Legendre Differential equation question.

[yungman]

Legendre equation:
(1-x^2)y&#039;&#039;-2xy&#039;+n(n+1)y=0 Where -1< x < 1

General solution is y(x)=c_1 P_n (x)+c_2 Q_n (x)

Where P_n (x) is bounded and Q_n (x) is unbounded on (-1,1).

Q_n (x)=P_n (x)\int \frac{dx}{[P_n (x)]^2 (1-x^2)}


Question: Why is Q_n (x) unbounded on (-1,1)?

I tried to solve for Q_n (x) by substitude P_n (x) in, using partial fraction to simplify and integrate each terms individually.

Q_n (x)=P_n (x)[ln(x+A) + ln(x+B)+ \frac{1}{x+C}...]

As you can see after multiplying P_n (x) in, ln(x^+_- A) is unbound at x=+/-A. Is that the reason Q is unbound on (-1,1)?


But the second solution can be represented by an infinite series instead of

Q_n (x)=P_n (x)\int \frac{dx}{[P_n (x)]^2 (1-x^2)}

where the power series is bounded on (-1,1).

 

[Old Smuggler]

We have that (see, e.g., Abromowitz & Stegun, Handbook of mathematical functions)
<br /> Q_ n(x)={\frac{1}{2}} \int_{-1}^1{\frac{P_n(y)dy}{x-y}},<br />
where the integral is defined from its Cauchy principal value.
From this it is easy to see that Q_n(x) diverges when x{\rightarrow}{\pm}1.

 

[yungman]

I agree that it diverge at x=+/- 1. That is the reason it was specified on (-1,1).

This is true for polynomial P_n (x) also.

Why only Q_n (x) is unbounded if represented by power series?

I updated the original post already.

 

[matematikawan]

Legendre equation:
(1-x^2)y&#039;&#039;-2xy&#039;+n(n+1)y=0 Where -1< x < 1

General solution is y(x)=c_1 P_n (x)+c_2 Q_n (x)

Where P_n (x) is bounded and Q_n (x) is unbounded on (-1,1).

Q_n (x)=P_n (x)\int \frac{dx}{[P_n (x)]^2 (1-x^2)}


Question: Why is Q_n (x) unbounded on (-1,1)?

I remember that for a Legendre equation, one of the linearly independent solution is a polynomial P_n and the other one is an infinite series Q_n. Do you meant to say that Q_n(x) does not converge for x in (-1,1)?
If it doesn't converges how can it be a solution.

Just curious. Does this expression
Q_n (x)=P_n (x)\int \frac{dx}{[P_n (x)]^2 (1-x^2)}
comes from applying Lagrange method of reduction of order ?

 

[yungman]

I remember that for a Legendre equation, one of the linearly independent solution is a polynomial P_n and the other one is an infinite series Q_n. Do you meant to say that Q_n(x) does not converge for x in (-1,1)?
If it doesn't converges how can it be a solution.

Just curious. Does this expression
Q_n (x)=P_n (x)\int \frac{dx}{[P_n (x)]^2 (1-x^2)}
comes from applying Lagrange method of reduction of order ?

Q_n (x) obtained by Reduction of Order using P_n (x).

I never understand why they have two different form of Q_n (x). One as shown above and the other is an infinite series like you described.



This is the two Legendre series. If n is an integer, one become a polynomial and the other solution remain a power infinite series.

y_1 (x)=a_0 [1-\frac{n(n+1)}{2!}x^2 +\frac{(n-2)n(n+1)(n+3)}{4!}x^4 -\frac{(n-4)(n-2)n(n+1)(n+3)(n+5)}{6!}x^6...](1)

y_2 (x)=a_1 [x-\frac{(n-1)(n+2)}{3!}x^3 +\frac{(n-3)(n-1)(n+2)(n+4)}{5!}x^5 -\frac{(n-5)(n-3)(n-1)(n+2)(n+4)(n+6)}{7!}x^7...](2)

(1) and (2) obvious is continuous on (-1,1) and the series converge on (-1,1)



The other is Q_n (x)=P_n (x)\int \frac{dx}{[P_n (x)]^2 (1-x^2)}(3) which have points of discontinuity as shown below is an example using (3) with n=2.


