Let's solve both problems using the **method of eigenfunction expansion**. --- ### (a) \( u_t - k u_{xx} = x \), \( 0 \leq x \leq \pi \) Boundary conditions: \( u(0, t) = 1 \), \( u(\pi, t) = 0 \), \( u(x, 0) = 0 \) #### **Step 1: Homogenize the Boundary Conditions** Let \( u(x, t) = v(x, t) + w(x) \), where \( w(x) \) is a steady-state function chosen so that \( v(0, t) = v(\pi, t) = 0 \). Set \( w(x) \) so that: - \( w(0) = 1 \) - \( w(\pi) = 0 \) Take \( w(x) = 1 - \frac{x}{\pi} \). Now, substitute into the PDE: \[ u_t - k u_{xx} = x \implies v_t - k v_{xx} = x - k w''(x) \] But \( w''(x) = 0 \), so \[ v_t - k v_{xx} = x \] with \[ v(0, t) = v(\pi, t) = 0, \quad v(x, 0) = -w(x) = \frac{x}{\pi} - 1 \] #### **Step 2: Expand Using Eigenfunctions** The eigenfunctions for Dirichlet BCs are \( \phi_n(x) = \sin(n x) \), with \[ v(x, t) = \sum_{n=1}^{\infty} A_n(t) \sin(nx) \] Substitute into the PDE: \[ \sum_{n=1}^{\infty} A_n'(t) \sin(nx) - k \sum_{n=1}^{\infty} A_n(t) n^2 \sin(nx) = x \] Expand \( x \) in terms of the eigenfunctions (Fourier sine series): \[ x = \sum_{n=1}^{\infty} f_n \sin(nx) \] where \[ f_n = \frac{2}{\pi} \int_{0}^{\pi} x \sin(nx) dx = \frac{2}{\pi} \left[ -\frac{x}{n} \cos(nx) \bigg|_0^{\pi} + \frac{1}{n} \int_0^{\pi} \cos(nx) dx \right] \] The calculation yields: \[ f_n = \frac{2}{\pi} \left( -\frac{\pi}{n} \cos(n\pi) + 0 \right) = \frac{2}{n} (-1)^{n+1} \] So the ODE system for each \( n \) is: \[ A_n'(t) + k n^2 A_n(t) = \frac{2}{n} (-1)^{n+1} \] This is a linear ODE; the integrating factor is \( e^{k n^2 t} \): \[ A_n(t) = e^{-k n^2 t} \left[ \int \frac{2}{n} (-1)^{n+1} e^{k n^2 t} dt + C_n \right] \] The particular solution is \[ A_{n,\text{part}}(t) = \frac{2}{k n^3} (-1)^{n+1} \] General solution: \[ A_n(t) = \frac{2}{k n^3} (-1)^{n+1} + C_n e^{-k n^2 t} \] Use initial condition \( v(x, 0) = \frac{x}{\pi} - 1 \): Expand \( v(x, 0) \) in sine series: \[ C_n = \text{Fourier sine coefficient of } \left( \frac{x}{\pi} - 1 \right) \] \[ C_n = \frac{2}{\pi} \int_0^\pi \left( \frac{x}{\pi} - 1 \right) \sin(nx) dx \] \[ = \frac{2}{\pi^2} \int_0^\pi x \sin(nx) dx - \frac{2}{\pi} \int_0^\pi \sin(nx) dx \] First term is as before: \[ \frac{2}{\pi^2} \cdot \frac{2}{n} (-1)^{n+1} = \frac{4}{\pi^2 n} (-1)^{n+1} \] Second term: \[ \frac{2}{\pi} \left( -\frac{1}{n} [\cos(n\pi) - 1] \right) = \frac{2}{\pi n} [1 - (-1)^n] \] So, \[ C_n = \frac{4}{\pi^2 n} (-1)^{n+1} - \frac{2}{\pi n} [1 - (-1)^n] \] #### **Final Answer for (a):** \[ u(x, t) = 1 - \frac{x}{\pi} + \sum_{n=1}^{\infty} \left[ \frac{2}{k n^3} (-1)^{n+1} + C_n e^{-k n^2 t} \right] \sin(nx) \] where \[ C_n = \frac{4}{\pi^2 n} (-1)^{n+1} - \frac{2}{\pi n} [1 - (-1)^n] \] --- ### (b) \( u_{tt} - c^2 u_{xx} = x(\pi - x) \), \( 0 \leq x \leq \pi \) Boundary conditions: \( u(0, t) = 0 \), \( u(\pi, t) = \sin t \), \( u(x, 0) = 0 \) #### **Step 1: Homogenize the Boundary Conditions** Let \( u(x, t) = v(x, t) + h(x, t) \), where \( h(x, t) \) satisfies the nonhomogeneous boundary: Set \( h(x, t) = \sin t \cdot \frac{x}{\pi} \), so: - \( h(0, t) = 0 \) - \( h(\pi, t) = \sin t \) So \( v(x, t) \) has homogeneous BCs: \[ v(0, t) = v(\pi, t) = 0 \] Now, \[ u_{tt} - c^2 u_{xx} = x(\pi - x) \] \[ v_{tt} + h_{tt} - c^2 (v_{xx} + h_{xx}) = x(\pi - x) \] So \[ v_{tt} - c^2 v_{xx} = x(\pi - x) - h_{tt} + c^2 h_{xx} \] Calculate \( h_{tt} = -\sin t \cdot \frac{x}{\pi} \), \( h_{xx} = 0 \): Thus, \[ v_{tt} - c^2 v_{xx} = x(\pi - x) + \sin t \cdot \frac{x}{\pi} \] with \[ v(0, t) = v(\pi, t) = 0, \quad v(x, 0) = -h(x, 0) = 0, \quad v_t(x, 0) = -h_t(x, 0) = -\cos 0 \cdot \frac{x}{\pi} = -\frac{x}{\pi} \] #### **Step 2: Expand Using Eigenfunctions** Expand \( v(x, t) = \sum_{n=1}^{\infty} A_n(t) \sin(nx) \). Expand right side in sine series: \[ x(\pi - x) = \sum_{n=1}^{\infty} F_n \sin(nx) \] where \[ F_n = \frac{2}{\pi} \int_0^\pi x(\pi - x) \sin(nx) dx \] Similarly, \( \sin t \cdot \frac{x}{\pi} \) expands as before with coefficients \( \frac{2}{n} (-1)^{n+1} \sin t \). So, \[ A_n''(t) + c^2 n^2 A_n(t) = F_n + \frac{2}{n} (-1)^{n+1} \sin t \] This is a nonhomogeneous ODE, solution is: \[ A_n(t) = \text{(particular solution for each term)} + C_{1n} \cos(c n t) + C_{2n} \sin(c n t) \] The particular solution for \( F_n \) is \( \frac{F_n}{c^2 n^2} \). For \( \frac{2}{n} (-1)^{n+1} \sin t \), use undetermined coefficients: Try \( a \sin t + b \cos t \): \[ A_n'' + c^2 n^2 A_n = \frac{2}{n} (-1)^{n+1} \sin t \] \[ (-a) \sin t - b \cos t + c^2 n^2 (a \sin t + b \cos t) = \frac{2}{n} (-1)^{n+1} \sin t \] \[ (a(-1 + c^2 n^2)) \sin t + b (-1 + c^2 n^2) \cos t = \frac{2}{n} (-1)^{n+1} \sin t \] So \( b = 0 \), \( a = \frac{2}{n} \frac{(-1)^{n+1}}{c^2 n^2 - 1} \) (assuming \( c^2 n^2 \neq 1 \)). Initial conditions \( v(x, 0) = 0 \), \( v_t(x, 0) = -\frac{x}{\pi} \): Expand \( -\frac{x}{\pi} \) in sine series, as before. #### **Final Answer for (b):** \[ u(x, t) = \sin t \frac{x}{\pi} + \sum_{n=1}^{\infty} \left[ \frac{F_n}{c^2 n^2} + \frac{2}{n} \frac{(-1)^{n+1}}{c^2 n^2 - 1} \sin t + C_{1n} \cos(c n t) + C_{2n} \sin(c n t) \right] \sin(nx) \] where \[ F_n = \frac{2}{\pi} \int_0^\pi x(\pi - x) \sin(nx) dx \] and \( C_{1n}, C_{2n} \) are found from the initial conditions. --- **If you want explicit evaluations of the integrals or the \( C_{1n}, C_{2n} \) coefficients, let me know!**

