# [SOLVED]Reduction of order

#### dwsmith

##### Well-known member
Legendre polynomials
I have one solution to $(1-x^2)y''-2xy'+\lambda^2 y=0$ and that is
$$\lambda^2a_0 - (2 - \lambda^2)xa_1 + 2a_2 + 6xa_3 + \sum_{k = 2}^{\infty}\left[a_k\left(\lambda^2 - k - k^2\right) + a_{k + 2}\left(k^2 + 3k + 2\right)\right]x^k = 0.$$
From this equation, we can conclude $\lambda^2a_0 + 2a_2 = 0$ and $-(2 - \lambda^2)xa_1 + 6xa_3 = 0$.
Therefore,
$$a_2 = -\frac{\lambda^2}{2}a_0\quad\text{and}\quad a_3 = \frac{2 - \lambda^2}{6}a_1.$$
Finally, we have $a_k\left(\lambda^2 - k - k^2\right) + a_{k + 2}\left(k^2 + 3k + 2\right) = 0$, i.e.
$$a_{k + 2} = \frac{k(k + 1) - \lambda^2}{(k + 2)(k + 1)}a_k.$$
Use the method of reduction of order to derive the following formula for the Legendre functions of the 2nd kind:
$$Q_n(x) = P_n(x)\int\frac{1}{[P_n(x)]^2(1 - x^2)}dx, \quad n = 0,1,2,\ldots$$
I basically need to solve
$$y_2 = y_1\int\frac{\exp\left(\int\frac{2x}{1-x^2}dx\right)}{y_1^2}$$
where $y_1$ is my solution above, correct?

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi dwsmith, Legendre polynomials
I have one solution to $(1-x^2)y''-2xy'+\lambda^2 y=0$ and that is
$$\lambda^2a_0 - (2 - \lambda^2)xa_1 + 2a_2 + 6xa_3 + \sum_{k = 2}^{\infty}\left[a_k\left(\lambda^2 - k - k^2\right) + a_{k + 2}\left(k^2 + 3k + 2\right)\right]x^k = 0.$$

From this equation, we can conclude $\lambda^2a_0 + 2a_2 = 0$ and $-(2 - \lambda^2)xa_1 + 6xa_3 = 0$.
Therefore,
$$a_2 = -\frac{\lambda^2}{2}a_0\quad\text{and}\quad a_3 = \frac{2 - \lambda^2}{6}a_1.$$
Finally, we have $a_k\left(\lambda^2 - k - k^2\right) + a_{k + 2}\left(k^2 + 3k + 2\right) = 0$, i.e.
$$a_{k + 2} = \frac{k(k + 1) - \lambda^2}{(k + 2)(k + 1)}a_k.$$
Note that,

$\lambda^2a_0 - (2 - \lambda^2)xa_1 + 2a_2 + 6xa_3 + \sum_{k = 2}^{\infty}\left[a_k\left(\lambda^2 - k - k^2\right) + a_{k + 2}\left(k^2 + 3k + 2\right)\right]x^k = 0$

$\Rightarrow \sum_{k = 0}^{\infty}\left[a_k\left(\lambda^2 - k - k^2\right) + a_{k + 2}\left(k^2 + 3k + 2\right)\right]x^k = 0$

Therefore,

$a_{k + 2} = \frac{k(k + 1) - \lambda^2}{(k + 2)(k + 1)}a_k\mbox{ for }k\geq 0$

Use the method of reduction of order to derive the following formula for the Legendre functions of the 2nd kind:
$$Q_n(x) = P_n(x)\int\frac{1}{[P_n(x)]^2(1 - x^2)}dx, \quad n = 0,1,2,\ldots$$
You have the differential equation,

$(1-x^2)y''-2xy'+\lambda^2 y=0$

with the solution,

$y_{1}(x)=P_{n}(x)$

>>Here<< you can find an example of how to perform the method of reduction of order. Try to find the second solution $$y_{2}(x)=Q_{n}(x)$$ using this method. Hope you can continue. Kind Regards,
Sudharaka.

#### dwsmith

##### Well-known member
>>Here<< you can find an example of how to perform the method of reduction of order. Try to find the second solution $$y_{2}(x)=Q_{n}(x)$$ using this method. Hope you can continue. Kind Regards,
Sudharaka.
I am looking through Zill's ODE book and it tells me to find the second solution using that formula I listed.

