This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision | Next revision Both sides next revision | ||
|
vpde_lecture25 [2020/03/31 08:49] trinh |
vpde_lecture25 [2020/03/31 08:55] trinh |
||
|---|---|---|---|
| Line 89: | Line 89: | ||
| end | end | ||
| </ | </ | ||
| + | |||
| + | ==== Implementation of the Fourier series ==== | ||
| + | |||
| + | Now we need to look to solve for the coefficients of our series by applying the boundary conditions. We have that | ||
| + | $$ | ||
| + | u(x, t) = \sum_{n=0}^\infty \sin\left(\frac{n\pi x}{L}\right) \left[ A_n \cos\left(\frac{n\pi ct}{L}\right) + B_n \sin\left(\frac{n\pi ct}{L}\right)\right] | ||
| + | $$ | ||
| + | |||
| + | Imposing the initial displacement, | ||
| + | $$ | ||
| + | u_0(x) = \sum_{n=0}^\infty A_n \sin\left(\frac{n\pi x}{L}\right), | ||
| + | $$ | ||
| + | |||
| + | which we recognise as the sine series for the odd $2L$-periodic extension of the function $u_0(x)$ originally defined on $[0, L]$. So the coefficients are (see theorem 12.7) | ||
| + | $$ | ||
| + | A_n = \frac{2}{L} \int_0^L u_0(x) \sin\left(\frac{n\pi x}{L}\right). | ||
| + | $$ | ||
| + | |||
| + | Imposing the initial velocity, we have | ||
| + | $$ | ||
| + | v_0(x) = \sum_{n=1}^\infty \left(\frac{n\pi c}{L}\right)B_n \sin\left(\frac{n\pi x}{L}\right) | ||
| + | $$ | ||
| + | |||
| + | Again we recognise this as the sine series, so we now need to equate | ||
| + | $$ | ||
| + | \left(\frac{n\pi c}{L}\right)B_n = \frac{2}{L} \int_0^L v_0(x) \sin\left(\frac{n\pi x}{L}\right). | ||
| + | $$ | ||