Certainly! Here's a detailed, step-by-step explanation for solving both problems using the method of eigenfunction expansion. --- ## Problem (a): **Equation:** \[ u_t - k u_{xx} = x, \quad 0 < x < \pi \] **Boundary conditions:** \[ u(0, t) = 1, \quad u(\pi, t) = 0 \] **Initial condition:** \[ u(x, 0) = 0 \] --- ### Step 1: Homogenize boundary conditions Since the boundary conditions are nonhomogeneous (values at boundaries are not zero), we first separate the solution into a steady part \( w(x) \) that satisfies the boundary conditions, and a transient part \( v(x, t) \): \[ u(x, t) = v(x, t) + w(x) \] Choose \( w(x) \) such that: \[ w(0) = 1, \quad w(\pi) = 0 \] A linear function satisfying this is: \[ w(x) = 1 - \frac{x}{\pi} \] which satisfies the boundary conditions. Then, the boundary conditions for \( v(x, t) \) are homogeneous: \[ v(0, t) = 0, \quad v(\pi, t) = 0 \] --- ### Step 2: Reformulate the PDE for \( v(x, t) \) Compute derivatives: \[ u_t = v_t, \quad u_{xx} = v_{xx} + w''(x) \] Since \( w(x) \) is linear, \( w''(x) = 0 \). The PDE becomes: \[ v_t - k v_{xx} = x - k w''(x) = x \] Initial condition: \[ u(x, 0) = 0 \implies v(x, 0) = -w(x) = -\left( 1 - \frac{x}{\pi} \right) = \frac{x}{\pi} - 1 \] --- ### Step 3: Eigenfunction expansion Because \( v(0, t) = v(\pi, t) = 0 \), expand \( v \) as: \[ v(x, t) = \sum_{n=1}^\infty A_n(t) \sin(n x) \] The right side \( x \) is expanded as a sine series: \[ x = \sum_{n=1}^\infty f_n \sin(n x) \] where \[ f_n = \frac{2}{\pi} \int_0^\pi x \sin(n x) dx \] **Calculating \( f_n \):** \[ f_n = \frac{2}{\pi} \left[ -\frac{x}{n} \cos(n x) \Big|_0^\pi + \frac{1}{n} \int_0^\pi \cos(n x) dx \right] \] \[ = \frac{2}{\pi} \left[ -\frac{\pi}{n} \cos(n \pi) + 0 \right] \] \[ = \frac{2}{\pi} \left( -\frac{\pi}{n} (-1)^n \right) = \frac{2}{n} (-1)^{n+1} \] --- ### Step 4: Derive ODEs for \( A_n(t) \) Substitute into PDE: \[ \sum_{n=1}^\infty A_n'(t) \sin(n x) + k \sum_{n=1}^\infty A_n(t) n^2 \sin(n x) = \sum_{n=1}^\infty f_n \sin(n x) \] Matching coefficients: \[ A_n'(t) + k n^2 A_n(t) = f_n = \frac{2}{n} (-1)^{n+1} \] which is a first-order linear ODE. --- ### Step 5: Solve for \( A_n(t) \) Solution: \[ A_n(t) = e^{-k n^2 t} \left[ \int_0^t e^{k n^2 s} \frac{2}{n} (-1)^{n+1} ds + C_n \right] \] The integral: \[ \int_0^t e^{k n^2 s} ds = \frac{1}{k n^2} \left( e^{k n^2 t} - 1 \right) \] So, \[ A_n(t) = e^{-k n^2 t} \left[ \frac{2}{n} (-1)^{n+1} \frac{e^{k n^2 t} - 1}{k n^2} + C_n \right] \] Simplify: \[ A_n(t) = \frac{2}{k n^3} (-1)^{n+1} \left( 1 - e^{-k n^2 t} \right) + C_n e^{-k n^2 t} \] --- ### Step 6: Apply initial conditions to find \( C_n \) At \( t=0 \): \[ v(x, 0) = \sum_{n=1}^\infty A_n(0) \sin(n x) = \frac{x}{\pi} - 1 \] Express \( \frac{x}{\pi} - 1 \) as sine series: \[ \frac{x}{\pi} - 1 = \sum_{n=1}^\infty C_n \sin(n x) \] Using Fourier sine coefficients: \[ C_n = \frac{2}{\pi} \int_0^\pi \left( \frac{x}{\pi} - 1 \right) \sin(n x) dx \] This integral has been computed earlier, yielding the coefficients: \[ C_n = \frac{4}{\pi^2 