I'm looking for results (or some ideas) on the following kind of pseudo-differential evolution equation:
$$ \frac{\partial u(t,x)}{\partial t} = \int_{-\infty}^{t} B(t-s,x)\, A(x,D_{x})u(s,x)\,ds \; ;\quad u(0,x) = u_{0}(x) $$
where $A$ is a $\Psi DO$ with symbol in $S^{m}(\mathbb{R}^{n}\times \mathbb{R}^{n})$ (let's say elliptic) and $B$ is a causal smooth function in $(t,x)$, decaying in time.
In particular, I am interested in construction of parametrices and propagation of singularities.
To get some feeling for the problem, we make the following simplifying assumptions:
$B$ is independent of $x$, say $B \in \mathcal S(\overline{\mathbb R}_+)$.
$A$ admits a (holomorphic) functional calculus.
There is a (holomorphic) function $h\colon \sigma(A)\to\overline{\mathbb H}_-$ such that $\bigl(i\tau - \lambda\,\hat{B}(\tau)\bigr)\bigr|_{\,\tau=h(\lambda)} =0$ for $\lambda\in\sigma(A)$, where $\hat{B}(\tau)=\int_0^\infty e^{-it\tau}B(t)\,dt$ is the Fourier transform of $B$.
Then it is readily seen that $$ u(t) = e^{ith(A)} u_0 $$ is a solution of the original problem. Notice that, in general, one can only solve backwards in time.
There are several difficult points with this approach, the two obvious ones being:
The range of $i\tau\,/\,\hat{B}(\tau)$ might miss parts of the spectrum of $A$. This puts restrictions on the initial value $u_0$.
There might be several choices for the function $h$. This leads to non-uniqueness of the solution $u$.
Here is an example: Take $B(t) = H(t)\,e^{-t}$. Then $\hat{B}(\tau)= \frac{1}{1+i\tau}$. Choose $A\in \operatorname{Op}S^2$ with principal symbol $|\xi|^2$ and $\sigma(A)\subset\{\lambda\in{\mathbb C}\mid \Re \lambda>0\}$. Then $h(\lambda) = \frac{i}2\left(1-\sqrt{1+4\lambda}\right)$. It follows that $h(A)\in \operatorname{Op}S^1$ with principal symbol $-i\,|\xi|$. This implies that the operators $e^{ith(A)}$ for $t<0$ belong to $\operatorname{Op}S^{-\infty}$, i.e., they are regularizing, and there is no propagation of singularities.
ADDED: I'm not aware of any holomorphic functional calculus for pseudodifferential operators - with $m>0$ and not necessarily having $A$ be normal (i.e., $AA^*=A^*A$ as unbounded operators in $L^2$) - which does what is required here.
Indeed, $\hat B$ extends to a $C^\infty$ function in $\overline{\mathbb H}_-$ which is holomorphic in $\mathbb H_-$, with $$ \hat B(\tau) \sim \sum_{k\geq 0} \frac{B^{(k)}(+0)}{(i\tau)^{k+1}} \quad \text{as $\tau\to \infty$ in $\overline{\mathbb H}_-$} $$ (and this asymptotic expansion can be differentiated any number of times). Therefore, assuming $B(+0)\neq0$, one has $$ \frac{i\tau}{\hat{B}(\tau)} = -\,\frac{\tau^2}{B(+0)} + O(\tau) \quad \text{as $\tau\to\infty$ in $\overline{\mathbb H}_-$.} $$ So, one would expect $h(A)$ to belong to $\operatorname{Op}S^{m/2}$ and to have principal symbol $$ -i\sqrt{B(+0)} \ a_m^{1/2}(x,\xi), $$ where $a_m(x,\xi)$ denotes the principal symbol of $A$.
Of course, $h(\lambda)$ need not be an algebraic function of $\lambda$ - as was the case in the example above - and then in order to prove such a result one cannot directly appeal to known facts about complex powers of (hypoelliptic) pseudodifferential operators.
Still, as $$ h(\lambda) \sim \sum_{l\geq0} c_l\lambda^{1/2-l} \quad\text{as $\lambda\to\infty$} $$ in a suitable sector depending on the choice of $B$ (again, this asymptotic expansion can be differentiated any number of times), with $c_0 = -i\sqrt{B(+0)}$ as just seen and $c_l$ for $l\geq0$ computable in terms of $B(+0),\dots, B^{(l)}(+0)$, there is a good chance that such a result holds. See also this previous post.
ADDED: Suppose that $e^{at}B\in \mathcal S(\overline{\mathbb R}_+)$ holds for some $a>0$. Then $\hat{B}$ extends to a $C^\infty$ function on $ia+\overline{\mathbb H}_-=\{\tau\in \mathbb B\mid\Im\tau\leq a\}$ which is holomorphic on $ia+\mathbb H_-$ (and still has an asymptotic expansion as $\tau\to\infty$, as above).
Here is another example using this observation: Let $B(t)=-\,H(t)\,e^{-t}$ and $A\in\operatorname{Op}S^2$ with principal symbol $|\xi|^2$. Suppose that $\sigma(A)\subset \{\lambda\in{\mathbb C}\mid (\Im \lambda)^2<\Re\lambda\}$. Then $i\tau/\hat{B}(\tau)=\tau^2-i\tau$ and there are the two choices $h_\pm(\lambda)= \frac12\left(i\pm \sqrt{4\lambda-1}\right)$ for $h$. Furthermore, $h_\pm(A) \in \operatorname{Op}S^1$ with principal symbol $\pm|\xi|$. To specify a unique solution $u$ requires a second initial condition $u_t(0)=u_1$. Contrary to the situation considered before, the solution $$ u(t) = e^{-t/2}\left(\cos\left(t\sqrt{A-1/4}\right)u_0 + \frac{\sin\left(t\sqrt{A-1/4}\right)}{\sqrt{A-1/4}}\left(\frac{u_0}2+u_1\right)\right) $$ is defined for all $t\in\mathbb R$. Moreover, now one has propagation of singularities.
