Question

The temperature in an infinite cylindrical rod of radius $a$ satisfies the conditions (a) ${\mathrm{\nabla }}^{2}T={\alpha }^{-1}\frac{\partial T}{\partial t}(\alpha =\frac{K}{C\rho })$ (b) $T=0$ at $t=0$ (c) $T=T_0\cos \varphi$ at $\rho =a$ Find $T(\rho ,\varphi ,t)$ for $t>0$

Short answer

The solution is $T(\rho ,\theta ,t)=T_0\cos (\theta )\text{(}{C}_1{I}_1(\gamma \rho )+C_2{K}_1(\gamma \rho )$ e^{-\lambda t}\), where $\gamma =\sqrt{\frac{\lambda }{\alpha }}$.

Step-by-step solution

- Understanding the Problem

The problem involves solving the heat equation in cylindrical coordinates for an infinite cylindrical rod with radius $a$. The heat equation is given together with initial and boundary conditions.

- Setup the Problem in Cylindrical Coordinates

The heat equation in cylindrical coordinates for temperature $T(\rho ,\theta ,t)$ is $$abl{a}^{2}T=\frac{1}{\alpha }\frac{\partial T}{\partial t},$$ where $$abl{a}^{2}T=\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial T}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^{2}T}{\partial {\theta }^2}.$$

- Apply the Initial Condition

The initial condition is given by $T(\rho ,\theta ,0)=0$. Therefore, initially, the temperature everywhere inside the rod is zero.

- Apply the Boundary Condition

The boundary condition is $T(\rho =a,\theta ,t)=T_0\cos \theta$. This means that at the surface of the cylinder ($\rho =a$), the temperature follows a cosine function with respect to the angular variable $\theta$.

- Assume a Separable Solution

Assume a solution of the form $$T(\rho ,\theta ,t)=R(\rho )\mathrm{\Theta }(\theta )e^{-\lambda t}.$$ Insert this into the heat equation.

- Solve the Angular Part

The angular part gives $$\frac{1}{\mathrm{\Theta }}\frac{d^2\mathrm{\Theta }}{d{\theta }^2}=-m^2.$$ The general solution to this is $$\mathrm{\Theta }(\theta )=A\cos (m\theta )+B\sin (m\theta ).$$ Given the boundary condition $T(a,\theta )=T_0\cos \theta$, we choose $m=1$ and $A=T_0,B=0$, so $$\mathrm{\Theta }(\theta )=T_0\cos \theta .$$

- Solve the Radial Part

The radial part gives $$R^{″}(\rho )+\frac{1}{\rho }{R}^{\prime }(\rho )-(\frac{m^2}{{\rho }^2}+\lambda \alpha )R(\rho )=0.$$ The general solution for $m=1$ is $$R(\rho )=C_1{I}_1(\gamma \rho )+C_2{K}_1(\gamma \rho ),$$ where $\gamma =\sqrt{\frac{\lambda }{\alpha }}$ and $I_1$ and $K_1$ are modified Bessel functions.

- Apply the Boundary Conditions to Find Constants

Applying the boundary conditions $R(a)=T_0$, and considering the behavior of Bessel functions at the origin and at large $\rho$, the solution simplifies.

- Combine Solutions

Combine the solutions for the angular, radial, and time parts. Substitute back $$T(\rho ,\theta ,t)=R(\rho )\mathrm{\Theta }(\theta )e^{-\lambda t}.$$

Key concepts

Boundary Conditions

Boundary conditions are essential in solving partial differential equations (PDEs) like the heat equation. They specify the value of the solution on the boundary of the domain. For this problem, the boundary condition is given by

• $T(\rho =a,\theta ,t)=T_0\text{cos}(\theta )$. This describes the temperature distribution at the surface of the cylindrical rod, where $a$ is the radius of the cylinder. The temperature at the boundary is a function of the angular position $\theta$ but is constant in time.

'Boundary conditions help us determine the particular form of the solution. They act as constraints that a valid solution must satisfy. Verifying that the solution satisfies the boundary condition is crucial for an accurate model.

Initial Conditions

Initial conditions specify the state of the system at the beginning of the observation. In this problem, the initial condition is given by

• $T(\rho ,\theta ,0)=0$. This means that at time $t=0$, the entire cylinder's temperature is zero.

'Initial conditions are used to determine the constants of integration when solving differential equations. They ensure that the solution is aligned with the physical situation at the start of the observation.

Separation of Variables

Separation of variables is a mathematical method used to solve PDEs, where the solution is assumed to be a product of functions, each depending on only one of the coordinates. For this problem, we assume a solution of the form

• $T(\rho ,\theta ,t)=R(\rho )\text{Θ}(\theta )e^{-\text{λ}t}$. Here, $R(\rho )$ is a function of the radial coordinate $\rho$, $\text{Θ}(\theta )$ is a function of the angular coordinate $\theta$, and $e^{-\text{λ}t}$ is the time-dependent part.

'By substituting this assumed form into the heat equation, we separate the variables, leading to simpler ordinary differential equations (ODEs) that are easier to solve. Each part of the assumed product solution corresponds to one of the coordinates.

Bessel Functions

Bessel functions arise in problems with cylindrical symmetry, such as this one. In our solution process, we encounter a differential equation whose solutions are the modified Bessel functions $I_1$ and $K_1$, given by

• $R(\rho )=C_1{I}_1(\text{γ}\rho )+C_2{K}_1(\text{γ}\rho )$. Here, $I_1$ and $K_1$ are the modified Bessel functions of the first and second kind respectively, and $\text{γ}=\text{√(λ/α)}$.

'Modified Bessel functions are especially useful in physical applications because they behave well at the origin and at infinity. In this particular case, the behavior of $I_1(\text{γ}\rho )$ near the origin and the boundary condition at $\rho =a$ determine the constants which simplifies our solution.