91Ó°ÊÓ

The wave equation is a fundamental equation in mathematical physics and engineering. It describes how waves, such as sound or light, propagate through a medium. In its standard form for one-dimensional wave propagation with wave speed c, it is written as:

\[ \frac{\partial^{2} \phi}{\partial t^{2}} - c^2 \frac{\partial^{2} \phi}{\partial x^{2}} = 0 \]

In this exercise, the wave speed is unity (c = 1), so the wave equation simplifies to:

\[ \frac{\partial^{2} \phi}{\partial t^{2}} - \frac{\partial^{2} \phi}{\partial x^{2}} = 0 \]

This equation states that the second time derivative of the function \( \phi (x, t) \) balances the second spatial derivative. The function \( \phi(x, t) \) represents the wave's displacement at position x and time t.

Boundary conditions are essential for solving partial differential equations like the wave equation, as they provide additional information that helps determine a unique solution. In this exercise, there are four specific boundary conditions given:

\[ \begin{array}{cc} \phi(x, t=0)=0, & \frac{\partial \phi}{\partial t}(x, t=0)=0 \ \phi(x=0, t)=0, & \phi(x=\ell, t)=1 \end{array} \]

These conditions can be interpreted as follows:

• At time t=0, the displacement \( \phi(x, 0) \) of the wave is zero everywhere along x.


• The initial velocity \( \frac{\partial \phi}{\partial t}(x, 0) \) is also zero at t=0.


• At position x=0, the displacement \( \phi(0, t) \) is always zero for any time t. This might represent a fixed boundary.


• At position x=\( \ell \), the displacement \( \phi(\ell, t) \) is always 1 for any time t, indicating a boundary that is held at a constant displacement.

An ordinary differential equation (ODE) is a relation that contains functions of only one independent variable and its derivatives. Solving ODEs is a critical step in solving partial differential equations. When the Laplace transform is applied to the wave equation, it transforms it into an ODE:

\[ \frac{\partial^{2} \hat{\Phi}(x,s)}{\partial x^{2}} - s^2 \hat{\Phi}(x, s) = 0 \]

This equation is linear and homogeneous, with a standard solution involving hyperbolic functions of the form:

\[ \hat{\Phi}(x,s) = A \sinh(sx) + B \cosh(sx) \]

The constants A and B are determined using the boundary conditions provided, turning the ODE into a specific solution.

The Laplace transform is a powerful tool used to convert differential equations into algebraic equations. For a function \( f(t) \), its Laplace transform \( \mathcal{L}[f(t)] \) is defined as:

\[ \hat{f}(s) = \mathcal{L}[f(t)] = \int_{0}^{\infty} e^{-st} f(t) dt \]

In the context of the wave equation, applying the Laplace transform with respect to time (t) converts the wave equation into an ODE. This simplifies solving the spatial part of the problem. After applying the Laplace transform to the wave equation and utilizing initial conditions, our transformed equation looks like:

\[ s^2 \hat{\Phi}(x, s) = \frac{\partial^{2} \hat{\Phi}(x, s)}{\partial x^{2}} \]

This facilitates solving for \( \hat{\Phi}(x, s) \).

The inverse Laplace transform is used to convert the solution back to the time domain. If \( \hat{F}(s) \) is the Laplace transform of \( f(t) \), the inverse Laplace transform is:

\[ f(t) = \mathcal{L}^{-1}[\hat{F}(s)] \]

In this problem, after finding the Laplace-transformed solution:

\[ \hat{\Phi}(x, s) = \frac{\sinh(sx)}{s \sinh(s \ell)} \]

we need to apply the inverse Laplace transform to get back to \( \phi(x, t) \):

\[ \phi(x, t) = \frac{x}{\ell} + \sum_{n=1}^{\infty} \frac{2(-1)^{n}}{n \pi} \sin \left( \frac{n \pi x}{\ell} \right) \cos \left( \frac{n \pi t}{\ell} \right) \]

This final expression represents the displacement of the wave in the original time and space domains.