Frobenius method

Definition

Let $L(y)=y^{\prime \prime }p(x)y^{\prime }+q(x)y=0⊛$ have a regular singular point at $x_0=0$ $\Rightarrow x^{2}L(y)=x^2{y}^{\prime \prime }+x(xp(x))y^{\prime }+(x^{2}q(x))y=0$ $\Rightarrow x^{2}L(y)=x^2{y}^{\prime \prime }+x(\sum _{n=0}^{\infty }{p}_n{x}^n)y^{\prime }+(\sum _{n=0}^{\infty }{q}_n{x}^n)y=0$

Since we can express coefficients as power series form, above expression satisfies $x^2{y}^{\prime \prime }+x{p}_0{y}^{\prime }+q_{0}y\approx 0$^[the corresponding Euler-Cauchy equation] around 0

So, let’s assume the solution $y=x^r\sum _{n=0}^{\infty }{a}_n{x}^n$ with $a_0\ne 0$ form like a solving process of Euler-Cauchy Equations There exists at least one solution of the form $y=y_1=x^r\sum _{n=0}^{\infty }{a}_n{x}^n$^[Frobenius type solution]

Assume $y=x^r\sum _{n=0}^{\infty }{a}_n{x}^n$ and fit in $⊛$ Then, we get $L(y)=a_{0}F(r)x^r+\sum _{n=1}^{\infty }\{F(r+n)a_n+\sum _{k=0}^{n-1}{a}_k(r+k)p_{n-k}+q_{n-k}\}{x}^{r+n}=0$ where $F(r)=r(r-1)+p_{0}r+q_0=0$ is called indicial equation of the ODE

Now, we can find $a_n$ from $a_0,a_1,\dots ,a_{n-1}$ if $F(n+r)\ne 0$ So, $a_n$ is a function of $r$

Considering the coefficient of $x^r$^[first non-zero coefficient], we obtain the indicial equation $F(r)=r(r-1)+p_{0}r+q_0=0$

If $r_1>r_2$ (real), $r_1-r_2\notin \mathbb{N}$

$\exists y_2$ s.t. $y_1=x^{r_1}\sum _{n=0}^{\infty }{a}_n{x}^n,y_2=x^{r_2}\sum _{n=0}^{\infty }{b}_n{x}^n$

If $r_1=r_2$ (real)

$y_1=x^{r_1}\sum _{n=0}^{\infty }{a}_n{x}^n,y_2=y_1\ln x+x^{r_1}\sum _{n=1}^{\infty }{b}_n{x}^n$

For euler-cauchy equation, $L(x^r)=(r-r_1{)}^2{x}^r=0$ at $r=r_1$,
$L(x^r\ln x)=L(\frac{\partial }{\partial r}{x}^r)=\frac{\partial }{\partial r}L(x^r)=2(r-r_1)x^r+(r-r_1{)}^2{x}^r\ln x{|}_{r=r_1}=0$

For general, $y_2=\frac{\partial }{\partial r}(x^r\sum _{n=0}^{\infty }{a}_n(r)x^n){|}_{r=r_1}=x^{r_1}\ln x\sum _{n=0}^{\infty }{a}_n{x}^n+x^{r_1}\sum _{n=1}^{\infty }\underset{b_n}{\underbrace{a_n^{\prime }(r_1)}}{x}^n$

If $r_1-r_2\in \mathbb{N}$

$y_1=x^{r_1}\sum _{n=0}^{\infty }{a}_n{x}^n,y_2=a{y}_1\ln x+x^{r_2}\sum _{n=0}^{\infty }{c}_n{x}^n$

If $r_1,r_2$ are complex

$\text{Re}(y_1),\text{Im}(y_1)$ are solutions