Section 9.10 Bounded Perturbations
Semigroup thoery is particulalry powerful if we consider nonhomogeneous equations, where the leading order differential operator is perturbed by an operator of lower order. There is a huge variety of perturbation results for semigroups \cite{pazy,lunardi,Engel} and here, we will focus on the example of bounded perturbations. We consider
\begin{equation*}
u_t = Au + Bu,
\end{equation*}
where \((A,D(A))\) is a generator and \(B\) is bounded. We want to show that \((A+B, D(A))\) is also a generator, and possibly get a relation of \(e^{(A+B)t}\) to \(e^{At}\text{.}\) The corresponding proof will use a switch of equivalent norms, which we prepare first: Proposition 9.10.1.
Let \(A\) be a linear operator on a Banach space \(X\) with \((0,\infty)\in \rho(A)\text{.}\) If for each \(\lambda\in(0,\infty)\)
\begin{equation*}
\|\lambda^n R_\lambda(A)^n\|\leq M, \qquad \mbox{for all } n\geq 0,
\end{equation*}
then there exists an equivalent norm \(|\cdot |_\diamond\) such
\begin{equation*}
\|x\|\leq |x|_\diamond \leq M\|x\|, \qquad \mbox{ for all } x\in X,
\end{equation*}
and
\begin{equation*}
|\lambda^n R_\lambda(A)^n|_\diamond \leq 1.
\end{equation*}
Proof.
\begin{equation*}
\|x\|_\mu := \sup_{n\geq 0} \|\mu^n R_\mu(A)^n x \|.
\end{equation*}
Then for \(n=0\) we get \(\|x\|\) such that
\begin{equation}
\|x\|\leq \|x\|_\mu \leq M\|x\|\label{munorm}\tag{9.10.1}
\end{equation}
and
\begin{equation*}
\|\mu R_\mu(A) x\|_\mu = \sup_{n\geq 0} \|\mu^n R_\mu(A)^n \mu R_\mu(A) x\| \leq \|x\|_\mu.
\end{equation*}
Hence the operator norm
\begin{equation*}
\|\mu R_\mu(A)\|_\mu \leq 1.
\end{equation*}
For \(0<\lambda\leq \mu\) we set \(y=R_\lambda(A) x\) and we get
\begin{align*}
R_\mu(A) (x-(\mu-\lambda)y) \amp=\amp R_\mu(A) \Bigl(x+(A-\mu I) y -\underbrace{(A-\lambda I) y}_{=x} \Bigr)\\
\amp=\amp R_\mu(A) (A-\mu I) y\\
\amp=\amp y.
\end{align*}
Evaluating \(y\) in the \(\mu\)-norm we get
\begin{align*}
\|y\|_\mu \amp=\amp \left\|\mu R_\mu(A) \frac{x-(\mu-\lambda)y}{\mu} \right \|_\mu\\
\amp\leq \amp \left\|\frac{x}{\mu} -\frac{\mu-\lambda}{\mu} y \right\|_\mu\\
\amp\leq \amp \frac{1}{\mu} \|x\|_\mu +\left(1-\frac{\lambda}{\mu} \right) \|y\|_\mu .
\end{align*}
This implies that
\begin{equation}
\frac{\lambda}{\mu} \|y \|_\mu \leq \frac{1}{\mu}\|x\|_\mu, \tag{9.10.2}
\end{equation}
leading to
\begin{equation*}
\lambda \|y\|_\mu \leq \|x\|_\mu .
\end{equation*}
Hence also the operator norm
\begin{equation*}
\|\lambda R_\lambda(A) \|_\mu \leq 1, \qquad \mbox{for all } 0<\lambda\leq \mu.
\end{equation*}
Now we take the limit \(\mu\to \infty\) and define the corresponding limit norm
\begin{equation*}
|x|_\diamond := \lim_{\mu\to\infty}\|x\|_\mu = \lim_{\mu\to \infty} \sup_{n\geq 0} \|\mu^n R_\mu(A)^n x\|
\end{equation*}
and
\begin{equation*}
|\lambda R_\lambda(A)|_{\diamond} \leq 1.
\end{equation*}
Taking the limit \(\mu\to\infty\) in ((9.10.1)) we get the equivalence of the \(|\cdot|_\diamond\) norm.Theorem 9.10.2.
Assume \((A,D(A))\) is a generator on \(X\) of a strongly continuous semigroup \(\{T(t)\}\) with exponential growth bound
\begin{equation*}
\|T(t)\| \leq M e^{\omega t}.
\end{equation*}
If \(B\) is a bounded linear operator on \(X\text{,}\) then \((A+B, D(A))\) is generator of a strongly continuous semigroup \(\{S(t)\}\text{,}\) denoted by \(S(t) = e^{(A+B)t}\text{,}\) with exponential bound
\begin{equation*}
\|S(t) \|\leq M e^{(\omega + M\|B\|) t}.
\end{equation*}
Proof.
\begin{equation*}
|T(t) |_\diamond \leq e^{\omega t}, \qquad |R_\lambda(A)|_\diamond \leq \frac{1}{\lambda-\omega}, \quad \lambda>\omega.
