10.7.1. Wave Equation with a Free End

Let us consider the one-dimensional wave equation:

∂²u / ∂t² = c² * ∂²u / ∂x²

With boundary conditions:

• u(0,t) = 0,
• lim (x→∞) u(x,t) = 0,
• u(x,0) = f(x),
• ∂u/∂t (x,0) = 0.

In this section, we introduce the fundamental wave equation, which describes how waves propagate through a medium. The notation used indicates how the displacement (u) changes with respect to time (t) and space (x). Here, c represents the speed of the wave. The boundary conditions specify that at the starting point (x=0), there is no displacement. Additionally, we require that as we move infinitely far from the origin, the displacement approaches zero. The initial state of the system, given by f(x), and the rate of change of displacement at time t=0, are also defined, indicating the wave's initial conditions.

Imagine holding one end of a rope tightly while the other end is free. If you create a wave by flicking the rope, the point where you flick it corresponds to u(0,t)=0—there's no movement at your hand. As you observe, the wave will travel along the rope but diminish as you move further away from your hand, similar to the boundary condition lim (x→∞) u(x,t)=0.

Apply the Fourier Sine Transform:

∂²U / ∂t² = −c²s²U

This is a second-order ODE in t.

In this step, we apply the Fourier Sine Transform to the displacement function to convert our spatial problem into a frequency domain problem. This transforms the wave equation into a second-order ordinary differential equation (ODE) with respect to time. The term s relates to the frequency components of the original function, which are now being analyzed over time. The Fourier transform helps us manage the wave problem by simplifying it into a form that can be solved for U(s,t), which represents the transformed function in the frequency domain.

Think of tuning a violin. The sound produced corresponds to the frequency of the strings vibrating. By applying the Fourier Sine Transform, we are effectively 'tuning' our equation, breaking down the wave into different frequency components, similar to finding the unique notes a violin string can produce.

U(s,t) = A(s)cos(cst) + B(s)sin(cst).

Once we have a second-order ordinary differential equation, we can express its solution in terms of two arbitrary functions, A(s) and B(s). These functions correspond to the two types of oscillatory motion: one that describes how the wave behaves in terms of cosine (A(s)) and the other in terms of sine (B(s)). This general form solves the wave equation and indicates that the wave can oscillate between different states based on these functions.

Just like how musical notes can mix to create melodies, the wave equation's solution can produce various wave patterns through the combination of the cosine and sine functions. The sounds from a musical instrument depend on the fundamental frequency (A(s)) and its harmonics (B(s))—together they make a harmonious tune, just as the wave equation forms a complex wave profile.

Initial conditions give:

• U(s,0) = F{s}{f(x)} = A(s),
• ∂U/∂t (t=0) = 0 ⇒ B(s) = 0.

To find the specific solutions A(s) and B(s), we incorporate the initial conditions into the general solution. The first condition allows us to express A(s) in terms of the Fourier transform of the original function f(x), while the second condition, stating that the initial rate of displacement is zero, leads to B(s) being determined as zero. As a result, the oscillatory behavior of the wave can be completely described using only A(s).

Imagine starting your day lying still in bed (where your initial rate of movement is zero). When you finally get up (which corresponds to U(s,0)), your position reflects where you began (A(s)). The absence of movement at that exact moment (B(s)=0) suggests that while you are initially stationary, your motion will change based on external influences as you move through your morning routine.

Thus,

U(s,t) = F{f(x)} · cos(cst) ⇒ u(x,t) = F{f(x)} cos(cst) sin(sx) ds.

The final result allows us to express the displacement u at position x and time t in terms of the integral of the transformed initial function F{f(x)}. Here, we see how the original function influences the solution at future times via the cosine term oscillating with time and the sine term filtering through the effects of position. This integral solution gives a comprehensive view of how the initial shape of the wave evolves over time.

Consider throwing a pebble into a still pond. The initial splash represents your initial condition, F{f(x)}, and as the ripples spread out, the combination of traveling waves (through the cosine and sine interaction) describes how the disturbance propagates through the water, just like our calculated u(x,t) depicts how the wave moves across a medium over time.

--