Let's break the solution into **two steps** with explanations after each, then give the **final answer** at the end. --- ## **Step 1: Formulate the Solution Using Separation of Variables** ### **Explanation** Given the PDE: \[ \frac{\partial T}{\partial t} = \frac{\partial^2 T}{\partial x^2} \] with boundary conditions: \[ \frac{\partial T}{\partial x}(, t) = , \quad T(1, t) = \] and initial condition: \[ T(x, ) = f(x) \] Since we have a Neumann BC at \( x = \) and Dirichlet BC at \( x = 1 \), the **eigenfunctions** are: \[ \phi_n(x) = \cos\left(\left(n - \frac{1}{2}\right)\pi x\right) \] with corresponding eigenvalues: \[ \lambda_n = \left(n - \frac{1}{2}\right)^2 \pi^2 \] The general solution is: \[ T(x, t) = \sum_{n=1}^\infty A_n \cos\left(\left(n - \frac{1}{2}\right)\pi x\right) e^{-\lambda_n t} \] with coefficients: \[ A_n = 2 \int_^1 f(x) \cos\left(\left(n - \frac{1}{2}\right)\pi x\right) dx \] --- ## **Step 2: Write \( f(x) \) and Find the Coefficients \( A_n \)** ### **Explanation** Given \( f(x) \) is the pulse: \[ f(x) = \begin{cases} & < x < \frac{1}{2} - \frac{w}{2} \\ \frac{T_}{w} & \frac{1}{2} - \frac{w}{2} < x < \frac{1}{2} + \frac{w}{2} \\ & \frac{1}{2} + \frac{w}{2} < x < 1 \end{cases} \] Plug this into the formula for \( A_n \): \[ A_n = 2 \int_{1/2 - w/2}^{1/2 + w/2} \frac{T_}{w} \cos\left(\left(n - \frac{1}{2}\right)\pi x\right) dx \] Change of variables: Let \( y = x - 1/2 \), so as \( x \) goes from \( 1/2 - w/2 \) to \( 1/2 + w/2 \), \( y \) goes from \( -w/2 \) to \( w/2 \): \[ A_n = 2 \cdot \frac{T_}{w} \int_{-w/2}^{w/2} \cos\left(\left(n - \frac{1}{2}\right)\pi (y + 1/2)\right) dy \] Use the cosine addition formula: \[ \cos(a(y + 1/2)) = \cos(ay)\cos(a/2) - \sin(ay)\sin(a/2) \] Integrate, noting that the sine terms will cancel over symmetric limits, and the cosine term integrates to: \[ \int_{-w/2}^{w/2} \cos(ay) dy = \frac{2 \sin(a w/2)}{a} \] So: \[ A_n = 2 \frac{T_}{w} \cos\left(\left(n - \frac{1}{2}\right)\frac{\pi}{2}\right) \cdot \frac{2 \sin\left(\left(n - \frac{1}{2}\right)\frac{\pi w}{2}\right)}{\left(n - \frac{1}{2}\right)\pi} \] Simplify: \[ A_n = \frac{4 T_}{w \pi (n - 1/2)} \cos\left(\left(n - \frac{1}{2}\right)\frac{\pi}{2}\right) \sin\left(\left(n - \frac{1}{2}\right)\frac{\pi w}{2}\right) \] --- ## **Final Answer** \[ \boxed{ T(x, t) = \sum_{n=1}^\infty \frac{4 T_}{w \pi (n - 1/2)} \cos\left(\left(n - \frac{1}{2}\right)\frac{\pi}{2}\right) \sin\left(\left(n - \frac{1}{2}\right)\frac{\pi w}{2}\right) \cos\left(\left(n - \frac{1}{2}\right)\pi x\right) e^{-\left(n - \frac{1}{2}\right)^2 \pi^2 t} } \] This formula gives the solution for \( T(x,t) \) in terms of the initial pulse function and the specified boundary conditions. You can use this expression to plot \( T(x, t) \) as requested. --- **Explanation Recap**: - **Step 1:** Used separation of variables and boundary conditions to set up the solution and eigenfunctions. - **Step 2:** Incorporated the pulse initial condition, calculated the Fourier coefficients, and gave the explicit final form.

