Can ##a_1## and ##a_2## be zero in the Frobenius' method solution for this DE?

[BearY]

Homework Statement

Use Forbenius' method to solve this DE:
$$ 5x^2y''+xy'+(x^3-1)y=0$$

Homework Equations

Seek power series solution in the form $y=\sum _{n=0}^{\infty } a_n x^{n+r}$, $a_0\neq0$

The Attempt at a Solution

Sub in the ansatz y, get $$ \sum _{n=0}^{\infty }a_n[5(n+r)(n+r+1)+n+r-1]x^n + \sum _{n=0}^{\infty }a_nx^{n+r+3} = 0 $$
Align the powers of x, split first sum to combine with second sum:$$ \sum _{n=0}^{\infty }\{a_{n+3}[5(n+r+3)(n+r+2)+n+r+3-1]+a_n\}x^{n+3} + a_0[5r(r-1)+r-1]+a_1[r(r+1)r+r]x+a_2[5(r+2)(r+1)+r+1]x^2=0$$
This is where I get stuck, I am not sure about how to handle $a_1$ and $a_2$ term or how to find solutions of $r$ with them being there. I can't find out if $a_1$ and $a_2$ can be zero or not, whether or why it matters.
Thanks in advance.

 

[Ray Vickson]

Homework Statement

Use Forbenius' method to solve this DE:
$$ 5x^2y''+xy'+(x^3-1)y=0$$

Homework Equations

Seek power series solution in the form $y=\sum _{n=0}^{\infty } a_n x^{n+r}$, $a_0\neq0$

The Attempt at a Solution

Sub in the ansatz y, get $$ \sum _{n=0}^{\infty }a_n[5(n+r)(n+r+1)+n+r-1]x^n + \sum _{n=0}^{\infty }a_nx^{n+r+3} = 0 $$
Align the powers of x, split first sum to combine with second sum:$$ \sum _{n=0}^{\infty }\{a_{n+3}[5(n+r+3)(n+r+2)+n+r+3-1]+a_n\}x^{n+3} + a_0[5r(r-1)+r-1]+a_1[r(r+1)r+r]x+a_2[5(r+2)(r+1)+r+1]x^2=0$$
This is where I get stuck, I am not sure about how to handle $a_1$ and $a_2$ term or how to find solutions of $r$ with them being there. I can't find out if $a_1$ and $a_2$ can be zero or not, whether or why it matters.
Thanks in advance.

I have not checked your work, but assuming it is correct we can say:
(1) either $a_0=0$ or $5r(r-1)+r-1 = 0$ (or both).
(2) either $a_1=0$ or $r^2(r+1)+r = 0$ (or both).
(3) either $a_2 = 0$ or $5(r+1)(r+2)+ r+1=0$ (or both).

From (1): if $a_0 \neq 0$ then $(5r+1)(r-1) = 0$, so either $r = 1$ or $r = -1/5$. Thus, if $r = 1$ or $r = -1/5$ then $a_0$ is arbitrary and $a_1 = a_2 = 0$. If $r \neq 1, -1/5$ then $a_0 = 0$.

Look at the other cases in a similar way. That will give you a candidate list of possible $r$ values and corresponding cases for $a_0,a_1,a_2 .$ Then you need to figure out how to eliminate some of those possibilities.

Note added in edit: I guess we can eliminate the case $a_0 = 0$, because $a_0 x^r$ is the lowest-power term in $y$, so cannot have a zero coefficient. In other words, we do have $a_0 \neq 0$, so we can say for sure that either $r = 1$ or $r = -1/5$, and also that $a_1 = a_2 = 0$ (assuming your work was correct).

 

[fresh_42]

Two polynomials $p(x) = \sum a_nx^n\, , \,q(x)= \sum b_nx^n$ are equal if all $a_n=b_n$ and if $q(x)=0$ then all $b_n=0$. Now which coefficients at position $0 , 1, 2$ are left on the left hand side? Don't forget, that $a_0 \neq 0$ if you solve for $r$. And do you know something about $r$, e.g. $r > 0$ or can it be arbitrary?

 

[BearY]

I have not checked your work, but assuming it is correct we can say:
(1) either $a_0=0$ or $5r(r-1)+r-1 = 0$ (or both).
(2) either $a_1=0$ or $r^2(r+1)+r = 0$ (or both).
(3) either $a_2 = 0$ or $5(r+1)(r+2)+ r+1=0$ (or both).

From (1): if $a_0 \neq 0$ then $(5r+1)(r-1) = 0$, so either $r = 1$ or $r = -1/5$. Thus, if $r = 1$ or $r = -1/5$ then $a_0$ is arbitrary and $a_1 = a_2 = 0$. If $r \neq 1, -1/5$ then $a_0 = 0$.

Look at the other cases in a similar way. That will give you a candidate list of possible $r$ values and corresponding cases for $a_0,a_1,a_2 .$ Then you need to figure out how to eliminate some of those possibilities.

That is to say since $a_0$ term must be 0, $a_0 \neq 0$, r can only be the solutions to $a_0$ term equal to zero. If the other two coefficients happen to be arbitrary constants that don't have to be 0, they can be included in the coefficients of the general solution of the DE, right?

 

[BearY]

Two polynomials $p(x) = \sum a_nx^n\, , \,q(x)= \sum b_nx^n$ are equal if all $a_n=b_n$ and if $q(x)=0$ then all $b_n=0$. Now which coefficients at position $0 , 1, 2$ are left on the left hand side? Don't forget, that $a_0 \neq 0$ if you solve for $r$. And do you know something about $r$, e.g. $r > 0$ or can it be arbitrary?

