Solving Heat Equation With Fourier Integrals: A Step-by-Step Guide

Hey guys! Today, we're diving deep into the fascinating world of partial differential equations and tackling a classic problem: solving the heat equation using Fourier integrals. This method is super powerful for dealing with problems defined on unbounded domains, and it's a cornerstone technique in mathematical physics and engineering. So, buckle up, and let's get started!

Understanding the Heat Equation and Boundary Conditions

First, let's break down the problem we're trying to solve. We're given the heat equation:

∂u/∂t = ∂²u/∂x²

This equation describes how temperature (u) changes over time (t) in a given region, where x represents the spatial coordinate. Think of it like modeling how heat diffuses along a metal rod. Now, we have some specific conditions that make our problem unique:

• u(0, t) = 0: This is a boundary condition, and it tells us that the temperature at x = 0 is always zero. Imagine the end of our metal rod being held in an ice bath – it's always kept at freezing!
• u(x, 0) = e^(-x): This is an initial condition, and it describes the temperature distribution at time t = 0. In our case, the temperature starts off exponentially decreasing as we move away from x = 0.
• x > 0, t > 0: This specifies the domain of our problem – we're only interested in positive values of x and t.

Our goal is to find the function u(x, t) that satisfies the heat equation and these conditions. This function will tell us the temperature at any point x and any time t.

The Magic of Fourier Integrals

So, how do we solve this beast? This is where the Fourier integral comes in. The Fourier integral is a powerful tool for representing functions as a continuous superposition of sines and cosines. It's like breaking down a complex musical chord into its individual notes, but for functions! For problems defined on semi-infinite domains (like our x > 0), we often use the Fourier sine transform, which is particularly useful when we have a condition like u(0, t) = 0.

Applying the Fourier Sine Transform

The Fourier sine transform of a function u(x, t) with respect to x is defined as:

U(ω, t) = ∫[0 to ∞] u(x, t) * sin(ωx) dx

where ω is the frequency variable. Essentially, we're transforming our function from the spatial domain (x) to the frequency domain (ω). This might seem abstract, but it simplifies the problem significantly.

Let's apply the Fourier sine transform to our heat equation. We'll need to use some properties of the transform, specifically how it handles derivatives. The key property we'll use is:

Fourier Sine Transform{∂²u/∂x²} = -ω²U(ω, t) + ωu(0, t)

Since u(0, t) = 0, this simplifies to:

Fourier Sine Transform{∂²u/∂x²} = -ω²U(ω, t)

Now, let's transform both sides of our heat equation:

Fourier Sine Transform{∂u/∂t} = Fourier Sine Transform{∂²u/∂x²}

This becomes:

∂U(ω, t)/∂t = -ω²U(ω, t)

Look at that! We've transformed our partial differential equation into an ordinary differential equation in t. This is a huge win! This ordinary differential equation is much easier to solve.

Solving the Ordinary Differential Equation

The equation ∂U(ω, t)/∂t = -ω²U(ω, t) is a simple first-order linear ODE. We can solve it using standard techniques (like separation of variables). The general solution is:

U(ω, t) = A(ω) * e^(-ω²t)

where A(ω) is an arbitrary function of ω. To find A(ω), we need to use our initial condition. Let's apply the Fourier sine transform to u(x, 0) = e^(-x):

U(ω, 0) = ∫[0 to ∞] e^(-x) * sin(ωx) dx

This integral can be solved using integration by parts (twice!) or by looking it up in a table of integrals. The result is:

U(ω, 0) = ω / (1 + ω²)

Now, we know that U(ω, 0) = A(ω), so:

A(ω) = ω / (1 + ω²)

Plugging this back into our solution for U(ω, t), we get:

U(ω, t) = (ω / (1 + ω²)) * e^(-ω²t)

The Inverse Fourier Sine Transform: Bringing it Back Home

We've now found the solution in the frequency domain. But we want the solution u(x, t) in the spatial domain. To do this, we need to apply the inverse Fourier sine transform. The inverse Fourier sine transform is defined as:

u(x, t) = (2/π) ∫[0 to ∞] U(ω, t) * sin(ωx) dω

Substituting our expression for U(ω, t), we get:

u(x, t) = (2/π) ∫[0 to ∞] ((ω / (1 + ω²)) * e^(-ω²t)) * sin(ωx) dω

This integral is a bit tricky to solve directly. It often requires techniques from complex analysis or the use of specialized integral tables. However, the result is a well-known integral, and its solution can be expressed in terms of the complementary error function, erfc(z), which is defined as:

erfc(z) = (2/√π) ∫[z to ∞] e^(-u²) du

After evaluating the integral (which is beyond the scope of this explanation but can be found in advanced mathematical texts), we obtain the final solution:

u(x, t) = (1/2) * (e^(-x) * erfc(x / (2√t) - √t) - e^(x) * erfc(x / (2√t) + √t))

Interpreting the Solution

Whoa! That's a mouthful of a solution! But what does it actually mean? This function u(x, t) describes the temperature distribution along our