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Section 9.7 The Hille-Yosida Theorem

The Hille-Yosida Theorem is the first result that will answer the question: When is an operator a generator of a \(C^0\)-semigroup?

Step 1: Item 1. is necessary (semigroup \(\Longrightarrow\) Item 1.) We already know from Theorem 9.3.5 that \(D(A)\) needs to be dense and \(A\) needs to be closed. Hence item 1. is necessary.

Step 2: Item 2. is necessary (semigroup \(\Longrightarrow\) Item 2.) Assume \(\{T(t)\}\) is a semigroup with the exponential bound Theorem 9.7.1. We define a strange looking integral operator, called \(I_n(\lambda)\text{,}\) and we will show that it corresponds to \((-R_\lambda(A))^n\text{.}\) The integral representation will then allow us to prove the desired resolvent estimate. Here we go
\begin{equation*} I_n(\lambda) x := \frac{1}{(n-1)!}\int_0^\infty t^{n-1} e^{-\lambda t} T(t) x \, dt, \qquad \lambda\in \CC, \mbox{Re}\lambda>\omega, n\geq 1. \end{equation*}
We will show that
\begin{equation*} I_n(\lambda) = (-R_\lambda(A))^n\qquad \mbox{and} \qquad \|I_n(\lambda)\|\leq \frac{M}{(\mbox{Re}\lambda-\omega)^n }. \end{equation*}
The inequality is straight-forward by using the exponential bound and integration by parts \((n-1)\)-times:
\begin{align} \|I_n(\lambda) \| \amp\leq \amp \frac{M}{(n-1)!}\int_0^\infty \left| t^{n-1} e^{-\lambda t} e^{\omega t} \right| \, dt\nonumber\tag{9.7.2}\\ \amp=\amp \frac{M}{(n-1)!}\int_0^\infty t^{n-1} \left| e^{(\omega-\lambda) t} \right| \, dt\nonumber\tag{9.7.3}\\ \amp=\amp \frac{M}{(n-1)!}\int_0^\infty t^{n-1} e^{(\omega-\mbox{Re}\lambda) t} \, dt\nonumber\tag{9.7.4}\\ \amp=\amp \frac{M(-1)^{n-1}}{(\omega-\mbox{Re}\lambda)^{(n-1)}}\int_0^\infty e^{(\omega-\mbox{Re}\lambda) t} \, dt\nonumber\tag{9.7.5}\\ \amp=\amp \frac{M(-1)^{n-1}}{(\omega-\mbox{Re}\lambda)^{n}} (-1)\tag{9.7.6}\\ \amp=\amp \frac{M}{(\mbox{Re}\lambda - \omega)^n}.\tag{9.7.7} \end{align}
For \(x\in D(A)\) and \(n>1\) we find
\begin{align} I_n(\lambda) A x \amp=\amp \frac{1}{(n-1)!} \int_0^\infty t^{n-1} e^{-\lambda t} T(t) A x \, dt \nonumber\notag\\ \amp=\amp \frac{1}{(n-1)!} \int_0^\infty t^{n-1} e^{-\lambda t}\frac{d}{dt} (T(t) x) \, dt \nonumber\notag\\ \amp=\amp - \frac{1}{(n-2)!} \int_0^\infty t^{n-2} e^{-\lambda t} T(t) x \, dt + \frac{\lambda}{(n-1)!} \int_0^\infty t^{n-1} e^{-\lambda t} T(t) x \, dt \nonumber\notag\\ \amp=\amp -I_{n-1}(\lambda)x + \lambda I_n(\lambda) x.\tag{9.7.8} \end{align}
Similarly, for \(n=1\) we get
\begin{align} I_1(\lambda) Ax \amp=\amp \int_0^\infty e^{-\lambda t} T(t) A x \, dt \nonumber\tag{9.7.9}\\ \amp=\amp \int_0^\infty e^{-\lambda t} \frac{d}{dt} ( T(t) x) \, dt \nonumber\tag{9.7.10}\\ \amp=\amp - e^0 T(0) x +\lambda \int_0^\infty e^{-\lambda t} T(t) x \, dt \nonumber\tag{9.7.11}\\ \amp=\amp - x +\lambda I_1(\lambda) x.\tag{9.7.12} \end{align}
Hence we get recursive relations between the \(I_n\text{.}\) From (Display Mathematics [STRUCT].[NUM]) we find
\begin{equation*} I_n(\lambda) (Ax-\lambda x) = - I_{n-1}(\lambda) x \end{equation*}
and since \(T\) and \(A\) commute, we also have
\begin{equation*} (Ax-\lambda x) I_n(\lambda) = - I_{n-1}(\lambda) x. \end{equation*}
Moreover
\begin{equation*} I_1(\lambda)(A-\lambda I) = - I = (A-\lambda I) I_1(\lambda). \end{equation*}
This implies that
\begin{equation} R_\lambda(A)x = (A-\lambda I)^{-1}x = - I_1(\lambda)x = -\int_0^\infty e^{-\lambda t} T(t) x dt.\label{ResolventLaplace}\tag{9.7.13} \end{equation}
This equation is an important relation in its own right, as it describes the resolvent as a Laplace transform of the semigroup!