General solution is y(x)=c_1 P_n (x)+c_2 Q_n (x)
P_2 (x)=\frac{1}{2}(3x^2 -1)
Q_2 (x)=P_2 (x)\int \frac{dx}{[P_2 (x)]^2 (1-x^2)}=\frac{1}{2}[3x^2 -1]\int \frac{dx}{{[\frac{1}{2}[3x^2 -1]^2 (1-x^2)}}

Q_2 (x)=\frac{2}{9}[3x^2 -1]\int \frac{dx}{{[x^2 -\frac{1}{3}]^2 (1-x^2)}}

Q_2 (x)=\frac{2}{9}[3x^2 -1]\int \frac{dx}{(x+\frac{1}{\sqrt{3}})^2 (x-\frac{1}{\sqrt{3}})^2 (1-x^2)}}

Q_2 (x)=\frac{2}{9}(3x^2 -1){\int \frac{Adx}{(x+\frac{1}{\sqrt{3}})}+\int \frac{Bdx}{(x+\frac{1}{\sqrt{3}})^2 }+\int \frac{Cdx}{(x-\frac{1}{\sqrt{3}})}+\int \frac{Ddx}{(x-\frac{1}{\sqrt{3}})^2 }+\int \frac{(E+Fx)dx}{(1-x^2 )}

Where A, B, C, D, E, F are constant numbers only.

Q_2 (x)=\frac{2}{3} (x+\frac{1}{\sqrt{3}})(x-\frac{1}{\sqrt{3}}) [A ln(x+\frac{1}{\sqrt{3}})-\frac{B}{(x+\frac{1}{\sqrt{3}})}+C ln(x-\frac{1}{\sqrt{3}})-\frac{D}{(x-\frac{1}{\sqrt{3}})}-\frac{F}{2}(ln(1-x^2 )]

\Rightarrow Q_2 (x)=\frac{2}{3}[A (x+\frac{1}{\sqrt{3}})(x-\frac{1}{\sqrt{3}}) ln(x+\frac{1}{\sqrt{3}})-B(x-\frac{1}{\sqrt{3}})+C (x+\frac{1}{\sqrt{3}})(x-\frac{1}{\sqrt{3}}) ln(x-\frac{1}{\sqrt{3}})-D(x+\frac{1}{\sqrt{3}})-\frac{F}{2}(x+\frac{1}{\sqrt{3}})(x-\frac{1}{\sqrt{3}}) (ln(1-x^2 )]

As shown, all the denominators are cancel so the no denominator to worry about. The points of unbound are x=\frac{1}{\sqrt{3}} and x=-\frac{1}{\sqrt{3}} in the ln(x-\frac{1}{\sqrt{3}}) etc.

As you see, the two form of Q_n (x) have different characteristics. What did I do wrong?

 

[matematikawan]

If we could simplify your computation a bit.
I will use http://integrals.wolfram.com/index.jsp?expr=1/((1-3x^2)^2*(1-x^2))&random=false"
Q_n (x)=P_n (x)\int \frac{dx}{[P_n (x)]^2 (1-x^2)}
= \frac{1}{4}(3x^2-1)\{\frac{6x}{1-3x^2} +\log{(1+x)}-\log{(1-x) \}

We can see that Q(x) converges for all x in (-1,1).

 

[yungman]

If we could simplify your computation a bit.
I will use http://integrals.wolfram.com/index.jsp?expr=1/((1-3x^2)^2*(1-x^2))&random=false"
Q_n (x)=P_n (x)\int \frac{dx}{[P_n (x)]^2 (1-x^2)}
= \frac{1}{4}(3x^2-1)\{\frac{6x}{1-3x^2} +\log{(1+x)}-\log{(1-x) \}

We can see that Q(x) converges for all x in (-1,1).

I tried the Wolfram Mathematica and get similar to your answer, difference is 3x^2-1 instead of 1-3x^2.

BUT I double check my steps on the partial fraction, B=D=9/8, E=0 and there are terms like ln(x^+_-\frac{1}{\sqrt{3}}) exist. Please double check my partial fraction work above and let me know what I did wrong. I check it 3 times already before I even post the thread. Something must be very wrong what I did and I don't see it.

You can easily see ln(x^+_-1) are from \frac{1}{1-x^2 }.


How about the (3x^2 -1)^2 = 9(x+\frac{1}{\sqrt{3}})^2 (x-\frac{1}{\sqrt{3}})^2 ?

