Let's solve each part step by step with explanations: --- ## **(a)** Given: \[ u_t = 4k u_{xx}, \quad 0 < x < 2\pi, \; t > 0 \] with boundary conditions: \[ u_x(0,t) = u_x(2\pi,t) = 0 \] and initial condition: \[ u(x,0) = x(2\pi - x)^2 \] ### **Step 1: Separation of Variables and Eigenfunctions** We seek solutions of the form \( u(x,t) = X(x)T(t) \). Plug into the PDE: \[ X(x)T'(t) = 4k X''(x)T(t) \implies \frac{T'(t)}{4k T(t)} = \frac{X''(x)}{X(x)} = -\lambda \] So, \[ T(t) = e^{-4k\lambda t} \] \[ X''(x) + \lambda X(x) = 0 \] with Neumann BCs: \( X'(0) = X'(2\pi) = 0 \). The eigenfunctions and eigenvalues are: \[ X_n(x) = \cos\left(\frac{n x}{2\pi}\right), \quad \lambda_n = \left(\frac{n}{2\pi}\right)^2, \quad n = 0, 1, 2, \ldots \] **Explanation:** We separated variables and found the eigenfunctions satisfying the given Neumann boundary conditions. --- ### **Step 2: Expanding Initial Condition and Writing Solution** Expand \( u(x,0) = x(2\pi - x)^2 \) in terms of the eigenfunctions: \[ u(x,0) = \sum_{n=0}^\infty A_n \cos\left(\frac{n x}{2\pi}\right) \] where \[ A_n = \frac{1}{\pi} \int_{0}^{2\pi} x(2\pi - x)^2 \cos\left(\frac{n x}{2\pi}\right) dx \] So the solution is: \[ u(x,t) = \sum_{n=0}^\infty A_n e^{-4k\lambda_n t} \cos\left(\frac{n x}{2\pi}\right) \] where \( \lambda_n = \left(\frac{n}{2\pi}\right)^2 \). **Explanation:** We wrote the initial condition as a Fourier cosine series and constructed the solution as a sum of eigenfunctions multiplied by their exponential decays. --- ### **Final Answer for (a):** \[ \boxed{ u(x,t) = \sum_{n=0}^\infty A_n e^{-k n^2 t / \pi^2} \cos\left(\frac{n x}{2\pi}\right), \quad A_n = \frac{1}{\pi} \int_{0}^{2\pi} x(2\pi-x)^2 \cos\left(\frac{n x}{2\pi}\right) dx } \] --- ## **(b)** Given: \[ u_t = k u_{xx}, \quad -\infty < x < \infty, \; t > 0 \] \[ u(x,0) = f(x), \quad \text{where} \quad f(x) = \begin{cases} x, & 0 \leq x \leq 4 \\ 0, & x < 0 \text{ or } x > 4 \end{cases} \] ### **Step 1: Use Fundamental Solution (Heat Kernel)** The general solution for the heat equation on the whole line is: \[ u(x,t) = \int_{-\infty}^{\infty} f(\xi) \frac{1}{\sqrt{4\pi k t}} e^{-\frac{(x-\xi)^2}{4k t}} d\xi \] **Explanation:** We use the heat kernel (fundamental solution) to represent the solution as a convolution with the initial condition. --- ### **Step 2: Plug in Initial Condition** Since \( f(\xi) = \xi \) for \( 0 \leq \xi \leq 4 \), and zero otherwise: \[ u(x,t) = \int_{0}^{4} \xi \frac{1}{\sqrt{4\pi k t}} e^{-\frac{(x-\xi)^2}{4k t}} d\xi \] **Explanation:** We restrict the integration to the support of \( f(\xi) \), simplifying the integral. --- ### **Final Answer for (b):** \[ \boxed{ u(x,t) = \int_0^4 \frac{\xi}{\sqrt{4\pi k t}} \exp\left(-\frac{(x-\xi)^2}{4 k t}\right) d\xi } \] Let me know if you need further simplification or explicit computation of the coefficients or integrals!