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Consider a vibrating string with fixed ends, having length $a$, tension $T$, linear density $\sigma$, (longitudinal) coordinate $x$, and transverse displacement $y(x,t)$.

I'm given the following Lagrangian: $$ L = \int_0^a \left[ \frac{\sigma}{2}\left(\frac{\partial y}{\partial t} \right)^2 - \frac{T}{2}\left( \frac{\partial y}{\partial x} \right)^2 \right] dx$$

and, assuming a Fourier series solution of the form: $$y(x,t) = \sqrt{\frac{2}{a}} \sum_{n=1}^\infty \sin\left(\frac{n\pi x}{a}\right) q_n(t)$$

I'm supposed to be able to re-write the Lagrangian as the discrete sum

$$L = \sum_{n=1}^\infty \left[ \frac{\sigma}{2} \dot{q}_n^2 -\frac{T}{2}\left( \frac{n\pi}{a} \right)^2 q_n^2 \right].$$

At first glance, I was expecting a "plug and chug" but realized that I'd have to expand a product of infinite Fourier series to calculate the squared partial derivatives. Should I expand the series, and hope most terms will cancel in the integration? Or is there a shortcut for calculating squares of odd Fourier series like this?

Source: working an old problem set I found online.

1 Answers 1

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The calculation does not require any laborious calculation and expansion. You don't really have to expand the product but just write it in a way that is short but correct. For this, introduce a new summing index, 'k'(say). Then, the squares (of the derivatives of y) are simply the double summation over 'n' and 'k' of sin(npix/a)sin(kpix/a) along with relevant constants and q's as coefficients. Plug them into the Lagrangian, take the summations outside the integral in order to integrate first. Then, due to orthonormality of sine and cosine functions terms, only those terms survive that have k=n. In the end, lagrangian is left with a single summation over 'n' and is the of the required form.