I have solved the following BVP (Border Value Problem):
$$y'''' = -P, y(0) = y(L) = 0, y'(0) = y'(L) = 0$$
Where $L=4 , P=24$
The DE describing it is: $y(x) = -x^2(x-4)^2$
This apparently is intended to model a horizontal beam.
I know that the maximum displacement takes place when $x = L/2$, however, I am unsure of how to prove this.
Any suggestions?
Thanks!
If we integrate the BVP four times, we have:
$$y(x) = -x^4 + c_1 x^3 + c^2 x^2 + c_3 x + c_4$$
Using the ICs, we find:
$$\tag 1 y(x) = -x^2(x-4)^2$$
A plot of $y(x)$ shows:
We now want to find where the maximum displacement takes place, so we take the derivative of $(1)$, set it equal to zero and find the critical points of interest. We have:
$$y'(x) = -4 x \left(x^2-6 x+8\right) = -4 x(x-2) (x-4) = 0 \implies x = 0, 2, 4$$
$x = 0$ and $x = 4$ are the boundary conditions and we have no displacement. At $x = 2$, we have the maximum displacement since this is the maximum of $|y(x)|$, so:
$$L = 4 , x = 2 \implies x = \dfrac{L}{2} = \dfrac{4}{2}$$