Results 1 to 4 of 4

Math Help - lenear dependance on a segment

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    lenear dependance on a segment5

    y''+\frac{1}{x}y'+(1-\frac{a^2}{x^2})y=0
    y1(1)=3 y2(1)=-8.1
    y1(1)=3 y2'(1)=5.4

    i need to deside:
    A.
    y1 and y2 are lenear dependant on (0,+infinity)
    B.
    y1 and y2 are not lenear dependant on (0,+infinity)
    C.
    we cant know if y1 and y2 are dependant only by a single point


    i don tknow how to solve this equation
    in a similar question i was told to use y=x^a
    but here its not working
    i am no get a certain a for which the equation is zero
    Last edited by transgalactic; November 27th 2010 at 04:01 AM. Reason: missed something
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    The solutions of the linear second order DE...

    \displaystyle x^{2}\ y^{''} + x\ y^{'} + (x^{2} - a^{2})\ y=0 (1)

    ... are called Bessel functions of order a. The parameter a can be real or integer [or even complex...] and the approach is different in the two cases. Curiously enough the case a 'non integer' is more easy to be 'attached' and the two independent solution are of the type \displaystyle J(x)= x^{\pm a}\ \sum_{n=0}^{\infty} \alpha_{n}\ x^{n} ...

    Kind regards

    \chi \sigma
    Last edited by chisigma; November 27th 2010 at 05:18 AM. Reason: orthography...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    what two cases
    how to eproach this one
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    The second order linear ordinary DE...

    \displaystyle y^{''} + \frac{y^{'}}{x} + (1-\frac{\nu^{2}}{x^{2}})=0 (1)

    ... is known as 'Bessel equation'. As I said in the previous post, the case in which \nu is not an integer is easier to manage, so that we start with it. First we observe that (1) is second order linear so that its general solution is of the type...

    \displaystyle y(x)= c_{1}\ \phi(x) + c_{2}\ \psi(x) (2)

    ... where \phi(*) and \psi(*) are two independent solution of (1). We can start with the hypothesis that a solution of (1) can be written as...

    \displaystyle y(x)= x^{\nu}\ \lambda(x)= x^{\nu}\ \sum_{n=0}^{\infty} a_{n}\ x^{n} (3)

    ... where \lambda(*) is an analytic function, so that (1) becomes...

    \displaystyle \sum_{n=0}^{\infty} a_{n}\ x^{n}  = - \sum_{n=2}^{\infty} n\ (n-1)\ a_{n}\ x^{n-2} - (1+2\ \nu)\ \sum_{n=1}^{\infty} n\ a_{n}\ x^{n-2} (4)

    In can be easily verified that all the a_{n} with n odd are nul, so that \lambda(*) is an even function. The coefiicient with n even are computed recursively starting from a_{0} which is arbitrary...

    \displaystyle a_{0}= -4\ (1+\nu)\ a_{2} \implies a_{2} = - \frac{a_{0}}{4\ (1+\nu)}

    \displaystyle a_{2}= -4\ 2\ \ (2+\nu)\ a_{4} \implies a_{4} = - \frac{a_{2}}{4\ 2\  (2+\nu)}

    ...

    \displaystyle a_{2n}= -4\ (n+1)\ (n+1+\nu)\ a_{2n+2} \implies a_{2 (n+1)} = - \frac{a_{2n}}{4\ (n+1)\ (n+1+\nu)} (5)

    ... so that we can write...

    \displaystyle y(x)= a_{0}\ x^{\nu}\ \{1+ \sum_{n=1}^{\infty} (-1)^{n} \frac{(\frac{x}{2})^{2n}}{n!\ (1+\nu)\ (2+\nu) ... (n+\nu)}\} (6)

    Some considerations...

    a) in (6) there is a term a_{0} and that's justified by the fact that a solution multiplied by a constant is also a solution...

    b) in (1) there is \nu^{2} and that means two different solution like (6), one with \nu >0 and the other with \nu<0...

    The b) permits us to whrite...

    \displaystyle y(x)= c_{1}\ J_{\nu} (x) + c_{2}\ J_{-\nu} (x) (7)

    At this point we have to identify a_{0}. At this scope we define the 'factorial function' as ...

    \displaystyle x!= \int_{0}^{\infty} t^{x}\ e^{-t}\ dt (8)

    Among the properties of this function the following is very important...

    \displaystyle (1+x)!= (1+x)\ x! (9)

    ... so that we can set a_{0}= \frac{1}{\nu!} for \nu >0 and a_{0}= \frac{1}{(-\nu)!} for \nu <0 obtaining...

    \displaystyle J_{\nu} = x^{\nu}\ \sum_{n=0}^{\infty} (-1)^{n} \frac{(\frac{x}{2})^{2n}}{n!\ (n+\nu)!}

    \displaystyle J_{-\nu} = x^{-\nu}\ \sum_{n=0}^{\infty} (-1)^{n} \frac{(\frac{x}{2})^{2n}}{n!\ (n-\nu)!} (10)

    It is important to remember that all that is true only if \nu isn't an integer. Finally an example of the J functions for \nu=\frac{1}{2}...



    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. infinit number of solutions (lenear akgebra)
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: April 24th 2011, 02:56 PM
  2. second order lenear equation 10
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: November 27th 2010, 06:28 AM
  3. lenear dependance proof
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 8th 2010, 02:39 AM
  4. Linear Dependance
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: January 22nd 2010, 01:44 AM
  5. is there lenear euqation that..
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: November 1st 2009, 10:28 PM

Search Tags


/mathhelpforum @mathhelpforum