Answer to Question #205551 in Differential Equations for Kumar Aditya

Question

 A string is stretched and fastened to two points “” a part and the points of the string are given 

   initial velocity v, v =. Find the displacement at a distance at time.

Answer 1

$\text{ The displacement }\mathrm{y}(\mathrm{x}, \mathrm{t}) \text{ is given by the equation }\\ \frac{\delta^{2} y}{\delta t^{2}}=a^{2} \frac{\delta^{2} y}{\delta x^{2}}-----------(1)\\ \text{The suitable solution of }(1) \text{ is given by :}\\ y(x, t)=( Acoslx + Bsinlx )( Ccoslat + Dsinlat )------(2)\\ \text{}\\ \text{The boundary conditions are:}\\ \text{(i) } y(0, t)=0, for t \geq 0\\ \text{(ii) } y(\ell, t)=0, for t \geq 0 .\\ \text{(iii) }y(x, 0)=0 \forall 0<x<i\\ \text{(i v) }\left[\frac{\delta y}{\delta t}\right]_{t=0}=g(x)=\left\{\begin{array}{ll}\frac{C x}{l} & 0<x<l \\ \frac{C(2 l-x)}{l} & l<x<2 l\end{array}\right.\\ \text{}\\ \text{Using }(i) \text{and }(ii) \text{in }(2), \text{we get :}\\ 0=A( Ccoslat +D sinlat ), for all t \geq 0 .\\ \text{Therefore, }\quad A=0\\ \text{Hence equation (2) becomes :}\\ y(x, t)=B sinlx ( Ccoslat + Dsinlat )----------(3)\\ \text{} \\ \text{Using (ii) in (3), we get :}$

$$0=B sinl\ell ( Ccoslat +D sinlat ), \forall t \geq 0, \text{ which gives } l \ell=n \pi\\ \text{Hence, }l=\frac{n \pi}{\ell}, n \text{ being an integer. }\\ \text{Thus }\\ y(x, t)=B \sin \frac{n \pi x}{l}\left[C \cos \frac{n \pi a t}{l}+D \sin \frac{n \pi a t}{l}\right]---------(4)\\ \text{Using (iii) in (4), we get }\\ 0=B \sin \frac{n \pi x}{l} C\\ \text{Therefore, }C=0\\ \text{Hence }y(x, t)=B \sin \frac{n \pi x}{l} \cdot D \sin \frac{n \pi a t}{l}=B_{1} \sin \frac{n \pi x}{l} \cdot D \sin \frac{n \pi a t}{l}\\ \text{Where }B_{1}=B D\\ \text{The most general solution is :}\\ y(x, t)=\sum_{n=1}^{\infty} B_{n} \sin \frac{n \pi x}{l} \cdot \sin \frac{n \pi a t}{l}-------(5)\\ \text{Differentiating (5) partially w.r.t t, we get }\\ \frac{\delta y}{\delta t}=\sum_{n=1}^{\infty} B_{n} \sin \frac{n \pi x}{l} \cos \frac{n \pi a t}{l} \cdot \frac{n \pi a}{l}\\ \text{}\\ \text{ Using condition (iv) in the above equation, we get: }$$

