0
$\begingroup$

For the ODE $$3z^2u''+8zu'+(z-2)u=0$$ construct a series solution of the form $$\sum_{k=0}^{\infty} A_kz^{k+r}$$ that is bounded as $z\rightarrow 0$. Take $A_0=1$ and compute explicitly the terms up to and including the one with $k=2$.

Now, I have determined that $$(3r(r-1)+8r-2)A_0z^r+\sum_{k=1}^{\infty} (3(k+r)(k+r-1)+8(k+r)-2)A_k+A_{k-1})z^{k+r}=0$$

Hence the indicial equation is $3r^2+5r-2=0$, with roots $r=\frac{1}{3}, -2$. The recurrence relation is $$A_k=-\frac{A_{k-1}}{3(k+r)(k+r-1)+8(k+r)-2}, \ \ k\geq 1$$ But how do I know which value for $r$ gives an unbounded solution or not?

$\endgroup$

1 Answer 1

1
$\begingroup$

You have, with $A_0=1$, terms $z^r$ in the series. For $r=-2$ that is unbounded at $z=0$, for $r=1/3$ you can convince yourself that there is a positive radius of convergence.

$\endgroup$
6
  • $\begingroup$ How do you know that $r=-2$ is unbounded? Can the ratio test confirm this? $\endgroup$
    – user557493
    Aug 26, 2018 at 11:00
  • $\begingroup$ Tell me how $z^{-2}$ can in any way be bounded around $z=0$. You do not need the other terms for divergence. $\endgroup$ Aug 26, 2018 at 11:08
  • $\begingroup$ I don't quite understand this idea of bounded and unbounded. For instance, I know the set $S=\{(x,y)\in\mathbb{R}^2\;|\; x^2+y^2=1\}$ is bounded as we can take a ball, $B(O,2)$, such that $S\cap B(O,2)=S$. I don't understand how $z^{-2}$ is unbounded at $z=0$. $\endgroup$
    – user557493
    Aug 26, 2018 at 11:20
  • $\begingroup$ What is $\max |z^{-2}|$ for $|z|<\delta$? Or any upper bound for the function values? $\endgroup$ Aug 26, 2018 at 12:53
  • 1
    $\begingroup$ Yes, exactly that. $\endgroup$ Aug 27, 2018 at 5:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy