - \lambda R_\lambda(A) \amp=\amp I\\ A R_\lambda(A) \amp=\amp I+\lambda R_\lambda(A). \end{align*}
Then for an analytic semigroup, we have
\begin{align*} A e^{At} \amp=\amp \frac{i}{2\pi} \ointop_\gamma e^{\lambda t} A R_\lambda(A) d\lambda\\ \amp=\amp \frac{i}{2\pi} \ointop_\gamma e^{\lambda t} (I+\lambda R_\lambda(A)) d\lambda\\ \amp=\amp \frac{i}{2\pi} \ointop_\gamma e^{\lambda t}\lambda (A-\lambda I)^{-1} d\lambda\\ \amp=\amp \frac{i}{2\pi} \ointop_{\gamma'} e^{\lambda'}\frac{\lambda'}{t} \left(A-\frac{\lambda'}{t} I\right)^{-1} \frac{d\lambda'}{t} , \end{align*}
where for \(t>0\) we used the substitution \(\gamma':=\gamma t\) and \(\lambda'=\lambda t\text{.}\) This leads to the estimate
\begin{align*} |Ae^{At}| \amp\leq \amp \frac{1}{2\pi t} \ointop_{\gamma'} |e^{\lambda'}| \left|\frac{\lambda'}{t}\right| \frac{1}{|\omega - \frac{\lambda'}{t}|} d\lambda'\\ \amp=\amp \frac{1}{2\pi t} \ointop_{\gamma'} |e^{\lambda'}| \left|\frac{\lambda'}{t}\right|\frac{\omega - \frac{\lambda'}{t} }{|\omega-\frac{\lambda'}{t}|} \frac{1}{\omega-\frac{\lambda'}{t}} d\lambda'\\ \amp=\amp \frac{1}{2\pi t} e^{\omega t} \left|\frac{\omega t}{t}\right|\\ \amp=\amp \frac{\omega}{2\pi} \frac{e^{\omega t}}{t} , \end{align*}
where we used Cauchy's integral formula in the second to last step.

Subsection 9.12.1 Perturbations of Analytic Semigroups

Consider
\begin{equation} \dot x = (A+B) x\label{perturbeins}\tag{9.12.1} \end{equation}
and define
\begin{equation*} y(t) = x(t) e^{-bt}. \end{equation*}
Then
\begin{equation*} \dot y = \dot x e^{-bt} - bx e^{-bt} = (A+B) x e^{-bt} - bx e^{-bt} = (A+\tilde B) y, \end{equation*}
with
\begin{equation*} \tilde B = B-bI. \end{equation*}
This means that \(x(t)\) solves ((9.12.1)) if and only if \(y\) solves \(\dot y = (A+\tilde B) y\text{,}\) and \(e^{(A+B)t}\) is an analytic semigroup if and only if \(e^{(A+\tilde B)t}\) is an analytic semigroup. Now, \(\|\tilde B x\|\leq a\|Ax\|\text{,}\) hence, without loss of generality, we can assume \(b=0\text{,}\) and use \(B\) instead of \(\tilde B\text{.}\)

Since \(A\) generates an analytic semigroup, we have
\begin{equation*} \|R_\lambda(A)\|\leq \frac{M_\ep}{|\lambda|}, \qquad \mbox{ for all }\lambda \in\Sigma_{\frac{\pi}{2}+\delta-\ep}, \end{equation*}
where \(\Sigma_{\frac{\pi}{2}+\delta-\ep}\) is a sector as defined above, for some small \(\ep>0\text{.}\)

Then
\begin{equation*} \|BR_\lambda(A) x \|\leq a \|A R_\lambda(A) x\| \end{equation*}
and since \(AR_\lambda(A) = I + \lambda R_\lambda(A) \text{,}\) we find
\begin{equation*} \|B R_\lambda(A) x\|\leq a \left|1+\lambda \frac{M_\ep}{|\lambda|} \right| \|x\| \leq a(1+M_\ep) \|x\|. \end{equation*}
Hence \(BR_\lambda(A)\) is bounded and for \(a<(1+M_\ep)^{-1}\) we have
\begin{equation*} \| BR_\lambda(A) \|<1. \end{equation*}
Then \(I+BR_\lambda(A)\) is invertible and we write
\begin{align*} A+B-\lambda I \amp=\amp (I+B((A-\lambda I)^{-1})(A-\lambda I)\\ (I+B R_\lambda(A))^{-1} (A+B-\lambda I) \amp=\amp A-\lambda I\\ R_\lambda(A) (I+B R_\lambda (A))^{-1} \amp=\amp R_\lambda(A+B). \end{align*}
This manipulation implies that
\begin{equation*} \rho(A)\subset \rho(A+B), \end{equation*}
which means that \(A+B\) is sectorial as well and we use the same angle \(\delta\) as for \(A\text{.}\) Moreover,
\begin{equation*} \|R_\lambda(A+B)\|\leq \| R_\lambda(A) (I+B R_\lambda(A))^{-1} \|\leq \frac{\tilde M}{|\lambda|}. \end{equation*}
Hence \(A+B\) is also sectorial and it generates an analytic semigroup defined by the Cauchy integral representation.

