So I have the following operator, which is not very nice...

\begin{equation} A w = \partial_{xx} w - a \partial_x w- b x^2 w =0, \end{equation} with the following decay conditions $w \longrightarrow 0$ as $x \longrightarrow \pm \infty$.

It is our goal to be able to write mild solutions to this problem, so we need to show that this operator generates a semigroup. The domain is the real line, so and I want to check the criteria of the Hille-Yosida theorem (density of the domain of operator $A$, $A$ is closed, $\lambda \in \rho(A)$ and$ || R(\lambda, A) || \leq \frac{1}{\lambda})$.

As we are on the real line, the domain is dense and the operator is closed (Right?). I have no idea how do the resolvent stuff! Help please! xx


The statement "As we are on the real line, the domain is dense and the operator is closed" is not correct. The "domain" refers to the domain of the operator A, not to the spacial domain where the differential equation is posed.

Recall that from the mathematical viewpoint an operator is not just the formal expression on the right hand side of the equation. You need to define it on an appropriate functional space. A good choice is usually a Sobolev space, and should take into account the boundary conditions. But for your example, this is not trivial (and it is an important step in solving your problem). The differential operator A is usually an unbounded lineal operator in that space, and what you really need to check is that A is a closed operator there.

I recommend you that you read the book by Brezis on functional analysis where simpler examples are given, before trying this more difficult one.

The condition of the resolvent usually means that you need to get apriori bounds (again, in the functional space of choice) for the associated elliptic problem. The Hille-Yosida theorem essentially reduces solving a (linear) parabolic problem to getting those estimates. I hope that my comments be of help !

  • $\begingroup$ Thank you! I am having a look now at the book! :) $\endgroup$ – Catherine Drysdale Feb 1 '18 at 14:29
  • $\begingroup$ Why is it that the boundary conditions complicate the choice of space? Kindest regards, $\endgroup$ – Catherine Drysdale Feb 1 '18 at 14:56
  • $\begingroup$ You need to put them into the space somehow. In the case of a bounded domain $\Omega$, the usual choices are the Sobolev space $H^1(\Omega)$ for Neumann boundary conditions, or it subset $H^1_0(\Omega)$ for Dirichlet boundary condtions. Hence my guess would be using $H^1(R)$. However, in this example, the real problem is that coefficient $x^2$ is not bounded, so that it is not clear that the weak formulation makes sense in that space. $\endgroup$ – Pablo De Napoli Feb 1 '18 at 15:03
  • $\begingroup$ Would you mind giving an example or resource? x $\endgroup$ – Catherine Drysdale Feb 1 '18 at 18:13
  • $\begingroup$ The case of a the heat and wave equations in a bounded domain are treated in the book by Brezis I've mention. Also look for the "internet seminars" in semigroup theory. $\endgroup$ – Pablo De Napoli Feb 1 '18 at 18:20

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.