12 - Frobenius Series

Frobenius Series

$$y(x)=(x-x_0{)}^{\alpha }\sum _{k=0}^{\infty }{a}_k(x-x_0{)}^k=\sum _{k=0}^{\infty }{a}_k(x-x_0{)}^{\alpha +k}$$

Indicial Equation

From Substituting Frobenius Series to Standard Form at a Regular Singular Point then

Indicial Equation is

$$F(\alpha ):=\alpha (\alpha -1)p_0\alpha +q_0=0$$

where the roots ${\alpha }_1$ and ${\alpha }_2$ are the indicial exponents (can be complex)

Generally it is ordered such that $\mathrm{Re}({\alpha }_1)\ge \mathrm{Re}({\alpha }_2)$

Proof

Assume $a_0\ne 0$ by choosing largest such $a$

$$\begin{array}{rl}\sum _{k=0}^{\infty }(\alpha +k)(\alpha +k-1)a_k(x-x_0{)}^{\alpha +k-2}+\sum _{k=0}^{\infty }& \sum _{j=0}^{\infty }(\alpha +k)p_j{a}_k(x-x_0{)}^{\alpha +k+j-2}\\ & +\sum _{k=0}^{\infty }\sum _{j=0}^{\infty }{q}_k{a}_k(x-x_0{)}^{\alpha +k+j-2}=0\end{array}$$

then equate coefficients

At lowest power $(x-0{)}^{\alpha -2}$ then

$$[\alpha (\alpha -1)+p_0\alpha +q_0]a_0=0$$

As $a_0$ is non-zero then define

$$F(\alpha )=\alpha (\alpha -1)p_0\alpha +q_0$$

Recurrence Relation for Frobenius Coefficients

Let $a_k$ be the coefficients of the Frobenius Series
Let $F$ be defined from the Indicial Equation

$$F(\alpha +k)a_k=-\sum _{j=0}^{k-1}[(\alpha +j)p_{k-j}+q_{k-j}]a_j$$

Finding the First Solution in the form of Frobenius Series

From Proof of Indicial Equation then
Continuing equating coefficients of powers of $(x-x_0)$ then

Coefficient of $(x-x_0{)}^{k+\alpha -2}$ satisfy

$$F(\alpha +k)a_k=-\sum _{j=0}^{k-1}[(\alpha +j)p_{k-j}+q_{k-j}]a_j$$

Generating the first series solution then let

$$\alpha ={\alpha }_1$$

Since $F$ is a quadratic function with roots at ${\alpha }_1$ and ${\alpha }_2$ then

$$F({\alpha }_1+k)\ne 0\text{ for any integer }k\ge 1$$

Thus

$$a_k=-\frac{1}{F({\alpha }_1+k)}\sum _{j=0}^{k-1}[(\alpha +j)p_{k-j}+q_{k-j}]a_j$$

Hence the coefficients $a_1,a_2,\cdots$ can be solved and thus obtain solution

$$y_1(x)=(x-x_0{)}^{{\alpha }_1}\sum _{k=0}^{\infty }{a}_k(x-x_0{)}^k$$

Hence at least one solution can be expressed as a Frobenius Series
with indicial exponent $\alpha ={\alpha }_1$

Finding the Second Solution - Case $I$: ${\alpha }_1-{\alpha }_2\notin \mathbb{Z}$

As ${\alpha }_1$ and ${\alpha }_2$ do not differ by an integer (and not equal)

Then

$$F({\alpha }_2+k)\ne 0\text{ for all }k\ge 1$$

Hence can solve using the second value of exponent $\alpha ={\alpha }_2$

Let the coefficients of the second solution be denoted as $b_n$

Then $b_k$ satisfy recurrence relation

$$b_k=-\frac{1}{F({\alpha }_2+k)}\sum _{j=0}^{k-1}[({\alpha }_2+j)p_{k-j}+q_{k-j}]b_j$$

Hence the second solution (also a Frobenius Series) with indicial exponent ${\alpha }_2$

$$y_2(x)=(x-x_0{)}^{{\alpha }_2}\sum _{k=0}^{\infty }{b}_k(x-x_0{)}^k$$

Finding the Second Solution - Case $II$: ${\alpha }_1={\alpha }_2$

Need to multiply by logs to get second solution in the form of

$$y_2(x)=y_1(x)\log (x-x_0)+(x-x_0{)}^{{\alpha }_1}\sum _{k=0}^{\infty }{b}_k(x-x_0{)}^k$$

where $y_1$ is the first solution

Note that form can be derived using derivatives method

Finding the Second Solution - Case $III$: ${\alpha }_1-{\alpha }_2$ is a positive integer

