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

Let $w(x,t)$ be the function describing deflection of the string. The function satisfies wave equation

$$w_{tt}=a^2w_{xx}$$

and from the condition of the question we have

$$w(0,t)=w(1,t)=0$$

as the ends are fixed and initial conditions are

$$w(x,0)=0$$

$$w_t(x,0)=u(x)$$

In order to solve the PDE we’ll use separation of variables, so $w(x,t)=X(x)T(t)$. If we put this in the equation, we’ll obtain $\frac{T''}{a^2T}=\frac{X''}{X}=-\lambda$. So we have Sturm-Liouville problem $\begin{cases} X''+\lambda X=0\\ X(0)=X(1)=0 \end{cases}$. The problem has infinitely many non-trivial solutions $X_n=\sin(\pi n x)$ and eigenvalues $\lambda_n=\pi^2n^2$. Then $T_n''+ \lambda_n a^2 T_n=0$ hence $T_n=C_{1n}\cos(\pi n at)+C_{2n}\sin(\pi n a t)$ where $C_{1n}, C_{2n}$ are arbitrary constants. So

$$w(x,t)=\sum_{n=1}^{\infty}T_n(t)X_n(x)=\sum_{n=1}^{\infty}(C_{1n}\cos(\pi n at)+C_{2n}\sin(\pi n a t))\sin(\pi n x)$$

In order to find $C_{1n}, C_{2n}$ we’ll use initial conditions. $w(x,0)=\sum_{n=1}^{\infty}C_{1n}\sin(\pi n x)=0$, so all $C_{1n}=0$

$$w_t(x,0)=\sum_{n=1}^{\infty}C_{2n}\pi n a\sin(\pi n x)=u(x)$$

let’s expand $u(x)$ into Fourier sine series on interval $0<x<1$.

$$u(x)=\sum_{n=1}^{\infty}b_n \sin(\pi n x)$$

$$b_n=2 \int_0^{1} u(x) \sin(\pi n x)\, dx=\frac{4\sin(\frac{\pi n}{2})}{\pi^2 n^2}$$

So

$$u(x)=\sum_{n=1}^{\infty}\frac{4\sin(\frac{\pi n}{2})}{\pi^2 n^2} \sin(\pi n x)$$

and equating coefficients in the sums we obtain $C_{2n}=\frac{4\sin(\frac{\pi n}{2})}{\pi^3n^3a}$. So

$$w(x,t)=\sum_{n=1}^{\infty}\frac{4\sin(\frac{\pi n}{2})}{\pi^3n^3a}\sin(\pi n a t)\sin(\pi n x)$$

or it can be rewritten as 

$$w(x,t)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{4}{\pi^3(2n-1)^3a}\sin(\pi (2n-1) a t)\sin(\pi (2n-1) x)$$