Starting with \(I_1\) we then argue recursively to find
\begin{equation*} I_n(\lambda) = (-R_\lambda(A))^n. \end{equation*}

Step 3: Item 1. and 2. are sufficient (Item 1.+2. \(\Longrightarrow\) semigroup). Now we come back to one of the ideas that we discussed in the introduction to this chapter; the idea to define an operator exponential as an inverse exponential limit
\begin{equation*} T_n(t) := \left(I-\frac{t}{n}A \right)^{-n}. \end{equation*}
We show \(T_n\to T\text{.}\) We rewrite \(T_n(t)\) as
\begin{equation*} T_n(t) =\left(\frac{t}{n}\right)^{-n} \left(-R_{\frac{n}{t}} (A) \right)^n, \end{equation*}
which is bounded by Item 2. for \(\frac{n}{t}>\omega\text{.}\) This is particularly bounded for large \(n\) and small \(t\text{.}\) To show that \(T_n(0)=I\text{,}\) it is sufficient to show that
\begin{equation*} \lim_{t\to 0} \left(I-\frac{t}{n}A\right)^{-1} = I, \end{equation*}
since the \(n\)-time iteration would then also be the identity. We use the relation for \(u\in D(A)\)
\begin{equation*} u = \left(I-\frac{t}{n} A \right)^{-1}\left( I-\frac{t}{n} A\right) u = \left(I-\frac{t}{n} A\right)^{-1} u - \frac{t}{n} \left( I-\frac{t}{n} A\right)^{-1} A u. \end{equation*}
Then
\begin{equation*} \left\| \left(I-\frac{t}{n} A\right)^{-1} u - u \right\| = \frac{t}{n} \left\|\left(I-\frac{t}{n} A\right)^{-1} A u \right\| \leq C t \|Au\| \to 0, \end{equation*}
for \(t\to 0\text{,}\) since \(\|Au\|\) is bounded for \(u\in D(A)\text{.}\) Hence \(T_n(0) u= u\) for a dense subset \(D(A)\text{,}\) and since \(T_n(0)\) is continuous, we have \(T_n(0)=I\) on \(X\text{.}\) \\

Next we want to show that indeed
\begin{equation*} T(t) = \lim_{n\to \infty} \left(I-\frac{t}{n} A\right)^{-n}. \end{equation*}
We compute
\begin{equation} \frac{d}{dt} T_n(t) = -n \left(I-\frac{t}{n} A\right)^{-n-1}\left(-\frac{A}{n}\right) = A\left(I-\frac{t}{n} A\right)^{-n-1}.\label{HYvier}\tag{9.7.14} \end{equation}
Again we argue for a dense subset, but this time for \(D(A^2)\subset D(A)\subset X\text{,}\) where it can be shown that \(D(A^2)\) is also dense by repeating the arguments in Theorem (Theorem 9.3.3). Let \(u\in D(A^2)\text{.}\) We show that \(\{T_n\}\) form a Cauchy sequence.
\begin{align} \amp\amp T_n(t) u - T_m(t) u \nonumber\tag{9.7.15}\\ \amp=\amp \int_0^t \frac{d}{ds} \left( T_m(t-s) T_n(s) u \right) ds \nonumber\tag{9.7.16}\\ \amp=\amp \int_0^t \left(-\dot T_m(t-s) T_n(s) u + T_m(t-s) \dot T_n(s) u\right)ds\nonumber\tag{9.7.17}\\ \amp=\amp \int_0^t \left(-A\left(I-\frac{t-s}{m} A\right)^{-m-1} \left(I-\frac{s}{n} A\right)^{-n} u +\left(I-\frac{t-s}{m}A\right)^{-m} A \left(I-\frac{s}{n} A\right)^{-n-1} u \right) ds \nonumber\tag{9.7.18}\\ \amp=\amp \int_0^t \left(I-\frac{t-s}{m} A\right)^{-m-1} \left(I-\frac{s}{n} A\right)^{-n-1} \left[-A\left(I-\frac{s}{n}A\right)+ \left(I-\frac{t-s}{m} A\right)A \right]u ds \nonumber\tag{9.7.19}\\ \amp=\amp \int_0^t \underbrace{\left(I-\frac{t-s}{m} A\right)^{-m-1}}_{\mbox{bounded}} \underbrace{\left(I-\frac{s}{n} A\right)^{-n-1}}_{\mbox{bounded}} \left[\frac{s}{n} - \frac{t-s}{m} \right] A^2 u \; ds.\tag{9.7.20} \end{align}
Then
\begin{align*} \|T_n(t) u - T_m(t) u\|\amp\leq \amp C\|A^2 u\|\int_0^t \frac{s}{n} + \frac{t-s}{m} ds\\ \amp=\amp \frac{C t^2}{2} \left[ \frac{1}{n} +\frac{1}{m} \right] \|A^2 u\|. \end{align*}
The last term is bounded, since \(u\in D(A^2)\) and it goes to zero for \(n,m\to\infty\text{.}\) Hence
\begin{equation*} T(t) =\lim_{n\to \infty} T_n(t) \end{equation*}
exists.

