10.6.1. Application: Heat Equation in a Semi-Infinite Rod

Consider a semi-infinite rod x ∈ [0,∞) initially at zero temperature, and for t > 0, the end x = 0 is held at a constant temperature T.

This chunk introduces the physical setup of the problem. We have a rod that stretches infinitely in the positive x-direction, where we initially assume the entire rod is at a temperature of zero. At some point, we start heating the very beginning of the rod (x = 0) to a temperature T, which will cause heat to travel along the rod over time. Understanding this setup is crucial as it establishes the conditions under which we will analyze heat conduction.

Think of a long metal rod partially submerged in hot water, where only the end of the rod touching the water heats up. Over time, the heat transfers along the rod, illustrating similar principles to our semi-infinite rod problem.

The governing equation is the heat conduction equation: ∂u/∂t = α² ∂²u/∂x², x > 0, t > 0.

This chunk presents the mathematical formulation describing how temperature changes over time (∂u/∂t) and how it changes along the rod (∂²u/∂x²). Here, u represents the temperature at a position x and time t, while α² is a constant that reflects the material properties of the rod. This equation is a partial differential equation (PDE) that enables us to model the physical process of heat conduction mathematically.

Imagine turning on a hair straightener. The straightener gets hot, and that heat gradually moves through the metal plates. The equation describes how the temperature in the plate evolves both over time and from one end to the other.

Subject to:

• u(0,t) = T,
• u(x,0) = 0,
• lim (x→∞) u(x,t) = 0.

Here, we define specific conditions that apply to our problem: At the boundary (x=0), the temperature is held constant at T. Initially (at time t=0), the entire rod is at zero temperature, and as we move infinitely far from the heated end, the temperature approaches zero. These conditions are critical as they define how we will solve our governing equation and find the temperature distribution over time.

Think of holding one end of a thick blanket over a heater while the other end is outside in the cold. Initially, the entire blanket feels cold (zero temperature), but as you keep it near the heater, one end gets hot (temperature T), while far away (as you move along the blanket), the warmth fades back to coldness.

We take the Fourier Cosine Transform with respect to x: Let U(s,t)=F {u(x,t)}= (2/π) ∫₀^∞ u(x,t)cos(sx)dx.

To solve the heat equation, we apply the Fourier Cosine Transform, which is useful for problems defined over semi-infinite domains. This transform helps convert the partial differential equation into an ordinary differential equation (ODE) in the time domain, simplifying the calculations. U(s,t) represents the transformed version of the temperature function u(x,t), and the integral computes how much of each cosine function contributes to the overall temperature distribution.

Imagine using a prism to decompose light into different colors. In our case, the Fourier transform is like a prism that decomposes the temperature profile into simpler components, allowing easier analysis.

Using properties of derivatives under transforms: F{ ∂²u/∂x²} = −s²U(s,t), ∂U/∂t = −α²s²U(s,t).

This chunk provides the transition to an ordinary differential equation. The Fourier transform of the second spatial derivative introduces a term involving s² (the frequency variable), while the Fourier transform of the time derivative results in a straightforward time derivative. This leads to a first-order linear ordinary differential equation for U(s,t), which is easier to solve compared to the original PDE.

Think of transforming a complex task into a series of easier steps. Just like how breaking down a complicated recipe into simple instructions makes cooking easier, transforming the temperature equation into an ODE simplifies finding the solution.

This is a first-order linear ODE in t with solution: U(s,t)=A(s)e^{−α²s²t}. From initial condition u(x,0)=0 ⇒ U(s,0)=0 ⇒ A(s)=0, but this contradicts the boundary condition.

We find that the solution for U(s,t) is of an exponential form, which is typical for heat equations. However, applying the initial condition directly leads to a contradiction regarding the constant A(s). This situation indicates that we cannot simply apply the initial condition without properly addressing the steady-state boundary condition at the same time.

Consider starting a plant from a seed. If you plant an initial seed in 'zero' conditions (no water, no sun), while wanting it to grow immediately, it doesn't work. In our mathematical scenario, we need to find a suitable condition to connect initial feelings with final results.

To fix this, we transform the non-homogeneous boundary via suitable substitution (e.g., Duhamel’s principle or separation of variables), which leads to:

u(x,t)=T erfc(√(x/(2αt))).

To resolve the issues with the boundary conditions, we apply techniques like Duhamel's principle or separation of variables, which help us incorporate the effect of the constant temperature boundary condition and lead us to the final solution using the complementary error function. This solution provides temperature at any point of the rod over time with respect to the specified boundary conditions.

It's like using special gardening techniques to ensure each plant (here, the solution) receives the right combination of conditions to thrive. We adapt our calculations to produce a result that respects both the initial and boundary conditions.

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