n} (-1)^{n+1} - \frac{2}{\pi n} [1 - (-1)^n] \] --- ## **Final solution for (a):** \[ \boxed{ u(x, t) = 1 - \frac{x}{\pi} + \sum_{n=1}^\infty \left[ \frac{2}{k n^3} (-1)^{n+1} \left( 1 - e^{-k n^2 t} \right) + C_n e^{-k n^2 t} \right] \sin(n x) } \] with \[ C_n = \frac{4}{\pi^2 n} (-1)^{n+1} - \frac{2}{\pi n} [1 - (-1)^n] \] --- ## **Problem (b):** **Equation:** \[ u_{tt} - c^2 u_{xx} = x(\pi - x) \] **Boundary conditions:** \[ u(0, t) = 0, \quad u(\pi, t) = \sin t \] **Initial conditions:** \[ u(x,0) = 0, \quad u_t(x,0) = 0 \] --- ### Step 1: Homogenize boundary conditions Since \( u(\pi, t) \neq 0 \), define: \[ u(x, t) = v(x, t) + h(x, t) \] where \( h(x, t) \) satisfies the boundary conditions: \[ h(0, t) = 0, \quad h(\pi, t) = \sin t \] Pick: \[ h(x, t) = \frac{x}{\pi} \sin t \] which satisfies: \[ h(0, t) = 0, \quad h(\pi, t) = \sin t \] The residual \( v(x, t) \) then satisfies: \[ v(0, t) = v(\pi, t) = 0 \] --- ### Step 2: PDE for \( v(x, t) \) Calculate derivatives: \[ u_{tt} = v_{tt} + h_{tt}, \quad u_{xx} = v_{xx} + h_{xx} \] Since \( h(x, t) = \frac{x}{\pi} \sin t \): \[ h_{tt} = -\frac{x}{\pi} \sin t \] \[ h_{xx} = 0 \] Plug into PDE: \[ v_{tt} + h_{tt} - c^2 (v_{xx} + 0) = x(\pi - x) \] which simplifies to: \[ v_{tt} - c^2 v_{xx} = x(\pi - x) + \frac{x}{\pi} \sin t \] Initial conditions: \[ v(x, 0) = u(x, 0) - h(x, 0) = 0 - 0 = 0 \] \[ v_t(x, 0) = u_t(x, 0) - h_t(x, 0) = 0 - \frac{x}{\pi} \cos 0 = - \frac{x}{\pi} \] --- ### Step 3: Eigenfunction expansion Expand \( v \): \[ v(x, t) = \sum_{n=1}^\infty A_n(t) \sin(n x) \] Expand RHS: \[ x(\pi - x) + \frac{x}{\pi} \sin t = \sum_{n=1}^\infty F_n \sin(n x) + \left( \sum_{n=1}^\infty G_n \sin(n x) \right) \sin t \] where: \[ F_n = \frac{2}{\pi} \int_0^\pi x(\pi - x) \sin(n x) dx \] and \[ G_n = \frac{2}{\pi} \int_0^\pi \left( -\frac{x}{\pi} \right) \sin(n x) dx = - \frac{2}{\pi^2} \int_0^\pi x \sin(n x) dx \] The integral for \( F_n \) is similar to earlier calculations; for \( G_n \): \[ G_n = - \frac{2}{\pi^2} \times \frac{2}{n} (-1)^{n+1} = - \frac{4}{\pi^2 n} (-1)^{n+1} \] --- ### Step 4: ODEs for \( A_n(t) \) The PDE reduces to: \[ A_n''(t) + c^2 n^2 A_n(t) = F_n + G_n \sin t \] with initial conditions: \[ A_n(0) = 0, \quad A_n'(0) = - \frac{2}{\pi} \int_0^\pi x \sin(n x) dx \] which can be computed similarly. The solution: \[ A_n(t) = \text{homogeneous solution} + \text{particular solution} \] homogeneous: \[ A_{n,hom}(t) = C_{1n} \cos(c n t) + C_{2n} \sin(c n t) \] particular: \[ A_{n,part}(t) = \frac{F_n}{c^2 n^2} + \frac{G_n}{c^2 n^2 - 1} \sin t \] assuming \( c^2 n^2 \neq 1 \). --- ## **Final solution for (b):** \[ \boxed{ u(x, t) = \frac{x}{\pi} \sin t + \sum_{n=1}^\infty \left[ \frac{F_n}{c^2 n^2} + \frac{G_n}{c^2 n^2 - 1} \sin t + C_{1n} \cos(c n t) + C_{2n} \sin(c n t) \right] \sin(n x) } \] where: \[ F_n = \frac{2}{\pi} \int_0^\pi x(\pi - x) \sin(n x) dx \] \[ G_n = - \frac{4}{\pi^2 n} (-1)^{n+1} \] and \( C_{1n}, C_{2n} \) are determined from initial conditions. --- ## **Summary:** - For **both problems**, the key steps are: 1. **Homogenize boundary conditions** by subtracting or adding a steady function. 