\noindent Step 4: It remains to show that this \(T(t)\) is indeed our semigroup and that \(A\) is its infinitesimal generator.

We know already that \(\lim_{t\to 0} T_n(t) = I\text{,}\) hence \(T(t)\) is strongly continuous. For the semigroup property we use equation (Display Mathematics [STRUCT].[NUM]) for \(n=m\text{,}\) without the integral, to obtain
\begin{equation*} \frac{d}{ds} \left( T_n(t-s) T_n(s) u\right) = \frac{2s-t}{n} \left(I-\frac{t-s}{n} A\right)^{-n-1} \left( I-\frac{s}{n} A\right)^{-n-1} A^2 u. \end{equation*}
In the limit as \(n\to \infty\) this becomes
\begin{equation*} \frac{d}{ds} T(t-s) T(t) = 0 . \end{equation*}
Hence \(T(t-s)T(s)\) is independent of \(s\text{.}\) Then it equals the value at \(s=0\text{,}\) which is \(T(t)\text{.}\)

Finally, taking the limit as \(n\to \infty\) in ((9.7.14)) leads to
\begin{equation*} \frac{d}{dt} T(t) = A T(t). \end{equation*}

\noindent The proof of the Hille-Yosida Theorem gave us two important formulas

Remark 9.7.3.
The formula for the semigroup (Corollary 9.7.2) is reminiscent to the discrete implicit Euler scheme for a numerical solver of \(\dot u = A u\text{.}\) Consider a small time increment \(\Delta t = \frac{t}{n}\text{.}\) Then the implicit Euler scheme is
\begin{equation*} \frac{u(t+\Delta t) - u(t) }{\Delta t} = A u(t+\Delta t), \end{equation*}
which gives
\begin{equation*} \left(\frac{1}{\Delta t} I - A\right) u(t+\Delta t) = \frac{u(t)}{\Delta t}. \end{equation*}
This is written as
\begin{equation*} u(t+\Delta t) = (I-\Delta t A)^{-1} u(t). \end{equation*}
An iteration of this scheme and taking \(\Delta t \to 0\) gives us something like (Corollary 9.7.2).
After all this work, we reward ourselves with a beautiful definition.
Definition 9.7.4.
If \(T(t)\) is a strongly continuous semigroup with generator \(A\text{,}\) then we write
\begin{equation*} T(t) = e^{At} . \end{equation*}

Remark 9.7.5.
In the semigroup triangle Figure 9.3 we are able to add the connection from \(A\) to \(T\) and also the connection from \(T\) to \(R_\lambda\text{.}\) The only missing link is the arrow from \(R_\lambda\) to \(T\text{.}\) This can be filled in for analytic semigroups, which we develop in Section Section 9.11.

\tikzstyle{semigr} = [ellipse, draw, fill=red!30, text width=3cm, text centered, minimum height=4em] \tikzstyle{gener} = [ellipse, draw, fill=yellow!40, text width=3cm, text centered, minimum height=4em] \tikzstyle{resol} = [ellipse, draw, fill=blue!20, text width=3cm, text centered, minimum height=4em] \tikzstyle{arrow} =[thick,->, >=stealth]

\begin{equation*} \{T(t)\}_{t\geq 0} \end{equation*}
Generator
\begin{equation*} (A, D(A)) \end{equation*}
Resolvent
\begin{equation*} R_\lambda(A) \end{equation*}

\(\displaystyle T(t) = e^{At}\)\(\displaystyle Ax = \lim_{t\to 0^+} \frac{T(t)x - x}{t} \)\(\displaystyle R_\lambda(A) = -\int_0^\infty e^{-\lambda t} T(t) dt \)\(R_\lambda(A) = (A-\lambda I)^{-1}\)\(A = R_\lambda(A)^{-1} + \lambda I\)semigroup triangle
Figure 9.7.6.