solve one dimensional heat diffusion equation by laplace transform

To solve the one-dimensional heat diffusion equation using the **Laplace transform**, let's walk through the steps.

### Problem Definition

The one-dimensional heat equation is given by:

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

where:
- \( u(x,t) \) is the temperature distribution along the rod.
- \( \alpha^2 \) is the thermal diffusivity constant.
- \( x \) is the spatial coordinate, and \( t \) is the time variable.

Assume the following initial and boundary conditions:
- Initial condition: \( u(x,0) = f(x) \).
- Boundary conditions: \( u(0,t) = 0 \) and \( u(L,t) = 0 \) (for a rod of length \( L \)).

### Step 1: Apply the Laplace Transform

We will apply the Laplace transform in time to the heat equation. The Laplace transform of \( u(x,t) \) is defined as:

\[
\mathcal{L}\{u(x,t)\} = \hat{u}(x,s) = \int_0^\infty u(x,t) e^{-st} \, dt
\]

Taking the Laplace transform of both sides of the heat equation:

\[
\mathcal{L}\left\{ \frac{\partial u(x,t)}{\partial t} \right\} = \mathcal{L}\left\{ \alpha^2 \frac{\partial^2 u(x,t)}{\partial x^2} \right\}
\]

Using the properties of Laplace transforms:
\[
s \hat{u}(x,s) - u(x,0) = \alpha^2 \frac{\partial^2 \hat{u}(x,s)}{\partial x^2}
\]

Substitute the initial condition \( u(x,0) = f(x) \):

\[
s \hat{u}(x,s) - f(x) = \alpha^2 \frac{\partial^2 \hat{u}(x,s)}{\partial x^2}
\]

This is a second-order ordinary differential equation (ODE) in \( x \):

\[
\frac{\partial^2 \hat{u}(x,s)}{\partial x^2} - \frac{s}{\alpha^2} \hat{u}(x,s) = - \frac{f(x)}{\alpha^2}
\]

### Step 2: Solve the Homogeneous ODE

First, solve the homogeneous equation:

\[
\frac{\partial^2 \hat{u}(x,s)}{\partial x^2} - \frac{s}{\alpha^2} \hat{u}(x,s) = 0
\]

The general solution to this second-order differential equation is:

\[
\hat{u}_h(x,s) = A(s) \sinh\left( \frac{\sqrt{s}}{\alpha} x \right) + B(s) \cosh\left( \frac{\sqrt{s}}{\alpha} x \right)
\]

### Step 3: Apply Boundary Conditions

Now, apply the boundary conditions:
- \( u(0,t) = 0 \Rightarrow \hat{u}(0,s) = 0 \)
  - This implies \( B(s) = 0 \), because \( \hat{u}(0,s) = A(s) \sinh(0) + B(s) \cosh(0) = B(s) \).

Thus, the solution simplifies to:

\[
\hat{u}(x,s) = A(s) \sinh\left( \frac{\sqrt{s}}{\alpha} x \right)
\]

- \( u(L,t) = 0 \Rightarrow \hat{u}(L,s) = 0 \)
  - This gives \( A(s) \sinh\left( \frac{\sqrt{s}}{\alpha} L \right) = 0 \).
  - For non-trivial solutions, \( A(s) = 0 \) is not allowed, so \( \sinh\left( \frac{\sqrt{s}}{\alpha} L \right) \neq 0 \).

Thus, \( A(s) \) can be determined from the initial condition, but the homogeneous solution approach is clear for now.

### Step 4: Invert the Laplace Transform

Once we obtain \( \hat{u}(x,s) \), the final step is to invert the Laplace transform and find \( u(x,t) \). This can involve partial fraction decomposition and applying known Laplace transform pairs.

Would you like me to continue with a specific example or clarify any of the steps above?

### Follow-up Questions:
1. How would you modify the solution for different boundary conditions?
2. What changes if we have a non-zero initial condition?
3. Can we solve this equation for infinite domain conditions (i.e., \( L \to \infty \))?
4. How would you apply the Fourier transform in conjunction with the Laplace method?
5. What physical systems can be modeled with variations of this heat equation?

### Tip: 
Laplace transforms are especially useful for solving PDEs with initial conditions because they convert time derivatives into algebraic equations, simplifying the problem structure.

Explore a detailed step-by-step solution to the one-dimensional heat diffusion equation using the Laplace transform. Learn about key concepts like partial differential equations, boundary value problems, and Laplace transforms. This advanced-level guide is ideal for university students studying heat transfer or applied mathematics.

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