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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!

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At a free end there can be no force imparted. You also need to distinguish between a supported end and a fixed end. A supported end forces the end to stay in position, so $u=0$, but u' can be non-zero. A fixed end forces the end to stay in position and the bar does not bend, so u=0, u'=0. As a free end may move, you may have $u \neq 0$, but as there is no force you must have u''=0. u' at a free end is set by the other conditions and can be non-zero (think of a bar with one end fixed and horizontal and the other end free-it will sag, so u' is non-zero at the free end). The two ends of the bar may have different conditions, so apply the correct one at $0$ and $L$.

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    @Heber: good question. It is a typo an you are right $u'(L/2)=0$ in that case. – 2011-05-24