Question

Consider a uniform rod of length \(L\) with an initial temperature given by \(u(x, 0)=\) \(\sin (\pi x / L), 0 \leq x \leq L\) Assume that both ends of the bar are insulated. (a) Find the temperature \(u(x, t) .\) (b) What is the steady-state temperature as \(t \rightarrow \infty\) ? (c) Let \(\alpha^{2}=1\) and \(L=40 .\) Plot \(u\) versus \(x\) for several values of \(t\). Also plot \(u\) versus \(t\) for several values of \(x .\) (d) Describe briefly how the temperature in the rod changes as time progresses.

Short answer

In summary, the temperature distribution of the rod changes over time as follows: 1. At the initial state (t = 0), the rod has a sinusoidal temperature distribution with the maximum temperature at the center (x = 20). 2. As time progresses, the amplitude of the temperature distribution decreases exponentially due to the heat transfer across the rod and the insulation at its ends. 3. The temperature decreases faster near the ends of the rod (x = 0 and x = 40) compared to the center of the rod (x = 20). 4. Eventually, as time approaches infinity, the rod reaches a steady-state temperature distribution, which is constant and equal to 0. Thus, the rod cools down over time and reaches a constant temperature distribution when it achieves thermal equilibrium.

Step-by-step solution

(a) Find the temperature u(x, t) using separation of variables

We'll assume that the temperature distribution in the rod can be described by the heat equation: $$ \frac{\partial u}{\partial t} = \alpha^2 \frac{\partial^2 u}{\partial x^2} $$ with the boundary conditions \(\frac{\partial u}{\partial x} \big|_{x=0} = \frac{\partial u}{\partial x} \big|_{x=L} = 0\) (since both ends of the bar are insulated) and the initial condition (IC) \(u(x, 0) = \sin(\frac{\pi x}{L})\). Let \(u(x, t) = X(x) \cdot T(t)\). Using separation of variables on the heat equation, we have: $$ \frac{T'(t)}{\alpha^2 T(t)} = \frac{X''(x)}{X(x)} $$ Since both sides of the equation depend on different variables, they must be equal to a constant. We'll denote this constant by \(-\lambda^2\): $$ T'(t) = -\alpha^2 \lambda^2 T(t) $$ $$ X''(x) = -\lambda^2 X(x) $$ Now, we'll solve these two ODEs separately. For the ODE in X(x): $$ X''(x) = -\lambda^2 X(x) $$ The general solution to this ODE is given by: $$ X(x) = A \cos(\lambda x) + B \sin(\lambda x) $$ Applying the boundary conditions (BCs), we get the following: $$ \frac{dX}{dx} \big|_{x=0} = -A \lambda \sin(\lambda x) + B \lambda \cos(\lambda x) = 0 $$ Let x=0, $$ -A \lambda = 0 \Rightarrow A = 0 $$ Now, the updated function for X(x) becomes: $$ X(x) = B \sin(\lambda x) $$ Now, the other boundary condition: $$ \frac{dX}{dx} \big|_{x=L} = B \lambda \cos(\lambda L) = 0. $$ Since B ≠ 0, we have: $$ \lambda L = n\pi,\,\, n = 1, 2, 3, ... $$ Therefore, $$ X_n(x) = B_n \sin(\frac{n\pi x}{L}) $$ Where \(n = 1, 2, 3, ...\) Now let's solve the ODE in T(t): $$ T'(t) = -\alpha^2 \lambda^2 T(t) $$ The general solution to this ODE is given by: $$ T(t) = C \exp(-\alpha^2 \lambda^2 t) $$ Therefore, the general solution for u(x, t) is given by: $$ u(x, t) = \sum_{n=1}^{\infty} B_n \sin\Big(\frac{n\pi x}{L}\Big) \cdot C \exp\Big(-\alpha^2 (\frac{n\pi}{L})^2 t\Big) $$ Now let's apply the initial condition: $$ u(x, 0) = \sin\Big(\frac{\pi x}{L}\Big) = \sum_{n=1}^{\infty} B_n \sin\Big(\frac{n\pi x}{L}\Big) $$ As we can see, only the first term (\(n=1\)) gives us the correct initial condition: $$ B_1 = 1,\,\,\,\, B_n = 0,\,\, n = 2, 3, ... $$ Thus, the temperature distribution is given by: $$ u(x, t) = \sin\Big(\frac{\pi x}{L}\Big) \cdot \exp\Big(-\alpha^2 (\frac{\pi}{L})^2 t\Big) $$

(b) Calculate the steady-state temperature as \(t \rightarrow \infty\)

To find the steady-state temperature, we take the limit as \(t \rightarrow \infty\): $$ u(x, \infty) = \lim_{t \rightarrow \infty} \sin\Big(\frac{\pi x}{L}\Big) \cdot \exp\Big(-\alpha^2 (\frac{\pi}{L})^2 t\Big) = 0 $$ As the exponential term goes to zero, the steady-state temperature distribution is constant and equal to 0.