#### dwsmith

##### Well-known member
Doing $y_2 = y_1v$. I would have to take the derivative of that series twice? That is going to be bit messy I imagine.
$$y_2 = v\sum_{k = 0}^{\infty}\left[a_k\left(\lambda^2 - k - k^2\right) + a_{k + 2}\left(k^2 + 3k + 2\right)\right]x^k$$
$$y_2' = v'\sum_{k = 0}^{\infty}\left[a_k\left(\lambda^2 - k - k^2\right) + a_{k + 2}\left(k^2 + 3k + 2\right)\right]x^k+ v\sum_{k = 1}^{\infty}\left[a_k\left(\lambda^2k - k^2 - k^3\right) + a_{k + 2}\left(k^3 + 3k^2 + 2k\right)\right]x^{k-1}$$
\begin{alignat}{3}
y_2'' & = & v''\sum_{k = 0}^{\infty}\left[a_k\left(\lambda^2 - k - k^2\right) + a_{k + 2}\left(k^2 + 3k + 2\right)\right]x^k\\
& + & 2v'\sum_{k = 1}^{\infty}\left[a_k\left(\lambda^2k - k^2 - k^3\right) + a_{k + 2}\left(k^3 + 3k^2 + 2k\right)\right]x^{k-1}\\
& + \ &v\sum_{k = 2}^{\infty}\left[a_k\left(\lambda^2(k^2-k) - k^4 + k^2\right) + a_{k + 2}\left(k^4 + 2k^3 -k^2-2k\right)\right]x^{k-2}
\end{alignat}
Is there a better way to do this? I mean plugging this into the Legendre equation is going to be ridiculous.

I plugged it into Mathematica and obtained
\begin{alignat}{2}
(1-x^2)\left(\sum_{k = 2}^{\infty}\left[a_k\left(\lambda^2(k^2-k) - k^4 + k^2\right) + a_{k + 2}\left(k^4 + 2k^3 -k^2-2k\right)\right]x^{k-2}\right) & -\\
2\left(\sum_{k = 1}^{\infty}\left[a_k\left(\lambda^2k - k^2 - k^3\right) + a_{k + 2}\left(k^3 + 3k^2 + 2k\right)\right]x^{k-1}\right)(xv+(x^2-1)v' & +\\
\left(\sum_{k = 0}^{\infty}\left[a_k\left(\lambda^2 - k - k^2\right) + a_{k + 2}\left(k^2 + 3k + 2\right)\right]x^k\right)(\lambda^2v-2xv'+(1-x^2)v'')& =\\
& 0
\end{alignat}
However, Paul's notes says the v parameter should be eliminated. I let Mathematica do all the work and the v parameter is still present.

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#### Sudharaka

##### Well-known member
MHB Math Helper
I think you are in a confusion as to what the solution to the differential equation is. It should be,

$y_{1}=P_{n}(x)=\sum_{k=0}^{\infty}a_{k}x^k\mbox{ where }\lambda^2=n(n+1)$

You will find >>this<< helpful in understanding how to solve the Legendre's differential equation using Frobenius method.

So in the method of reduction of order you have to take,

$y_{2}(x)=v(x)\sum_{k=0}^{\infty}a_{k}x^k$

#### dwsmith

##### Well-known member
I think you are in a confusion as to what the solution to the differential equation is. It should be,

$y_{1}=P_{n}(x)=\sum_{k=0}^{\infty}a_{k}x^k\mbox{ where }\lambda^2=n(n+1)$

You will find >>this<< helpful in understanding how to solve the Legendre's differential equation using Frobenius method.

So in the method of reduction of order you have to take,

$y_{2}(x)=v(x)\sum_{k=0}^{\infty}a_{k}x^k$
That is what I did original to get my first solution. Now I am looking for a second solution.

