12.3. Solution of the Heat Equation by Separation of Variables

Assume solution in separable form:

𝑢(𝑥,𝑡) = 𝑋(𝑥)𝑇(𝑡)

In this first step, we assume that the temperature distribution, denoted as u(x,t), can be split into two independent functions: one that depends only on position (X(x)) and another that depends only on time (T(t)). This assumption simplifies the problem by reducing it from a function of two variables into two functions of one variable each.

Think of it like baking a cake. Instead of considering the entire cake as a single piece, you can look at the ingredients (like flour, eggs) separately while mixing them. Once you mix them, you have a cake! Here, we mix position and time separately to find the overall solution.

Substitute into the heat equation:

𝑑𝑇 𝑑2𝑋 𝑋(𝑥) = 𝛼2𝑇(𝑡) 𝑑𝑡 𝑑𝑥2

Divide both sides by 𝛼2𝑋𝑇:

1 𝑑𝑇 1 𝑑2𝑋 = = −𝜆 𝛼2𝑇(𝑡) 𝑑𝑡 𝑋(𝑥)𝑑𝑥2

After making the assumption of separable solutions, we substitute the expression u(x,t) = X(x)T(t) into the One-Dimensional Heat Equation. This substitution allows us to rearrange the equation so that one side depends entirely on time, while the other side depends solely on position. By dividing each side appropriately, we introduce a constant (-λ), which leads us to two separate ordinary differential equations, one for time and one for space.

Imagine you are trying to balance two different scales – one for the weight of fruits (position) and another for their freshness over time (time). By isolating each scale, we can analyze how much fruit we have and how fresh it is separately before deciding how to preserve or sell them.

We now get two ordinary differential equations:

• Time equation:

𝑑𝑇 + 𝛼2𝜆𝑇 = 0 ⇒ 𝑇(𝑡) = 𝐴𝑒−𝛼2𝜆𝑡

𝑑𝑡

• Spatial equation:

𝑑2𝑋 + 𝜆𝑋 = 0 ⇒ 𝑋(𝑥)= 𝐵sin(√𝜆𝑥)+ 𝐶cos(√𝜆𝑥)

𝑑𝑥2

From the separation of variables, we derive two ordinary differential equations (ODEs). The first is a time-dependent equation that has an exponentially decaying solution, indicating how temperature decreases over time. The second equation depends purely on the spatial component and yields solutions involving sine and cosine functions, which are characteristic of oscillatory behaviors in spatial distributions.

Think of a pendulum clock. The time it takes to swing back and forth (time equation) can be described by how quickly it slows down due to friction (exponential decay), while the angle it covers on each side (spatial equation) is determined by its physical properties, like length and weight (sine and cosine functions).

Apply boundary conditions to find allowed eigenvalues and eigenfunctions.

Example (Dirichlet BCs):

𝑢(0,𝑡) = 0 ⇒ 𝑋(0) = 0 ⇒ 𝐶 = 0

𝑢(𝐿,𝑡) = 0 ⇒ 𝑋(𝐿)= 0 ⇒ 𝐵sin(√𝜆𝐿) = 0

Non-trivial solution if:

𝑛𝜋 2 √𝜆𝐿 = 𝑛𝜋 ⇒ 𝜆 = ( )

𝑛 𝐿

To find specific solutions to our ODEs, we apply the boundary conditions relevant to our physical setup. For Dirichlet boundary conditions, we set the temperature at the ends of the rod to zero (u(0,t)=0 and u(L,t)=0). This leads us to conclude that the constant C must be zero and gives us an important condition that allows us to solve for the eigenvalues (λ) and associated eigenfunctions (X(x)). In particular, we find that the allowed values of λ come from specific integer multiples (n) that relate to the length of the rod.

Imagine sealing both ends of a water hose. Just like how the water can't go beyond those points, we apply similar constraints to the temperatures at the boundaries of our rod, helping us determine valuable characteristics (like wavelengths) of how heat behaves inside the rod.

Then:

𝑛𝜋𝑥 −𝛼2(𝑛𝜋 )2 𝑡

𝑆(𝑥) = sin( ), 𝑇 (𝑡) = 𝑒 𝐿

𝑛 𝐿 𝑛

Final Solution:

∞

𝑢(𝑥,𝑡) = ∑𝐵 sin( )𝑒 𝐿

𝑛 𝐿

𝑛=1

Where 𝐵 are Fourier coefficients determined using the initial condition 𝑢(𝑥,0) = 𝑓(𝑥).

After applying the boundary conditions and solving the ordinary differential equations, we arrive at the final solutions for temperature distribution in terms of a series of sine functions and exponential decay terms. The solution represents the temperature u(x,t) as an infinite sum of sinusoidal functions multiplied by time-varying exponential terms. The coefficients (B) are calculated using the initial temperature distribution, giving us a comprehensive view of how the temperature evolves over time throughout the rod.

Visualize a piano. Each key produces a different sound (sine wave) depending on how hard you strike it (initial condition). When you press a key, that note can last for a certain duration before fading away (exponential decay). Like playing different keys together to create a symphony, our final solution combines all the different sine waves into a beautifully complex melody representing the temperature changes within the rod over time.

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