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The problem is this: $\begin{cases} U_t = 3U_{xx}, \quad 0 < x < 2\pi, \\ U(0,t) = U(2\pi,t) = 0, & \\ U(x,0) = 2 \sin x + 5 \sin 3x \end{cases}$ I want to express this as an infinite series but I'm not sure how to express the coefficient. It seems that given its form I shouldn't be too complicated.

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    Fixed it. Thanks2012-10-22

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$U(x,t) = \sum_{n=1}^{\infty}f_n(t) \sin(n x)$ This gives us $\sum_{n=1}^{\infty}f_n'(t) \sin(n x) = -3 \sum_{n=1}^{\infty}n^2 f_n(t) \sin(n x)$ Hence, $f_n'(t) + 3n^2 f_n(t) = 0 \implies f_n(t) = f_n(0) \exp(-3n^2t)$ $f_n(0) = \begin{cases}2 & n=1\\ 5 & n=3 \\ 0 & \text{otherwise}\end{cases}$ Hence, $U(x,t) = 2 \exp(-3t) \sin(x) + 5 \exp(-27t) \sin(3x)$

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    Perfect. This is what I got as well2012-10-22