Let's tackle ** (i)** step-by-step. --- ## **Step 1: Find the General Solution** The PDE is: \[ \frac{\partial T}{\partial t} = \frac{\partial^ T}{\partial x^2} \] with BCs: \[ T(, t) = T(1, t) = \] and initial condition: \[ T(x, ) = f(x) \] where \( f(x) \) is a pulse function. ### **Separation of Variables and Fourier Series** The general solution (using eigenfunction expansion for Dirichlet BCs) is: \[ T(x, t) = \sum_{n=1}^\infty B_n \sin(n \pi x) e^{-(n \pi)^2 t} \] where the coefficients \( B_n \) are found from the initial condition: \[ T(x, ) = f(x) = \sum_{n=1}^\infty B_n \sin(n \pi x) \] \[ B_n = 2 \int_{}^{1} f(x) \sin(n \pi x) dx \] **Explanation:** We use separation of variables, given the homogeneous Dirichlet BCs, leading to a sine series expansion for the solution. The time decay comes from the heat equation eigenvalues. --- ## **Step 2: Apply the Pulse Initial Condition** The pulse function is: \[ f(x) = \begin{cases} & < x < \frac{1}{2} - \frac{w}{2} \\ \frac{T_}{w} & \frac{1}{2} - \frac{w}{2} < x < \frac{1}{2} + \frac{w}{2} \\ & \frac{1}{2} + \frac{w}{2} < x < 1 \end{cases} \] So, \[ B_n = 2 \int_{}^{1} f(x) \sin(n \pi x) dx = 2 \int_{\frac{1}{2} - \frac{w}{2}}^{\frac{1}{2} + \frac{w}{2}} \frac{T_}{w} \sin(n \pi x) dx \] Evaluate the integral: \[ B_n = \frac{2T_}{w} \int_{\frac{1}{2} - \frac{w}{2}}^{\frac{1}{2} + \frac{w}{2}} \sin(n \pi x) dx \] \[ = \frac{2T_}{w} \left[ -\frac{1}{n\pi} \cos(n\pi x) \right]_{\frac{1}{2} - \frac{w}{2}}^{\frac{1}{2} + \frac{w}{2}} \] \[ = -\frac{2T_}{n\pi w} \left( \cos\left(n\pi \left(\frac{1}{2} + \frac{w}{2}\right)\right) - \cos\left(n\pi \left(\frac{1}{2} - \frac{w}{2}\right)\right) \right) \] Use the identity \( \cos(a + b) - \cos(a - b) = -2 \sin a \sin b \): Let \( a = n\pi/2 \), \( b = n\pi w/2 \): \[ = -\frac{2T_}{n\pi w} \left( -2 \sin\left(\frac{n\pi}{2}\right) \sin\left(\frac{n\pi w}{2}\right) \right) \] \[ = \frac{4 T_}{n\pi w} \sin\left(\frac{n\pi}{2}\right) \sin\left(\frac{n\pi w}{2}\right) \] So, \[ \boxed{ T(x, t) = \sum_{n=1}^\infty \left[ \frac{4 T_}{n\pi w} \sin\left(\frac{n\pi}{2}\right) \sin\left(\frac{n\pi w}{2}\right) \right] \sin(n\pi x) e^{-(n\pi)^2 t} } \] **Explanation:** We substitute the piecewise initial condition and evaluate the integral for the Fourier coefficients, simplifying using trigonometric identities. --- ## **Final Answer** \[ \boxed{ T(x, t) = \sum_{n=1}^\infty \left[ \frac{4 T_}{n\pi w} \sin\left(\frac{n\pi}{2}\right) \sin\left(\frac{n\pi w}{2}\right) \right] \sin(n\pi x) e^{-(n\pi)^2 t} } \] This formula gives the temperature at any point and time for the given initial pulse.

# Solution to the Heat Equation Problem ## **(i) General Solution** ### **Step 1: Determine the General Solution** The heat equation is given by: \[ \frac{\partial T}{\partial t} = \frac{\partial^2 T}{\partial x^2} \] with boundary conditions: \[ T(0, t) = T(1, t) = 0 \] and initial condition: \[ T(x, 0) = f(x) \] where \( f(x) \) is a pulse function. #### **Separation of Variables and Fourier Series** Using separation of variables, the general solution can be expressed as: \[ T(x, t) = \sum_{n=1}^\infty B_n \sin(n \pi x) e^{-(n \pi)^2 t} \] where the coefficients \( B_n \) are determined from the initial condition: \[ B_n = 2 \int_0^1 f(x) \sin(n \pi x) dx \] **Explanation:** This approach involves using the method of separation of variables, which is suitable given the homogeneous boundary conditions. The solution comprises a series of sine functions that decay over time. --- ### **Step 2: Apply the Initial Condition** The pulse function is defined as: \[ f(x) = \begin{cases} 0 & \text{for } x < \frac{1}{2} - \frac{w}{2} \\ \frac{T_0}{w} & \text{for } \frac{1}{2} - \frac{w}{2} < x < \frac{1}{2} + \frac{w}{2} \\ 0 & \text{for } x > \frac{1}{2} + \frac{w}{2} \end{cases} \] Calculating \( B_n \): \[ B_n = 2 \int_{\frac{1}{2} - \frac{w}{2}}^{\frac{1}{2} + \frac{w}{2}} \frac{T_0}{w} \sin(n \pi x) dx \] \[ = \frac{2T_0}{w} \left[ -\frac{1}{n\pi} \cos(n \pi x) \right]_{\frac{1}{2} - \frac{w}{2}}^{\frac{1}{2} + \frac{w}{2}} \] \[ = -\frac{2T_0}{n\pi w} \left( \cos\left(n\pi \left(\frac{1}{2} + \frac{w}{2}\right)\right) - \cos\left(n\pi \left(\frac{1}{2} - \frac{w}{2}\right)\right) \right) \] Using the identity: \[ \cos(a + b) - \cos(a - b) = -2 \sin a \sin b \] we find: \[ = \frac{4 T_0}{n\pi w} \sin\left(\frac{n\pi}{2}\right) \sin\left(\frac{n\pi w}{2}\right) \] Thus, the temperature solution is: \[ \boxed{ T(x, t) = \sum_{n=1}^\infty \left[ \frac{4 T_0}{n\pi w} \sin\left(\frac{n\pi}{2}\right) \sin\left(\frac{n\pi w}{2}\right) \right] \sin(n\pi x) e^{-(n\pi)^2 t} } \] **Explanation:** We evaluate the Fourier coefficients based on the piecewise definition of the initial condition, allowing us to construct the series solution. --- ## **Final Answer** The complete solution for the temperature distribution is: \[ \boxed{ T(x, t) = \sum_{n=1}^\infty \left[ \frac{4 T_0}{n\pi w} \sin\left(\frac{n\pi}{2}\right) \sin\left(\frac{n\pi w}{2}\right) \right] \sin(n\pi x) e^{-(n\pi)^2 t} } \] This formulation provides the temperature at any point \( x \) and time \( t \) for the specified initial pulse shape.