Section 9.8 The Lumer-Phillips Theorem
A special case arises when the constant \(M_\gamma\) =1. In this case we only need the resolvent estimate for \(n=1\text{:}\)
\begin{equation}
\|R_\lambda(A)\|\leq \frac{1}{\lambda-\omega}\tag{9.8.1}
\end{equation}
and the rest follows through iteration
\begin{equation}
\|R_\lambda(A)^n\|\leq \frac{1}{(\lambda-\omega)^n}.\tag{9.8.2}
\end{equation}
Corollary 9.8.1.
The operator \(A\) is a generator of a strongly continuous semigroup \(\{T(t)\}\) with
\begin{equation*}
\|T(t)\| \leq e^{\omega t},
\end{equation*}
if - \(D(A)\) is dense and \(A\) is closed.
- For all \(\lambda>\omega\) we have\begin{equation*} \|R_\lambda(A) \|\leq \frac{1}{\lambda-\omega} . \end{equation*}
Definition 9.8.2.
The semigroup \(\{T(t)\}\) is called a quasi contraction if \(\|T(t) \|\leq e^{\omega t}\text{,}\) and it is called a contraction if \(\|T(t)\|\leq 1\) (i.e. \(\omega=0\)).Theorem 9.8.3.
Let \(A\) be a linear operator on a Hilbert space. Assume- \(D(A)\) is dense.
- For all \(x \in D(A)\text{,}\) \(\mbox{Re}(x, Ax) \leq \omega(x,x) \) for some \(\omega\geq 0\text{.}\)
- There exists a \(\lambda_0>\omega\) such that \(A-\lambda_0I \) is onto.
\begin{equation*}
\|e^{At} \|\leq e^{\omega t} .
\end{equation*}
Proof.
\begin{equation*}
\|(A-\lambda I) x \| \|x\| \geq \mbox{Re} (x, (\lambda I - A) x) \geq (\lambda - \omega) (x,x),
\end{equation*}
which implies
\begin{equation*}
\|(A-\lambda I) x\|\geq (\lambda-\omega) \|x\|.
\end{equation*}
Hence \(A-\lambda I\) is bounded below, away from \(0\text{,}\) and by Corollary Corollary 8.2.9 the resolvent \(R_\lambda(A)\) exists and
\begin{equation*}
\|R_\lambda(A)\|\leq \frac{1}{\lambda-\omega}.
\end{equation*}
Hence for any \(\lambda_0>\omega\) the map \(A-\lambda_0 I\) is onto. Then \(R_{\lambda_0}(A)\) has range \(X\) and is continuous. This implies that \(A-\lambda_0 I\) is closed, which means that \(A\) is closed. By the Corollary Corollary 9.8.1 to the Hille-Yosida Theorem, \(A\) generates a strongly continuous semigroup \(T(t)\) with
\begin{equation*}
\|T(t) \|\leq e^{\omega t}.
\end{equation*}
Corollary 9.8.4.
Let \(A\) be a linear operator on a Hilbert space. Assume- \(D(A)\) is dense.
- For all \(x \in D(A)\text{,}\) \(\mbox{Re}(x, Ax) \leq \omega(x,x) \) for some \(\omega\geq 0\text{.}\)
- The resolvent set \(\rho(A)\cap (\omega,\infty) \neq \emptyset\text{.}\)
\begin{equation*}
\|e^{At} \|\leq e^{\omega t} .
\end{equation*}
Example 9.8.5.
\begin{equation}
\omega >\max\{\mbox{Re}(\lambda), \lambda\in \sigma(A)\}.\label{spbd}\tag{9.8.3}
\end{equation}
Let \(\{\phi_i\}_i\) denote an orthonormal basis of eigenvectors. Then
\begin{equation*}
(x, Ax) = \sum_{i=1}^\infty \lambda_i (x,\phi_i) (x,\phi_i) \leq \omega \sum_{i=1}^\infty (x,\phi_i)(x,\phi_i) = \omega \|x\|^2.
\end{equation*}
Then for \(\lambda_0>\omega\) the resolvent \(R_{\lambda_0}(A)\) exists and \(A-\lambda_0 I\) is onto. By the Lumer-Phillips theorem \(A\) generates a quasi contraction semigroup. Note here that the spectral bound ((9.8.3)) is essential.Example 9.8.6.
\begin{equation*}
D(A) = \{u\in H^1(0,1), u(1) =0\}.
\end{equation*}
\(D(A)\) is dense in \(L^2(0,1)\text{,}\) since \(H^1\) is dense in \(L^2\) and functions in \(L^2\) are only defined up to a set of measure zero. Hence the condition on one of the boundary points is not relevant. We compute the spectrum of \(A\text{.}\) For \(\varphi\in L^2(0,1)\) we like to solve the resolvent equation \((A-\lambda I) u = \varphi\text{.}\) This is written as an ODE
\begin{equation*}
u' = \lambda u +\varphi, \qquad u(1) =0,
\end{equation*}
which is a linear ODE that can be solved for any \(\lambda\in \CC\text{.}\) Hence \((A-\lambda I)^{-1}\) exists for all \(\lambda\in \CC\text{,}\) and we can chose a spectral bound of \(\omega=0\text{.}\) To apply Lumer-Phillips we need one more estimate
\begin{equation*}
(u,Au) = \int_0^1 u(x) u'(x) dx = - \frac{1}{2} (u(0))^2 \leq 0 .
\end{equation*}
Hence \(A\) generates a contraction semigroup \(T(t)\) with
\begin{equation*}
\|T(t) \|\leq 1.
\end{equation*}
