Answer to Question #286502 in Differential Equations for Ashok

Question

A taut string of length 2l , fastened at both ends, is disturbed from its position of equilibrium by imparting to each of its points on an initial velocity of magnitude k(2lx-x^2). Find the displacement function y(x,t)

Answer 1

The displacement $y(x,t)$ of any point '$x$ ' of the string at any time '$t$ ' is given by

$\frac{\partial²y}{\partial t²}=a²\frac{\partial²y}{\partial x^{2}}$ $---(1)$

We have to solve equation (1) satisfying the following boundary conditions.

$y(0,t)=0,$ $for$ $t≥0$ $----(2)$

$y(2l,t)=0,$ $for$ $t≥0$ $----(3)$

$y(x,0)=0,$ $for$ $0≤x≤2l$ $----(4)$

$\frac{\delta y}{\delta t}(x,0)=k(2lx-x²),$ $for$ $0≤x≤2l$ $----(5)$

The suitable solution of Eq (1), consistent with the vibration of the string,is

$y(x,t)=(A\cos px + B\sin px)$

$(C\cos pat+D\sin pat)$ $----(6)$

Using boundary conditions (2) in (6), we have

$A(C\cos pat+D\sin pat)=0$ $for$ $all$ $t≥0$

$A=0$

Using boundary conditions (3) in (6), we have

$B\sin 2lp(C\cos pat+D\sin pat)=0$

$for$ $all$ $t≥0$

Either $B=0$ or $\sin 2lp=0$

If we assume that $B=0,$ we get a trivial solution.

$\sin 2lp=0$

$2lp=nπ$

$p=\frac{nπ}{2l}$

Where $n=0,1,2,...\infty$

Using boundary conditions (4) in (6), we have

$B\sin px.C=0$

$for$ $0≤x≤2l$

As $B≠0,$ we get $C=0$

Using these values of $A,$ $p,$ $C$ in (6), the solution reduces to

$y(x,t)=k\sin \frac{nπ}{2l}\sin \frac{nπat}{2l}$

$----(7)$

where $n=0,1,2,3...\infty$

The most general solution of Eq.(1) is

$y(x,t)=\sum_{n=1}^{\infty}\lambda_{n}\sin \frac{nπ}{2l}\cos \frac{nπat}{2l}$ $----(8)$

Differentiating both sides of (8) partially with respect to t, we have

$\frac{\delta y}{\delta t} (x,t)=$ $\sum_{n=1}^{\infty}(\frac{nπa}{2l}.\lambda_{n})\sin \frac{nπx}{2l}\cos \frac{nπat}{2l}$ $----(9)$

Using boundary condition (5) in (9), we have

$\sum_{n=1}^{\infty}(\frac{nπa}{2l}\lambda_{n})\sin \frac{nπx}{2l}=k(2lx-x²)$

$for$ $0≤x≤2l$

$=\sum_{n=1}^{\infty}b_{n}\sin \frac{nπx}{2l}$

Which is Fourier half-range sine series of $k(2lx-x²)$ in $(0,2l).$

Comparing like terms,we get

$\frac{nπa}{2l}.\lambda_{n}=b_n=$

$\frac{2}{2l}\int_{0}^{2l}f(x)\sin \frac{nπx}{l}dx,$ by Euler's formula

$=\frac{2}{2l}\int_{0}^{2l}k(2lx-x²)\sin \frac{nπx}{2l}dx$

$=\frac{2k}{n\pi a}[\,(2lx-x^{2})[\,\frac{-cos\frac{n\pi x}{2l}}{\frac{n\pi}{2l}}]\,-(2l-2x[\,\frac{-sin\frac{n\pi x}{2l}}{\frac{n^{2}\pi^{2}}{4l^{2}}}]\,\\+(-2)[\,\frac{cos\frac{n\pi x}{2l}}{\frac{n^{3}\pi^{3}}{8l^{3}}}]\,]\,_{0}^{2l}$

$=\frac{32kl^{3}}{n^{4}\pi^{4}a}[1-(-1)^{n}]$

$=\begin{cases}0,\hspace{0.4cm}if\hspace{0.1cm} n \hspace{0.1cm}is\hspace{0.1cm} even\\\\\\ \frac{32kl^{3}}{n^{4}\pi^{4} a},\hspace{0.4cm}if\hspace{0.1cm} n \hspace{0.1cm}is\hspace{0.1cm} odd\end{cases}$

Using this value of $\lambda_n$ in (8), the required solution is

$y(x,t)=$ $\frac{64kl³}{π⁴a}\sum_{n=1}^{\infty}\frac{1}{(2n-1)⁴}\sin \frac{(2n-1)πx}{2l}\cos \frac{(2n-1)πat}{2l}$