2
$\begingroup$

Let $(A,\mathcal{D}(A))$ be an infinitesimal generator of a strongly continuous semigroup $(T(t))_{t\ge 0}$ on a Banach space $X$ and define on $\mathcal{X} := X \times X$ the operator matrix $$\mathcal{A}=\left( \begin{array}{cc} 0 & A \\ A & 0 \\ \end{array} \right)$$

with domain $\mathcal{D}(\mathcal{A}) := \mathcal{D}(A) \times D(A).$

I want to know if $\mathcal{A}$ generates a strongly continuous semigroup on the product space $\mathcal{X}$.

$\endgroup$
2
  • 2
    $\begingroup$ I seems to me that your matrix is diagonal rather than anti-diagonal (as suggested in the title). Do you mean $\mathcal{A} = \begin{pmatrix} 0 & A \\ A & 0 \end{pmatrix}$? $\endgroup$ Sep 26 '19 at 12:16
  • $\begingroup$ yeah exactly thank you that's what I meant. (The diagonal case is direct) $\endgroup$ Sep 26 '19 at 12:30
3
$\begingroup$

The answer is no in general.

For a counterexample, let $A$ be your favourite semigroup generator that has a sequence of eigenvalues $(\lambda_n) \subseteq \mathbb{R}$ such that $\lambda_n \to -\infty$ (for instance, let $A$ be the Dirichlet or Neumann Laplace operator on $L^2(0,1)$).

If $f_n$ is an eigenvector for $\lambda_n$, then $(f_n,-f_n) \in \mathcal{D}(\mathcal{A})$ and $\mathcal{A}(f_n,-f_n) = -\lambda_n (f_n,-f_n)$. Hence, $\mathcal{A}$ has a sequence of eigenvalues that converges to $\infty$ and thus, it cannot be a semigroup generator.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.