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Consider the homogeneous wave equation $$ u_{t t}=a u_{x x} $$ on the whole real line \(x \in \mathbb{R}\). Since the spatial domain is unbounded in both directions, we only need to specify the initial conditions $$ u(x, 0)=u_{0}(x), \quad u_{t}(x, 0)=v_{0}(x), \quad x \in \mathbb{R} $$ Verify the solution formula $$ \begin{aligned} u(x, t) &=\frac{1}{2}\left[u_{0}(x+\sqrt{a} t)+u_{0}(x-\sqrt{a} t)\right]+\frac{1}{2 \sqrt{a}} \int_{x-\sqrt{a} t}^{x+\sqrt{a} t} v_{0}(s) d s \\\ & \equiv u_{1}(x, t)+u_{2}(x, t) \end{aligned} $$ This is called D'Alembert's formula. The part \(u_{1}(x, t)\) represents the propagation of the initial displacement for zero initial velocity, whereas the other part \(u_{2}(x, t)\) describes the vibrations caused by the initial velocity with zero initial displacement. The quantity \(\sqrt{a}\) can be interpreted as the wave propagation speed.

Short Answer

Expert verified
Both components of D'Alembert's formula satisfy the wave equation individually, confirming the given solution.

Step by step solution

01

Substitute the Solution into the Wave Equation

Verify that the provided formula\[u(x, t) = \frac{1}{2}(u_{0}(x+\sqrt{a} t) + u_{0}(x-\sqrt{a} t)) + \frac{1}{2 \sqrt{a}} \int_{x-\sqrt{a} t}^{x+\sqrt{a} t} v_{0}(s) d s\]satisfies the wave equation\[u_{tt} = a u_{xx}.\]Begin by calculating \(u_{tt}\) and \(u_{xx}\) for the given formula.
02

Calculate \(u_{1}(x, t)\) Components

First, find the second partial derivatives with respect to time and space for \(u_{1}(x,t) = \frac{1}{2}(u_{0}(x+\sqrt{a} t) + u_{0}(x-\sqrt{a} t))\). Calculate the first time derivative:\[\frac{\partial}{\partial t} u_{1}(x, t) = \frac{1}{2} (\sqrt{a} u_{0}'(x+\sqrt{a}t) - \sqrt{a} u_{0}'(x-\sqrt{a}t)) = \frac{\sqrt{a}}{2} (u_{0}'(x+\sqrt{a}t) - u_{0}'(x-\sqrt{a}t)).\]
03

Calculate the Second Time Derivative

Continue by taking the second time derivative of \(u_{1}(x, t)\):\[\frac{\partial^2}{\partial t^2} u_{1}(x,t) = \frac{\sqrt{a}}{2} (\sqrt{a} u_{0}''(x+\sqrt{a}t) + \sqrt{a} u_{0}''(x-\sqrt{a}t)) = \frac{a}{2} (u_{0}''(x+\sqrt{a}t) + u_{0}''(x-\sqrt{a}t)).\]
04

Calculate the Second Spatial Derivative

Now, compute the second spatial derivative of \(u_{1}(x, t)\):\[\frac{\partial^2}{\partial x^2} u_{1}(x, t) = \frac{1}{2} (u_{0}''(x+\sqrt{a}t) + u_{0}''(x-\sqrt{a}t)).\]
05

Compare the Derivatives

Compare the second time derivative and the scaled second spatial derivative:\[\frac{\partial^2}{\partial t^2} u_{1}(x,t) = a \frac{\partial^2}{\partial x^2} u_{1}(x,t).\]Thus, \(u_{1}(x,t)\) satisfies the wave equation.
06

Verify \(u_{2}(x,t)\)

Determine the first time derivative for \(u_{2}(x,t)\):\[\frac{\partial}{\partial t} u_{2}(x, t) = \frac{1}{2\sqrt{a}}(v_{0}(x+\sqrt{a} t)\sqrt{a} - v_{0}(x-\sqrt{a} t)\sqrt{a}) = \frac{1}{2} (v_{0}(x+\sqrt{a} t) - v_{0}(x-\sqrt{a} t)).\]
07

