A long string, for which the transverse wave velocity is \(c\), is given a displacement specified by some function \(\eta=\eta_{0}(x)\) that is localized near the middle of the string. The string is released at \(t=0\) with zero initial velocity. Find the equations for the traveling waves that are produced and make a sketch showing the waves at several instants of time with \(t \geq 0\). Hint: Find two waves traveling in opposite directions that together satisfy the initial conditions.

The wave equation governs the behavior of waves in a medium (like the string in this problem). In one dimension, the wave equation is: \[ \frac{\partial^2 \eta}{\partial t^2} = c^2 \frac{\partial^2 \eta}{\partial x^2}\] where \(\eta\) is the displacement function, \(c\) is the wave velocity, and \(x\) and \(t\) are the spatial and time coordinates, respectively.

The general solution of the wave equation represents waves traveling in both positive and negative \(x\) directions. So, we can write the general solution as a superposition of both rightward and leftward traveling waves: \[ \eta(x, t) = f(x - ct) + g(x + ct)\] where \(f\) and \(g\) are arbitrary functions that represent the shapes of the rightward and leftward traveling waves, respectively.

We are given the initial displacement function \(\eta\) as \(\eta_0(x)\) and the initial velocity is zero. We can use these initial conditions to find the functions \(f\) and \(g\). First, we'll apply the initial displacement condition: \[ \eta(x, 0) = \eta_0(x) = f(x) + g(x)\] Next, we'll apply the initial velocity, which is the first derivative of \(\eta\) with respect to time, \(\frac{\partial \eta}{\partial t}\). The initial velocity is zero, so we have: \[ \frac{\partial \eta}{\partial t}(x,0) = 0 = -cf'(x) + cg'(x)\] Now, we have a system of two equations with two unknown functions, \(f(x)\) and \(g(x)\). We can solve this system by using the method of characteristics. We need to add and subtract the equations to eliminate either \(f\) or \(g\). Adding both equations, we eliminate \(g(x)\) and get: \[ f(x) = \frac{1}{2}[\eta_0(x) + \int_{-x}^{x}c\frac{d\eta_0(u)}{du}du]\] Subtracting both equations, we eliminate \(f(x)\) and get: \[ g(x) = \frac{1}{2}[\eta_0(x) - \int_{-x}^{x}c\frac{d\eta_0(u)}{du}du]\] Finally, substitute \(f(x)\) and \(g(x)\) back into the general solution: \[ \eta(x, t) = \frac{1}{2}[\eta_0(x - ct) + \int_{-x+ct}^{x-ct}c\frac{d\eta_0(u)}{du}du] + \frac{1}{2}[\eta_0(x + ct) - \int_{-x-ct}^{x+ct}c\frac{d\eta_0(u)}{du}du]\] To visualize the waves at several instants of time, you can plot \(\eta(x, t)\) as a function of \(x\) for different values of \(t\geq0\).