Question

Let \( u(x, t) \) be the solution of the wave equation \[ \frac{\partial^2 u}{\partial t^2} = \frac{\partial^2 u}{\partial x^2}, \quad 0<x<\pi, \, t>0, \] with the initial conditions \[ u(x, 0) = \sin x + \sin 2x + \sin 3x, \quad \frac{\partial u}{\partial t}(x, 0) = 0, \quad 0<x<\pi, \] and the boundary conditions \[ u(0, t) = u(\pi, t) = 0, \quad t \geq 0. \] Then, the value of \( u \left( \frac{\pi}{2}, \pi \right) \) is

• A. \( -\frac{1}{2} \)
• B. 0
• C. \( \frac{1}{2} \)
• D. 1

Hint:

For problems involving the wave equation, the solution often takes the form of a sum of sines and cosines. The coefficients are determined by the initial and boundary conditions.

Answer 1

The Correct Option is B

We are given the wave equation with initial and boundary conditions. To solve for \( u \left( \frac{\pi}{2}, \pi \right) \), we can use the general solution for the wave equation, which is a sum of sine and cosine functions. The solution for the wave equation with these boundary conditions can be expressed as: \[ u(x, t) = \sum_{n=1}^{\infty} \left( A_n \cos(nt) + B_n \sin(nt) \right) \sin(nx). \] Step 1: Initial conditions
Using the initial condition \( u(x, 0) = \sin x + \sin 2x + \sin 3x \), we see that: \[ u(x, 0) = \sum_{n=1}^{\infty} A_n \sin(nx). \] Comparing this with \( \sin x + \sin 2x + \sin 3x \), we deduce that: \[ A_1 = 1, \, A_2 = 1, \, A_3 = 1, \quad \text{and} \quad A_n = 0 \, \text{for} \, n>3. \] Also, the condition \( \frac{\partial u}{\partial t}(x, 0) = 0 \) implies that all \( B_n = 0 \). Step 2: General solution
Thus, the solution becomes: \[ u(x, t) = \cos(t) \sin(x) + \cos(2t) \sin(2x) + \cos(3t) \sin(3x). \] Step 3: Evaluate at \( x = \frac{\pi}{2} \) and \( t = \pi \)
Now, we substitute \( x = \frac{\pi}{2} \) and \( t = \pi \) into the solution: \[ u\left( \frac{\pi}{2}, \pi \right) = \cos(\pi) \sin\left( \frac{\pi}{2} \right) + \cos(2\pi) \sin(\pi) + \cos(3\pi) \sin\left( \frac{3\pi}{2} \right). \] Since \( \sin(\frac{\pi}{2}) = 1 \), \( \sin(\pi) = 0 \), and \( \sin(\frac{3\pi}{2}) = -1 \), we get: \[ u\left( \frac{\pi}{2}, \pi \right) = (-1)(1) + (1)(0) + (-1)(-1) = -1 + 1 = 0. \] Step 4: Conclusion
Thus, the value of \( u\left( \frac{\pi}{2}, \pi \right) \) is \( 0 \).