If ${\alpha }_1-{\alpha }_2=N$ where $N>0$ is an integer

Trouble when $k=N$

Case $III(a)$

For $k=N$

If right-hand side of Recurrence Relation for Frobenius Coefficients is non-zero then

Contradiction so standard Frobenius Solution method does not work
Second solution is same form as Case II so

$$y_2(x)=y_1(x)\log (x-x_0)+(x-x_0{)}^{{\alpha }_1}\sum _{k=0}^{\infty }{b}_k(x-x_0{)}^k$$

Note that the indicial exponent for second series is ${\alpha }_2$

Case $III(b)$

For $k=N$

If right-hand side of Recurrence Relation for Frobenius Coefficients is zero then

There is no contradiction and any value of $b_N$ will work

Hence second solution has Frobenius For

$$y_2(x)=(x-x_0{)}^{{\alpha }_2}\sum _{k=0}^{\infty }{b}_k(x-x_0{)}^k$$

where $b_0$ can be chosen to be $b_0=1$ (WLOG) and $b_N$ is arbitrary

As ${\alpha }_2+N={\alpha }_1$ then changing $b_N$ corresponds to adding multiples of $y_1$

Finding the Second Solution - Case $II$ Proof - Derivative Method

As ${\alpha }_1$ is a double rood of $F(\alpha )$
Without loss of generality, let $a_0=1$

Suppose solve for $a_1,a_2,\cdots$ with arbitrary $\alpha$(so $F(\alpha )$ is not generally equal to zero)
Thus each coefficient $a_k$ is a function of $\alpha$

Let

$$y(x;\alpha )=(x-x_0{)}^{\alpha }+\sum _{k=1}^{\infty }{a}_k(\alpha )(x-x_0{)}^{k+\alpha }$$

By Recurrence Relation then coefficient of $(x-x_0{)}^{\alpha +k-2}$ in $Ly$ is zero for all $k\ge 1$ hence

$$Ly(x;\alpha )=(x-x_0{)}^{\alpha -2}F(\alpha )$$

As $F({\alpha }_1)=0$ then $Ly(x;{\alpha }_1)=0$ thus

$$y_1(x)=y(x;{\alpha }_1)=\sum _0^{\infty }{a}_k({\alpha }_1)(x-x_0{)}^{{\alpha }_1+k}$$

is a solution

Differentiate $y$ with respect to $\alpha$ then set $\alpha ={\alpha }_1$
Since $L$ has no dependence on $\alpha$ then

$$\begin{array}{rl}L\left[\frac{\partial }{\partial \alpha }y(x;\alpha )\right]& =\frac{\partial }{\partial \alpha }[Ly(x;\alpha )]\\ & =\frac{\partial }{\partial \alpha }[(x-x_0{)}^{\alpha -2}F(\alpha )]\\ & =(x-x_0{)}^{\alpha -2}\log (x-x_0)F(\alpha )+(x-x_0{)}^{\alpha -2}{F}^{\prime }(\alpha )\end{array}$$

Since ${\alpha }_1$ is a double root of $F$, then right hand side is zero at $\alpha ={\alpha }_1$ hence

$$y_2(x)=\frac{\partial }{\partial \alpha }y(x;\alpha ){\mid }_{\alpha ={\alpha }_2}$$

satisfies $L{y}_2=0$

Explicitly calculating

$$\begin{array}{rl}\frac{\partial }{\partial \alpha }y(x;\alpha )& =\frac{\partial }{\partial \alpha }\left(\sum _{k=0}^{\infty }{a}_k(\alpha )(x-x_0{)}^{\alpha +k}\right)\\ & =\log (x-x_0)\sum _{k=0}^{\infty }{a}_k(\alpha )(x-x_0{)}^{\alpha +k}+\sum _{k=0}^{\infty }{a}_k^{\prime }(x-x_0{)}^{\alpha +k}\end{array}$$

Setting $\alpha ={\alpha }_1$

$$y_2(x)=\log (x-x_0)y_1(x)+\sum _{k=0}^{\infty }{b}_k(x-x_0{)}^{{\alpha }_1+k}$$

where $b_k=a_k^{\prime }({\alpha }_1)$