Heat equation with neumann boundary conditions

By: Prof. Dr. Fazal Rehman Shamil | Last updated: December 1, 2024

[latex]
\[
\textbf{Heat Equation with Neumann Boundary Conditions}
\]

\[
\text{The heat equation is given as:}
\]
\[
\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}, \quad x \in [0, L], \, t > 0
\]

\[
\text{Neumann Boundary Conditions:}
\]
\[
\frac{\partial u}{\partial x}\bigg|_{x=0} = 0, \quad \frac{\partial u}{\partial x}\bigg|_{x=L} = 0
\]

\[
\text{Initial Condition:}
\]
\[
u(x, 0) = f(x), \quad 0 \leq x \leq L
\]

\[
\text{Solution Process:}
\]

\[
\text{1. Assume a separable solution: } u(x, t) = X(x)T(t)
\]

\[
\text{2. Substitute into the heat equation:}
\]
\[
X(x) \frac{dT}{dt} = \alpha T(t) \frac{d^2X}{dx^2}
\]

\[
\text{3. Separate variables:}
\]
\[
\frac{1}{T(t)} \frac{dT}{dt} = \alpha \frac{1}{X(x)} \frac{d^2X}{dx^2} = -\lambda
\]

\[
\text{4. Solve the spatial equation:}
\]
\[
\frac{d^2X}{dx^2} + \frac{\lambda}{\alpha} X = 0
\]

\[
\text{5. Apply Neumann Boundary Conditions to find eigenvalues and eigenfunctions.}
\]

\[
\text{6. Solve the temporal equation:}
\]
\[
\frac{dT}{dt} + \lambda T = 0 \quad \text{which leads to } T(t) = C e^{-\lambda t}
\]

\[
\text{7. Combine the solutions:}
\]
\[
u(x, t) = \sum_{n=0}^\infty C_n X_n(x) e^{-\lambda_n t}
\]

\[
\text{The coefficients } C_n \text{ are determined using the initial condition.}
\]