Answer to Question #92950 in Differential Equations for Subhasis Padhy

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):

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