Question

Use the result of Exercise 63 to find a solution of $$ u_{t t}=a^{2} u_{x x}, \quad-\infty< x <\infty $$ that satisfies the given initial conditions. $$ u(x, 0)=x^{2}, \quad u_{t}(x, 0)=1, \quad-\infty

Short answer

Short answer: The solution to the one-dimensional wave equation with the given initial conditions is $u(x, t) = x^2 + \frac{1}{2}a^2t^2 + t$.

Step-by-step solution

Recall the result of Exercise 63

The d'Alembert's solution for the one-dimensional wave equation is given by: $$ u(x, t) = \frac{1}{2}[f(x + at) + f(x - at)] + \frac{1}{2a}\int_{x-at}^{x+at} g(y) \, dy $$ where f(x) and g(x) are functions defined by the initial conditions \(u(x,0) = f(x)\) and \(u_t(x,0) = g(x)\).

Identify the initial conditions

For this problem, we are given the initial conditions: $$ u(x, 0) = x^2 \quad \Rightarrow \quad f(x) = x^2 $$ $$ u_t(x, 0) = 1 \quad \Rightarrow \quad g(x) = 1 $$

Substitute the initial conditions into the d'Alembert's solution

Now, we will substitute the functions f(x) and g(x) obtained from the initial conditions into the d'Alembert's solution: $$ u(x, t) = \frac{1}{2}[(x+at)^2 + (x-at)^2] + \frac{1}{2a}\int_{x-at}^{x+at} 1 \, dy $$

Simplify the expression

We will now simplify the expression for u(x,t) by combining the terms: $$ u(x, t) = x^2 + \frac{1}{2}a^2t^2 + \frac{1}{2a}\int_{x-at}^{x+at} 1 \, dy $$

Solve the integral

In this step, we will evaluate the integral: $$ \int_{x-at}^{x+at} 1 \, dy = y\Big|_{x-at}^{x+at} = (x+at) - (x-at) = 2at $$

Combine the results

Finally, we will substitute the result of the integral back into the expression for u(x,t) and obtain the final solution: $$ u(x, t) = x^2 + \frac{1}{2}a^2t^2 + \frac{1}{2a}(2at) = x^2 + \frac{1}{2}a^2t^2 + t $$ So, the solution of the given wave equation with the specified initial conditions is: $$ u(x, t) = x^2 + \frac{1}{2}a^2t^2 + t $$

Key concepts

Wave Equation

The wave equation is a mathematical model that describes the propagation of waves, such as sound waves or light waves, through a medium. In its simplest form, the wave equation for one-dimensional space is expressed as
\[ u_{tt} = a^2 u_{xx} \]
where \( u(x, t) \) represents the displacement of the wave at position \( x \) and time \( t \), \( u_{tt} \) is 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 \( a \) is the speed at which the waves travel through the medium. This partial differential equation (PDE) is fundamental in fields like physics and engineering, capturing how waves behave in a given space and time.

Initial Conditions

Initial conditions are essential to solving partial differential equations like the wave equation because they specify the state of the system at the onset \( t = 0 \). In other words, they tell us the starting shape of the wave and how it is initially moving. For instance, the initial conditions
\[ u(x, 0) = x^2, \quad u_t(x, 0) = 1 \]
from our exercise define the initial displacement of the wave to be the shape of a parabola and its initial velocity to be uniform across all points in space. These conditions enable us to tailor general solutions, such as d'Alembert's solution, to the specifics of the problem we need to solve. Without initial conditions, we would have an infinite number of potential solutions to the wave equation, lacking any physical significance.

Partial Differential Equations

Partial differential equations (PDEs) are equations that involve multiple independent variables and their respective partial derivatives. They are utilized to formulate problems involving functions of several variables and are indispensable in the natural sciences and engineering. An example of a PDE is the wave equation we discussed earlier. PDEs, like the wave equation, are challenging because they require a detailed understanding of calculus, especially the concepts of differentiation and integration with respect to different variables. PDEs can have infinitely many solutions, so we need additional information, the 'initial conditions', to find the one that corresponds to the physical problem at hand. Solving a PDE often involves integrating functions to find the general form of the solution, and then applying the initial conditions to find the specific solution that fits the scenario being analyzed.