1. ## Boundary Value problem

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:
$y'' = f(x) (0 \leq x \leq 1)$
$y'(0) = 0, y'(1) = 0$
does not in general have a solution.

I used the LDE solution to be $y = Asin(x)+Bcos(x)$.
So, solving for both boundary conditions:
$y'(0) = Acos(0)-Bsin(0) =0$ Therefore $A=0$.
$y'(1) = Acos(1)-Bsin(1) = -Bsin(1) = 0$ Therefore $B=0$
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 $f(x)$ for there to be a solution. If choose the boundary conditions of $0\leq x \leq \pi$, and making the right boundary condition to be $y'(\pi)=0$, I'll get a solution of the form $y=Bcos(x)$.

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

2. 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:
$y'' = f(x) (0 \leq x \leq 1)$
$y'(0) = 0, y'(1) = 0$
does not in general have a solution.

I used the LDE solution to be $y = Asin(x)+Bcos(x)$.
So, solving for both boundary conditions:
$y'(0) = Acos(0)-Bsin(0) =0$ Therefore $A=0$.
$y'(1) = Acos(1)-Bsin(1) = -Bsin(1) = 0$ Therefore $B=0$
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 $f(x)$ for there to be a solution. If choose the boundary conditions of $0\leq x \leq \pi$, and making the right boundary condition to be $y'(\pi)=0$, I'll get a solution of the form $y=Bcos(x)$.

Am I close? Or have I got the wrong idea here.
So you find that:

$y(x)=0$

is a solution to your linear BVP.

Consider:

$y'(x)=\int_0^xf(\xi)\; d\xi$

CB

3. Ah, so you're saying I found a solution to the BVP? Then how would I go about showing this. I have no idea. If I make choose the general solution to be of the form Ax + B, then I'll get A = 0 for both boundary cases. Meaning the answer to the BVP is y(x) = B...

4. Originally Posted by Silverflow
Ah, so you're saying I found a solution to the BVP? Then how would I go about showing this. I have no idea. If I make choose the general solution to be of the form Ax + B, then I'll get A = 0 for both boundary cases. Meaning the answer to the BVP is y(x) = B...
That does not make much sense to me. Forget the linear case and look at what i SAID IN THE SECOND PART OF MY LAST POST.

cb

5. Oh, I understand now, sorry. Thanks for that.

6. Sorry, for this question, we know that there is a solution to the nonhomogeneous BVP iff the equation given by CaptainBlack is equal to 0 but integrating from 0 to 1. But how does this help?

7. Originally Posted by TheFirstOrder
Sorry, for this question, we know that there is a solution to the nonhomogeneous BVP iff the equation given by CaptainBlack is equal to 0 but integrating from 0 to 1. But how does this help?
The boundary values are $y'(0)=0$ and $y'(1)=0$ and

$y'(x)=\int_0^x f(\xi)\; d\xi \ \ \ \ \ (1)$

is the first derivative of a solution satisfying the $y'(0)=0$ condition, the other condition then requires that:

$y'(1)=\int_0^1 f(\xi)\; d\xi=0$

So the boundary value problem cannot have a solution if

$\int_0^1 f(\xi)\; d\xi \ne 0$

If this integral is zero then the solution is:

$y(x)=\int_0^x \int_0^u f(\xi)\; d\xi \; du + c$

We note that $y'(1)=0$ is not a condition independed of $y'(0)=0$, which is why in solution to the boundary value problem we still have an arbitary constant.

CB

8. Thanks! Makes sense now