2. **Expand the remaining problem** in eigenfunctions (sine functions for Dirichlet BCs). 3. **Express RHS** as a sine series. 4. **Solve the resulting ODEs** for the time-dependent coefficients. 5. **Apply initial conditions** to determine integration constants. 6. **Write the complete solution** by combining all parts. This method converts PDEs into manageable ordinary differential equations, leveraging the orthogonality and completeness of eigenfunctions. --- If you need further clarification on any step or specific integral evaluations, feel free to ask!

Certainly! Here's a clear, step-by-step explanation in words for solving **problem (b)** using the method of eigenfunction expansion: --- ### Explanation for Problem (b): **Equation:** \[ u_{tt} - c^2 u_{xx} = x(\pi - x) \] defined for \( 0 < x < \pi \), with boundary conditions: \[ u(0, t) = 0, \quad u(\pi, t) = \sin t \] and initial conditions: \[ u(x, 0) = 0, \quad u_t(x, 0) = 0 \] --- ### Step 1: Homogenizing the boundary conditions Since the boundary condition at \( x = \pi \) is not zero (it's \( \sin t \)), we first create a function that satisfies this boundary condition, so we can work with a problem with homogeneous (zero) boundary conditions. We do this by defining a new function: \[ u(x, t) = v(x, t) + h(x, t) \] where \( h(x, t) \) is chosen such that: \[ h(0, t) = 0, \quad h(\pi, t) = \sin t \] A simple function satisfying these boundary conditions is: \[ h(x, t) = \frac{x}{\pi} \sin t \] because at \( x=0 \), \( h=0 \), and at \( x=\pi \), \( h=\sin t \). This transformation means our new function \( v(x, t) \) now has boundary conditions: \[ v(0, t) = 0, \quad v(\pi, t) = 0 \] --- ### Step 2: Deriving the PDE for \( v(x, t) \) Next, we substitute \( u = v + h \) into the original PDE. Since \( h \) depends on \( x \) and \( t \), derivatives are: - \( u_{tt} = v_{tt} + h_{tt} \) - \( u_{xx} = v_{xx} + h_{xx} \) Calculate: - \( h_{tt} = - \frac{x}{\pi} \sin t \) (since \( h \) is linear in \( x \), only \( \sin t \) varies with \( t \)) - \( h_{xx} = 0 \) (since \( h \) is linear in \( x \)) Substituting into the PDE gives: \[ v_{tt} + h_{tt} - c^2 (v_{xx} + h_{xx}) = x(\pi - x) \] which simplifies to: \[ v_{tt} - c^2 v_{xx} = x(\pi - x) + \frac{x}{\pi} \sin t \] Note the right side now includes the original source term \( x(\pi - x) \) plus an additional term involving \( \sin t \). The initial conditions become: \[ v(x, 