Continue with Second Time Derivative

Continue by taking the second time derivative for \(u_{2}(x, t)\):\[\frac{\partial^2}{\partial t^2} u_{2}(x,t) = \frac{\sqrt{a}}{2} (v_{0}'(x+\sqrt{a}t) + v_{0}'(x-\sqrt{a}t)).\]Calculate the first spatial derivative:\[\frac{\partial}{\partial x} u_{2}(x, t) = \frac{1}{2 \sqrt{a}}(v_{0}(x+\sqrt{a} t) - v_{0}(x-\sqrt{a} t)).\]
08

Complete Second Spatial Derivative

Finally, compute \(u_{2}(x,t)\)'s second spatial derivative:\[\frac{\partial^2}{\partial x^2} u_{2}(x,t) = \frac{\sqrt{a}}{2} (v_{0}'(x+\sqrt{a}t) + v_{0}'(x-\sqrt{a}t)).\]Again, we find that:\[\frac{\partial^2}{\partial t^2} u_{2}(x,t) = a \frac{\partial^2}{\partial x^2} u_{2}(x,t).\]
09

Combine and Validate

Since both components \(u_{1}(x,t)\) and \(u_{2}(x,t)\) satisfy the wave equation individually, their sum, \(u(x,t)\), also satisfies it:\[\frac{\partial^2}{\partial t^2} u(x,t) = a \frac{\partial^2}{\partial x^2} u(x,t).\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

homogeneous wave equation
The homogeneous wave equation is a type of partial differential equation that models the propagation of waves through a medium. It has the form:
\[ u_{tt} = a u_{xx} \]
Here, \( u \) is a function that depends on both time \( t \) and spatial coordinate \( x \). The notation \( u_{tt} \) represents the second partial derivative of \( u \) with respect to time, and \( u_{xx} \) is the second partial derivative of \( u \) with respect to space. The constant \( a \) determines the characteristics of the wave’s speed and behavior.

In simple terms, the equation states that the acceleration of the wave at any point (i.e., second derivative with respect to time) is proportional to the curvature of the wave at that point (i.e., second derivative with respect to space). The homogeneous aspect means there are no external forces or sources generating or altering the wave as it propagates.
D'Alembert's formula
D'Alembert's formula provides a direct method to solve the wave equation for an unbounded domain. The formula is given by: \[ u(x, t) = \frac{1}{2}\big[u_{0}(x+\beta t) + u_{0}(x-\beta t)\big] + \frac{1}{2 \beta} \bigg[ \bigintssss_{x-\beta t}^{x+\beta t} v_{0}(s) \, \text{d}s \bigg]\ \ \]
Here, \( \beta = \sqrt{a} \), where \( a \) is the wave speed from the wave equation.
In simpler terms, D'Alembert's formula decomposes the wave into two parts:
  • The first part represents how the initial displacement \( u_{0}(x) \) evolves over time.
    It’s the sum of the initial wave shape shifted left and right by a factor of the wave speed over time.
  • The second part describes how initial velocities \( v_{0}(x) \) affect the wave. It considers contributions along a line from \( x - \sqrt{a} t \) to \( x + \sqrt{a} t \), weighted by the inverse of the wave speed. This integral measures the cumulative effect of initial velocities over space and time.
To sum up, D'Alembert’s formula elegantly combines initial displacement and velocity to describe wave dynamics over time.
partial derivatives
Partial derivatives are fundamental in understanding how functions change with respect to multiple variables. When addressing the wave equation, you encounter derivatives with respect to both time \( t \) and space \( x \).
  • A partial derivative with respect to time, denoted as \( u_{t} \) or \( \frac{\partial u}{\partial t} \), measures how the function \( u(x, t) \) changes as time changes, while keeping \( x \) constant.
  • Similarly, a partial derivative with respect to space, denoted as \( u_{x} \) or \( \frac{\partial u}{\partial x} \), indicates how \( u(x, t) \) varies as \( x \) changes, while fixing time \( t \).

When higher-order partial derivatives are involved, such as \( u_{tt} \) or \( u_{xx} \), they provide deeper insight:
  • The second-time derivative, \( u_{tt} \), describes the acceleration or the rate of change of velocity over time.
  • The second-space derivative, \( u_{xx} \), depicts the curvature or how steeply the wave's shape changes in space.
Both of these higher-order partial derivatives are critical in solving and understanding wave behavior, as they encapsulate the dynamic (time-related) and structural (space-related) aspects of the wave.
wave propagation speed
The wave propagation speed determines how fast waves travel through a medium. In the context of the homogeneous wave equation \( u_{tt} = a u_{xx} \), the speed \( \sqrt{a} \) plays a crucial role.

