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Is there something analogous to the regularity results of the implicit function theorem, but for pseudodifferential operators? I'm looking for something to the effect of, "under certain conditions, the solution operator to a family of pseudodifferential equations is itself pseudodifferential".

In detail, suppose we have a family of invertible symmetric operators $H(q): X \rightarrow X$ (parameterized by $q$ in function space $Q$), each of which maps a function space $X$ to itself. Further, suppose each $H(q)$ has representation as a pseudo-differential operator so that $$H(q)u = \int e^{i x \xi} P(x,\xi;q) \hat{u}(\xi) d\xi$$ for some symbol $P$ satisfying $$\partial^\alpha_\xi \partial^\beta_x P(x,\xi;q) \le C_{\alpha,\beta}(1 + |\xi|^2)^{\frac{1}{2}(m-|\alpha|)}$$ ($m$ is the order of the pseudodifferential operator). Also let $P$ satisfy whatever other requirements in terms of $q$ that are needed.

Finally, let $S:Q \rightarrow X$ be the solution operator to the equation $$H(q)u=w,$$ for a fixed right hand side $w$. Ie., to find $S(q)$, solve the equation $H(q)u=w$ for $u$, and set $S(q):=u$. Are there any conditions under which $S$ is also pseudodifferential, and if so can we say anything about it's symbol?

Nick Alger
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1 Answers1

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I figured out on my own that the answer is affirmative, under the condition that the derivative of the constraint, $[H'_q(q)\delta q]u$, is pseudodifferential in $\delta q$.

By the implicit function theorem, the linearization of the solution operator, $S'$ to the equation $H(q)u=w$ is given by, $$S'(q)\delta q = H(q)^{-1} [H'_q(q) \delta q]u$$

Assuming $H(q)$ is invertible (otherwise it doesn't make much sense to talk about solving $H(q)u=w$), we have that $H(q)^{-1}$ is also pseudodifferential with symbol $1/P$ to leading order, where $P$ was the symbol of $H(q)$.

Thus when $[H'_q(q)\delta q]u$ is pseudodifferential in $\delta q$, we have that $S'$ is the composition of pseudodifferential operators, and so is itself pseudodifferential.

Nick Alger
  • 19,977