## **Problem Overview** We have a heat conduction problem described by the following partial differential equation (PDE): \[ \frac{\partial T}{\partial t} = \frac{\partial^2 T}{\partial x^2} \] with specific boundary and initial conditions. We need to find the solution for \( T(x, t) \) and analyze the behavior of the temperature at specific points. --- ## **Step 1: Formulate the Solution Using Separation of Variables** ### **Boundary Conditions** 1. Neumann BC at \( x = 0 \): \(\frac{\partial T}{\partial x}(0, t) = 0\) 2. Dirichlet BC at \( x = 1 \): \(T(1, t) = 0\) ### **Initial Condition** Given by the pulse function \( f(x) \). ### **General Solution Using Separation of Variables** Assuming a solution of the form \( T(x, t) = X(x)T(t) \), we can rewrite the PDE as: \[ \frac{1}{T}\frac{dT}{dt} = \frac{1}{X}\frac{d^2X}{dx^2} = -\lambda \] This leads to two ordinary differential equations (ODEs): 1. **For \( T(t) \)**: \[ \frac{dT}{dt} + \lambda T = 0 \quad \Rightarrow \quad T(t) = T_0 e^{-\lambda t} \] 2. **For \( X(x) \)**: \[ \frac{d^2X}{dx^2} + \lambda X = 0 \] The boundary conditions yield eigenfunctions: \[ X_n(x) = \cos\left(\left(n - \frac{1}{2}\right)\pi x\right) \] with eigenvalues: \[ \lambda_n = \left(n - \frac{1}{2}\right)^2 \pi^2 \] ### **Complete Solution** The general solution becomes: \[ T(x, t) = \sum_{n=1}^\infty A_n \cos\left(\left(n - \frac{1}{2}\right)\pi x\right) e^{-\left(n - \frac{1}{2}\right)^2 \pi^2 t} \] ### **Explanation** We have separated the variables and derived the general form of the solution using boundary conditions. This allows us to express the solution as a series. --- ## **Step 2: Calculate Coefficients \( A_n \)** ### **Using the Pulse Function** The function \( f(x) \) is defined as follows: \[ f(x) = \begin{cases} 0 & 0 < x < \frac{1}{2} - \frac{w}{2} \\ \frac{T_0}{w} & \frac{1}{2} - \frac{w}{2} < x < \frac{1}{2} + \frac{w}{2} \\ 0 & \frac{1}{2} + \frac{w}{2} < x < 1 \end{cases} \] ### **Finding Coefficients** The coefficients \( A_n \) are computed using: \[ A_n = \frac{2}{1} \int_0^1 f(x) \cos\left(\left(n - \frac{1}{2}\right)\pi x\right) dx \] Substituting \( f(x) \): \[ A_n = 2 \int_{\frac{1}{2} - \frac{w}{2}}^{\frac{1}{2} + \frac{w}{2}} \frac{T_0}{w} \cos\left(\left(n - \frac{1}{2}\right)\pi x\right) dx \] ### **Integration** Using a change of variables \( y = x - \frac{1}{2} \): \[ A_n = 2 \frac{T_0}{w} \int_{-\frac{w}{2}}^{\frac{w}{2}} \cos\left(\left(n - \frac{1}{2}\right)\pi \left(y + \frac{1}{2}\right)\right) dy \] Applying the cosine addition formula: \[ \cos\left(a(y + \frac{1}{2})\right) = \cos(ay)\cos\left(\frac{a}{2}\right) - \sin(ay)\sin\left(\frac{a}{2}\right) \] Integrating the cosine term yields: \[ A_n = \frac{4 T_0}{w \pi (n - \frac{1}{2})} \sin\left(\left(n - \frac{1}{2}\right)\frac{\pi w}{2}\right) \] ### **Explanation** By substituting the pulse function into the formula for \( A_n \) and performing the integration, we can determine the coefficients that will shape our solution. --- ## **Final Answer** The solution for \( T(x, t) \) can be expressed as: \[ \boxed{ T(x, t) = \sum_{n=1}^\infty A_n \cos\left(\left(n - \frac{1}{2}\right)\pi x\right) e^{-\left(n - \frac{1}{2}\right)^2 \pi^2 t} } \] Where: \[ A_n = \frac{4 T_0}{w \pi (n - \frac{1}{2})} \sin\left(\left(n - \frac{1}{2}\right)\frac{\pi w}{2}\right) \] ### **Convergence Behavior** At the points \( x = \frac{1}{2} - \frac{w}{2} \) and \( x = \frac{1}{2} + \frac{w}{2} \), the solution converges to the average temperature influenced by the initial pulse function. This formula gives a comprehensive understanding of the temperature distribution over time in the rod under the given conditions.