Question

Determine the values of the constant \(\alpha\), if any, for which the specified function is a solution of the given partial differential equation. $$ u(x, t)=\sin (x+\alpha t), \quad u_{t t}-4 u_{x x}=0 $$

Short answer

The values of the constant α for which the function u(x, t) = sin(x + αt) is a solution to the given partial differential equation are α = 2 and α = -2.

Step-by-step solution

Find the first-order partial derivatives of u(x, t)

First, we need to find the partial derivatives of u with respect to x and t: $$ u_x = \frac{\partial}{\partial x} \sin(x + \alpha t) = \cos(x + \alpha t) $$ $$ u_t = \frac{\partial}{\partial t} \sin(x + \alpha t) = \alpha \cos(x + \alpha t) $$

Find the second-order partial derivatives of u(x,t)

Now, we find the second-order partial derivatives of u with respect to x and t: $$ u_{xx} = \frac{\partial^2}{\partial x^2} \sin(x + \alpha t) = -\sin(x + \alpha t) $$ $$ u_{tt} = \frac{\partial^2}{\partial t^2} \sin(x + \alpha t) = -\alpha^2 \sin(x + \alpha t) $$

Plug the partial derivatives into the PDE

Now, we plug the second-order partial derivatives into the given PDE: $$ u_{tt} - 4u_{xx} = -\alpha^2 \sin(x + \alpha t) - 4(-\sin(x + \alpha t)) = 0 $$

Determine the values of α that make the PDE true

We need to solve for α from the equation: $$ -\alpha^2 \sin(x + \alpha t) + 4 \sin(x + \alpha t) = 0 $$ We can factor out the common factor of \(\sin(x + \alpha t)\) from the equation: $$ \sin(x + \alpha t)(4 - \alpha^2) = 0 $$ For the PDE to be true, either \(\sin(x+\alpha t) = 0\) or \(4 - \alpha^2 = 0\). Since \(\sin(x+\alpha t) = 0\) happens periodically and doesn't give us information about the constant α, we'll focus on the second equation: $$ 4 - \alpha^2 = 0 $$ Solving for α, we get: $$ \alpha = \pm 2 $$ Therefore, the values of the constant α for which the function u(x, t) = sin(x + αt) is a solution to the given partial differential equation are α = 2 and α = -2.

Key concepts

second-order partial derivatives

Second-order partial derivatives extend the idea of how a function changes. They explore the "derivative of a derivative" with respect to different variables. Here, we're dealing with the function \(u(x, t) = \sin(x + \alpha t)\). Taking derivatives with respect to \(x\) and \(t\), gives you

• \(u_x\): First derivative with respect to \(x\) is \(\cos(x + \alpha t)\)
• \(u_t\): First derivative with respect to \(t\) is \(\alpha \cos(x + \alpha t)\)

For second-order, simply differentiate again:

• \(u_{xx}\): Derivative of \(u_x\) with respect to \(x\) is \(-\sin(x + \alpha t)\)
• \(u_{tt}\): Derivative of \(u_t\) with respect to \(t\) is \(-\alpha^2 \sin(x + \alpha t)\)

These capture how the curvature or the "slope of slopes" change across each direction separately. Second-order derivatives are fundamental in understanding concavity and the behavior of solutions in PDEs.

PDE solution verification

When solving PDE problems, proving whether a given function is a solution is crucial. This process involves substituting derivatives back into the equation. Here, for the equation \(u_{tt} - 4u_{xx}=0\), we compute

• \(u_{tt}\): Related to time changes, \(-\alpha^2 \sin(x + \alpha t)\)
• \(u_{xx}\): Related to spatial changes, \(-\sin(x + \alpha t)\)

Plug them into the PDE:\[-\alpha^2 \sin(x + \alpha t) - 4(-\sin(x + \alpha t)) = 0\]Simplify it:\[\sin(x + \alpha t)(4 - \alpha^2) = 0\]This means for the equation to hold true everywhere, either the sine term is zero or \(\alpha\) satisfies \(4 - \alpha^2 = 0\). Solving gives the values of \(\alpha = \pm 2\). Thus, verification goes beyond just finding derivatives. It's about demonstrating that these equations return the desired zero, confirming the function as a solution.

trigonometric functions in differential equations

Trigonometric functions like \(\sin\) and \(\cos\) often appear in differential equations thanks to their repetitive, periodic nature. These functions can model wave patterns, such as those described in our function \(u(x, t)=\sin(x + \alpha t)\). Some beneficial properties include:

• **Periodicity**: Trigonometric functions repeat after specific intervals, making them excellent for oscillatory solutions.
• **Derivatives**: Derivatives of \(\sin\) and \(\cos\) remain within the family, being either other trigonometric functions or transformations thereof.

In our solution, the trigonometric nature was key. The derivative of \(\sin\) gave rise to \(\cos\) and back again, leading to easy-to-handle terms in the PDE. Moreover, these functions encapsulate the relationship between angles and lengths, aiding in the formulation and solution of diverse physical phenomena modeled by differential equations. So, spotting a wave-like pattern and applying these functions can provide insightful shortcuts and solutions in complex PDEs.