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Thread: power series solution of second order ode

  1. #1
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    power series solution of second order ode

    I am supposed to find a series solution for y''=y^2.

    My problem is with the y^2. I've started with the following but then I get stuck:

    y= sum(anx^n) and y''=sum(n*n-1)anx^n-2 and then I put these into the equation.

    With the y^2 I end up with the multiplication of two series...I don't know if this is correct, and I don't know how to deal with the product of two series....
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  2. #2
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    Quote Originally Posted by sarah232 View Post
    I am supposed to find a series solution for y''=y^2.

    My problem is with the y^2. I've started with the following but then I get stuck:

    y= sum(anx^n) and y''=sum(n*n-1)anx^n-2 and then I put these into the equation.

    With the y^2 I end up with the multiplication of two series...I don't know if this is correct, and I don't know how to deal with the product of two series....
    $\displaystyle y''=\sum_{n \geq 2}n(n-1)a_nx^{n-2}=\sum_{n \geq 0} (n+1)(n+2)a_{n+2}x^n.$ we also have $\displaystyle y^2=\sum_{n \geq 0}b_n x^n,$ where $\displaystyle b_n = \sum_{k=0}^n a_k a_{n-k}.$ so you must have: $\displaystyle (n+1)(n+2)a_{n+2}=\sum_{k=0}^n a_k a_{n-k}. \ \ \ (*)$

    so using $\displaystyle (*)$ we need to find all $\displaystyle a_n, \ n \geq 2,$ in terms of $\displaystyle a_0,a_1.$ for n = 0, $\displaystyle (*)$ gives us: $\displaystyle 2a_2=a_0^2$ and thus $\displaystyle a_2 = \frac{a_0^2}{2}$ and for n = 1, $\displaystyle (*)$ gives us $\displaystyle 3a_3 = a_0a_1$ and so $\displaystyle a_3 = \frac{a_0a_1}{3}.$

    for n = 2, you'll get $\displaystyle a_4 = \frac{a_0^3 + a_1^2}{12},$ etc. things get much messier for larger values of $\displaystyle n.$
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    wow...that was fast. I've been working on this problem for 2 days. I have more questions if you are willing!

    I am supposed to show that the solution of x' = 1 + z^4 + x^2
    with x(0)=0 cannot be bounded for all z. I'm guessing I'm supposed to show that it doesn't satisfy a Lipschitz condition but I have no idea how to do that.

    Thanks again...where have you been all my life.
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    Quote Originally Posted by sarah232 View Post
    wow...that was fast. I've been working on this problem for 2 days. I have more questions if you are willing!

    I am supposed to show that the solution of x' = 1 + z^4 + x^2
    with x(0)=0 cannot be bounded for all z. I'm guessing I'm supposed to show that it doesn't satisfy a Lipschitz condition but I have no idea how to do that.

    Thanks again...where have you been all my life.
    is $\displaystyle x$ assumed to be a function of $\displaystyle z$?
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    I guess so. It doesn't say anything else in the question but the theory says I'm dealing with differential equations of the form x'=f(x,z) with initial point x(zo)=x0
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    Hello,
    Quote Originally Posted by sarah232 View Post
    wow...that was fast. I've been working on this problem for 2 days. I have more questions if you are willing!

    I am supposed to show that the solution of x' = 1 + z^4 + x^2
    with x(0)=0 cannot be bounded for all z. I'm guessing I'm supposed to show that it doesn't satisfy a Lipschitz condition but I have no idea how to do that.

    Thanks again...where have you been all my life.
    It is better to ask new questions in new threads...

    Anyway, let f(z,x) such that : $\displaystyle x'=f(z,x)=1+z^4+x^2$
    Now you have to prove that f is Lipschitz wrt to x, uniformly to z.

    So let's consider x and y. Does there exist a constant C such that $\displaystyle |f(z,x)-f(z,y)|\leq C \cdot |x-y|$ ?

    From the mean value theorem, we know that there exists $\displaystyle \xi \in \mathbb{R}$ such that :
    $\displaystyle |f(z,x)-f(z,y)| = \left|\frac{\partial f(z,\xi)}{\partial \xi}\right| \cdot |x-y|$

    So if you can't bound that partial derivative, then it's not a Lipschitz function.

    $\displaystyle \frac{\partial f(z,\xi)}{\partial \xi}=2 \xi$

    And over $\displaystyle \mathbb{R}$, this is not bounded.
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