4
$\begingroup$

How can I get the (Volterra) operator from an equation of the type

$$u''(x)+xu'(x)+u(x)=0\text{ ?}$$

I know that there is a general way of doing it, if you could point me at the proper book I'd be thankful!

$\endgroup$
  • $\begingroup$ It is simpler to solve this equations than to understand what you are asking about $\endgroup$ – Norbert Sep 23 '12 at 16:51
  • $\begingroup$ Ok maybe I explain myself a little bad. Fredholm operators are helpful in order to solve differential equations, because you can reduce the problem to one of the tipe $u(x)=T(u(x))$ where $T$ is the operator, that is, the answer will be the eigenvalues of $T$. The question is adressed to people who know a little about Fredholm and Volterra operators. $\endgroup$ – Miguel Sep 23 '12 at 17:40
2
$\begingroup$

I understand that you want to rewrite the differential equation in terms of an integral (Volterra-type) operator. The resulting operator $T$ will be Hilbert-Schmidt, hence compact, hence $I-T$ is Fredholm.

Introducing $v=u'$, we get the system of 1st order equations $u'=v$, $v'=-u-xv$. Using the initial values $(u_0,v_0)$, we rewrite the IVP as a system $$u(t)=u_0+\int_0^t v(s)\,ds, \qquad v(t)=v_0+\int_0^t [-u(s)-xv(s)]\,ds$$ The desired operator $T$ takes the vector-valued function $(u,v)$ and produces $$t\mapsto \left(u_0+\int_0^t v(s)\,ds, v_0+\int_0^t [-u(s)-xv(s)]\,ds\right)$$

$\endgroup$

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.