# Step-by-Step Solution Given PDE: \[ u_{xx} + u_{xt} - 6u_{tt} = 0 \] with initial conditions: \[ u(x,0) = \phi(x), \quad u_t(x,0) = \psi(x) \] **Hint:** Factor the operator as we did for the wave equation. --- ## Step 1: Write the Equation in Operator Form Let’s use operator notation: \[ u_{xx} + u_{xt} - 6u_{tt} = 0 \] \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial x \partial t} - 6 \frac{\partial^2 u}{\partial t^2} = 0 \] --- ## Step 2: Factor the Operator Let \( D_x = \frac{\partial}{\partial x} \) and \( D_t = \frac{\partial}{\partial t} \): \[ u_{xx} + u_{xt} - 6u_{tt} = (D_x^2 + D_x D_t - 6 D_t^2)u = 0 \] Try to factor: \[ D_x^2 + D_x D_t - 6 D_t^2 = (D_x + a D_t)(D_x + b D_t) \] Expand: \[ (D_x + a D_t)(D_x + b D_t) = D_x^2 + (a+b) D_x D_t + ab D_t^2 \] Match coefficients: - \( D_x^2 \): Coefficient is 1 (matches) - \( D_x D_t \): \( a+b = 1 \) - \( D_t^2 \): \( ab = -6 \) Solve for \( a \) and \( b \): \[ a + b = 1 \\ ab = -6 \] This is a quadratic: \[ a^2 - a - 6 = 0 \implies (a-3)(a+2) = 0 \implies a=3, b=-2 \] So, \[ D_x^2 + D_x D_t - 6 D_t^2 = (D_x + 3 D_t)(D_x - 2 D_t) \] --- ## Step 3: Write the Factored PDE \[ (D_x + 3 D_t)(D_x - 2 D_t) u = 0 \] Let \( v = (D_x - 2 D_t)u \). Then, \[ (D_x + 3 D_t) v = 0 \] --- ## Step 4: Solve the First-Order PDEs ### 4.1. Solve \( (D_x - 2 D_t)u = 0 \) This is: \[ u_x - 2u_t = 0 \] This is a transport equation. Its general solution is: \[ u(x, t) = F(x + 2t) \] for some function \( F \). ### 4.2. Now, solve \( (D_x + 3 D_t)v = 0 \) \[ v_x + 3v_t = 0 \] General solution: \[ v(x, t) = G(x - 3t) \] Recall \( v = (D_x - 2 D_t)u \): \[ v(x, t) = u_x - 2u_t = G(x - 3t) \] --- ## Step 5: Find the General Solution We have: \[ u(x, t) = F(x + 2t) \] and \[ u_x - 2u_t = G(x - 3t) \] Let’s try a general solution of the form: \[ u(x, t) = f(x + 2t) + g(x - 3t) \] Calculate: \[ u_x = f'(x + 2t) + g'(x - 3t) \] \[ u_t = 2f'(x + 2t) - 3g'(x - 3t) \] So, \[ u_x - 2u_t = f'(x + 2t) + g'(x - 3t) - 2[2f'(x + 2t) - 3g'(x - 3t)] = f'(x+2t) + g'(x-3t) - 4f'(x+2t) + 6g'(x-3t) = -3f'(x+2t) + 7g'(x-3t) \] Set this equal to \( G(x-3t) \): But for the original equation, \( G(x-3t) \) can be arbitrary, so the general solution is: \[ u(x, t) = f(x + 2t) + g(x - 3t) \] --- ## Step 6: Apply Initial Conditions Given: \[ u(x, 0) = f(x) + g(x) = \phi(x) \] \[ u_t(x, 0) = 2f'(x) - 3g'(x) = \psi(x) \] Now solve for \( f \) and \( g \): From \( u(x, 0) \): \[ f(x) + g(x) = \phi(x) \tag{1} \] From \( u_t(x, 0) \): \[ 2f'(x) - 3g'(x) = \psi(x) \tag{2} \] Differentiate (1): \[ f'(x) + g'(x) = \phi'(x) \tag{3} \] Now solve (2) and (3) for \( f'(x) \) and \( g'(x) \): Multiply (3) by 3: \[ 3f'(x) + 3g'(x) = 3\phi'(x) \] Add to (2): \[ [2f'(x) - 3g'(x)] + [3f'(x) + 3g'(x)] = \psi(x) + 3\phi'(x) \\ 5f'(x) = \psi(x) + 3\phi'(x) \implies f'(x) = \frac{1}{5}[\psi(x) + 3\phi'(x)] \] From (3): \[ g'(x) = \phi'(x) - f'(x) = \phi'(x) - \frac{1}{5}[\psi(x) + 3\phi'(x)] = \frac{5\phi'(x) - (\psi(x) + 3\phi'(x))}{5} = \frac{2\phi'(x) - \psi(x)}{5} \] Integrate: \[ f(x) = \int f'(x) dx = \frac{1}{5} \int [\psi(x) + 3\phi'(x)] dx = \frac{1}{5}\left( \int \psi(x) dx + 3\phi(x) \right) \] \[ g(x) = \int g'(x) dx = \frac{1}{5} \int [2\phi'(x) - \psi(x)] dx = \frac{1}{5} (2\phi(x) - \int \psi(x) dx) \] --- ## Step 7: Write the Final Solution \[ \boxed{ u(x, t) = f(x+2t) + g(x-3t) } \] where \[ f(x) = \frac{1}{5}\left( \int \psi(x) dx + 3\phi(x) \right) \] \[ g(x) = \frac{1}{5} \left(2\phi(x) - \int \psi(x) dx \right) \] Or, **explicitly**: \[ \boxed{ u(x, t) = \frac{1}{5}\left[ \int \psi(x+2t) dx + 3\phi(x+2t) \right] + \frac{1}{5}\left[ 2\phi(x-3t) - \int \psi(x-3t) dx \right] } \] where the indefinite integrals are with respect to the argument. --- ## **Summary of Logic** 1. **Factor the PDE** as in the wave equation. 2. **Solve the resulting first-order PDEs** using the method of characteristics. 3. **Combine to get the general solution** in terms of arbitrary functions. 4. **Apply initial conditions** to determine these functions. 5. **Write the final solution** in explicit form. Let me know if you'd like an example with specific \( \phi \) and \( \psi \)!