# Indicial equation to Legendre's equation

1. Feb 7, 2005

### meteorologist1

Could someone show me how to find the indicial equation and the indicies relative to any regular singular point of the Legendre equation:

(1 - z^2)w'' - 2zw' + kw = 0

Thank you.

2. Feb 7, 2005

### dextercioby

Compare it to the standard form of the Laplace equation for the first kind Gauss hypergeometric series and extract the indices.Consult the website of wolfram for more details especially on the irregular solution and the points in the complex plane where the latter is not valid.

Daniel.

3. Feb 8, 2005

### Galileo

Assume a series solution:

$$w(z)=\sum_{n=0}^{\infty}a_nz^n$$

plug it into the equation, collect powers. Note that the coefficient of each power must vanish. Use this to get a recursive relation for the coefficients.

4. Feb 8, 2005

### HallsofIvy

The "singular points" are where the leading coefficient, 1- z2= 0: z= 1 and z= -1.

To find the "indicial" equation, try solutions of the form $\Sigma a_n (z-1)^{n+c}$ and $\Sigma a_n(z+1)^{n+c}$. Gather the lowest power and determine c so that a0 is NOT 0. (That's the "indicial equation" and those values of c are the "indices".

5. Feb 8, 2005

### saltydog

Well, me too:

According to Rainville & Bedient, need to consider the series in the form:

$$w(z)=\sum_{n=0}^{\infty}a_nz^{n+c}$$

If I then plug this into the ODE, I get:

$$a_n=\frac{n^2-5n+4cn-7c+6-k}{n^2+2cn-n+c^2-c} a_{n-2}$$

with:
$$n\geq 2$$

(first two terms arbitrary)
Need to check this though.

For the n=0 term however, I get the following indicial equation:

$$c^2-c=0$$

or c=0,1

Need to verify this and also back-substitute the resulting solution for confidence in my opinion.

Last edited: Feb 8, 2005
6. Feb 8, 2005

### meteorologist1

HallsofIvy, using your method, I get what saltydog got: c (c - 1) = 0. So I'm confident that that's the indicial equation for Legendre. Thanks.

7. Feb 8, 2005

### saltydog

I think my above statements are not correct.

Rather: To solve an equation "about the point x0" means to obtain solutions expressed in terms of powers of (x-xo).

Thus for:

$$(1-x^2)y^{''}-2xy^{'}+ky=0$$

and an solution is desired about the point x0=1, make the substitution v=x-1 and thus obtain:

$$(2v-v^2)y^{''}-2(v-1)y^{'}+ky=0$$

Really, I think if the indicial equation for a solution about the point xo=1 is desired, it should be determined from the equation above in v. I'm not sure but how about if this is for your work in school and you find out, you report it back here?

Thanks,
Salty

8. Feb 8, 2005

### meteorologist1

Yes, this is problem I'm doing for school. I will report you back the solution.

9. Feb 9, 2005

### saltydog

I determined the generating function about arbitrary point x0 for this ODE in terms of power series:

$$w(z)=\sum_{n=0}^{\infty}a_n(z+x_0)^n}$$

This is needed if a plot outside of (-1,1) is desired since the power series converges within a radius only up to the closest singularity.

Since this ODE has singularities at x=-1 and x=1, it's no supprise that the plots (see attached graph) have asymptotes at these values. I selected arbitrary values for the constants a0, a1, and k to generate the plots.

$$a_n=\frac{n^2-3n+2-k}{(1-x_0^2)(n^2-n)} a_{n-2}+ \frac{2x_0(n^2-2n+1)}{(1-x_0^2)(n^2-n)} a_{n-1}$$

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Last edited: Feb 9, 2005