Question

Let \( u(x,t) \) be the solution of the initial-value problem \[ \frac{\partial^2 u}{\partial t^2} - 9 \frac{\partial^2 u}{\partial x^2} = 0, \quad x \in \mathbb{R}, \quad t>0, \quad u(x, 0) = e^x, \quad \frac{\partial u}{\partial t}(x, 0) = \sin x. \] Then, the value of \( u\left(\frac{\pi}{2}, \frac{\pi}{6}\right) \) is:

• A. \( \frac{1}{2} \left( e^{\pi} - \frac{1}{3} \right) \)
• B. \( \frac{1}{2} \left( e^{\pi} + \frac{1}{3} \right) \)
• C. \( \frac{1}{2} \left( e^{\pi} + \frac{5}{3} \right) \)
• D. \( \frac{1}{2} \left( e^{\pi} - \frac{5}{3} \right) \)

Hint:

For wave equations, use the general solution \( u(x, t) = f(x - 3t) + g(x + 3t) \), and apply the initial conditions to find the specific form of the solution.

Answer 1

The Correct Option is C

The given partial differential equation is the wave equation: \[ \frac{\partial^2 u}{\partial t^2} - 9 \frac{\partial^2 u}{\partial x^2} = 0. \] This equation has the general solution: \[ u(x, t) = f(x - 3t) + g(x + 3t), \] where \( f \) and \( g \) are determined by the initial conditions. 
Step 1: Apply the initial conditions The initial conditions are: \[ u(x, 0) = e^x, \quad \frac{\partial u}{\partial t}(x, 0) = \sin x. \] From \( u(x, 0) = f(x) + g(x) = e^x \), we get: \[ f(x) + g(x) = e^x. \] From \( \frac{\partial u}{\partial t}(x, 0) = -3f'(x) + 3g'(x) = \sin x \), we get: \[ -3f'(x) + 3g'(x) = \sin x \quad \Rightarrow \quad f'(x) - g'(x) = -\frac{1}{3} \sin x. \] Integrating this equation, we find: \[ f(x) - g(x) = \int -\frac{1}{3} \sin x \, dx = \frac{1}{3} \cos x + C. \] Step 2: Solve for \( f(x) \) and \( g(x) \) We now have the system: \[ f(x) + g(x) = e^x, \quad f(x) - g(x) = \frac{1}{3} \cos x + C. \] Solving these equations gives: \[ f(x) = \frac{e^x + \frac{1}{3} \cos x + C}{2}, \quad g(x) = \frac{e^x - \frac{1}{3} \cos x - C}{2}. \] Substitute these into the solution for \( u(x, t) \): \[ u(x, t) = \frac{e^{x-3t} + \frac{1}{3} \cos(x-3t) + C}{2} + \frac{e^{x+3t} - \frac{1}{3} \cos(x+3t) - C}{2}. \] Step 3: Evaluate at \( x = \frac{\pi}{2} \) and \( t = \frac{\pi}{6} \) Now, substitute \( x = \frac{\pi}{2} \) and \( t = \frac{\pi}{6} \) into the expression for \( u(x,t) \) to find: \[ u\left(\frac{\pi}{2}, \frac{\pi}{6}\right) = \frac{1}{2} \left( e^{\pi} + \frac{5}{3} \right). \] Thus, the correct answer is: \[ \boxed{(C) \frac{1}{2} \left( e^{\pi} + \frac{5}{3} \right)}. \]