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Question

Find u(x, t) for the string of length :=1 and c²=1 when the initial velocity is zero and the inital deflection with small k(say, 0.01) is as follows. Sketch or graph u(x, t). kx²(1-x)

Solution

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Recall that the solution of the wave equation

2ut2=c22ux2\frac{\partial^2 u}{\partial t^2}=c^2\frac{\partial^2u}{\partial x^2}

with

u(0,t)=0u(L,t)=0 for all t0u(x,0)=f(x)ut(x,0)=g(x) for all 0xL\begin{align*} u(0,t)&=0 &&u(L,t)=0 &&\text{ for all } t\geqq 0 \\[7pt] u(x,0)&=f(x) &&u_t(x,0)=g(x) &&\text{ for all } 0\leqq x \leqq L \end{align*}

is given by

u(x,t)=n=1(Bncos(λnt)+Bnsin(λnt))sin(nπxL)u(x,t)= \sum^{ \infty }_{n=1}\left( B_n\cos(\lambda_n t)+B_n^*\sin(\lambda_nt) \right) \sin( \frac{n\pi x}{L} )

where λn=cnπL\lambda_n= \frac{cn\pi}{L}, Bn=2L0Lf(x)sin(nπxL)dxB_n= \frac{2}{L} \int_{0}^{L} f(x)\sin( \frac{n\pi x}{L} ) \,dx and

Bn=2cnπ0Lg(x)sin(nπxL)dxB_n^*= \frac{2}{cn\pi} \int_{0}^{L} g(x)\sin( \frac{n\pi x}{L} ) \,dx

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