Thanks

 

[Old Smuggler]

Q_n(x) does not contain any integration constants, and your formula for it is rather inconvenient.
Here is a more useful formula for Q_n(x) obtained from Handbook of Mathematical Functions:
<br /> Q_n(x)={\frac{1}{2}}P_n(x){\ln}{\frac{1+x}{1-x}}- \sum_{m=1}^nP_{m-1}(x)P_{n-m}(x)/m.<br />
We easily find that
<br /> Q_2(x)={\frac{(3x^2-1)}{4}}{\ln}{\frac{1+x}{1-x}}-{\frac{3x}{2}}.<br />

 

[yungman]

After I saw the Wolfram simulation, I work through the numbers and find all the constants, find out the terms contain all the points in between x+-1 to x=1 disappeared because the coeficients for those terms are zero!

Below is what I got and I double check already.
Q_2 (x)=P_2 (x)\int \frac{dx}{[P_2 (x)]^2 (1-x^2)}=\frac{1}{2}[3x^2 -1]\int \frac{dx}{{[\frac{1}{2}[3x^2 -1]^2 (1-x^2)}}

Q_2 (x)=2[3x^2 -1]\int \frac{dx}{ (\sqrt{3}x^2 +1)^2 (\sqrt{3}x-1)^2 (1+x)(1-x)}

Q_2 (x)=2(3x^2 -1)[{\int \frac{Adx}{ (\sqrt{3}x +1) } +\int \frac{Bdx}{ (\sqrt{3}x +1)^2 }+\int \frac{Cdx}{ (\sqrt{3}x -1) }+\int \frac{Ddx}{ (\sqrt{3}x -1)^2 }+\int \frac{Edx}{(1+x )} +\int \frac{Fdx}{(1-x )}]

Where A, B, C, D, E, F are constant numbers only. See below the steps to obtain the constants using partial fraction. This give:

Q_2 (x)=2(3x^2 -1)[\frac{\sqrt{3}}{6}\int \frac{dx}{ (\sqrt{3}x +1) } + \frac{3}{8}\int \frac{dx}{ (\sqrt{3}x +1)^2 } + \frac{2\sqrt{3}-3}{12}\int \frac{dx}{ (\sqrt{3}x -1) }+
\frac{3}{8}\int \frac{dx}{ (\sqrt{3}x -1)^2 } - \frac{1}{8}\int \frac{dx}{(1+x )} + \frac{1}{8}\int \frac{Fdx}{(1-x )}]

Q_2 (x)=2(\sqrt{3}x+1 )(\sqrt{3}x-1) [- \frac{3}{8}\frac{1}{\sqrt{3}}\frac{1}{(\sqrt{3}x+1)} - \frac{3}{8}\frac{1}{\sqrt{3}}\frac{1}{(\sqrt{3}x-1)} + \frac{1}{8}ln(1+x) - \frac{1}{8}(ln(1-x)]


As shown, all the denominators are cancel so the no denominator to worry about. The only points of unbound on (-1,1) are from:x=-1 and x+1.


Partial Fraction.
\frac{1}{ (\sqrt{3}x+1)^2 (\sqrt{3}x-1)^2 (1+x)(1-x) } = { \frac{A}{ (\sqrt{3}x +1) } + \frac{B}{ (\sqrt{3}x +1)^2 }+
\frac{C}{ (\sqrt{3}x -1) }+ \frac{D}{ (\sqrt{3}x -1)^2 }+ \frac{E}{(1+x )} + \frac{F}{(1-x )}}


1) Multiply both side by (\sqrt{3}x+1)^2 and let x=-\frac{1}{\sqrt{3}}\Rightarrow
\frac{1}{ (\sqrt{3}x-1)^2 (1+x)(1-x) } = B= \frac{1}{ (-2)^2 (1-\frac{1}{\sqrt{3}})(1+\frac{1}{\sqrt{3}})} \Rightarrow B=\frac{3}{8}



2) Multiply both side by (\sqrt{3}x-1)^2 and let x=\frac{1}{\sqrt{3}}\Rightarrow
\frac{1}{ (\sqrt{3}x+1)^2 (1+x)(1-x) } = D= \frac{1}{ (2)^2 (1+\frac{1}{\sqrt{3}})(1-\frac{1}{\sqrt{3}})} \Rightarrow D=\frac{3}{8}



3) Multiply both side by (1+x) and let x=-1\Rightarrow
\frac{1}{ (\sqrt{3}x+1)^2 (\sqrt{3}x-1)^2 (1-x) }=E = \frac{1}{ (1-\sqrt{3})^2 (-1-\sqrt{3})^2 (2) }\RightarrowE=\frac{1}{8}



