(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Consider an elastic string of length L whose ends are held fixed. The string is set in motion with no initial velocity from an initial position u(x, 0) = f (x). Assume that the parameter alpha = 1. Find the displacement u(x,t) for the given initial position f(x)

3. The attempt at a solution

[tex]

u(x,t) = X(x)T(t) \Rightarrow \frac{X''}{X} = \frac{T''}{T} = -\lambda\\

X = C_1 cos(\sqrt{\lambda}x) + C_2 sin(\sqrt{\lambda}x)\\X(0) = C_1 = 0\\X(L) = C_2 sin(\sqrt{\lambda}L) = 0 \Rightarrow \sqrt{\lambda}L = n\pi \forall n\geq 1 \Rightarrow \lambda = (\frac{n \pi}{L})^2\\

T''+ \lambda T = 0 \Rightarrow T = C_1 cos(\frac{n \pi t}{L}) + C_2 sin(\frac{n \pi t}{L})\\

\Rightarrow u_n(x,t) = A_n cos(\frac{n \pi t}{L})sin(\frac{n \pi x}{L}) + B_n sin(\frac{n \pi t}{L}) sin(\frac{n \pi x}{L})\\

\Rightarrow u(x,t) = \sum _1 ^\infty A_n cos(\frac{n \pi t}{L})sin(\frac{n \pi x}{L}) + B_n sin(\frac{n \pi t}{L}) sin(\frac{n \pi x}{L})\\

u(x,0) = 1 = \sum _1 ^\infty A_n cos(\frac{n \pi (0)}{L})sin(\frac{n \pi x}{L}) + B_n sin(\frac{n \pi (0)}{L}) sin(\frac{n \pi x}{L}) = \sum _1 ^\infty A_n sin(\frac{n \pi x}{L})\\

\frac{\partial u(x,0)}{\partial t} = \sum _1 ^\infty B_n \frac{n \pi}{L} sin(\frac{n \pi x}{L}) = 0\\

[/tex]

So this forces all of the B_n to zero. We now solve the A_n

[tex]

A_n = 2 \int _0 ^L f(x)sin(\frac{n \pi x}{L}) dx = 2 \int_{\frac{L}{2} - 1} ^{\frac{L}{2} + 1} sin(\frac{n \pi x}{L}) dx = \frac{4Lsin(\frac{n \pi}{2})sin(\frac{n \pi}{L} )}{\pi n}

[/tex]Thus, my solution should be:

[tex]

u(x,t) = \sum_1 ^\infty \frac{4Lsin(\frac{n \pi}{2})sin(\frac{n \pi}{L} )}{\pi n} sin(\frac{n \pi x}{L})cos(\frac{n \pi t}{L})

[/tex]

However, this answer is off by a factor of L from what the book has. Can anyone help me find my mistake?

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