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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!

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It is simpler to solve this equations than to understand what you are asking about –  no identity Sep 23 '12 at 16:51
    
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. –  Miguel Sep 23 '12 at 17:40

1 Answer 1

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 $$t\mapsto \left(u_0+\int_0^t v(s)\,ds, v_0+\int_0^t [-u(s)-xv(s)]\,ds\right)$$

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