Water at \(100^{\circ}\) is flowing through a long pipe of radius 1 rapidly enough so that we may assume that the temperature is \(100^{\circ}\) at all points. At \(t=0\), the water is turned off and the. surface of the pipe is maintained at \(40^{\circ}\) from then on (neglect the wall thickness of the pipe). Find the temperature distribution in the water as a function of \(r\) and \(t\). Note that you need only consider a cross section of the pipe.

Heat transfer in materials is often modeled by the heat equation. This is a partial differential equation (PDE) that describes how temperature changes over time within a given region. For our cylindrical problem, the heat equation in terms of cylindrical coordinates (considering radial symmetry) is:
\[ \frac{\partial T}{\partial t} = \alpha \left( \frac{\partial^2 T}{\partial r^2} + \frac{1}{r} \frac{\partial T}{\partial r} \right) \]
Here, T is the temperature, and \alpha is the thermal diffusivity, which encapsulates how quickly heat diffuses through the material. The term r represents the radial position from the center of the pipe.
Given the pipe scenario, the temperature distribution depends solely on the radial distance r and time t. Initial temperature is uniform (100°C) and boundary conditions change the scenario as time progress, impacting the heat diffusion.

Boundary conditions play a critical role in solving the heat equation. They specify the temperature at the edges of the domain over which the PDE is being solved. For the cylindrical pipe:

• Initial Condition: At time t=0, the temperature everywhere in the water is \(100^{\circ}C\).
• Boundary Condition: At the pipe surface (\(r=a\)), the temperature is maintained at \(40^{\circ}C\) after t=0.
• Symmetry Condition: At the center of the pipe (\(r=0\)), the temperature gradient is zero, i.e., \( \frac{\partial T}{\partial r} (0, t) = 0 \).

These conditions help establish constraints that any potential solution must satisfy.

When solving the heat equation in cylindrical coordinates, Bessel functions naturally arise due to the radial component of the geometry. We derive a Bessel differential equation during the separation of variables step. Specifically, the equation is:
\[ r^2 \frac{d^2 R}{dr^2} + r \frac{dR}{dr} + \lambda r^2 R = 0 \]
Here, \(R(r)\) represents the radial part of the temperature distribution, and \(\lambda\) is a separation constant. The general solution to this equation involves the Bessel functions of the first kind, \(J_0\), and the second kind, \(Y_0\):
\[ R(r) = A J_0(\sqrt{\lambda} r) + B Y_0(\sqrt{\lambda} r) \]
Since \(Y_0\) is singular at \(r=0\), we discard it (thus, \(B=0\)), leaving:
\[ R(r) = A J_0(\sqrt{\lambda} r) \]
The Bessel function properties help us handle circular symmetry in PDEs efficiently.

To solve the heat equation, we use a technique called separation of variables. This involves assuming a solution where time and space can be separated, i.e.,
\[ T(r, t) = R(r) T(t) \]
Plugging this form into our heat diffusion equation and separating the variables yields two ordinary differential equations (ODEs):
\[ \frac{1}{\alpha} \frac{1}{T} \frac{dT}{dt} = \frac{1}{R} \left( \frac{d^2 R}{dr^2} + \frac{1}{r} \frac{dR}{dr} \right) = -\lambda \]
Here, \(\lambda\) is the separation constant. The resulting ODEs are:

• For the radial component: \[ r^2 \frac{d^2 R}{dr^2} + r \frac{dR}{dr} + \lambda r^2 R = 0 \]
• For the time component: \[ \frac{dT}{dt} + \lambda \alpha T = 0 \]

By solving these ODEs separately and combining the solutions while applying initial and boundary conditions, we obtain a complete solution for T(r, t), the desired temperature distribution.