I'm trying to apply the Frobenius method to a differential equation where the indicial roots differ by an integer value (which should produce a pair of identical series solutions). I run into trouble when divining the recurrence relation for the solution's coefficients.


Assume the solution has the form $y=\sum \limits_{n=0}^\infty C_n x^{n+r}$

$\sum \limits_{n=0}^{\infty}(n+r)(n-1+r) C_n x^{n-1+r} + \sum \limits_{n=0}^{\infty}-(n+r)C_n x^{n+r} + \sum \limits_{n=0}^{\infty}C_n x^{n+r} $

Adjust the sums so each $x$ has the same exponent, using $k=n-1$ for the first sum and $k=n$ for the other two sums

$ \sum \limits_{k=-1}^{\infty}(k+1+r)(k+r) C_n x^{k+r} + \sum \limits_{k=0}^{\infty}-(k+r)C_k x^{k+r} + \sum \limits_{k=0}^{\infty}C_k x^{k+r} $

Adjust the sums so each one starts at 0

$ r(r-1)C_{-1}x^{r-1}+ \sum \limits_{k=0}^{\infty}(k+1+r)(k+r) C_{k+1} x^{k+r} + \sum \limits_{k=0}^{\infty}-(k+r)C_k x^{k+r} + \sum \limits_{k=0}^{\infty}C_k x^{k+r} $

$ r(r-1)C_{-1}x^{r-1}+ \sum \limits_{k=0}^{\infty}\bigg [(k+1+r)(k+r) C_{k+1} -(k+r)C_k + C_k \bigg ]x^{k+r} $

The indicial roots are $r=0$, $r=1$

The recurrence relation is

$C_{k+1} = { {(k+r-1)C_k}\over {(k+r+1)(k+r)} }$

For $r=1$,

$C_{k+1} = { { k C_k}\over {(k+2)(k+1)} }$

$C_1 = 0$

$C_2 = { C_1 \over {3 \cdot 2} }$

$C_3 = { C_2 \over {4 \cdot 3} }$

For $r=0$,

$C_{k+1} = { { (k-1) C_k}\over {(k+1)(k)} }$

$C_1 = {-1 \over {0}}$

$C_2 = 0$

$C_3 = { C_2 \over {3 \cdot 2} }$

$C_4 = { C_2 \over {4 \cdot 3} }$

My recurrence relations have sustained mortal wounds somehow. Is it okay to shift the sum index to start at a negative value? Are my substitutions to equalize the exponents okay? Has my algebra packed up and departed without me?


There should be no $C_{-1}$ occurring, or it should be understood that this additional coefficient is automatically zero. In fact, the equation after index shift should read as $$\small 0=r(r-1)C_0x^{r-1}+ \sum_{k=0}^{\infty}(k+1+r)(k+r) C_{k+1} x^{k+r} + \sum_{k=0}^{\infty}-(k+r)C_k x^{k+r} + \sum_{k=0}^{\infty}C_k x^{k+r} $$ As the indicial roots not only have an integer difference, but are integers themselves, the solutions for the larger root are contained in the solutions of the smaller root. Thus it is sufficient to consider $r=0$ with the equations $k(k+1)C_{k+1}=(k-1)C_k$ which for $k=0,1,2,3$ read as \begin{align} 0·C_1&=-C_0\\ 2C_2&=0·C_1\\ 6C_3&=C_2\\ 12C_4&=2C_3 \end{align} etc. giving the solutions $y=C_1x$. Search for other solutions in the form $y(x)=xu(x)$, $$ 0=x(xu''+2u')-x(xu'+u)+xu=x^2u''+(2x-x^2)u' $$ resulting in $(\ln|u'|)'=1-\frac2x$, $$ u'(x)=C\frac{e^x}{x^2}\\ u(x)=D+C\Bigl(-\frac1x+\ln|x|+\frac12x+…+\frac{x^{n-1}}{(n-1)n!}+…\Bigr) $$ so that $$ y(x)=xu(x)=Dx+C\Bigl(-1+x\ln|x|+\frac12x^2+…+\frac{x^n}{(n-1)n!}+…\Bigr) $$ contains the term $x\ln|x|$ that can not be developed into a shifted power series around $x=0$.

You should again check the assumptions on the applicability of the Frobenius method, there should be something about a "regular singularity".


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