Answer to Question #92952 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

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 

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