# The Method of Frobenius - Find roots of indicial EQ and 1st terms of series solution

1. Jul 19, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

Apply the method of Frobenius to find the roots of the indicial equation to show that $c_1\,=\,c_2\,=\,0$.

The equation in question is a 2nd order DE that was https://www.physicsforums.com/showthread.php?t=177492".

$$z\,\frac{d^2T}{dx^2}\,+\,\frac{z}{x}\,\frac{dT}{dx}\,-\,z\,y\,=\,0$$

If you look at the thread where this equation was derived, we assume that $z\,=\,x^2$.

$$x^2\,\frac{d^2T}{dx^2}\,+\,\frac{x^2}{x}\,\frac{dT}{dx}\,-\,x^2\,y\,=\,0$$

2. Relevant equations

http://en.wikipedia.org/wiki/Frobenius_method" [Broken]

3. The attempt at a solution

There is a regular, singular point at $x_0\,=\,0$. We seek a solution of the form

$$\sum_0^\infty\,a_n\,x^{n\,+\,r}$$

Differentiating that sum twice and substituting into the steady-state heat balance equation above and bringing the x's into the sums

$$\sum_0^\infty\,(n\,+\,r)(n\,+\,r\,+1)\,a_n\,x^{n\,+\,r}\,+\,\sum_0^\infty\,(n\,+\,r)\,a_n\,x^{n\,+\,r\,-1}\,-\,\sum_0^\infty\,a_n\,x^{n\,+\,r\,+\,2}\,+\,T_a\,x^2\,=\,0$$

Note the last term, it arises from using $y\,=\,T\,-\,T_a$ from the https://www.physicsforums.com/showthread.php?t=177492". I really don't know what to do with it, I don't think it is constant, but how do I combine into sums and how do I deal with the constant $T_a$? I am going to eliminate it, but don't know why!

Now, changing the indicies to combine the summations

$$\sum_0^\infty\,\left[(n\,+\,r)(n\,+\,r\,+\,1)\,a_n\,+\,(n\,+\,r\,+1)\,a_{n\,+\,1}\,-\,a_{n\,-\,2}\right]\,x^{n\,+\,r}\,=\,0$$

Now, I get the indicial equation

$$(r^2\,-\,r)\,a_0\,+\,(r\,+\,1)\,a_1\,=\,0$$

$$r^2\,-\,r\,=\,0\,\longrightarrow\,r^2\,-\,r\,=\,0\,\longrightarrow\,r\,=\,0,\,1$$

But the roots are both supposed to be zero, what did I do wrong?

Last edited by a moderator: May 3, 2017
2. Jul 19, 2007

### HallsofIvy

Staff Emeritus
So the equation is really
$$x^2\frac{d^2T}{dx^2}+ x\frac{dT}{dx}- x^y= 0$$
and with y= T-Ta, that's
$$x^2\frac{d^2T}{dx^2}+ x\frac{dT}{dx}- x^2T- x^2Ta= 0$$

No, don't change the indicies yet. You should never have any "a" except a0 in the indicial equation. Going back to the equation before you changed the indicies, observe that the lowest power occurs when n= 0 (that's always true) in the middle sum. Then you get $a_0 r x^r$ and that is the ONLY way you get xr in the entire sum. In order that that be 0 (so the entire sum will be identically 0) is to have either a0= 0 or r= 0. The whole point of Frobenious' method is to choose r so that the first term, a0 is NOT 0. We must have r= 0.

As far as the "Tax2" term is concerned, include Ta in the equation for the coefficient of x2 but no other power.

Last edited by a moderator: May 3, 2017