(c) Plot u(x, t) and u(t, x) for given values of α and L

Given \(\alpha^2 = 1\) and \(L = 40\), we have: $$ u(x, t) = \sin\Big(\frac{\pi x}{40}\Big) \cdot \exp\Big(- (\frac{\pi}{40})^2 t\Big) $$ To plot the temperature distribution, we can create two plots: 1. u(x) vs x for different values of t: This can be done by varying x from 0 to 40 and plotting the temperature distribution for different values of t, such as t = 0 s, 10 s, 50 s, 100 s, and 500 s. 2. u(t) vs t for different values of x: This can be done by varying t from 0 s to 500 s and plotting the temperature distribution for different values of x such as x = 10, 20, 30, and 40. Note: This task requires plotting which cannot be performed within the text-based environment. However, you can use any plotting software or programming language (like Python, MATLAB, or Wolfram Alpha) to visualize the graphs described above.

(d) Describe briefly how the temperature in the rod changes as time progresses

As time progresses, the temperature distribution \(u(x, t)\) shows that the rod cools down and approaches a steady-state temperature of 0. Initially, the rod has a sinusoidal temperature distribution, with a maximum temperature at the center of the rod (x = 20). As time passes, the amplitude of the temperature distribution decreases exponentially due to the exp(- \(\alpha^2(\frac{\pi}{L})^2 t\)) term. The rod cools down faster near the ends than at the center, as evident from the decreasing value of the temperature distribution when the position x is close to 0 and 40. In the steady-state, the rod has a constant temperature distribution of 0.

Key concepts

Separation of Variables

The method of separation of variables is a standard approach to solving partial differential equations (PDEs), such as the heat equation. By assuming the solution can be written as the product of two functions, each depending on one variable, separation of variables transforms the PDE into ordinary differential equations (ODEs).

For example, consider a heat equation problem where the temperature distribution, denoted as \(u(x,t)\), depends on spatial position \(x\) and time \(t\). The solution \(u(x,t) = X(x) \times T(t)\) splits the equation into separate parts, where \(X\) is a function of \(x\) only and \(T\) is a function of \(t\) only. Solving these ODEs you may find specific solutions, which consider the physical constraints of the system.

Boundary Conditions

The boundary conditions of a problem specify the behavior of a system at the boundaries of the domain. For thermal problems, such as the heat equation, boundary conditions are crucial as they include information about heat transfer at the surface. Insulated ends, as in the given exercise, are represented mathematically by Neumann boundary conditions which dictate that the spatial derivative of the temperature distribution, \( \frac{\text{d}X}{\text{d}x} \), at the boundaries \(x = 0\) and \(x = L\) is zero. These conditions lead to a set of possible solutions, specifically sine and cosine functions, which are used to match the initial temperature distribution.

Steady-State Temperature

The concept of steady-state temperature refers to a condition where the temperature distribution no longer changes with time. In the context of the heat equation for an insulated rod, the steady-state is reached when the time variable \(t\) approaches infinity. As a result of the heat equation’s time-dependent term, which includes an exponential decay based on time, the temperature distribution ultimately becomes zero everywhere along the rod. This happens since the exponential term goes to zero as time goes to infinity, indicating the rod has equilibrated with its surroundings.

Initial Temperature Distribution

The term initial temperature distribution refers to the starting temperature profile of an object. For the heat equation, it is essential to specify how the temperature varies throughout the domain at \(t = 0\). In the given exercise, a sine function represents the initial condition for the temperature distribution along the length of the rod. This form of the initial condition is particularly amenable to the method of separation of variables, as sine functions naturally emerge from the solution of the spatial part of the heat equation under insulated boundary conditions. The initial condition directly influences the constants found in the final expression of the temperature distribution, which satisfies both the heat equation and the given boundary and initial conditions.