Let's carefully solve the given initial-boundary value problem (IBVP) using the eigenfunction expansion method. --- ### **Problem Statement:** \[ u_{tt} = u_{xx} + x(x+1)t, \quad 0 < x < 1, \quad t > 0 \] with boundary conditions: \[ u(0,t) = 0, \quad u(1,t) = 0, \] and initial conditions: \[ u(x,0) = 0, \quad u_t(x,0) = 0. \] --- ### **Step 1: Decompose the problem** The PDE has a nonhomogeneous term: \( x(x+1)t \). To use the eigenfunction expansion, we decompose the solution \( u(x,t) \) into: \[ u(x,t) = v(x,t) + w(x,t), \] where: - \( v(x,t) \) solves the *homogeneous* PDE with the same boundary conditions, - \( w(x,t) \) is a particular solution that accounts for the nonhomogeneous term. --- ### **Step 2: Find a particular solution \( w(x,t) \)** Since the nonhomogeneous term is separable as \( x(x+1) t \), a natural approach is to look for a particular solution of the form: \[ w(x,t) = A(x) t, \] where \( A(x) \) is to be determined. Plug into the PDE: \[ w_{tt} = 0, \] \[ w_{xx} = A''(x) t, \] \[ \text{Right side} = x(x+1) t. \] The PDE becomes: \[ 0 = A''(x) t + x(x+1) t, \] \[ \Rightarrow A''(x) + x(x+1) = 0, \] which is an ODE for \( A(x) \): \[ A''(x) = -x(x+1). \] --- ### **Step 3: Solve for \( A(x) \)** Integrate twice: \[ A''(x) = -x^2 - x. \] First integration: \[ A'(x) = -\int (x^2 + x) dx + C_1 = -\left( \frac{x^3}{3} + \frac{x^2}{2} \right) + C_1, \] Second integration: \[ A(x) = -\int \left( \frac{x^3}{3} + \frac{x^2}{2} \right) dx + C_1 x + C_2. \] Compute: \[ A(x) = -\left( \frac{x^4}{12} + \frac{x^3}{6} \right) + C_1 x + C_2. \] --- ### **Step 4: Define \( w(x,t) \)** Thus, \[ w(x,t) = A(x) t = \left( -\frac{x^4}{12} - \frac{x^3}{6} + C_1 x + C_2 \right) t. \] To satisfy the boundary conditions \( u(0,t) = u(1,t) = 0 \), we require: \[ w(0,t) = 0, \] \[ w(1,t) = 0. \] Check: \[ w(0,t) = \left( 0 - 0 + 0 + C_2 \right) t = C_2 t, \] \[ w(1,t) = \left( -\frac{1}{12} - \frac{1}{6} + C_1 \cdot 1 + C_2 \right) t. \] Set these to zero for all \( t \): \[ C_2 = 0, \] \[ -\frac{1}{12} - \frac{1}{6} + C_1 + 0 = 0, \] \[ C_1 = \frac{1}{12} + \frac{1}{6} = \frac{1}{12} + \frac{2}{12} = \frac{3}{12} = \frac{1}{4}. \] **Therefore:** \[ A(x) = -\frac{x^4}{12} - \frac{x^3}{6} + \frac{1}{4} x, \] and \[ w(x,t) = \left( -\frac{x^4}{12} - \frac{x^3}{6} + \frac{1}{4} x \right) t. \] --- ### **Step 5: Homogeneous problem for \( v(x,t) \)** Now, define: \[ u(x,t) = v(x,t) + w(x,t), \] where \( v(x,t) \) satisfies: \[ v_{tt} = v_{xx}, \] with boundary conditions: \[ v(0,t) = 0, \quad v(1,t) = 0, \] and initial conditions: \[ v(x,0) = u(x,0) - w(x,0) = 0 - 0 = 0, \] \[ v_t(x,0) = u_t(x,0) - w_t(x,0) = 0 - A(x) = -A(x). \] Since \( w(x,0) = 0 \), and \( w_t(x,0) = A(x) \). --- ### **Step 6: Eigenfunction expansion for \( v(x,t) \)** The homogeneous wave equation: \[ v_{tt} = v_{xx}, \] with boundary conditions: \[ v(0,t) = v(1,t) = 0, \] has eigenfunctions: \[ \phi_n(x) = \sin(n \pi x), \quad n=1,2,\dots, \] with eigenvalues: \[ \lambda_n = (n \pi)^2. \] Express: \[ v(x,t) = \sum_{n=1}^\infty T_n(t) \sin(n \pi x). \] --- ### **Step 7: Derive ODEs for \( T_n(t) \)** Plug into the PDE: \[ \sum_{n=1}^\infty T_n''(t) \sin(n \pi x) + \sum_{n=1}^\infty T_n(t) \lambda_n \sin(n \pi x) = 0, \] which reduces to: \[ T_n''(t) + (n \pi)^2 T_n(t) = 0. \] Initial conditions: \[ v(x,0) = 0 \Rightarrow \sum T_n(0) \sin(n \pi x) = 0 \Rightarrow T_n(0) = 0, \] \[ v_t(x,0) = -A(x) \Rightarrow \sum T_n'(0) \sin(n \pi x) = -A(x). \] Project \( -A(x) \) onto \( \sin(n \pi x) \): \[ T_n'(0) = - \frac{2}{1} \int_0^1 A(x) \sin(n \pi x) dx. \] --- ### **Step 8: Compute the Fourier coefficients \( c_n \)** Define: \[ c_n = \int_0^1 A(x) \sin(n \pi x) dx, \] then: \[ T_n'(0) = -2 c_n, \] and \[ T_n(t) = \frac{\sin(n \pi t)}{n \pi} \cdot (-2 c_n), \] since initial displacement is zero, and initial velocity is \( T_n'(0) \). **Solution for \( T_n(t) \):** \[ T_n(t) = \frac{-2 c_n}{n \pi} \sin(n \pi t). \] --- ### **Step 9: **Final solution** Putting everything together: \[ u(x,t) = v(x,t) + w(x,t), \] where \[ v(x,t) = \sum_{n=1}^\infty T_n(t) \sin(n \pi x) = \sum_{n=1}^\infty \left( \frac{-2 c_n}{n \pi} \sin(n \pi t) \right) \sin(n \pi x), \] and \[ w(x,t) = \left( -\frac{x^4}{12} - \frac{x^3}{6} + \frac{1}{4} x \right) t. \] --- ### **Step 10: Summary and Conclusion** - **Particular solution \( w(x,t) \):** Found explicitly to handle the nonhomogeneous term. - **Homogeneous solution \( v(x,t) \):** Expanded in eigenfunctions, with initial conditions projected to compute \( T_n(t) \). - **Total solution:** Sum of the homogeneous and particular solutions. --- ### **Remarks:** - The eigenfunction expansion method effectively reduces the PDE to a set of ODEs. - The particular solution accounts for the nonhomogeneous forcing term. - The initial conditions determine the coefficients for the eigenfunction expansion. --- **This completes the step-by-step eigenfunction expansion solution to the IBVP.**