#### dwsmith

##### Well-known member
That is what I did original to get my first solution. Now I am looking for a second solution.
Making that change I still have
$$(1-x^2)\left(\sum_{k = 2}^{\infty}k(k-1)a_kx^{k-2}\right) - 2\left(\sum_{k = 1}^{\infty}ka_kx^{k-1}\right)(xv+(x^2-1)v') + \left(\sum_{k = 0}^{\infty}a_kx^k\right)(\lambda^2v-2xv'+(1-x^2)v'') = 0$$

#### Sudharaka

##### Well-known member
MHB Math Helper
Making that change I still have
$$(1-x^2)\left(\sum_{k = 2}^{\infty}k(k-1)a_kx^{k-2}\right) - 2\left(\sum_{k = 1}^{\infty}ka_kx^{k-1}\right)(xv+(x^2-1)v') + \left(\sum_{k = 0}^{\infty}a_kx^k\right)(\lambda^2v-2xv'+(1-x^2)v'') = 0$$
Simplify this and equate the coefficients of similar powers of $$x$$ to zero; similar to what you have done in your first post to find the recurrence relation.

#### dwsmith

##### Well-known member
Simplify this and equate the coefficients of similar powers of $$x$$ to zero; similar to what you have done in your first post to find the recurrence relation.
\begin{align}
a_0(\lambda^2v-2v'x+v'')+a_1(v''+2v'+vx(2+\lambda^2)+a_2(2+v'x)+6a_3+\\
\sum_{n=2}^{\infty}[a_{n+2}(n^2+3n+2)+a_{n+1}2v'(n+1)+a_n(v''+v( \lambda^2 -2n)-n(n-1))+a_{n-1}2v'n+a_{n-2}v'']x^n
\end{align}

Is this correct (^)?
If so,
$$a_2 = \frac{2v'-v''-\lambda^2v}{2+v'x}a_0$$
and
$$a_3=-a_1\frac{v''+2v'+vx(\lambda^2+2)}{6}.$$

#### Sudharaka

##### Well-known member
MHB Math Helper
\begin{align}
a_0(\lambda^2v-2v'x+v'')+a_1(v''+2v'+vx(2+\lambda^2)+a_2(2+v'x)+6a_3+\\
\sum_{n=2}^{\infty}[a_{n+2}(n^2+3n+2)+a_{n+1}2v'(n+1)+a_n(v''+v( \lambda^2 -2n)-n(n-1))+a_{n-1}2v'n+a_{n-2}v'']x^n
\end{align}

Is this correct (^)?
If so,
$$a_2 = \frac{2v'-v''-\lambda^2v}{2+v'x}a_0$$
and
$$a_3=-a_1\frac{v''+2v'+vx(\lambda^2+2)}{6}.$$
The recurrence relations should not be dependent on $$x$$. #### dwsmith

##### Well-known member
The recurrence relations should not be dependent on $$x$$. What? That solution is following what you said.

#### Sudharaka

##### Well-known member
MHB Math Helper
What? That solution is following what you said.
The recurrence relations you have obtained have $$x$$ in them,

$a_2 = \frac{2v'-v''-\lambda^2v}{2+v'{\color{red}x}}a_0$

$a_3=-a_1\frac{v''+2v'+v{\color{red}x}(\lambda^2+2)}{6}.$

Anyway, thinking about this problem further I came to note that there is no need to substitute for $$P_{n}(x)$$ since the answer is given in terms of $$P_{n}(x)$$. So you can take,

$y_{2}(x)=v(x)P_{n}(x)$

and perform the reduction of order. Let me guide you through the first steps. The differential equation is,

$(1-x^2)y''-2xy'+\lambda^2 y=0$

We know that $$y_{1}(x)=P_{n}(x)$$. Let, $$y_{2}(x)=Q_{n}(x)=v(x)P_{n}(x)$$. Then,

$(1-x^2)[v(x)P_{n}(x)]''-2x[v(x)P_{n}(x)]'+\lambda^2 v(x)P_{n}(x)=0$

Using, $$(1-x^2)P''_{n}(x)-2xP'_{n}(x)+\lambda^2 P_{n}(x)=0$$ you can obtain,

$2(1-x^2)v'(x)P'_{n}(x)+(1-x^2)v''(x)P_{n}(x)-2xv'(x)P_{n}(x)=0$

$\Rightarrow 2(1-x^2)v'(x)P_{n}(x) P'_{n}(x)+(1-x^2)v''(x)P_{n}^{2}(x)-2xv'(x)P_{n}^{2}(x)=0$

$\Rightarrow \frac{d}{dx}\left[(1-x^2)P_{n}^{2}(x)\right]v'(x)+[(1-x^2)P_{n}^{2}(x)]v''(x)=0$

Now you should be able to solve for $$v(x)$$ by separation of variables.