Subsection 9.12.2 Regularity of Mild Solutions

We consider the inhomogeneous initial value problem
\begin{align} \dot u(t) \amp=\amp A u(t) + f(t),\tag{9.12.2}\\ u(0) \amp=\amp u_0, \nonumber\tag{9.12.3} \end{align}
where \(A\) is a sectorial generator. We know already that \(e^{At} u_0\) is analytic in \(t\) for \(t>0\text{.}\) Moreover \(e^{At} u_0 \in D(A^n)\) for each \(n>0\) and
\begin{equation*} \|A e^{At} u_0 \|\leq C\frac{\|u_0\|}{t}. \end{equation*}
Hence we prove a technical lemma:
By Theorem [cross-reference to target(s) "6.34" missing or not unique] we chose constants \(M,C>0\) such that
\begin{equation*} \|e^{At}\|\leq M, \qquad \|Ae^{At} \|\leq \frac{C}{t}\qquad \mbox{ for all } t\in (0,T]. \end{equation*}
Then
\begin{equation*} \left\|\int_0^t A e^{A(t-s)} (f(s) - f(t)) ds \right\| \leq \int_0^t \frac{C}{t-s} L (t-s)^\theta ds = \frac{C L t^\theta}{\theta}, \end{equation*}
where \(L\) is the H\"older constant of \(f\text{.}\) This implies \(w\in D(A)\text{.}\) To show the H\"older continuity in \(t\) we compute
\begin{align*} \|A e^{At} - Ae^{As} \| \amp=\amp \left\|\int_s^t A^2 e^{A\tau} d\tau \right\|\\ \amp\leq\amp \int_s^t \|A^2 e^{A\tau} \| d\tau\\ \amp=\amp \int_s^t \|A e^{\frac{A\tau}{2}} A e^{\frac{A\tau}{2}}\| d\tau\\ \amp\leq\amp \int_s^t \frac{2C}{\tau}\frac{2C}{\tau} d\tau\\ \amp=\amp 4 C^2 \frac{t-s}{st}. \end{align*}
Then
\begin{align*} Aw(t+h) - A w(t) \amp=\amp A\int_0^{t+h} e^{A(t+h-s)} (f(s) - f(t+h)) ds-\int_0^t e^{A(t-s)} (f(s) - f(t)) ds\\ \amp=\amp \underbrace{A\int_0^t\left(e^{A(t+h-s)} - e^{A(t-s)} \right) (f(s) - f(t)) ds}_{I_1}\\ \amp\amp + \underbrace{ A \int_0^t e^{A(t+h-s)} (f(t) - f(t+h)) ds}_{I_2}\\ \amp\amp + \underbrace{A\int_t^{t+h} e^{A(t+h-s)} (f(s) - f(t+h)) ds}_{I_3}. \end{align*}
We begin with \(I_1\text{,}\) where we use the substitution \(s=t-h\tau\) in the third step:
\begin{align*} \|I_1\| \amp\leq \amp \int_0^t 4 C^2 \frac{(t+h-s)-(t-s) }{(t+h-s)(t-s)} L (t-s)^\theta ds\\ \amp=\amp 4 C^2 L h \int_0^t (t+h-s)^{-1} (t-s)^{\theta-1} ds\\ \amp=\amp 4 C^2 L h \int_0^{\frac{t}{h}} \frac{h^{\theta-1} \tau^{\theta-1}}{1+\tau} h d\tau \quad = \quad 4 C^2 h^\theta \int_0^{\frac{t}{h}} \frac{\tau^{\theta-1}}{1+\tau} d\tau\\ \amp\leq\amp 4 C^2 h^\theta \frac{\tau^\theta}{\theta} \Big|_0^{\frac{t}{h}} \quad \leq \quad C h^\theta. \end{align*}
For \(I_2\) we find
\begin{align*} \|I_2\| \amp \leq \amp \left\|A\int_0^t e^{A(t+h-s)} (f(t)-f(t+h)) ds \right\|\\ \amp\leq\amp \left\|(e^{A(t+h)} - e^{At}) (f(t+h) - f(t) ) \right\|\\ \amp\leq \amp M h^\theta. \end{align*}
Finally \(I_3\) is estimated as
\begin{align*} \|I_3\|\amp\leq \amp \int_t^{t+h} \frac{C}{t+h-s} L(t+h-s)^\theta ds\\ \amp\leq\amp C \int_t^{t+h} (t+h-s)^{\theta-1} ds \quad = \quad \left|\frac{C}{\theta} (t+h-s)^\theta \Big|_t^{t+h} \right|\\ \amp=\amp \frac{C}{\theta} h^\theta. \end{align*}

