I have this eigenvalue problem:

$$-x^2u_{xx} + 2xu_x - 2u = \lambda x^2 u$$

for $0 < x < 1$, with boundary conditions $u_x(0) = 0, u(1) = u_x(1)$.

I'm trying to find a function $M(x)$, such that under the change of variable $u(x) = M(x)w(x)$, it can be transformed to

$$w_{xx} = - \lambda w$$

for $0 < x < 1$.

The solution says this:

With $u = Mw$, the ODE $-x^2 u'' + 2xu' - 2u = \lambda x^2u$ becomes

$$-x^2 (M'' -aM' - bM) + 2x(M' w + Mw') - 2Mw = \lambda x^2Mw$$

$$-x^2 Mw'' + (2xM - 2x^2 M')w' + (-x^2 M'' + 2xM' - 2M)w = \lambda x^2Mw$$

I was wondering if someone could please explain where the $a$ and $b$ came from? Thanks.


Seems like a typo to me, since $u=Mw$ means $u''=M''w +2M'w' +Mw''$ rather than $M'' -aM' -bM$.

  • $\begingroup$ Hmm, it seems unlikely to be a typo though? Putting in $a$ and $b$ seems deliberate. $\endgroup$ – Wyuw Aug 16 '18 at 23:31
  • 2
    $\begingroup$ @Wyuw Why do you think it is deliberate? If it was, shouldn't $a$ and $b$ appear in the original substitution $u = Mw$ and also in the second line? $\endgroup$ – Mattos Aug 16 '18 at 23:50
  • $\begingroup$ @Mattos Ahh you make a good point! Yes, I think you are correct. $\endgroup$ – Wyuw Aug 16 '18 at 23:53

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