Question

The solution of \(\frac{\partial^2 u}{\partial t^2} = 4\frac{\partial^2 u}{\partial x^2}\), \(t > 0\), \(-\infty < x < \infty\) satisfying the conditions \(u(x,0) = x\) and \(\frac{\partial u}{\partial t}(x,0) = 0\) is

• A. \(x\)
• B. \(\frac{x^2}{2} + t\)
• C. \(x + 2t\)
• D. \(\frac{t^2}{2} + x\)

Hint:

When initial velocity \(\frac{\partial u}{\partial t}(x,0) = 0\), the solution is simply the average of the initial wave profile shifted left and right.
If \(f(x)\) is linear, the time dependency \(t\) cancels out completely.

Answer 1

The Correct Option is A

Concept:
• The 1D wave equation is given by \(\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}\).
• D'Alembert's formula states that for initial conditions \(u(x,0) = f(x)\) and \(\frac{\partial u}{\partial t}(x,0) = g(x)\), the solution is: \[ u(x,t) = \frac{1}{2}[f(x - ct) + f(x + ct)] + \frac{1}{2c}\int_{x-ct}^{x+ct} g(\xi) \, d\xi \]

Step 1: Identify parameters from the problem statement
Comparing with the standard equation, \(c^2 = 4 \implies c = 2\).
The initial shape profile is \(f(x) = x\).
The initial velocity profile is \(g(x) = 0\).

Step 2: Substitute parameters into D'Alembert's formula
Since \(g(x) = 0\), the integral term becomes zero: \[ u(x,t) = \frac{1}{2}[f(x - 2t) + f(x + 2t)] \] Substitute \(f(x) = x\): \[ u(x,t) = \frac{1}{2}[(x - 2t) + (x + 2t)] \] \[ u(x,t) = \frac{1}{2}[2x] = x \]