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:

  1. $B$ is independent of $x$, say $B \in \mathcal S(\overline{\mathbb R}_+)$.

  2. $A$ admits a (holomorphic) functional calculus.

  3. 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.

  • $\begingroup$ Thanks @Ingo this is more or less what I was attempting. Could you give me some references on the holomorphic function calculus? (I already checked the books by Shubin and Grubb) – J.C. 10 mins ago $\endgroup$ – J.C. Mar 14 '15 at 21:54
  • $\begingroup$ @J.C. Could you please be more specific about your assumptions on A, B? There are simply too many possibilities. To demonstrate this, I've added another example. The essential difference between both examples is the sign of B (for t>0) while leaving A unaltered. $\endgroup$ – ifw Mar 15 '15 at 15:57

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.