A note on a Fourier sine series solution to a heat flow problem

A heat flow problem led to the following general series solution:

$u(x, t) = \sum_{n=0}^{\infty} b_n \exp\bigg[ -\bigg(\frac{(2n+1)\pi}{2L} \bigg)^2 kt \bigg] \sin\bigg(\frac{(2n+1)\pi}{2L}x\bigg) \qquad \qquad \qquad \qquad \qquad (1)$

An initial condition at $t = 0$ required $u(x, 0) = u_0$ , so it was necessary to find the coefficients $b_n$ in the following sine series:

$u_0 = \sum_{n=0}^{\infty} b_n \sin\bigg(\frac{(2n+1)\pi}{2L}x\bigg) \qquad \qquad \qquad \qquad \qquad (2)$

The sine functions here involve coefficients of $x$ of the form $\frac{(2n+1)\pi}{2L}$ rather than the usual $\frac{n \pi}{L}$ . They have period $4L$ and by looking at graphs it is easy to convince oneself that

$\int_{-2L}^{2L} \sin^2\bigg(\frac{(2n+1)\pi}{2L}x\bigg) dx = \int_{-2L}^{2L} \cos^2\bigg(\frac{(2n+1)\pi}{2L}x\bigg) dx$

$\int_{-2L}^{2L} \sin^2\bigg(\frac{(2n+1)\pi}{2L}x\bigg) dx$

$= \frac{1}{2} \int_{-2L}^{2L} \bigg[\sin^2\bigg(\frac{(2n+1)\pi}{2L}x\bigg) + \cos^2\bigg(\frac{(2n+1)\pi}{2L}x\bigg) \bigg] dx$

$= \frac{1}{2} \cdot 4L = 2L$

Also, orthogonality can be demonstrated in the usual ways (e.g., using the exponential form of the sine function):

$\int_{-2L}^{2L} \sin\bigg(\frac{(2n+1)\pi}{2L}x\bigg) \cdot \sin\bigg(\frac{(2m+1)\pi}{2L}x \bigg) dx = 0$

for $n \neq m$ . In fact, the functions $\sin\bigg(\frac{(2n+1)\pi}{2L}x \bigg)$ for $n = 0, 1, 2, \ldots$ form a complete, orthogonal set of basis functions and we can expand $u_0$ in this basis. To this end, we need to create an odd function involving $u_0$ of the form

$f(x) = \left\{ \begin{array}{rl} u_0 & \text{for } 0 < x < 2L \\ -u_0 & \text{for } -2L < x< 0 \end{array} \right.$

The coefficients of the Fourier sine series are then obtained as

$b_n = \frac{1}{2L} \int_{-2L}^{2L} f(x)\sin\bigg(\frac{(2n+1)\pi}{2L}x \bigg) dx = \frac{1}{L} \int_0^{2L} u_0 \sin\bigg(\frac{(2n+1)\pi}{2L}x \bigg) dx$

$= \frac{u_0}{L} \bigg[-\frac{2L}{(2n+1)\pi}\cos\bigg(\frac{(2n+1)\pi}{2L}x \bigg) \bigg]_0^{2L} = \frac{-2u_0}{(2n+1)\pi} \big(\cos((2n+1)\pi)-1\big)$

$= \frac{-2u_0}{(2n+1)\pi} \cdot (-2) = \frac{4u_0}{(2n+1)\pi}$

The full solution of the heat flow problem is thus

$u(x, t) = \frac{4u_0}{\pi}\sum_{n=0}^{\infty} \frac{1}{2n+1} \exp\bigg[ -\bigg(\frac{(2n+1)\pi}{2L} \bigg)^2 kt \bigg] \sin\bigg(\frac{(2n+1)\pi}{2L}x\bigg)$

A note on a Fourier sine series solution to a heat flow problem

Published by Dr Christian P. H. Salas

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