Inverse problems for memory kernels by Laplace transform methods (Q1578916)

The authors are concerned with the problem of recovering the kernel \(m\) in the following \(j\)-th order integro-differential equation \((j=1,2),\) related to the cylinder \(\Omega=D\times {\mathbb R}_+\), \(D\) being a bounded domain in \({\mathbb R}^N\) with a smooth boundary \(\partial D\): \[ \begin{gathered} \beta(x)D^j_tu(x,t) - \text{ div} (\gamma(x)\nabla u)(x,t)\\ + \int_0^t m(t-s)\text{ div} (\gamma(x)\nabla u)(x,s) ds = f(x,t),\qquad (x,t)\in \Omega,\end{gathered} \] subject to the initial condition \[ u(x,0) = \varphi(x),\qquad \forall x\in D,\quad \text{ if} j=1, \] \[ u(x,0) = \varphi(x),\qquad D_tu(x,0) = \psi(x),\qquad \forall x\in D,\quad \text{ if} j=2, \] and to either of the boundary conditions \[ u(x,t) = 0\;\;\text{ or\;} \;\lambda (x){\partial u\over \partial \nu}(x,t) + \mu(x)u(x,t) = 0,\quad (x,t)\in \partial D \times (0,T). \] Functions \(\beta\), \(\lambda\) and \(\mu\) are, respectively, continuous on \({\overline D}\) and \(\partial D\) as well as positive and non-negative, while \(\nu(x)\) stands for the outer normal to \(\partial D\) at \(x\). \noindent Moreover, to recover \(m\), either of the following additional measurements on the state function \(u\) is prescribed: \[ \Psi[u(\cdot,t)]=h(t),\qquad \forall t\in [0,T], \] or \[ \Psi[u(\cdot,t)]-\int_0^t m(t-s)\Psi[u(\cdot,s)] ds=h(t), \qquad \forall t\in [0,T], \] where \(\Psi\) is a given linear (continuous) functional acting on the spatial variables, only. Using Laplace transform techniques the authors reduce the problems to two fixed point equations for \(M\), the Laplace transform of \(m\), and solve them in a specific space of analytic functions under suitable assumptions on the Laplace transforms of the data \((f\beta){\widetilde {\;}}\), \({\widetilde \varphi}\), \({\widetilde h}\) and \({\widetilde \psi}\), if \(j=2\). Then such conditions are replaced with explicit (sufficient) conditions on the data \(f\), \(\varphi\), \(h\) and \(\psi\), if \(j=2\).