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[SOLVED] Reduction of order

dwsmith

Well-known member
Feb 1, 2012
1,673
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
Feb 5, 2012
1,621
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
Feb 1, 2012
1,673
>>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
Feb 1, 2012
1,673
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.
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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
Feb 1, 2012
1,673
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
Feb 1, 2012
1,673
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
Feb 5, 2012
1,621
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
Feb 1, 2012
1,673
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
Feb 5, 2012
1,621
\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
Feb 1, 2012
1,673

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
\[\Rightarrow 2\frac{d}{dx}\left[(1-x^2)P_{n}(x)\right]v'(x)+[(1-x^2)P_{n}(x)]v''(x)=0\]

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
Feb 5, 2012
1,621
$$
\left[(1-x^2)P_{n}(x)\right]' = [P - x^2P]' = P' - 2xP - x^2P' = P'(1-x^2) - 2xP
$$
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
Feb 1, 2012
1,673
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.
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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
Feb 1, 2012
1,673
From this, how can I find $P_0(x)$?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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
Feb 1, 2012
1,673
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
Feb 5, 2012
1,621