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Math Help - Pointers and help - solving an equation

  1. #1
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    Pointers and help - solving an equation

    I want to solve a differential equation, the exercise being, find two different independent solutions. (I don't know the proper English terms for this.)

    z^2 w'' - (z^2 + z)w' + (z+1)w = 0

    I start out with \displaystyle w = \sum_0^\infty a_n z^{n+m}. Differentiate to match each term in the equation and shove it in.

    \displaystyle \sum_0^\infty (n+m)(n+m-1)a_n z^{n+m} - \sum_0^\infty (n+m)a_n z^{n+m+1}-\sum_0^\infty (n+m)a_n z^{n+m} + \sum_0^\infty a_n z^{n+m+1} + \sum_0^\infty a_n z^{n+m} = 0

    \displaystyle \sum_0^\infty [(n+m)(n+m-1) -(n+m) - 1]a_n z^{n+m} = \sum_0^\infty [(n+m) -1]a_n  z^{n+m+1}

    Recursive series.

    \displaystyle (m+n-1)^2 a_n = (n+m -2)a_{n-1}

    Now I'm starting to get wobbly legs. I set n = 0, a_0 = 1 to see if I can pinpoint m. (m-1)^2 = 0 \implies m = \pm 1

    m = -1 \implies a_n = \frac{(n-3)}{(n-2)^2} a_{n-1} results in a series a_n = \infty, n \geq 2. Can I render a solution from that even if it diverges?

    \displaystyle m = 1 \implies a_n = \frac{(n-1)}{n^2} a_{n-1} results in a series a_n = 0, n \geq 1 that would imply that one solution is w = z. Or..?

    Now I've sort of played all my cards here. What if I state a solution w = zu and solve the equation..? Or is there a simpler way to render an other solution here?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Given the linear second order 'incomplete' DE...

    \displaystyle z^{2}\ w^{''} -(z + z^{2})\ w^{'} + (1+z)\ w=0 (1)

    ... we are able to solve it from the knowledge of two independent solutions u and v. Because u and v are solutions of (1) You can write...

    \displaystyle z^{2}\ u^{''} -(z + z^{2})\ u^{'} + (1+z)\ u=0

    \displaystyle z^{2}\ v^{''} -(z + z^{2})\ v^{'} + (1+z)\ v=0 (2)

    ... so that subtracting with some easy step You obtain...

    \displaystyle (v\ u^{''} - u\ v^{''}) - \frac{1+z}{z}\ (v\ u^{'} - u\ v^{'}) =0 (3)

    Now setting  \varphi = v\ u^{'} - u\ v^{'} the (3) becomes...

    \displaystyle \varphi^{'} = \frac{1+z}{z}\ \varphi (4)

    ... the solution of which is very easy...

    \varphi= c_{1}\ z\ e^{z} (5)

    Deviding both terms of (3) by v^{2} and taking into account (5) You obtain...

    \displaystyle \frac{v\ u^{''} - u\ v^{''}}{v^{2}} = \frac{d}{dz} \frac{u}{v} = c_{1}\ \frac{z\ e^{z}}{v^{2}} \implies \frac{u}{v}= c_{2} + c_{1} \int \frac{z\ e^{z}}{v^{2}}\ dz \implies

    \displaystyle \implies u= c_{2}\ v + c_{1}\ v\ \int \frac{z\ e^{z}}{v^{2}}\ dz (6)

    Now it is easy to verify that v=z is solution of (1) so that the general solution of (1) is...

    \displaystyle w(z)= c_{2}\ z + c_{1}\ z\ \int \frac{e^{z}}{z}\ dz (7)

    The integral is (7) is a non elementary function known as 'exponential integral function'...



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  3. #3
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    Ok, the part where you solve the de i did slightly different.

    With my approach, w = zu I get u''+ (\frac{1}{z} + z) u' = 0. Multiply with e^{ln z - z} = z e ^{-z} to get u' = C z^{-1} e ^z.

    C = 1 should work as I need one solution, I already have w = z.

    \displaystyle u = \int \frac{e^z}{z} dz = \int \sum_{n=0}^\infty \frac{z^{n-1}}{n!} dz = \log z + \sum_{n=1}^\infty \frac{z^{n}}{n*n!} dz

    w_1 = z

    \displaystyle w_2 = zu = z \log z + \sum_{n=1}^\infty \frac{z^{n+1}}{n*n!} dz

    Edit: Back to my questions, is there a simpler way to get a solution, using the recursive formula? Can I render a solution if it diverges?
    Last edited by liquidFuzz; December 20th 2010 at 10:42 PM.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by liquidFuzz View Post
    ...

    \displaystyle u = \int \frac{e^z}{z} dz = \int \sum_{n=0}^\infty \frac{z^{n-1}}{n!} dz = \log z + \sum_{n=1}^\infty \frac{z^{n}}{n*n!} dz
    The definition of 'exponential integral function' is a little different as You can see in...

    Exponential Integral -- from Wolfram MathWorld

    Pratically is...

    \displaystyle Ei(z)= \ln z + \gamma + \sum_{n=1}^{\infty} \frac{z^{n}}{n\ n!} (1)

    ... where the constant \gamma is the Euler's constant...

    \displaystyle \gamma= \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k} - \ln n (2)

    That is not a marginal detail because the 'particular solution' is given by the product...

    \displaystyle z\ Ei(z)= z\ \ln z + \gamma\ z + \sum_{n=2}^{\infty} \frac{z^{n}}{(n-1)\ (n-1)!} (3)



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