Question

Find the deflection of the fixed end vibrating string of unit length corresponding to zero initial deflection and u(x) given below as the initial velocity.

x , 0 ≤ x < 1/2
u(x) =
1-x , 1/2 ≤ x < 1

Answer 1

A vibrating string of unit length {L = 1 and with fixed ends} satisfies the following:

$$PDE: \qquad \frac{\partial^2u}{\partial t^2}=c^2 \frac{\partial^2u}{\partial x^2}$$ $$BC: \qquad u(0,t)=u(1,t)=0$$ $$IC: \qquad u(x,0)=0, \\ u_t(x,0)=\begin{cases} x &\text{, } 0<=x<1/2 \\ 1-x &\text{, } 1/2<=x<1 \end{cases}$$

where c - const (is the speed of propagation of the wave in the string).

Separation of variables for this equation allows you to get a solution:

$$u(x,t)=\underset{n=1}{\overset{\infin}{\sum}}[A_n cos(n\pi ct)sin(n\pi x)+B_n sin(n\pi ct)sin(n\pi x)] \qquad (1)$$

where

$$A_n=2\underset{0}{\overset{1}{\int}}u(x,0) sin(n\pi x) dx=0 \qquad (2)$$

$$B_n=\frac{2}{n\pi c}\underset{0}{\overset{1}{\int}}u_t(x,0) sin(n\pi x) dx= \\ \frac{2}{n\pi c} \Bigg(\underset{0}{\overset{1/2}{\int}}x sin(n\pi x) dx+\underset{1/2}{\overset{1}{\int}}(1-x) sin(n\pi x) dx \Bigg)=\\ \frac{2}{n\pi c} \Bigg( (\frac{sin(n\pi x)-(n\pi x)cos(n\pi x)}{n^2\pi ^2})\underset{0}{\overset{1/2}{|}}-\frac{cos(n\pi x)}{n\pi}\underset{1/2}{\overset{1}{|}} -\\ (\frac{sin(n\pi x)-(n\pi x)cos(n\pi x)}{n^2\pi ^2})\underset{1/2}{\overset{1}{|}} \Bigg)=\\ \frac{2}{n^3\pi^3 c} \Bigg( 2sin(\frac{n\pi}{2})-sin(n\pi)\Bigg) =\\ \frac{4}{n^3\pi^3 c} sin(\frac{n\pi}{2}), \space n=1,2,3,... \qquad (3)$$

from (1),(2) and (3):

$$u(x,t)=\underset{n=1}{\overset{\infin}{\sum}} \frac{4}{n^3\pi^3 c} sin(\frac{n\pi}{2}) sin(n\pi ct)sin(n\pi x), \space n=1,2,3,...$$