Orthogonality Proof for R(x)*Wn*Yn(x) = -T(d^2)[Yn(x)]/dx

[sachi]

We have the eigenvalue equation
R(x)*(Wn*2)*Yn(x) = -T (d^2)[Yn(x)]/dx

T the tension is a constant

R(x) is the weight function, Wn is the eigenvalue of the nth normal mode and Yn is the nth normal mode eigenfunction. We have to show that the integral of (Yn(x)*Ym(x)*R(x)) between the limits of 0 and L is zero. we are also told that the streched string in question is held under tension T between rigid supports at x=0 and x=L.
I can complete most of the proof by intregation by parts, but I'm having trouble making a couple of terms disappear to make the proof work.
We end up with
[Yn'*Ym] - [Ym'Yn] having to equal zero, where the square brackets indicate that we are subtracting the value of the bracketed expression at L from its value at 0, and ' indicate a derivative wrt x. We are not explicitly told that Y or Y' goes to zero at 0 or L, only that the supports are "rigid". My argument is as follows:let us suppose that Y and it's first derivative are continuous at the boundaries. Therefore their values at 0 and L will be the same for all normal modes, therefore the expression above goes to zero as required. The problem is that I'm not sure about the justification for the first derivative to be zero (also, will all derivatives be zero? I suspect this will be so for a physically sensible model of a string.)

 

[lippyka]

The justification for the first derivative to be zero at the boundaries is that the assumption of "rigid supports" implies no displacement, and therefore no rate of change in displacement (or velocity) at the boundaries. Therefore all derivatives at the boundaries should be zero, not just the first derivative. Furthermore, if Y and its derivatives are continuous, then it follows that Y and its derivatives must also be zero at the boundaries.