$y_{t}(x, 0)=g(x)=\left\{\begin{array}{ll}\frac{C_{x}}{l} & 0<x<l \\ \frac{2 l-x}{l} & l<x<2 l\end{array}\right.\\ \text{Taking }\frac{C x}{l}, 0<x<l \text{we get: }\\ \frac{C x}{l}=\sum_{n=1}^{\infty} B_{n} \frac{n \pi a}{l} \cdot \sin \frac{n \pi x}{l}\\ i.e B_{n} \frac{n \pi a}{l}=\frac{2}{l} \int_{0}^{1} f(x) \cdot \sin \frac{n \pi x}{l} d x\\ i.e B_{n}=\frac{2}{n \pi a} \int_{0}^{1} f(x) \cdot \sin \frac{n \pi x}{l} d x=\frac{2}{n \pi a} \int_{0}^{1} \frac{C x}{l} \cdot \sin \frac{n \pi x}{l} d x=\frac{2}{n \pi a} \int_{0}^{1} \frac{C x}{l} d\left[\frac{-\cos \frac{n \pi z}{1}}{\frac{\pi T}{T}}\right]\\ =\frac{2}{n \pi a}\left\{\frac{C x}{l} d\left[\frac{-\cos \frac{n \pi}{l}}{\frac{\pi \pi}{T}}\right]-\frac{C}{l}\left[\frac{-\sin \frac{n \pi}{2}}{\frac{n^{2} \pi^{2}}{l}}\right]\right\}=\frac{2}{n \pi a}\left\{\frac{-2 \cos n \pi+2}{\frac{n^{3} \pi^{3}}{1^{3}}}\right\}=\frac{2}{n \pi a} \cdot \frac{2 l^{3}}{n^{3} \pi^{3}}\{1-\cos n \pi\}\\ i.e B_{n}=\frac{4 l^{3}}{n^{4} \pi^{4} a}\left\{1-(-1)^{n}\right\}\\ \text{or } B_{n}=\left\{\begin{array}{ll}\frac{8 l^{3}}{l} & \text { if } n \text { is odd } \\ 0 & \text { if } n \text { is even }\end{array}\right.\\ \therefore \text{ solution is:}\\ y(x, t)=\frac{8 l^{3}}{n^{4} a} \sum_{n=1}^{\infty} \frac{1}{(2 n-1)^{4}} \sin \frac{(2 n-1) \pi a t}{l} \sin \frac{(2 n-1) \pi x}{l}, \forall 0<x<l---> \bold{Answer}\\ \text{Taking } \frac{C(2 l-x)}{l}, l<x<2 l \text{ we get:}$

$\frac{2 C l-C x}{l}=\sum_{n=1}^{\infty} B_{n} \frac{n \pi a}{2 l} \cdot \sin \frac{n \pi x}{2 l}\\ i.e B_{n} \frac{n \pi a}{l}=\frac{2}{l} \int_{l}^{2 l} g(x) \cdot \sin \frac{n \pi x}{2 l} d x\\ i.e B_{n}=\frac{2}{n \pi a} \int_{l}^{2 l} g(x) \cdot \sin \frac{n \pi x}{2 l} d x=\frac{2}{n \pi a} \int_{l}^{2 l} \frac{2 C l-C x}{l} \cdot \sin \frac{n \pi x}{2 l} d x=\\ \frac{2}{n \pi a} \int_{l}^{2 l} \frac{2 C l-C x}{l} d\left[\frac{-\cos \frac{n \pi x}{2}}{\frac{n \pi x}{2 l}}\right]\\ =\frac{2}{n \pi a}\left\{\frac{2 C l-C x}{l} d\left[\frac{-\cos \frac{n \pi z}{2}}{\frac{n \pi}{2}}\right]-\frac{2 C l-C}{l}\left[\frac{-\sin \frac{\pi \pi}{2}}{\frac{n^{2} \pi 2}{22}}\right]\right\}=\frac{2}{n \pi a}\left\{\frac{-2 \cos n \pi+2}{\frac{n^{3} \pi^{3}}{4 !^{3}}}\right\}=\frac{2}{n \pi a} \cdot \frac{8 l^{3}}{n^{3} \pi^{3}}\{1- \cos n \pi\}\\ i.e B_{n}=\frac{16 l^{3}}{n^{4} \pi^{4} a}\left\{1-(-1)^{n}\right\}\\ or\\ B_{n}=\left\{\begin{array}{ll}\frac{32 l^{3}}{l} & \text { if } n \text { is odd } \\ 0 & \text { if } n \text { is even }\end{array}\right.\\ \therefore \text{ solution is :}\\ y(x, t)=\frac{32 l^{3}}{n^{4} a} \sum_{n=1}^{\infty} \frac{1}{(2 n-1)^{4}} \sin \frac{(2 n-1) \pi a t}{l} \sin \frac{(2 n-1) \pi x}{l}, \forall l<x<2 l---> Answer$