A string of length l has initial displacement ${\mathbf{\text{y}}}_{\mathbf{\text{0}}}\mathbf{=}\mathbf{\text{x}}\mathbf{(}\mathbf{\text{l}}\mathbf{-}\mathbf{\text{x}}\mathbf{)}$.Find the displacement as a function of x and t.

The initial displacement has been given as$y_0=x(1-x)$.

Laplace’s equation in cylindrical coordinates is,

$\begin{array}{c}{\mathbf{\nabla }}^{\mathbf{2}}\mathbf{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\left(}\mathbf{r}\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{r}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{\theta }}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{z}}^{\mathbf{2}}}\\ \mathbf{=}\mathbf{0}\end{array}$

And to separate the variable the solution assumed is of the form$\mathit{u}\mathbf{=}\mathit{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{\Theta }\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathit{Z}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$.

The one-dimensional wave equation.

$\frac{{\partial }^{2}y}{\partial x^2}=\frac{1}{v^2}\frac{{\partial }^{2}y}{\partial t^2}$ ….. (1)

Separate the variables so substitute$y=X\left(x\right)T\left(t\right)$into equation (1).

$\begin{array}{c}\frac{1}{X}\frac{d^{2}X}{d{x}^2}=\frac{1}{v^2}\frac{1}{T}\frac{d^{2}T}{d{t}^2}\\ =-k^2\end{array}$

Solve further and you have,

$\begin{array}{l}{X}^{\text{'}\text{'}}+k^{2}X=0\\ \ddot{T}+k^2{v}^{2}T=0\end{array}$

The solutions of the two previous equations.

$X=\left\{\begin{array}{l}\sin \left(kx\right)\\ \cos \left(kx\right)\end{array}\right.$

$T=\left\{\begin{array}{l}\sin \left(kvt\right)\\ \cos \left(kvt\right)\end{array}\right.$

Boundary conditions:

Since the string is fastened at x = 0 and x = l, there must be y = 0 for these values for x and all t. This means that only the role="math" localid="1664350634952" $\sin \left(kx\right)$factors is needed, and also select k so that $\sin \left(kl\right)=0$, so $k=\raisebox{1ex}{$n\pi $}\!\left/ \!\raisebox{-1ex}{$l$}\right.$. Given the initial conditions the combination of solutions is found.

$y=\sin (n\pi x/l)\cos (n\pi vt/l)$

So, given these basis functions, write the solution in the form of a series as below.

$y=\sum _{n=1}^{\infty }{b}_n\sin (n\pi x/l)\cos (n\pi vt/l)$

The coefficients $b_n$are to be determined so that at t = 0 you have,

$y_0=x(l-x)$.

Therefore,

$\begin{array}{c}{b}_n=\frac{2}{l}\underset{0}{\overset{l}{\int }}x(l-x)\sin (n\text{pix}/l)dx\\ =\frac{2}{{\pi }^3{n}^3}[-l^2{(-1)}^n+2{\left(l\right)}^2]\\ =\frac{8l^2}{{\left(\pi n\right)}^3}\text{;ifnisodd}\end{array}$

Now, replace $b_n$to find the displacement.

$y(x,t)=\frac{8l^2}{{\left(\pi n\right)}^3}\sum _{\text{nodd}}\frac{1}{n^3}\sin (n\pi x/l)\cos (n\pi vt/l)$

Hence, this is the required the solution.