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#### dwsmith

##### Well-known member
$\Rightarrow 2\frac{d}{dx}\left[(1-x^2)P_{n}(x)\right]v'(x)+[(1-x^2)P_{n}(x)]v''(x)=0$

Now you should be able to solve for $$v(x)$$ by separation of variables.
$$\left[(1-x^2)P_{n}(x)\right]' = [P - x^2P]' = P' - 2xP - x^2P' = P'(1-x^2) - 2xP$$
If I multiply 2 and v' in, we have
$$4P'x$$
which isn't right.

#### Sudharaka

##### Well-known member
MHB Math Helper
$$\left[(1-x^2)P_{n}(x)\right]' = [P - x^2P]' = P' - 2xP - x^2P' = P'(1-x^2) - 2xP$$
If I multiply 2 and v' in, we have
$$4P'x$$
which isn't right.
Sorry. There was a little error. I have edited my last post (#12). #### dwsmith

##### Well-known member
Sorry. There was a little error. I have edited my last post (#12). Then we have
$$\frac{v''}{v'} = -\frac{1}{2}\frac{[(1-x^2)P^2]'}{(1-x^2)P^2}\Rightarrow \ln v' = \frac{1}{2}\ln((1-x^2)P^2)$$
I don't see how to go from this to the solution.

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#### Sudharaka

##### Well-known member
MHB Math Helper
Then we have
$$\frac{v''}{v'} = -\frac{1}{2}\frac{[(1-x^2)P^2]'}{(1-x^2)P^2}\Rightarrow \ln v' = \frac{1}{2}\ln((1-x^2)P^2)$$
I don't see how to go from this to the solution.
From where did the fraction $$\frac{1}{2}$$ come from? Continuing from post #12 we have,

$\frac{d}{dx}\left[(1-x^2)P_{n}^{2}(x)\right]v'(x)+[(1-x^2)P_{n}^{2}(x)]v''(x)=0$

$\Rightarrow -\frac{v''(x)}{v'(x)}=\frac{\frac{d}{dx}\left[(1-x^2)P_{n}^{2}(x)\right]}{[(1-x^2)P_{n}^{2}(x)]}$

Integrate both sides and we get,

$\ln[v'(x)]=\ln\left[(1-x^2)P_{n}^{2}(x)\right]^{-1}$

$\Rightarrow v'(x)=\frac{1}{(1-x^2)P_{n}^{2}(x)}$

Integrate again to obtain,

$v(x)=\int \frac{1}{(1-x^2)P_{n}^{2}(x)}\,dx$

$\therefore Q_{n}(x)=v(x)P_{n}(x)=P_{n}(x)\int \frac{1}{(1-x^2)P_{n}^{2}(x)}\,dx$

#### dwsmith

##### Well-known member
From this, how can I find $P_0(x)$?

#### Sudharaka

##### Well-known member
MHB Math Helper
From this, how can I find $P_0(x)$?
You have,

$P_{n}(x)=\sum_{k=0}^{\infty}a_k x^k\mbox{ where }a_{k + 2} = \frac{k(k + 1) - [n(n+1)]^2}{(k + 2)(k + 1)}a_k$

Substitute $$n=0$$ and try to obtain a solution to the recurrence relation.

#### dwsmith

##### Well-known member
You have,

$P_{n}(x)=\sum_{k=0}^{\infty}a_k x^k\mbox{ where }a_{k + 2} = \frac{k(k + 1) - [n(n+1)]^2}{(k + 2)(k + 1)}a_k$

Substitute $$n=0$$ and try to obtain a solution to the recurrence relation.
How does one come up with $Q_0$ if $P_0$ is represented as a relation?

#### Sudharaka

##### Well-known member
MHB Math Helper
How does one come up with $Q_0$ if $P_0$ is represented as a relation?
Recall that,

$Q_{n}(x)=v(x)P_{n}(x)$