Solution of the Legendre's ODE using Frobenius Method

Question

This is the Legendre's differential equation given in my book:

$(1-x)^{2}\ddot{y}-2x\dot{y}+k(k+1)y=0$

I solved this equation by taking:

$y=x^{c}\{a_{0}+a_{1}x+a_{2}x^{2}+.....+a_{r}x^{r}+.....\}$

Therefore, each term in the equation becomes (I deliberately didn't use the sum notation to see it clearly):

$k(k+1)y=k(k+1)a_{0}x^{c}+k(k+1)a_{1}x^{c+1}+k(k+1)a_{2}x^{c+2}+.....+k(k+1)a_{r}x^{c+r}+.....$

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$-2x\dot{y}=-2ca_{0}x^{c}-2(c+1)a_{1}x^{c+1}-2(c+2)a_{2}x^{c+2}-.....-2(c+r)a_{r}x^{c+r}-.....$

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$\ddot{y}=(c-1)ca_{0}x^{c-2}+c(c+1)a_{1}x^{c-1}+(c+1)(c+2)a_{2}x^{c}+.....+(c+r-1)(c+r)a_{r}x^{c+r-2}+.....$

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$-x^{2}\ddot{y}=-(c-1)ca_{0}x^{c}-c(c+1)a_{1}x^{c+1}-(c+1)(c+2)a_{2}x^{c+2}-.....-(c+r-1)(c+r)a_{r}x^{c+r}-.....$

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The indicial equation is $(c-1)ca_{0}x^{c-2}=0$. Therefore, there are two solutions for $c=0$ and $c=1$. Since,

$(c+r+2)(c+r+1)a_{r+2}+k(k+1)a_{r}-2(c+r)a_{r}-(c+r)(c+r-1)a_{r}=0$

general recurrence equation is as follows:

$a_{r+2}=\frac{[(c+r-k)(c+r+k+1)]a_{r}}{(c+r+2)(c+r+1)}$

There is no problem so far. But my book gives the following answer for $c=0$:

$y=a_{0}\left\{ 1-\frac{k(k+1)}{2}x^{2}+\frac{k(k-2)(k+1)(k+3)}{4!}x^{4}-.....\right\}$

This is only possible for $a_1=0$. But I can't see that. I will be very glad if anyone can show me why $a_1$ should be equal to zero when $c=0$.

Answer:

I solved both for $c=0$ and $c=1$ as @Semiclassical suggested. For $c=0$, $a_1$ is indeterminate because of the term: $c(c+1)a_1x^{c-1}$ (It can be any value). Therefore, the solution is:

$u=a_{0}\left\{ 1-\frac{k(k+1)}{2!}x^{2}+\frac{k(k-2)(k+1)(k+3)}{4!}x^{4}-.....\right\} +a_{1}\left\{ x-\frac{(k-1)(k+2)}{3!}x^{3}+\frac{(k-3)(k-1)(k+2)(k+4)}{5!}x^{5}-.....\right\}$

For $c=1$, $a_1=0$ because of the term: $c(c+1)a_1x^{c-1}$, thus the solution is:

$w=a_{0}\left\{ x-\frac{(k-1)(k+2)}{3!}x^{3}+\frac{(k-3)(k-1)(k+2)(k+4)}{5!}x^{5}-.....\right\}$

As it is seen this is not a distinct solution. It is already available in the solution for $c=0$. Therefore, the solution for $c=0$ in my book is incomplete. Actually, it should give the complete solution as above. I learned that this happens when indicial roots differ by an integer value.

Answer 1

Hint: Write out the solution for $c=1$ (which necessarily has no constant term and leading term $a_0 x$) and compare with what you get for your solution at $c=0$ if you don't assume $a_1=0$.

Answer 2

When $c=0$, the recurrence doesn't determine $a_0$ or $a_1$. There is a two-parameter family of solutions, and you can use $a_0$ and $a_1$ as the parameters. So you can choose $a_0$ and $a_1$ to be whatever you like. It's convenient to take the two basic solutions to be the ones with $(a_0,a_1) = (1,0)$ and $(0,1)$; every solution is then a linear combination of these two.

When you have a regular singular point and the indicial roots differ by an integer, there can be a logarithmic term that complicates things, but that doesn't happen in this case ($x=0$ is not a singular point).