Question

Suppose \[ u(x, t) = \frac{1}{2} \left[ g(x + ct) + g(x - ct) \right] \] is a solution of the following initial value problem of the wave equation \[ u_{tt} = 9 u_{xx}, u(x, 0) = g(x), u_t(x, 0) = 0. \] Then the value of \( c^2 \) is \(\underline{\hspace{2cm}}\) (round off to 2 decimal places).

Hint:

In wave equations, the wave speed \( c \) is related to the coefficient of the second derivative with respect to space and time. For \( u_{tt} = 9 u_{xx} \), \( c^2 = 9 \).

Answer 1

Correct Answer: 9

We are given the wave equation \( u_{tt} = 9 u_{xx} \) and a solution \( u(x, t) = \frac{1}{2} \left[ g(x + ct) + g(x - ct) \right] \). We need to find the value of \( c^2 \).

Step 1: Apply the wave equation to the solution.
The wave equation requires that \( u_{tt} = 9 u_{xx} \), so we need to differentiate the given solution twice with respect to \( t \) and \( x \), respectively. First, differentiate \( u(x, t) \) with respect to \( t \): \[ u_t(x, t) = \frac{1}{2} \left[ g'(x + ct) \cdot c - g'(x - ct) \cdot (-c) \right] = \frac{c}{2} \left[ g'(x + ct) + g'(x - ct) \right]. \] Now, differentiate \( u_t(x, t) \) with respect to \( t \) again to find \( u_{tt} \): \[ u_{tt}(x, t) = \frac{c^2}{2} \left[ g''(x + ct) + g''(x - ct) \right]. \]

Step 2: Differentiate \( u(x, t) \) with respect to \( x \).
Now, differentiate \( u(x, t) \) with respect to \( x \): \[ u_x(x, t) = \frac{1}{2} \left[ g'(x + ct) - g'(x - ct) \right]. \] Now, differentiate \( u_x(x, t) \) with respect to \( x \) again to find \( u_{xx} \): \[ u_{xx}(x, t) = \frac{1}{2} \left[ g''(x + ct) + g''(x - ct) \right]. \]

Step 3: Substitute into the wave equation.
We substitute the expressions for \( u_{tt} \) and \( u_{xx} \) into the wave equation \( u_{tt} = 9 u_{xx} \): \[ \frac{c^2}{2} \left[ g''(x + ct) + g''(x - ct) \right] = 9 \cdot \frac{1}{2} \left[ g''(x + ct) + g''(x - ct) \right]. \] Canceling the common factor of \( \frac{1}{2} \) and solving for \( c^2 \), we get: \[ c^2 = 9. \] Thus, the value of \( c^2 \) is \( 9 \).