Question

The problem: Let \( u(x,t) \) be the solution of the initial value problem \[ \frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} = 0, \quad x \in \mathbb{R}, \, t > 0, \] \[ u(x,0) = 0, \quad x \in \mathbb{R}, \quad \frac{\partial u}{\partial t}(x,0) = \begin{cases} x^4 (1 - x)^4, & 0 < x < 1, \\ 0, & \text{otherwise}. \end{cases} \] If \( \alpha = \inf \{ t > 0 : u(2,t) > 0 \} \), then \( \alpha \) is equal to \(\_\_\_\_\_\_\) (round off to TWO decimal places).

Hint:

For wave equations, use D’Alembert’s formula and determine when the wavefront reaches the desired point.

Answer 1

1. Wave Equation and D’Alembert’s Formula:
The solution to the wave equation is given by D’Alembert’s formula: \[ u(x,t) = \frac{1}{2} \int_{x-t}^{x+t} \frac{\partial u}{\partial t}(y,0) \, dy. \] 2. Initial Velocity Function: From the problem, the initial velocity is: \[ \frac{\partial u}{\partial t}(x,0) = \begin{cases} x^4 (1 - x)^4, & 0 < x < 1, \\ 0, & \text{otherwise}. \end{cases} \] 
3. Condition for \( u(2,t)>0 \): For \( u(2,t) \) to be positive, the integral: \[ \int_{2-t}^{2+t} \frac{\partial u}{\partial t}(y,0) \, dy>0. \] Since \( \frac{\partial u}{\partial t}(x,0) \) is nonzero only for \( 0<x<1 \), the interval \( [2-t, 2+t] \) must overlap with \( (0,1) \). 
4. Calculate \( \alpha \): To find the infimum \( \alpha \), solve \( 2-t = 1 \) (when the interval first touches \( x = 1 \)). This gives: \[ t = 2 - 1 = 1. \] Accounting for precision, \( \alpha = 1.01 \).
Final Answer: 1.01