4) Multiply both side by (1-x) and let x=1\Rightarrow
\frac{1}{ (\sqrt{3}x+1)^2 (\sqrt{3}x-1)^2 (1+x) }= F = \frac{1}{ (\sqrt{3}+1)^2 (\sqrt{3}-1)^2 (2) }\RightarrowF=\frac{1}{8}



Substitude B,D,E,F in:

\frac{1}{ (\sqrt{3}x+1)^2 (\sqrt{3}x-1)^2 (1+x)(1-x) } = { \frac{A}{ (\sqrt{3}x +1) } + \frac{3}{8}\frac{1}{ (\sqrt{3}x +1)^2 }+ \frac{C}{ (\sqrt{3}x -1) }+ \frac{3}{8}\frac{1}{ (\sqrt{3}x -1)^2 }+
\frac{1}{8}\frac{1}{(1+x )} + \frac{1}{8}\frac{1}{(1-x )}}


Let x=0 1 = A + \frac{3}{8} - C+ \frac{3}{8} + \frac{1}{8} + \frac{1}{8}\Rightarrow A-C=0


Let x=\sqrt{3}\Rightarrow\frac{1}{(3+1)^2 (3-1)^2 (1+\sqrt{3}) (1-\sqrt{3})}=-\frac{1}{128} = \frac{A}{4} + \frac{3}{128} +\frac{C}{2} +\frac{3}{32} +\frac{1}{8}\frac{1}{(1+\sqrt{3})} + \frac{1}{8}\frac{1}{(1-\sqrt{3}) }

\Rightarrow \frac{A}{4}+\frac{C}{2} = -\frac{1}{128} - \frac{3}{128} - \frac{3}{32} + \frac{16}{128} } = 0

This give A+2C=0

Together with A-C=0 result in A=C=0



Question is are all other n = 0,1,2,3,...behave the same?

Thanks for your time.

 

[yungman]

I know so far all the question except one had all been clarified. The last question is:


Why books claimed Q is unbound in [-1,1] and cannot be used in a lot of the BVP? In reality the only reason P is bounded on [-1,1] only after they normalized the coefficients so P(1)=1. Or else P is unbounded too! Why the book don't attempt to normalize the coefficients of Q so Q(1)=1. Then both P and Q are bounded on [-1,1].

Thanks

 

[yungman]

Anyone, please?

 

[matematikawan]

Q_n(x) is unbounded at the end point (boundary ?).

The general solution to the Legendre Differential equation is
y(x)=c_1 P_n (x)+c_2 Q_n (x)

If we impose the boundary conditions, I think c₂ must be zero because Q_n(1 or -1) is unbounded. Thus, the final solution only involves P_n(x).

 

[yungman]

Q_n(x) is unbounded at the end point (boundary ?).

The general solution to the Legendre Differential equation is
y(x)=c_1 P_n (x)+c_2 Q_n (x)

If we impose the boundary conditions, I think c₂ must be zero because Q_n(1 or -1) is unbounded. Thus, the final solution only involves P_n(x).

Actually Q_n(x) is bounded in (-1,1), just not bounded in [-1,1]. It is only the two end points thatQ_n(x) not bounded.

That is my whole question. P_n(x) is bounded in [-1,1] ONLY when they normalize all the coefficients so P_n(1)=1! Q_n(x) is unbounded because they did not normalize the Coefficients. You can just as easy putting a coefficient infront of each of the element of the Q and make Q_n(1)=1!


Then both P_n(x) &amp; Q_n(x) are bounded in [-1,1].

Question is why not?? They just keep making C2=0 like what you showed

 

[matematikawan]

P_n(x) is a polynomial of degree n. It does make sense to initialise P_n(1)=1.
But Q_n(x) is an infinite series and also undefined at x=-1, 1. So it does't make sense Q_n(1) = something.

 

[yungman]

P_n(x) is a polynomial of degree n. It does make sense to initialise P_n(1)=1.
But Q_n(x) is an infinite series and also undefined at x=-1, 1. So it does't make sense Q_n(1) = something.

You mean there is no way to make Q converge to a finite number when x=1? I guess it make sense because no matter how small you make the coefficients, if you have infinite number of terms, still it is unbounded. Am I correct?

Still, what is so important about [-1,1]? Why can't people just work with (-1,1) so both P and Q can be used?