give me an exemplary solution for heat equation for a rod

Sure, let's go through an example of solving the heat equation for a rod. The heat equation is given by:

\[ \frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2} \]

where \( u(x,t) \) is the temperature at position \( x \) and time \( t \), and \( \alpha \) is the thermal diffusivity constant.

### Example Problem
Consider a rod of length \( L \) with the following initial and boundary conditions:
- Initial temperature distribution: \( u(x,0) = f(x) \)
- Boundary conditions: \( u(0,t) = 0 \) and \( u(L,t) = 0 \)

### Solution Method
We use the method of separation of variables, assuming a solution of the form \( u(x,t) = X(x)T(t) \).

1. **Separation of Variables:**
   Substitute \( u(x,t) = X(x)T(t) \) into the heat equation:

   \[ X(x) \frac{dT(t)}{dt} = \alpha T(t) \frac{d^2X(x)}{dx^2} \]

   Divide both sides by \( \alpha X(x)T(t) \):

   \[ \frac{1}{\alpha T(t)} \frac{dT(t)}{dt} = \frac{1}{X(x)} \frac{d^2X(x)}{dx^2} = -\lambda \]

   Here, \(-\lambda\) is a separation constant. This gives us two ordinary differential equations:

   \[ \frac{dT(t)}{dt} + \alpha \lambda T(t) = 0 \]

   \[ \frac{d^2X(x)}{dx^2} + \lambda X(x) = 0 \]

2. **Solve the Spatial Equation:**
   The spatial equation is:

   \[ \frac{d^2X(x)}{dx^2} + \lambda X(x) = 0 \]

   With boundary conditions \( X(0) = 0 \) and \( X(L) = 0 \). The solutions are:

   \[ X_n(x) = \sin\left(\frac{n\pi x}{L}\right), \quad \lambda = \left(\frac{n\pi}{L}\right)^2, \quad n = 1, 2, 3, \ldots \]

3. **Solve the Temporal Equation:**
   The temporal equation is:

   \[ \frac{dT(t)}{dt} + \alpha \left(\frac{n\pi}{L}\right)^2 T(t) = 0 \]

   The solution is:

   \[ T_n(t) = e^{-\alpha \left(\frac{n\pi}{L}\right)^2 t} \]

4. **Combine Solutions:**
   The general solution is a sum of solutions for each \( n \):

   \[ u(x,t) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{L}\right) e^{-\alpha \left(\frac{n\pi}{L}\right)^2 t} \]

5. **Determine Coefficients \( B_n \):**
   Use the initial condition \( u(x,0) = f(x) \):

   \[ f(x) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{L}\right) \]

   The coefficients \( B_n \) are found using the Fourier sine series:

   \[ B_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx \]

### Final Solution
Thus, the solution to the heat equation with the given initial and boundary conditions is:

\[ u(x,t) = \sum_{n=1}^{\infty} \left( \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx \right) \sin\left(\frac{n\pi x}{L}\right) e^{-\alpha \left(\frac{n\pi}{L}\right)^2 t} \]

This represents the temperature distribution along the rod over time.

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Would you like more details or have any questions? Here are five additional questions you might ask:

1. How do you solve the heat equation with different boundary conditions?
2. Can you explain the physical interpretation of the heat equation?
3. What happens to the solution as \( t \) approaches infinity?
4. How would the solution change for a rod with insulated ends?
5. Can you provide an example with a specific initial temperature distribution \( f(x) \)?

**Tip:** When solving partial differential equations, always check the boundary and initial conditions carefully to ensure they match the physical situation you are modeling.

Explore an in-depth solution to the heat equation for a rod, covering initial and boundary conditions, separation of variables, and Fourier sine series. Ideal for advanced high school students and college learners.

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