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)=\sin x, \quad u_{t}(x, 0)=a \cos x, \quad-\infty

Short answer

Answer: The specific solution to the given wave equation with the given initial conditions is \(u(x,t) = \sin(x-at)\).

Step-by-step solution

Write down the general solution from Exercise 63

The general solution of the wave equation \(u_{t t} = a^2 u_{x x}\) is given by: $$ u(x,t) = F(x-at) + G(x+at) $$ where F and G are arbitrary functions of their respective arguments.

Apply the initial condition \(u(x, 0) = \sin x\)

Using the initial condition \(u(x, 0) = \sin x\), we can write the solution as: $$ u(x,0) = F(x) + G(x) = \sin x $$

Apply the initial condition \(u_t(x, 0)=a \cos x\)

Using the initial condition \(u_t(x, 0)=a \cos x\), we can calculate the partial derivative with respect to t: $$ u_t(x,t) = -a F'(x-at) + a G'(x+at) $$ Now, by setting \(t=0\), we get: $$ u_t(x,0) = -a F'(x) + a G'(x) = a \cos x $$

Solve the system of equations for F and G

We now have the following system of equations: $$ F(x) + G(x) = \sin x \\ -a F'(x) + a G'(x) = a \cos x $$ With some manipulation, we can isolate F(x) in the first equation to obtain: $$ F(x) = \sin x - G(x) $$ Next, substitute this into the second equation and divide by \(a\): $$ F'(x) - G'(x) = \cos x $$

Integrate both sides of the above equation

Integrating both sides of this equation gives us: $$ F(x) - G(x) = \sin x + C $$ Here, C is the constant of integration, and we need to find its value.

Determine the value of C using the first equation

Comparing the first equation and the equation obtained in step 5, we can see that they are identical: $$ F(x) + G(x) = \sin x \\ F(x) - G(x) = \sin x + C $$ Adding both equations, we find: $$ 2F(x) = 2 \sin x + C $$ Thus, \(C = 0\). The equation in step 5 now becomes: $$ F(x) - G(x) = \sin x $$

Solve for F(x) and G(x)

From the equations \(F(x) + G(x) = \sin x\) and \(F(x) - G(x) = \sin x\), we can now solve for F(x) and G(x). Adding both equations gives: $$ 2F(x) = 2 \sin x $$ Therefore, \(F(x) = \sin x\). Similarly, subtracting the equations gives: $$ 2G(x) = 0 $$ Thus, \(G(x) = 0\).

Substitute the solutions for F(x) and G(x) into the general solution

Now, we substitute these values into the general solution given in step 1: $$ u(x,t) = F(x-at) + G(x+at) = \sin(x-at) + 0 $$ So, the solution to the given wave equation with the given initial conditions is: $$ u(x,t) = \sin(x-at) $$

Key concepts

Understanding Partial Differential Equations

Partial differential equations (PDEs) are mathematical equations that involve rates of change with respect to continuous variables. A PDE for a function of two or more independent variables contains partial derivatives with respect to those variables.

PDEs are used to formulate problems involving functions of several variables and they are especially useful in simulating physical phenomena such as sound, heat, fluid flow, and wave propagation. In the context of wave propagation, the wave equation is a second-order linear PDE that describes the distribution of a given variable (like the pressure or displacement of a wave) over space and time.

For example, the wave equation mentioned in the exercise, \( u_{tt} = a^2 u_{xx} \), is a pivotal PDE in physics where \( u(x,t) \) denotes the wave function dependent on space \( x \) and time \( t \) and \( a \) represents the wave speed. This equation is satisfied by any function \( u \) where the second temporal derivative is proportional to the second spatial derivative—signifying a balance between the rates of change of the function in time and space.

Initial Conditions and Their Importance

The solution to a PDE usually involves arbitrary functions or constants. To find a unique solution, initial conditions (or boundary conditions) are required. These are values of the function, or its derivatives, specified at the beginning of the problem.

In the context of the wave equation, initial conditions often represent the initial shape of the wave and the initial velocity distribution. For instance, the initial conditions provided in the exercise, \( u(x, 0) = \sin x \) and \( u_t(x, 0) = a \cos x \), reveal the wave's configuration and velocity at time \( t = 0 \). These conditions are used to determine the functions \( F(x) \) and \( G(x) \) that appear in the general wave equation solution. The need to satisfy both the equation and initial conditions simultaneously is a classic example of an initial value problem (IVP), which is a specific type of PDE problem.

Without initial conditions, we would have infinitely many solutions to the wave equation, and thus, the conditions allow us to zero in on the one solution that fits the physical context of the problem.

d’Alembert’s Formula and Solving the Wave Equation

d’Alembert’s formula is a specific solution to the one-dimensional wave equation when the initial conditions are given. The genius of d'Alembert was in recognizing that the one-dimensional wave equation can be factored into a product of two simpler first-order terms, allowing the solution to be written as a sum of two arbitrary functions.

In the exercise, the given wave equation solution is expressed using d'Alembert’s formula: \( u(x,t) = F(x-at) + G(x+at) \). Here, the two terms represent waves traveling in opposite directions. The functions \( F \) and \( G \) are determined by the initial shape and velocity of the wave through integration, which aligns with the given initial conditions.

The method of solving the wave equation using d'Alembert's formula includes steps like applying the given initial conditions, differentiating to find \( F \) and \( G \) (or their derivatives), and integrating or solving simultaneous equations to find explicit forms. These steps result in a specific wave function \( u(x,t) \) that satisfies both the wave equation and the initial conditions presented at the start, exemplifying d'Alembert's strategy of separating and then recombining wave components to achieve the solution.

In our exercise, the student found that \( G(x) = 0 \) and \( F(x) = \sin x \), yielding the final wave function \( u(x,t) = \sin(x - at) \), which represents a sine wave propagating in the positive x-direction.