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How do you put

$u'' +c u' +d =0$

into regular SL-form? Can not see how it's an eigenvalue problem without a first order term. But the theorem states EVERY second order operator can be put into SL form, right?

Benjamin Lindqvist
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  • What is the theorem you're talking about? – user37238 Jun 30 '14 at 14:27
  • By the way, where is the partial differential equation here? – user37238 Jun 30 '14 at 14:28
  • My textbook (folland) stated is as a theorem, or so I thought. So we CAN'T put it into standard form? – Benjamin Lindqvist Jun 30 '14 at 14:35
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    I don't know (at this point) the answer to your question : could you please say what are the variables with respect to you differentiate, what are the hypothesis on $c$ and $d$, what is the book the name of the book and where can we find the theorem, and where is the pde? – user37238 Jun 30 '14 at 14:53

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To put the ODE into a SLE form, we need to find $p$ and $q$ such that

$$ -(p u')' + q u = f \iff u'' + cu' +d = 0 $$

If we expand the SLE, we see

$$ - p u'' -p'u' + qu = f \iff u '' + \frac{ p'}{p}u' - \frac{q}{p} u = -\frac{f}{p} $$

By simple comparison we see

$$q \equiv 0, \quad \& \quad \frac{p'}{p} = c, \quad \& \quad \frac{f}{p} = d $$

Jeb
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