Answer to Question #202053 in Differential Equations for Shrenik Gadiya

Question

A string is stretched and fastened to two points  apart. Motion is started by displacing the string in the form  from which it is released at a time , the initial velocity is zero. To find the deflection

Answer 1

$\displaystyle \textsf{The displacement}\,\, y(x,t) \,\, \textsf{of the point of the string}\\ \textsf{at a distance}\,\, x \,\, \textsf{from the left end}\,\, 0\,\, \textsf{at time}\,\, t \\ \textsf{is given by the equation}\\ \frac{\partial^2 y}{\partial t^2} = a^2\frac{\partial^2 y}{\partial x^2}\\ \textsf{Since the ends of the string}\,\, x=0 \,\, \textsf{and}\,\, x = l \,\, \textsf{are fixed,}\\ \textsf{they do not undergo any displacement at any time.}\\ \textsf{Therefore,} y(0,t) = 0 \,\, \textsf{for}\,\, t \geq 0 \,\, \textsf{and} \,\, y(l,t) = 0 \,\, \textsf{for}\,\, t \geq 0\\ \textsf{Since the string is released from rest initially,} \textsf{that is, at}\,\, t=0, \,\, \textsf{the initial velocity of every point}\\ \textsf{of the string in the}\,\, y-\textsf{direction is zero.}\\ \textsf{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} \\ \textsf{Solving the PDE by Seperation of Variables,}\\ \textsf{Let}\,\, y = X(x)T(t)\\ \frac{\partial y}{\partial x} = X'(x)T(t)\\ \frac{\partial^2 y}{\partial x^2} = X''(x)T(t)\\ \frac{\partial y}{\partial t} = X(x)T'(t)\\ \frac{\partial^2 y}{\partial t^2} = X(x)T''(t)\\ X(x)T''(t) = a^2X''(x)T(t)\\ \frac{X''(x)}{X(x)} = \frac{T''(t)}{a^2T(t)} = K^2\\ \frac{X''(x)}{X(x)} = K^2\\ X''(x) - K^2X(x) = 0\\ \textsf{Solution of}\,\, X(x)\,\, \textsf{is}\\ X(x) = C_1\cos(Kx) + C_2\sin(Kx)\\ \frac{T''(t)}{a^2T(t)} = K^2\\ T''(t) - a^2 K^2T(t) = 0\\ T(t) = C_3\cos(aKt) + C_4\sin(aKt)\\ \textsf{Thus,}\,\,y(x,t) = \left(C_1\cos(Kx) + C_2\sin(Kx)\right)\left(C_3\cos(aKt) + C_4\sin(aKt)\right)\\ \textsf{By initial condition,}\,\,\frac{\partial y}{\partial t}(x,0) = 0\\ \frac{\partial y}{\partial t} = \left(C_1\cos(Kx) + C_2\sin(Kx)\right)\left(-C_3aK\sin(aKt) + C_4aK\cos(aKt)\right)\\ \frac{\partial y}{\partial t}(x,0) = C_4aK\left(C_1\cos(Kx) + C_2\sin(Kx)\right) = 0\\ \implies C_4 = 0.\\ \textsf{By the initial condition}\,\, y(0,t) = 0\\ y(x,t) = \left(C_1\cos(Kx) + C_2\sin(Kx)\right)\left(C_3\cos(aKt) + C_4\sin(aKt)\right) = 0\\ y(0,t) = C_1\left(C_3\cos(aKt) + C_4\sin(aKt)\right) = 0\\ \implies C_1 = 0\\ \therefore y(x,t) = C_2\sin(Kx)C_3\cos(aKt)\\ \textsf{By the initial condition}\,\, y(l,t) = 0\\ y(l,t) = C_2\sin(Kl)C_3\cos(aKt) = 0\\ \implies \sin(Kl) = 0\\ \sin(Kl) = \sin(n\pi)\hspace{1cm} \,\, \textsf{for}\,\, n \geq 1\\ K = \frac{n\pi}{l}\\ \therefore y(x,t) = C_2 C_3\sin\left(\frac{an\pi x}{l}\right)\cos\left(\frac{an\pi t}{l}\right) = \lambda\sin\left(\frac{an\pi x}{l}\right)\cos\left(\frac{an\pi t}{l}\right)\\ \textsf{The general solution is thus,}\\ y(x, t) = \sum_{n=1}^{\infty} \lambda_n \sin\left(\frac{an\pi x}{l}\right)\cos\left(\frac{an\pi t}{l}\right)\\ \textsf{Where,}\,\, \lambda_n \,\, \textsf{can be related to the Fourier} \\ \textsf{coefficients by incorporating the initial conditions.}\\ y(x, 0) = \sum_{n=1}^{\infty} \lambda_n \sin\left(\frac{an\pi x}{l}\right)\cos\left(\frac{an\pi \times 0}{l}\right) = \sum_{n=1}^{\infty} \lambda_n \sin\left(\frac{an\pi x}{l}\right)\\ \textsf{By Fourier's Half-range sine series, we derive}\,\,\lambda_n\\ \lambda_n = \frac{2a}{l}\int_0^{\frac{l}{a}} y(x,0) \sin\left(\frac{an\pi x}{l}\right) \mathrm{d}x$