We write the mild solution as
\begin{equation*} u(t) = e^{At} u_0 +\int_0^t e^{A(t-s)}(f(s)-f(t)) ds + \int_0^t e^{A(t-s)} f(t) ds. \end{equation*}
Upon applying \(A\) to the mild solution we find
\begin{equation} A u = \underbrace{A e^{At} u_0}_{C^{0,\theta}} + \underbrace{Aw}_{C^{0,\theta}} + A\int_0^t e^{A(t-s)} f(t) ds. \tag{9.12.4} \end{equation}
For the last term we have from the Fundamental Theorem for Semigroups Theorem 9.3.3 that
\begin{equation*} A\int_0^t e^{A(t-s)} f(t) ds = (e^{At}-I) f(t), \end{equation*}
and we know that \(I f(t) = f(t) \in C^{0,\theta}\text{.}\) We still need to check the term \(e^{At} f(t)\text{.}\) For \(t\geq \delta, h>0\) we estimate
\begin{align*} \|e^{A(t+h)} f(t+h) - e^{At} f(t) \| \amp\leq\amp \|e^{A(t+h)}\| \|f(t+h) - f(t) \| + \|e^{A(t+h)} - e^{At} \| \|f(t) \|\\ \amp\leq\amp C_1 h^\theta + C_2 \frac{h}{\delta} \quad \leq \quad C_3 h^\theta. \end{align*}
This proves that \(Au\in C^{0,\theta}([\delta, T],X) \text{,}\) and since \(u\) is a mild solution it also implies \(\dot u\in C^{0,\theta}([\delta, T],X) \text{,}\) and we have item 1. In the last inequality, we clearly see that \(\delta>0\) needs to be imposed.

To show continuity at time \(0\text{,}\) we consider for \(t\to 0\)
\begin{equation*} \|e^{At} f(t) - f(0) \|\leq \|e^{At} f(0)-f(0)\| + \|e^{At}\|\|f(t) - f(0)\|. \end{equation*}
The first term of the right hand side goes to zero since \(e^{At}\) is strongly continuous and the second term goes to zero since \(f(t)\) is H\"older continuous. Hence we proved item 2.

Finally, for item 3. we estimate
\begin{align*} \|(e^{A(t+h)} - e^{At}) f(t) \| \amp=\amp \left \|\int_t^{t+h} A e^{A\tau} f(t) d\tau \right\|\\ \amp\leq\amp \int_t^{t+h} \|A e^{A\tau}(f(t) - \underbrace{f(0)}_{=0}) \| d\tau\\ \amp\leq\amp c \int_t^{t+h} \frac{1}{\tau} t^\theta d\tau \quad \leq C \int_t^{t+h} \tau^{\theta-1} d\tau\\ \amp=\amp c ((t+h)^\theta-t^\theta) \quad \leq \quad c h^\theta. \end{align*}

The previous result can essentially be visualized as
\begin{equation*} \underbrace{\dot u}_{C^{0,\theta}} = \underbrace{Au}_{C^{0,\theta}} + \underbrace{f(t)}_{C^{0,\theta}}. \end{equation*}
For an analytic semigroup \(\dot u \) and \(Au\) are as good as \(f\text{.}\) It also means that
\begin{equation*} u\in C^1((0,T], C^{0,\theta}), \qquad u \in C((0,T), C^{2,\theta}). \end{equation*}
Hence \(u\) is twice continuously differentiable in \(x\) with H\"older continuous second derivative. This is called regularity.

Higher regularity results are possible with analytic semigroups and this is an entire research area in itself. See the excellent book by A. Lunardi \cite{lunardi}.