# Thread: IVP and Boundary Value Problem

1. ## IVP and Boundary Value Problem

Find an interval centered about x=0 for which the given IVP has a unique solution:
(x-2)y''+3y=x, y(0)=0, y'(0)=1

what exactly is the question asking?? Do I need to solve the IVP to find the boundaries?

2. Originally Posted by daskywalker
Find an interval centered about x=0 for which the given IVP has a unique solution:
(x-2)y''+3y=x, y(0)=0, y'(0)=1

what exactly is the question asking?? Do I need to solve the IVP to find the boundaries?
Well, you could solve it then prove that this solution is unique (not too difficult, I think, since it's a linear equation). Or you could use the existence and uniqueness theorem for linear equations that says that if $\displaystyle y''=P(x)+Q(x)y+R(x)y'$ where $\displaystyle P,Q,R$ are continous functions of $\displaystyle x$ on $\displaystyle a<x<b$ then there is a unique solution to this equation $\displaystyle ya,b) \rightarrow \mathbb{R}$ with the specified initial conditions on a point.

3. If we suppose that the solution we are searching is analitic in $\displaystyle x=0$ we can write...

$\displaystyle y(x)= \sum_{n=0}^{\infty} a_{n}\cdot x^{n}$

$\displaystyle y^{'}(x)= \sum_{n=1}^{\infty} n\cdot a_{n}\cdot x^{n-1}$

$\displaystyle y^{''} (x)= \sum_{n=2}^{\infty} n\cdot (n-1)\cdot a_{n}\cdot x^{n-2}$ (1)

From (1) and the 'initial conditions' we derive that is...

$\displaystyle a_{0}=0$

$\displaystyle a_{1}=1$ (2)

... and the DE can be written as...

$\displaystyle 3\cdot \sum_{n=2}^{\infty} a_{n}\cdot x^{n} + \sum_{n=2}^{\infty} n\cdot (n-1)\cdot a_{n}\cdot x^{n-1} -2\cdot \sum_{n=2}^{\infty} n\cdot (n-1)\cdot a_{n}\cdot x^{n-2} = -2\cdot x$ (3)

If we impose the identity (3) for all $\displaystyle n>1$ we obtain first...

$\displaystyle a_{2}=0$ (4)

... and for $\displaystyle n>2$ ...

$\displaystyle a_{n}= \frac{3\cdot a_{n-2} + (n-1)\cdot (n-2)\cdot a_{n-1}}{2\cdot n\cdot (n-1)}$ (5)

The conlusion is that the solution can be written as...

$\displaystyle y(x)= x + \sum_{n=3}^{\infty} \alpha_{n}\cdot (\frac{x}{2})^{n}$ (6)

... and because the sequence $\displaystyle \alpha_{n}$ in (6) is bounded the series in (6) will converge if $\displaystyle |x|<2$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$