Solve the Indicial Equation to Show c_1\,=\,c_2\,=\,0

[VinnyCee]

Homework Statement

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 $z\,=\,x^2$.

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

Homework Equations

http://en.wikipedia.org/wiki/Frobenius_method"

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?

 

[HallsofIvy]

Homework Statement

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 $z\,=\,x^2$.

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

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

Homework Equations

http://en.wikipedia.org/wiki/Frobenius_method"

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

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

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 x^r 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.

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

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

As far as the "T_ax²" term is concerned, include T_a in the equation for the coefficient of x² but no other power.