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While using the Frobenius method to solve a second order ODE of the form

$$y^{\prime \prime} + p(x)y^\prime + q(x)y = 0$$

if the roots of the indicial equation $(r(r-1)+p_0r+q_0=0)$ $r_1, r_2 (r_1>r_2)$ differ by an integer then we first obtain $y_1=\displaystyle \sum_{n=0}^{\infty} a_nx^{n+r_1}$as a solution and then use reduction of order to find $y_2$. The formula for reduction for order is

$$y_2 = Cy_1\int\dfrac{e^{-\int p(x)dx}}{y_1^2(x)} dx$$

as derived here. Now, apparently you can derive the form of $y_2$ rigorously by substituting $y_1=\displaystyle \sum_{n=0}^{\infty} a_nx^{n+r_1}$ in the above eqaution. However, I cant find this derivation anywhere. If anyone knows how to derive

$$y_2 = Cy_1\ln x + x^{r_2}\sum_{n=0}^{\infty}b_nx^n$$

using this method or if there is ANY source/textbook that does this, please let me know.

Aadhaar Murty
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  • When $r_1-r_2$ is a positive integer, the power series method may (e.g. Bessel's DE of order $\frac12$: $x^2y''+xy'+(x^2-\frac14)y$) or may not (e.g. Bessel's order $1$) produce a second solution. You don't necessarily get the $y_1\log x$. – user10354138 Sep 27 '21 at 15:16
  • Yeah but when you do, how do we arrive at the form of $y_2$? – Aadhaar Murty Sep 27 '21 at 17:47
  • See https://math.stackexchange.com/questions/3907666/intuition-behind-the-general-form-of-the-solution-in-frobenius-method for an exploration of this topic. The notation does not translate one-to-one, and there is something skewed in the first formula for the reduction-of-order here. – Lutz Lehmann Sep 28 '21 at 08:10

1 Answers1

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First, the formula of reduction of order is $y_2=\color\red{y_1}\int\frac{e^{-\int p(x)dx}}{y_1^2(x)}dx$.
Also note that we can ignore $C$ since the solution $c_1y_1+c_2y_2$ already contains general constants.
To derive the given formula, let's start with evaluating $\frac{1}{y_1^2(x)}=\frac{1}{x^{2r_1}(c_0+c_1x+c_2x^2+\cdots)}$.
Let $X=\frac{1}{c_0+c_1x+c_2x^2+\cdots}$, and observe that $X\cdot(c_0+c_1x+c_2x^2+\cdots)=1$.
By comparing the coefficient of each term, we can know that $X=\frac{1}{c_0}-\frac{c_1}{c_0^2}x+\frac{c_2c_0-c_1^2}{c_0^3}x^2+\cdots$.
So we can say that $\frac{1}{y_1^2(x)}=x^{-2r_1}(C_0+C_1x+C_2x^2+\cdots)$ for $C_0=\frac{1}{c_0},C_1=-\frac{c_1}{c_0^2},C_2=-\frac{c_2c_0-c_1^2}{c_0^3},\cdots$.
Now, observe that $p(x)=\frac{a_0}{x}+a_1+a_2x+\cdots$ so that $e^{-\int p(x)dx}=e^{-a_0\ln(x)-a_1x-\frac{a_1}{2}x^2\cdots}$
$=x^{-a_0}e^{-a_1x-\frac{a_1}{2}x^2\cdots}=x^{-a_0}(1+R(x)+\frac{1}{2}\{R(x)\}^2+\cdots)=x^{-a_0}(1+d_1x+d_2x^2+\cdots)$
where $R(x)=-a_1x-\frac{a_1}{2}x^2\cdots$, by using taylor series of $e^x$.
Thus we can write $\frac{e^{-\int p(x)dx}}{y_1^2(x)}=x^{-2r_1-a_0}(D_0+D_1x+D_2x^2+\cdots)$
$=D_0x^{-2r_1-a_0}+D_1x^{-2r_1-a_0+1}+\cdots+D_{2r_1+a_0-1}\frac{1}{x}+\cdots$.
Then $\int\frac{e^{-\int p(x)dx}}{y_1^2(x)}dx=D_{2r_1+a_0-1}\ln (x)+\left[\frac{D_0}{-2r_1-a_0+1}x^{-2r_1-a_0+1}+\frac{D_1}{-2r_1-a_0+2}x^{-2r_1-a_0+2}+\cdots\right]$.
Therefore $y_2=y_1(x)\ln (x)+x^{r_1}(E_0x^{-2r_1-a_0+1}+E_1x^{-2r_1-a_0+2}+\cdots)$.
Note that we divided RHS by $D_{2r_1+a_0-1}$.
Now, the proof ends by showing $-r_1-a_0+1\geq r_2$ and they differ by integer so that $(E_0x^{-r_1-a_0+1}+E_1x^{-r_1-a_0+2}+\cdots)$ can be written as $x^{r_2}\sum_{n=0}^\infty b_nx^n$.
Since $r_1$ and $r_2$ are the roots of the indicial equation $r^2-(1-a_0)r+b_0$, $r_1+r_2=1-a_0$.
Then $-r_1-a_0+1=r_2$, so they differ by an integer $0$, and the proof ends.
Moreover, we can know that $b_0 \neq 0$. (Your book may also mention this condition.)

