# power series solution of second order ode

• Apr 10th 2009, 09:45 AM
sarah232
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....
• Apr 10th 2009, 10:12 AM
NonCommAlg
Quote:

Originally Posted by sarah232
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 $\displaystyle n.$
• Apr 10th 2009, 10:33 AM
sarah232
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.
• Apr 10th 2009, 10:41 AM
NonCommAlg
Quote:

Originally Posted by sarah232
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 $\displaystyle z$?
• Apr 10th 2009, 10:45 AM
sarah232
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
• Apr 10th 2009, 10:46 AM
Moo
Hello,
Quote:

Originally Posted by sarah232
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.

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.