\end{equation*}
For \(\lambda> \omega+|B|_\diamond\) we find
\begin{equation*}
|BR_\lambda(A)|_\diamond \leq \frac{|B|_\diamond}{\lambda-\omega} < 1,
\end{equation*}
which implies that \(I + B R_\lambda(A) \) is invertible. We set
\begin{align*}
V\amp:=\amp R_\lambda(A) (I+ B R_\lambda (A))^{-1}\\
\amp=\amp R_\lambda(A) (I-(-BR_\lambda(A)))^{-1}\\
\amp=\amp \sum_{k=0}^\infty R_\lambda(A)(-BR_\lambda(A))^k,
\end{align*}
where we used the Neumann series in the last step. Now we claim that \(V=R_\lambda(A+B)\text{.}\) Indeed,
\begin{align*}
(A+B-\lambda I) V \amp=\amp \left[ (A-\lambda I) + B\right] R_\lambda(A)(I+BR_\lambda(A))^{-1}\\
\amp=\amp (I+ BR_\lambda(A))(I + BR_\lambda(A))^{-1}\\
\amp=\amp I.
\end{align*}
For the reverse case, we employ the above Neumann series
\begin{align*}
V(A+B-\lambda I)x \amp=\amp \sum_{k=0}^\infty R_\lambda(A) (-BR_\lambda(A))^k ((A-\lambda I) + B) x\\
\amp=\amp x + R_\lambda(A) Bx + \sum_{k=1}^\infty R_\lambda(A)(-BR_\lambda(A))^k ((A-\lambda I) + B) x\\
\amp=\amp x + R_\lambda(A) B x + \sum_{k=1}^\infty \underbrace{R_\lambda(A) (-B) R_\lambda(A)(-B) \cdots }_{k \mbox{ times}} R_\lambda(A)(A-\lambda I) x\\
\amp\amp + \sum_{k=1}^\infty \underbrace{R_\lambda(A) (-B) R_\lambda(A)(-B) \cdots }_{k \mbox{ times}} R_\lambda(A) B x\\
\amp=\amp x+ R_\lambda(A) Bx + \sum_{k=1}^\infty (-R_\lambda(A) B)^k x - \sum_{k=2}^\infty (-R_\lambda(A) B)^k x\\
\amp=\amp x.
\end{align*}
Then
\begin{align*}
|R_\lambda(A+B) |_\diamond \amp=\amp |V|_\diamond\\
\amp=\amp|R_\lambda(A)|_\diamond |(BR_\lambda(A) +I)^{-1} |_\diamond\\
\amp\leq\amp \frac{1}{\lambda-\omega} \left|\sum_{k=0}^\infty \left(-B R_\lambda(A)\right)^{-k} \right|_\diamond\\
\amp\leq\amp \frac{1}{\lambda-\omega} \sum_{k=0^\infty} |B|_\diamond^k |R_\lambda(A)|_\diamond^k\\
\amp \leq\amp \frac{1}{\lambda-\omega} \sum_{k=0}^\infty \frac{|B|^k_\diamond}{(\lambda-\omega)^k}\\
\amp=\amp \frac{1}{\lambda-\omega} \frac{1}{1-\frac{|B|_\diamond}{\lambda-\omega}}\\
\amp=\amp
\frac{1}{\lambda-\omega-|B|_\diamond}.
\end{align*}
Now we apply the Corollary Corollary 9.8.1 of the Hille-Yosida Theorem and conclude that \(A+B\) is a generator of a strongly continuous semigorup \(\{S(t)\}\) with
\begin{equation*}
|S(t)|_\diamond \leq e^{(\omega+|B|_\diamond)t},
\end{equation*}
which, upon reverting to the original norm, gives
\begin{equation*}
\|S(t) \|\leq M e^{(\omega + M\|B\|) t} .
\end{equation*}
Remark 9.10.3.
The semigroup \(\{S(t)\}\) solves the ODE \(\dot u = Au + Bu\) and it can be related to the semigroup \(\{T(t)\}\) with the variation of constants formula
\begin{equation*}
S(t) x = T(t) x +\int_0^t T(t-s) B S(s) x dx .
\end{equation*}
Remark 9.10.4.
The shift in norm is a very useful trick, since we get rid of the constant \(M\) in the Hille-Yosida theorem. This means, that in the \(|\cdot|_\diamond\)-norm we need the resolvent estimate only for \(n=1\) and not for all \(n\text{.}\) In fact, whenever such estimates arise in the next sections, we implicitly assume to have chosen the \(|\cdot|_\diamond\) norm to begin with.Example 9.10.5.
\begin{align*}
u_t \amp=\amp D\Delta u + \int_\Omega k(x,y) u(y,t) dy,\\
u|_{\partial\Omega} \amp=\amp 0.\\
k\amp\geq \amp0, \quad k\in L^2, \quad \int_\Omega k(x,y) dx =1.
\end{align*}
The operator \(D\Delta \) is generator of the heat-equation semigroup \(\{e^{D\Delta t}\}\) with Dirichlet boundary conditions. The integral operator \(Ku = \int_\Omega k(x,y) u(y,t) dy\) is a compact Hilbert-Schmidt operator, hence bounded. We directly apply the above perturbation result Theorem Theorem 9.10.2 and there exists a strongly continuous semigroup
\begin{equation*}
S(t)= e^{(D\Delta + K)t}
\end{equation*}
on \(L^2(\Omega)\) with
\begin{equation*}
\|S(t) \|\leq M e^{(\omega + M\|K\|) t} .
\end{equation*}
For each \(u_0\in D(A)\) and for each \(T>0\) we have
\begin{equation*}
S(t)u_0 \in C^1([0,T], L^2(\Omega))\cap C([0,T], D(A)).
\end{equation*}