I was following some online lecture relating to an elastic bar with length $L$ that obey the differential equation $\displaystyle \frac{d^{2}u}{dx^{2}} = f(x)$, where $f(x)$ is its own weight or some load. The professor during that online lecture states that $u(0)=0$, $u(L)= 0$ as a boundary condition if the bar is fixed on both ends (this part I got it). But if one end is free then boundary conditions will change to $u''(x) = f(x)$ and $\displaystyle \frac{du(0)}{dx} = 0$ (?). Anyone could tell me why is that the case? How come boundary conditions changes from $u(0)=u(L)=0$ fixed end to $u’(0) = 0 = u(L)$ free end? Thank you!
Boundary conditions of an elastic bar
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ordinary-differential-equations