0
$\begingroup$

Use the method of Frobenius, with the larger root of the indicial equation, to find the first three terms of the power series of the solution to $$2t^2x'' + tx' -(t+1)x=0.$$

My work:

Note that $t=0$ is a regular singular point. On solving the indicial equation, we get $r=1$ or $r=-\frac12$. We proceed by letting $$x(t) = t^r \sum_{k=0}^{\infty} a_k t^k$$ which, after plugging into the equation, yields $$(ra_1 - a_0)t^{r+1} - a_0 t^r + \sum_{n=r+2}^{\infty} [(2n^2-n-1)a_{n-r} - a_{n-r-1}]t^n=0.$$ However, I cannot proceed. The above with $r=1$ would imply that all coefficients are zero. The calculation should be fine as several of my friends also get the same answer. Should we conclude that the first three terms are all zero?

$\endgroup$
  • $\begingroup$ Using $r=1$, it seems to me that effectively all coefficients are equal to $0$. $x(t)=0$ is a solution of the differential equation. The story is totally different using $r=-\frac12$. $\endgroup$ – Claude Leibovici Mar 20 '15 at 9:50
  • $\begingroup$ there is nothing to balance $-tx$ in the last term? $\endgroup$ – abel Mar 20 '15 at 10:16
  • $\begingroup$ @abel : what do you mean by balancing $-tx$? $\endgroup$ – Nighty Mar 20 '15 at 10:20
  • $\begingroup$ that was an error. i don't have time now. i will take a look at it later. $\endgroup$ – abel Mar 20 '15 at 11:40
  • $\begingroup$ $t+(1/5)t^2+(1/70)t^3+\dots$. Try again plugging in. $\endgroup$ – GEdgar Mar 20 '15 at 12:35
1
$\begingroup$

you have $$2t^2x''+tx' - (t+1)x= 0 $$ now sub $$x = t^k + a_1t^{k+1} + a_2 t^{k+2} + \cdots\\ x' = kt^{k-1} + (k+1)a_1t^k + (k+2)a_2t^{k+1} + \cdots\\ x'' = (k-1)kt^{k-2} + k(k+1)a_1t^{k-1} + (k+1)(k+2)a_2t^{k} + \cdots$$ we get $$2t^2\left( (k-1)kt^{k-2} + k(k+1)a_1t^{k-1} + (k+1)(k+2)a_2t^{k} + \cdots\right) + t\left( kt^{k-1} + (k+1)a_1t^k + (k+2)a_2t^{k+1} + \cdots\right)- (t+1)\left( t^k + a_1t^{k+1} + a_2 t^{k+2} + \cdots\right) =0$$

equating the coefficient of $t^{k}$ we have

$$f(k)=2(k-1)k +k-1=(k-1)(2k+1)=0 \to k = 1, k = -\frac 12.$$

equating the coefficient of $t^{k+1}$ we have

$$2k(k+1)a_1+(k+1)a_1 -a_1-1=0\to a_1 = \frac 1{f(k+1)} $$ in the same way you find $$a_2 = \frac {a_1}{f(k+2)}, \cdots, a_{n+1}=\frac {a_n}{f(n+k)}, = n = 1, 2, \cdots $$

let us look at the series for the case $k = 0$

$$a_1 = \frac 1{f(2)} = \frac 1{1 \cdot 5}\\ a_2 = \frac {a_1}{f(3)}=\frac 1{1\cdot 2 \cdot 5 \cdot 7}\\ a_3 = \frac{a_2} {f(4)}=\frac 1{1\cdot 2 \cdot 3 \cdot 5 \cdot 7 \cdot 9} \\ \vdots$$

$\endgroup$
  • $\begingroup$ Thanks for replying. Indeed we are doing the same thing, except the fact that i am using summation... is that a reason for you not to use summation? $\endgroup$ – Nighty Mar 22 '15 at 16:01
  • $\begingroup$ @LeeKM, summation notation, i find, sometimes obscures what is going on. once i see the pattern, i may go back to the generic terms and the summation notation. i tend to make lots careless errors with the summation notation. that is the reason i prefer the longhand notation. $\endgroup$ – abel Mar 22 '15 at 16:04
  • $\begingroup$ When the goal is "find the first three terms", then probably summation notation is more trouble than it's worth. $\endgroup$ – GEdgar Mar 22 '15 at 16:53
1
$\begingroup$

Plugging into the equation: $$ a_0(2r+1)(r-1)t^r + (2a_1r^2+3a_1r-a_0)t^{r+1} +(2a_2r^2+7a_2r+5a_2-a_1)t^{r+2}+\dots $$
So with $r=1$ (or $r=-1/2$) the first term goes away. That is why you choose a root of the indicial equation. Anything else will have a first term there, leading to $a_0 = 0$. If you get $a_0=0$, it means you have made a mistake.

$\endgroup$

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.