ILLE2304
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  • I'm sure Im missing something really obvious but how did you compare the coefficients of each term in $X$? – Aadhaar Murty Sep 28 '21 at 15:04
  • @AadhaarMurty First, multiply $\frac{1}{c_0}$ to $c_0+c_1x+c_2^2+\cdots$. Then we have $1+\frac{c_1}{c_0}x+\frac{c_2}{c_0}x^2+\cdots$. To eliminate $\frac{c_1}{c_0}x$, add $-\frac{c_1}{c_0^2}x(c_0+c_1x+c_2x^2+\cdots)$. Now the remaining is $-\frac{c_2c_0-c_1^2}{c_0^2}x^2+\cdots$. So add $\frac{c_2c_0-c_1^2}{c_0^3}x^2(c_0+c_1x+c_2x^2+\cdots)$. By repeating this steps, we can eliminate all terms except $1$. And by merging the terms we have added, $\left(\frac{1}{c_0}-\frac{c_1}{c_0^2}x+\frac{c_2c_0-c_1^2}{c_0^3}x^2+\cdots\right)(c_0+c_1x+c_2x^2+\cdots)=1$. – ILLE2304 Sep 29 '21 at 11:21
  • This concept is very similar to the division of polynomials, but different from the point that it starts dividing from the lowest order, constant. – ILLE2304 Sep 29 '21 at 11:24
  • If you already are normalizing arbitrary constants, you could also set $a_0=1$ so that also $c_0=1$, reducing complexity from the formulas. /// You should avoid double use of variable names in close context, here $a_k$ for the coefficients in $y_1$ and then in $p$. /// One should not discount the possibility that $D_{2r_1+a_0-1}=0$, removing the logarithmic term /// (Use display math environments - align, multline - for long formulas, it is better readable.) – Lutz Lehmann Sep 30 '21 at 15:38
  • @okw1124 thanks now it's clear. I get the rest of the solution. Last question though : could you just elaborate on how $p(x)\cdot x = a_0 + a_1x + a_2x^2...$ – Aadhaar Murty Oct 01 '21 at 10:17
  • @AadhaarMurty : The usual form similar to an Euler-Cauchy equation is $x^2y''(x)+x\bar p(x)y'(x)+\bar q(x)y(x)=0$, where $\bar p$ and $\bar q$ are continuous in $x=0$, or optimally have a power series expansion. Then normalizing so that the leading coefficient is $1$ as formulated in the question you get $p(x)=\frac{\bar p(x)}{x}$ and $q(x)=\frac{\bar q(x)}{x^2}$. – Lutz Lehmann Oct 01 '21 at 11:22
  • @AadhaarMurty Sorry for confusing notations. As Lutz Lehmann said, $a_0,a_1,\cdots$ are just constants, and they are not related to $y_1$. I think you used $p(x)+\frac{p_0}{x}+p_1+\cdots$ since you wrote the indicial equation $r(r-1)+p_0r+q_0$. What we are saying is the same, only differ in the name of constants. – ILLE2304 Oct 01 '21 at 11:23
  • @AadhaarMurty Or if you want to ask why it is not $(x-x_0)p(x)$ but $xp(x)$, it is a critical question. Here, I assumed (as many books do) that the singular point is $x_0=0$, but in fact, it may not be zero. So it would be more accurate to say $(x-x_0)p(x)=a_0+a_1x+\cdots$, or in your notation, $(x-x_0)p(x)=p_0+p_1x+\cdots$. – ILLE2304 Oct 01 '21 at 11:26
  • okw1124 OK so just to reiterate : you have set $p^{\bar}(x) = a_0 + a_1x+a_2x^2.....$ in the general form by @Lutz Lehmann above and these coefficients are unrelated to $y_1$. – Aadhaar Murty Oct 01 '21 at 17:15
  • @AadhaarMurty Exactly right. – ILLE2304 Oct 01 '21 at 21:11
  • @okw1124 a lot of sources saying that the second solution when $|r_1 - r_2| \in Z$ is $(\frac {\partial y_1}{\partial r})_{r_2}. $ Can that somehow be derived from the result you obtained for $y_2$ – Aadhaar Murty Oct 03 '21 at 13:08
  • @AadhaarMurty You have to recall how we obtained the coefficients in $y_1(x)$. These constants are given by the recurrence relations about $r$. So we can say $y_1=\eta(x,r_1)$ where $\eta(x,r)=x^r \phi(x,r)$, as we consider the series part as a function of $x$ and $r$. Then $y_2=\left(\frac{\partial \eta}{\partial r}\right){r_1}=x^r \ln(x)\phi(x,r)+x^r \phi{r}(x,r)\vert_{r_1}=y_1\ln (x)+x^{r_1}\phi_{r}(x,r_1)$. Now we may express $x^{r_1}\phi_{r}(x,r_1)$ as a new series, $x^{r_2}\sum_{n=0}^\infty b_nx^n$. I think what you saw from other sources is about this theorem. – ILLE2304 Oct 03 '21 at 21:01