Originally Posted by

**Silverflow** Hi all,

Gotta BVP here that I'd like to run by someone to see if my thinking is correct.

Show that the following boundary value problem:

$\displaystyle y'' = f(x) (0 \leq x \leq 1)$

$\displaystyle y'(0) = 0, y'(1) = 0 $

does not in general have a solution.

I used the LDE solution to be $\displaystyle y = Asin(x)+Bcos(x)$.

So, solving for both boundary conditions:

$\displaystyle y'(0) = Acos(0)-Bsin(0) =0$ Therefore $\displaystyle A=0$.

$\displaystyle y'(1) = Acos(1)-Bsin(1) = -Bsin(1) = 0$ Therefore $\displaystyle B=0$

This, to me, shows that there is no solution of this form to the BVP. Is this sufficient to answer the question?

The next part asks to find a condition of $\displaystyle f(x)$ for there to be a solution. If choose the boundary conditions of $\displaystyle 0\leq x \leq \pi$, and making the right boundary condition to be $\displaystyle y'(\pi)=0$, I'll get a solution of the form $\displaystyle y=Bcos(x)$.

Am I close? Or have I got the wrong idea here.

Thanks for your time!