I want to ask about the observability inequality for the heat equation ( internal control), we consider the backward heat equation with Dirichlet boundary conditions:

\begin{array}{c} \varphi _{t}+\Delta \varphi =0\text{ in }\Omega \times (0,T) \\ \varphi =0\text{ on } \partial \text{}\Omega \times (0,T)\text{} \\ \varphi (T)=\varphi _{T}\text{ }% \end{array} The observability inequality in some books is given by: $$ \left\Vert \varphi (0)\right\Vert ^{2}\leq c\int_{\omega }\int_{0}^{T}\varphi (x,t)dtdx $$ and some times $$\left\Vert \varphi (T)\right\Vert ^{2}\leq c\int_{\omega }\int_{0}^{T}\varphi (x,t)dtdx$$ and $$\int_{\omega }\int_{T/4}^{3T/4}\varphi (x,t)dtdx\leq c\int_{\omega }\int_{0}^{T}\varphi (x,t)dtdx$$ To passe from the first to the second I guess that they made the substitution $s=T-t$, however, for the third it comes from the golobal Carleman estimates but I can not see the equivalence between them. Thank you.

  • 1
    $\begingroup$ First, you should have $\varphi^2$ inside all of the integrals instead of just $\varphi$. For the equivalence, you can read Theorem 2.44 of Coron, Control and Nonlinearity or Theorem 11.2.1 of Tucsnak and Weiss, Observation and Control for Operator Semigroups. Basically, this is Integration by parts, Riesz Theorem in Hilbert spaces and Closed Graph Theorem. I can put a full answer if tou want, but I can't promise it will we easier to understand than these references :) . $\endgroup$ – Armand Koenig Jun 22 '18 at 10:12

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