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Legendre Differential equation question.

  1. Jan 21, 2010 #1
    Legendre equation:
    [tex](1-x^2)y''-2xy'+n(n+1)y=0[/tex] Where -1< x < 1

    General solution is [tex]y(x)=c_1 P_n (x)+c_2 Q_n (x)[/tex]

    Where [tex]P_n (x)[/tex] is bounded and [tex]Q_n (x)[/tex] is unbounded on (-1,1).

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

    Question: Why is [tex]Q_n (x)[/tex] unbounded on (-1,1)?

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

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

    As you can see after multiplying [tex]P_n (x)[/tex] in, [tex] ln(x^+_- A)[/tex] 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

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

    where the power series is bounded on (-1,1).
    Last edited: Jan 22, 2010
  2. jcsd
  3. Jan 21, 2010 #2
    We have that (see, e.g., Abromowitz & Stegun, Handbook of mathematical functions)
    Q_ n(x)={\frac{1}{2}} \int_{-1}^1{\frac{P_n(y)dy}{x-y}},
    where the integral is defined from its Cauchy principal value.
    From this it is easy to see that [itex]Q_n(x)[/itex] diverges when [itex]x{\rightarrow}{\pm}1[/itex].
  4. Jan 21, 2010 #3
    I agree that it diverge at x=+/- 1. That is the reason it was specified on (-1,1).

    This is true for polynomial [tex]P_n (x)[/tex] also.

    Why only [tex]Q_n (x)[/tex] is unbounded if represented by power series?

    I updated the original post already.
    Last edited: Jan 22, 2010
  5. Jan 22, 2010 #4
    I remember that for a Legendre equation, one of the linearly independent solution is a polynomial Pn and the other one is an infinite series Qn. Do you meant to say that Qn(x) does not converge for x in (-1,1)?
    If it doesn't converges how can it be a solution. :confused:

    Just curious. Does this expression
    [tex]Q_n (x)=P_n (x)\int \frac{dx}{[P_n (x)]^2 (1-x^2)}[/tex]
    comes from applying Lagrange method of reduction of order ?
  6. Jan 22, 2010 #5

    [tex]Q_n (x)[/tex] obtained by Reduction of Order using [tex]P_n (x)[/tex].

    I never understand why they have two different form of [tex]Q_n (x)[/tex]. 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.

    [tex] 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..........][/tex](1)

    [tex] 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..........][/tex](2)

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

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

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

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

    [tex]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)}}[/tex]

    [tex]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 )}[/tex]

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

    [tex]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 )][/tex]

    [tex]\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 )][/tex]

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

    As you see, the two form of [tex]Q_n (x)[/tex] have different characteristics. What did I do wrong?
  7. Jan 23, 2010 #6
    Last edited by a moderator: Apr 24, 2017
  8. Jan 23, 2010 #7
    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 [tex]ln(x^+_-\frac{1}{\sqrt{3}})[/tex] 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 [tex]ln(x^+_-1)[/tex] are from [tex]\frac{1}{1-x^2 }[/tex].

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

    Last edited by a moderator: Apr 24, 2017
  9. Jan 25, 2010 #8
    [itex]Q_n(x)[/itex] does not contain any integration constants, and your formula for it is rather inconvenient.
    Here is a more useful formula for [itex]Q_n(x)[/itex] obtained from Handbook of Mathematical Functions:
    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.
    We easily find that
    Last edited: Jan 25, 2010
  10. Jan 26, 2010 #9
    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.
    [tex]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)}}[/tex]

    [tex]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)}[/tex]

    [tex]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 )}][/tex]

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

    [tex]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) }+ [/tex]
    [tex]\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 )}][/tex]

    [tex]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)][/tex]

    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.
    [tex]\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 }+[/tex]
    [tex] \frac{C}{ (\sqrt{3}x -1) }+ \frac{D}{ (\sqrt{3}x -1)^2 }+ \frac{E}{(1+x )} + \frac{F}{(1-x )}}[/tex]

    1) Multiply both side by [tex] (\sqrt{3}x+1)^2[/tex] and let [tex] x=-\frac{1}{\sqrt{3}}\Rightarrow[/tex]
    [tex]\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}[/tex]

    2) Multiply both side by [tex] (\sqrt{3}x-1)^2[/tex] and let [tex] x=\frac{1}{\sqrt{3}}\Rightarrow[/tex]
    [tex]\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}[/tex]

    3) Multiply both side by [tex] (1+x)[/tex] and let [tex] x=-1\Rightarrow[/tex]
    [tex] \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}[/tex]

    4) Multiply both side by [tex] (1-x)[/tex] and let [tex] x=1\Rightarrow[/tex]
    [tex] \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}[/tex]

    Substitude B,D,E,F in:

    [tex]\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 }+ [/tex]
    [tex]\frac{1}{8}\frac{1}{(1+x )} + \frac{1}{8}\frac{1}{(1-x )}}[/tex]

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

    Let x=[tex]\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}) }[/tex]

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

    This give [tex]A+2C=0[/tex]

    Together with [tex]A-C=0[/tex] result in [tex]A=C=0[/tex]

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

    Thanks for your time.
  11. Jan 29, 2010 #10
    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].

  12. Jan 31, 2010 #11
    Anyone, please?
  13. Feb 2, 2010 #12
    Qn(x) is unbounded at the end point (boundary ?).

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

    If we impose the boundary conditions, I think c2 must be zero because Qn(1 or -1) is unbounded. Thus, the final solution only involves Pn(x).
  14. Feb 3, 2010 #13

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

    That is my whole question. [tex]P_n(x)[/tex] is bounded in [-1,1] ONLY when they normalize all the coefficients so [tex]P_n(1)=1[/tex]!!! [tex]Q_n(x)[/tex] 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 [tex]Q_n(1)=1[/tex]!!!!

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

    Question is why not?? They just keep making C2=0 like what you showed
  15. Feb 3, 2010 #14
    Pn(x) is a polynomial of degree n. It does make sense to initialise Pn(1)=1.
    But Qn(x) is an infinite series and also undefined at x=-1, 1. So it does't make sense Qn(1) = something.
  16. Feb 3, 2010 #15
    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?
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