I beg your pardon, I am not sure what does being left on the left hand side mean. r is arbitary constant.

 

[Ray Vickson]

That is to say since $a_0$ term must be 0, $a_0 \neq 0$, r can only be the solutions to $a_0$ term equal to zero. If the other two coefficients happen to be arbitrary constants that don't have to be 0, they can be included in the coefficients of the general solution of the DE, right?

I have no idea what you are attempting to say; it makes no sense to me. Re-read my post---especially the part added in edit.

 

[fresh_42]

I beg your pardon, I am not sure what does being left on the left hand side mean. r is arbitary constant.

You have an equation $\text{ (LHS =) } p(x) = p_0 +p_1x + p_2x^2 + \ldots = 0 = 0\cdot 1 + 0 \cdot x + 0\cdot x^2 + \ldots \text{ (= RHS) }$. Now compare the positions at $x^n \, : \, p_0=0 \; , \; p_1=0\; , \; p_2=0\; \ldots \;p_n=0$. I wrote $p_i$ because those new coefficients are themselves polynomial expressions in the $a_j$, e.g. $p_0 = a_0 (5r^2-4r-1)$. This can be solved and you get two possible values for $r$. Take them and continue with $p_1 = a_1 \left( (1+r)\cdot (1+5r)-1 \right)$ and so on.

 

[BearY]

I have no idea what you are attempting to say; it makes no sense to me. Re-read my post---especially the part added in edit.

I am sorry, I meant to ask, in this question we can indeed be sure that $a_1 = a_2 = 0$. But what happens if a solution to $r$ allows other coefficients to be non-zero? Or is that impossible?

 

[fresh_42]

I am sorry, I meant to ask, in this question we can indeed be sure that $a_1 = a_2 = 0$.

Re-read post #3 which explains the comparison of two polynomials.

But what happens if a solution to $r$ allows other coefficients to be non-zero? Or is that impossible?

You cannot rule it out. All you know is $a_0 \neq 0$. E.g. I got $a_3 = -\dfrac{1}{63} a_0$ in one case, which is thus unequal to zero, too.

 

[BearY]

You have an equation $\text{ (LHS =) } p(x) = p_0 +p_1x + p_2x^2 + \ldots = 0 = 0\cdot 1 + 0 \cdot x + 0\cdot x^2 + \ldots \text{ (= RHS) }$. Now compare the positions at $x^n \, : \, p_0=0 \; , \; p_1=0\; , \; p_2=0\; \ldots \;p_n=0$. I wrote $p_i$ because those new coefficients are themselves polynomial expressions in the $a_j$, e.g. $p_0 = a_0 (5r^2-4r-1)$. This can be solved and you get two possible values for $r$. Take them and continue with $p_1 = a_1 \left( (1+r)\cdot (1+5r)-1 \right)$ and so on.

I understand it now. Thank you

Re-read post #3 which explains the comparison of two polynomials.

You cannot rule it out. All you know is $a_0 \neq 0$. E.g. I got $a_3 = -\dfrac{1}{63} a_0$ in one case, which is thus unequal to zero, too.

I think $a_3$ will be the result of the recurrence relationship given by the sum in the last step I wrote in #1. And so will be $a_6$ and so on. The value of $r$ only makes $a_1 = a_2 = 0$ right?

 

[fresh_42]

In the case $r=1$ I had $a_1=a_2=0$ as well. I think this is also true for the other value of $r$, but I haven't checked.

 

[Ray Vickson]

I am sorry, I meant to ask, in this question we can indeed be sure that $a_1 = a_2 = 0$. But what happens if a solution to $r$ allows other coefficients to be non-zero? Or is that impossible?

We need $a_1 = a_2 = 0$. Then we can get $a_3$ in terms of $a_0$, can get $a_4$ in terms of $a_1$---so is 0, etc. Altogether, the only nonzero values of $a_n$ for $n >0$ are $a_3, a_6, a_9, \ldots ,$ and in principle you can get them all in terms of $a_0.$

 

[BearY]

We need $a_1 = a_2 = 0$. Then we can get $a_3$ in terms of $a_0$, can get $a_4$ in terms of $a_1$---so is 0, etc. Altogether, the only nonzero values of $a_n$ for $n >0$ are $a_3, a_6, a_9, \ldots ,$ and in principle you can get them all in terms of $a_0.$

I am sorry for the vagueness, I meant other as $a_1$ and $a_2$, I don't know why I said other instead of that, doesn't save much typing anyway. I am just wondering what if for $a_0f(r)$ and $a_1g(r)$, the solution to $f(r) = 0$ and $g(r) = 0$ are the same. Or is it impossible?

 

[Ray Vickson]

I am sorry for the vagueness, I meant other as $a_1$ and $a_2$, I don't know why I said other instead of that, doesn't save much typing anyway. I am just wondering what if for $a_0f(r)$ and $a_1g(r)$, the solution to $f(r) = 0$ and $g(r) = 0$ are the same. Or is it impossible?

Obviously, if $a_0 \neq 0$ the solution of $a_0 f(r) = 0$ is the same as that of $f(r) = 0$. But if $a_0 = 0$ the equation $a_0 f(r) = 0$ contains no useful information about $r$.

So: go back are re-read past posts: we DO have $a_0 \neq 0$ (essentially by definition). That tells us enough about $r$ that we can say with absolute, 100% certainty, that $g(r) \neq 0$, FORCING $a_1 = 0$. End of story.