Question

Determine the values of the constant \(\alpha\), if any, for which the specified function is a solution of the given partial differential equation. $$ u(x, t)=2 \cos (x+\alpha t), \quad u_{t t}-u_{x x}+2 u=0 $$

Short answer

The values of the constant α that make the given function a solution to the PDE are α = ±√3.

Step-by-step solution

Calculate first partial derivatives

Compute the first partial derivatives \(u_t\) and \(u_x\) with respect to \(t\) and \(x\) respectively: $$ u_t(x,t) = \frac{\partial}{\partial t}(2\cos(x+\alpha t)) = -2\alpha\sin(x+\alpha t) \\ u_x(x,t) = \frac{\partial}{\partial x}(2\cos(x+\alpha t)) = -2\sin(x+\alpha t) $$

Calculate second partial derivatives

Now, compute the second partial derivatives \(u_{tt}\) and \(u_{xx}\): $$ u_{tt}(x,t) = \frac{\partial^2}{\partial t^2}(2\cos(x+\alpha t)) = -2\alpha^2 \cos(x+\alpha t) \\ u_{xx}(x,t) = \frac{\partial^2}{\partial x^2}(2\cos(x+\alpha t)) = -2\cos(x+\alpha t) $$

Plug the derivatives into the PDE

Substitute the calculated partial derivatives in the original PDE: $$ u_{tt}(x, t) - u_{xx}(x, t) + 2u(x, t) = -2\alpha^2 \cos(x+\alpha t) + 2\cos(x+\alpha t) + 2(2\cos(x+\alpha t)) = 0 $$

Solve for \(\alpha\)

Since the given function is nontrivial, the expression involving the trigonometric function must vanish at all points \((x,t).\) That is, we must find \(\alpha\) for which $$ -2\alpha^2 \cos(x+\alpha t) + 2\cos(x+\alpha t) + 4\cos(x+\alpha t)=0 $$ using the fact that \(\cos(x+\alpha t)\neq 0\). Now simplify the equation and solve for \(\alpha\): \begin{align*} (-2\alpha^2 + 2 + 4)\cos(x+\alpha t) &= 0 \\ (-2\alpha^2 + 6)\cos(x+\alpha t) &= 0 \\ -2\alpha^2 + 6 &= 0 \end{align*} Now we solve for \(\alpha\): \begin{align*} -2\alpha^2 &= -6 \\ \alpha^2 &= 3 \\ \alpha &= \pm\sqrt{3} \end{align*} Therefore, the values of \(\alpha\) that make \(u(x, t) = 2 \cos (x + \alpha t)\) a solution to the partial differential equation are \(\alpha = \pm\sqrt{3}\).

Key concepts

Boundary Value Problems

Understanding boundary value problems (BVPs) is critical when studying partial differential equations (PDEs). A BVP is a type of differential equation accompanied by a set of additional constraints called boundary conditions. These conditions specify values or behaviors of the solution to the PDE at the boundaries of the domain. For example, in physical problems, these boundaries might represent the edges of a string, the surface of a drum, or the walls of a container.

In the given exercise, the function to be tested is not explicitly presented within a BVP context; however, when working with PDEs, knowing if a function satisfies particular boundary conditions is essential for concluding whether it is the appropriate solution to a physical problem. The exercise might be part of a larger BVP, where validating a solution to the PDE is the first step before examining the boundary constraints.

First Partial Derivatives

First partial derivatives represent the rate of change of a function with respect to one of its variables, while keeping the other variables constant. This concept is central to the study of PDEs, as it reflects how a function behaves in a singular dimension, which is key for understanding multidimensional systems. When calculating first partial derivatives, as shown in the exercise, the notation \( u_t \) signifies the derivative of \( u \) with respect to time (\( t \) ), and \( u_x \) denotes the derivative with respect to space (\( x \) ).

These derivatives give insight into the dynamics of a system at a local level. For the function provided in the problem, \( u(x, t)=2 \cos (x+\alpha t) \), finding its first partial derivatives is the first step in verifying that it satisfies the given PDE.

Second Partial Derivatives

Second partial derivatives expand upon the concept of first derivatives by examining the rate at which the rate of change itself changes. These derivatives are crucial in PDEs, particularly when analyzing phenomena involving acceleration, curvature, or propagation of waves. For instance, in the studied exercise, \( u_{tt} \) and \( u_{xx} \) represent the second partial derivatives of \( u \) with respect to \( t \) and \( x \) respectively. These describe how the function's change rate varies over time and space.

For the trigonometric function \( u(x, t)=2 \cos (x+\alpha t) \), its second partial derivatives convey information on the acceleration of the wave and spatial curvature. Plugging these second derivatives back into the PDE and solving for the constant \( \alpha \) confirms whether \( u(x, t) \) can indeed be a solution to the problem, provided it also satisfies any associated boundary conditions.