Answer to Question #238079 in Differential Equations for Surya

Question

A String is stretched and fastened to two points l apart. Motion is started by
displacing the string in the form

Answer 1

$The \ displacement \ y(x, t) \ of \ the \ point \ of \ the \ string \ at \ a \\\ distance \ x \ from \ the \ left \ end \ 0 \ at \ time \ t \ is \ given\\ \ by \ the \ equation \frac{\partial^{2} y}{\partial t^{2}}=a^{2} \frac{\partial^{2} y}{\partial x^{2}} \\ Since \ the \ ends \ of \ the \ string \\ \ x=0 \ and \ x=l \ are \ fixed, \ they \ do \ not \ undergo\\ \ any \ displacement \ at \ any \ time.\\ Therefore, y(0, t)=0 \ for \ t \geq 0 \ and \ y(l, t)=0 \ for \ t \geq 0 \\ Since \ the \ string \ is \ released \ from \ rest \ initially, \\\ that \ is, \ at \ t= 0 , the \ initial \ velocity \ of \ every \ point \ of \ the\\ \ string \ in \ the \ y - direction \ is \ zero.\\ Hence, \ \frac{\partial y}{\partial t}(x, 0)=0 , for \ 0 \leq x \leq l\\ \frac{\partial^{2} y}{\partial t^{2}}=a^{2} \frac{\partial^{2} y}{\partial x^{2}}\\ Solving \ the \ PDE \ by \ Separation \ of \ Variables,\\$

$Let y=X(x) T(t) \frac{\partial y}{\partial x}=X^{\prime}(x) T(t) \\ \frac{\partial^{2} y}{\partial x^{2}}=X^{\prime \prime}(x) T(t) \\ \frac{\partial y}{\partial t}=X(x) T^{\prime}(t) \\ \frac{\partial^{2} y}{\partial t^{2}}=X(x) T^{\prime \prime}(t) \\ X(x) T^{\prime \prime}(t)=a^{2} X^{\prime \prime}(x) T(t) \\ \frac{X^{\prime \prime}(x)}{X(x)}=\frac{T^{\prime \prime}(t)}{a^{2} T(t)}=K^{2} \\ \frac{X^{\prime \prime}(x)}{X(x)}=K^{2} \\ X^{\prime \prime}(x)-K^{2} X(x)=0\\ Solution \,of \,X(x) \, is\, X(x)=C_{1} \cos (K x)+C_{2} \sin (K x) \\ \frac{T^{\prime \prime}(t)}{a^{2} T(t)}=K^{2} \\ T^{\prime \prime}(t)-a^{2} K^{2} T(t)=0$

$T(t)=C_{3} \cos (a K t)+C_{4} \sin (a K t)\\ Thus, \ y(x, t)=\left(C_{1} \cos (K x)+C_{2} \sin (K x)\right)\left(C_{3} \cos (a K t)+C_{4} \sin (a K t)\right) \\ By \ initial \ condition, \ \frac{\partial y}{\partial t}(x, 0)=0\\ \frac{\partial y}{\partial t}=\left(C_{1} \cos (K x)+C_{2} \sin (K x)\right)\left(-C_{3} a K \sin (a K t)+C_{4} a K \cos (a K t)\right) \\ \frac{\partial y}{\partial t}(x, 0)=C_{4} a K\left(C_{1} \cos (K x)+C_{2} \sin (K x)\right)=0 \\ \Longrightarrow C_{4}=0\\ By \ the \ initial \ condition \ y(0, t)=0 \\ y(x, t)=\left(C_{1} \cos (K x)+C_{2} \sin (K x)\right)\left(C_{3} \cos (a K t)+C_{4} \sin (a K t)\right)=0 \\ y(0, t)=C_{1}\left(C_{3} \cos (a K t)+C_{4} \sin (a K t)\right)=0 \\ \Longrightarrow C_{1}=0 \\ \therefore y(x, t)=C_{2} \sin (K x) C_{3} \cos (a K t)\\ By \ the \ initial \ condition \ y(l, t)=0 \\ y(l, t)=C_{2} \sin (K l) C_{3} \cos (a K t)=0 \\ \Longrightarrow \sin (K l)=0 \\ \sin (K l)=\sin (n \pi) \quad \text { for } n \geq 1\\$

$K=\frac{n \pi}{l} \\ \therefore y(x, t)=C_{2} C_{3} \sin \left(\frac{a n \pi x}{l}\right) \cos \left(\frac{a n \pi t}{l}\right)=\lambda \sin \left(\frac{a n \pi x}{l}\right) \cos \left(\frac{a n \pi t}{l}\right)\\ The \ general \ solution \ is \ thus,\\ y(x, t)=\sum_{n=1}^{\infty} \lambda_{n} \sin \left(\frac{a n \pi x}{l}\right) \cos \left(\frac{a n \pi t}{l}\right) Where, \ \lambda_{n} \ can \ be \ related \ to \ the \ Fourier \\ Coefficients \ by \ incorporating \ the \ initial \ conditions.\\ y(x, 0)=\sum_{n=1}^{\infty} \lambda_{n} \sin \left(\frac{a n \pi x}{l}\right) \cos \left(\frac{a n \pi \times 0}{l}\right)=\sum_{n=1}^{\infty} \lambda_{n} \sin \left(\frac{a n \pi x}{l}\right) \\ By \ Fourier's \ Half-range \ sine \ series, \ we \ derive \ \\ \lambda_{n}=\frac{2 a}{l} \int_{0}^{\frac{1}{a}} y(x, 0) \sin \left(\frac{a n \pi x}{l}\right) \mathrm{d} x$