Here are some key points to understand about wave propagation speed:
  • The constant \( a \) in the wave equation is related to the medium's properties, influencing how quickly waves can travel.
  • The physical wave speed \( \sqrt{a} \) dictates how far and how fast disturbances (initial displacements and velocities) spread through the medium.
  • A higher value of \( a \) means the wave travels faster.
  • In D'Alembert’s formula, \( \sqrt{a} \) shifts initial conditions forward and backward over time by this speed, essentially stretching or compressing the wave.
Understanding wave propagation speed is vital for applications like signal transmission, sound waves, and other physical phenomena where waves are present.

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Most popular questions from this chapter

For any constants \(k, \omega, c_{1}\), and \(c_{2}\), verify that the function $$ u(x, t)=e^{-k \omega^{2} t}\left(c_{1} \cos \omega x+c_{2} \sin \omega x\right) $$ satisfies the differential equation $$ u_{t}=k u_{x x} $$

Consider a heat conduction problem for a homogeneous rod of unit length. Assume the rod is sufficiently thin and its surface is thermally insulated. Then the temperature of the rod, \(u=u(x, t)\), satisfies the equation $$ u_{t}=a u_{x x}+f(x, t) $$ where \(a=k /(c \rho)\) with \(c\) the specific heat, \(\rho\) the density, \(k\) a thermal conductivity coefficient, and \(f(x, t)\) a scaled heat source function. Suppose the temperatures at the two ends of the rod are fixed at 0 , the initial rod temperature is \(0, a=1\), and \(f(x, t)=x(1-x) \sin \pi t .\) Compute and plot the temperature at \(t=1\) for several values of \(n_{t}\) and \(n_{x}\) until the numerical solution at \(x=1 / 2, t=1\) has three ahilized dioits

Verify that the function \(u(x, t)=e^{-t} \cos x\) solves the initial boundary value problem $$ \left\\{\begin{array}{l} u_{t}=u_{x x}, \quad 00 \\ u_{x}(0, t)=u_{x}(\pi, t)=0, \quad t \geq 0 \\ u(x, 0)=\cos x, \quad 0 \leq x \leq \pi \end{array}\right. $$ Note that here the boundary conditions are specified on the values of the spatial derivative. Such boundary conditions are examples of Neumann ones.

Consider the planar problem of small transverse vibrations of an elastic string of length \(L\), whose two ends are fixed at \(x=0\) and \(x=L\) on the \(x\) -axis. Let the string be subject to the action of a vertical force of density \(f(x, t)\) per unit mass. Assume the displacement vector is perpendicular to the \(x\) -axis and is small in size. Then the equation for the vertical displacement function, \(u(x, t)\), is $$ u_{t t}=a u_{x x}+f(x, t) $$ where \(a=T_{0} / \rho\), with \(T_{0}\), the tension of the string, and \(\rho\), the mass density of the string, both assumed constant. Suppose \(L=1, a=1\). Compute the vertical displacement for the following cases: (a) \(\quad f(x, t)=\sin \pi x, u(x, 0)=0\) and \(u_{t}(x, 0)=0 .\) (b) \(\quad f(x, t)=0, u(x, 0)=0\) and \(u_{t}(x, 0)=1\).

Modify the MaTLaB program Wave to solve the initial boundary value problem $$ \left\\{\begin{array}{l} \frac{\partial^{2} u}{\partial t^{2}}=\frac{\partial^{2} u}{\partial x^{2}}+\left(1+\frac{\pi^{2}}{4}\right) e^{-t} \cos \frac{\pi x}{2}, \quad x \in(0,1), t \in(0,1) \\ u(x, 0)=\cos \frac{\pi x}{2}, \quad u_{t}(x, 0)=-\cos \frac{\pi x}{2}, \quad x \in[0,1] \\ u_{x}(0, t)=u(1, t)=0, \quad t \in[0,1] \end{array}\right. $$ The true solution if \(u(x, t)=e^{-t} \cos (\pi x / 2)\). Compute the numerical solutions with \(n_{x}=n_{t}=5,10\), and 20 , and plot the error functions. Give the solution errors at \(t=1\) and \(x=i / 5,0 \leq i \leq 4\), and comment on the numerical results.

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