Let's solve the PDE \( u_{tt} = 9u_{xx} \) using the **Fourier sine series** step by step. --- ### **Step 1: Assume a solution using separation of variables** Assume \( u(x, t) = X(x)T(t) \). Plug into the PDE: \[ u_{tt} = 9u_{xx} \implies X(x)T''(t) = 9X''(x)T(t) \] Divide both sides by \( X(x)T(t) \): \[ \frac{T''(t)}{T(t)} = 9 \frac{X''(x)}{X(x)} = -\lambda \] where \( \lambda \) is a separation constant. - **Explanation:** We separate variables to reduce the PDE to two ODEs. --- ### **Step 2: Write general solution using the Fourier sine series** The spatial part is: \[ X''(x) + \frac{\lambda}{9} X(x) = 0 \] The general solution (assuming Dirichlet BCs, e.g., \( u(0,t)=u(L,t)=0 \)): \[ X_n(x) = \sin\left(\frac{n\pi x}{L}\right), \quad \frac{\lambda_n}{9} = \left(\frac{n\pi}{L}\right)^2 \implies \lambda_n = 9\left(\frac{n\pi}{L}\right)^2 \] The temporal part: \[ T_n''(t) + \lambda_n T_n(t) = 0 \implies T_n(t) = A_n \cos\left(3\frac{n\pi}{L}t\right) + B_n \sin\left(3\frac{n\pi}{L}t\right) \] So the full solution is: \[ u(x, t) = \sum_{n=1}^{\infty} \left[ A_n \cos\left(3\frac{n\pi}{L}t\right) + B_n \sin\left(3\frac{n\pi}{L}t\right) \right] \sin\left(\frac{n\pi x}{L}\right) \] - **Explanation:** The boundary conditions (typically Dirichlet) lead us to a sine Fourier series in space, and the time-dependent part is a combination of sines and cosines with frequencies determined by separation. --- ## **Final Answer** \[ \boxed{ u(x, t) = \sum_{n=1}^{\infty} \left[ A_n \cos\left(3\frac{n\pi}{L}t\right) + B_n \sin\left(3\frac{n\pi}{L}t\right) \right] \sin\left(\frac{n\pi x}{L}\right) } \] where \( A_n \) and \( B_n \) are determined by the initial conditions.