0) = u(x, 0) - h(x, 0) = 0 - 0 = 0 \] \[ v_t(x, 0) = u_t(x, 0) - h_t(x, 0) = 0 - \frac{x}{\pi} \cos 0 = - \frac{x}{\pi} \] --- ### Step 3: Eigenfunction expansion Since \( v \) has homogeneous boundary conditions (\( v(0,t)=v(\pi,t)=0 \)), we expand \( v \) as a series of sine functions: \[ v(x, t) = \sum_{n=1}^\infty A_n(t) \sin(n x) \] Similarly, the right side of the PDE (the source term) is expanded into sine series: \[ x(\pi - x) + \frac{x}{\pi} \sin t = \sum_{n=1}^\infty F_n \sin(n x) + \left( \sum_{n=1}^\infty G_n \sin(n x) \right) \sin t \] The coefficients \( F_n \) and \( G_n \) are obtained by integrating the respective functions against \( \sin(n x) \), leveraging the orthogonality of sine functions. --- ### Step 4: Solving for the coefficients \( A_n(t) \) Each coefficient \( A_n(t) \) satisfies a second-order ordinary differential equation: \[ A_n''(t) + c^2 n^2 A_n(t) = F_n + G_n \sin t \] This is a standard forced harmonic oscillator equation with a constant forcing \( F_n \) and a time-dependent forcing \( G_n \sin t \). The general solution includes: - **Homogeneous part:** involving constants determined by initial conditions - **Particular part:** accounting for the steady forcing terms The particular solution for the constant forcing \( F_n \) is simply \( \frac{F_n}{c^2 n^2} \). For the \( G_n \sin t \) forcing, a particular solution involves sine and cosine functions with coefficients depending on the parameters. --- ### Step 5: Applying initial conditions Using the initial conditions (\( v(x,0)=0 \), \( v_t(x,0) = -x/\pi \)), we compute the initial values of the coefficients \( A_n(0) \) and \( A_n'(0) \). These are obtained by projecting the initial displacement and velocity onto the sine basis, resulting in formulas for the constants \( C_{1n} \) and \( C_{2n} \). --- ### **Final solution:** Putting it all together, the original solution \( u(x,t) \) is reconstructed as: \[ u(x, t) = \frac{x}{\pi} \sin t + \sum_{n=1}^\infty \left[ \frac{F_n}{c^2 n^2} + \text{(terms involving } \sin t \text{ and initial conditions)} \right] \sin(n x) \] This expression combines the particular solution that accounts for the source term and the homogeneous eigenfunction expansion that satisfies the boundary conditions and initial data. --- ### **Summary:** - We transformed the problem to have homogeneous boundary conditions by subtracting a carefully chosen function. - We expanded the solution and source terms into sine series, leveraging their orthogonality. - We solved the resulting ordinary differential equations for each mode. - The complete solution is a sum of steady-state parts, oscillatory parts, and initial condition adjustments. --- If you'd like, I can help with specific calculations of the integrals or the coefficients!