Question

Consider a bar of length \(40 \mathrm{cm}\) whose initial temperature is given by \(u(x, 0)=x(60-\) \(x) / 30 .\) Suppose that \(\alpha^{2}=1 / 4 \mathrm{cm}^{2} / \mathrm{sec}\) and that both ends of the bar are insulated. (a) Find the temperature \(u(x, t) .\) (b) Plot \(u\) versus \(x\) for several values of \(t\). Also plot \(u\) versus \(t\) for several values of \(x\) (c) Determine the steady-state temperature in the bar. (d) Determine how much time must elapse before the temperature at \(x=40\) comes within 1 degree of its steady-state value.

Short answer

Question: Find the time it takes for the temperature at x=40 to be within 1 degree of its steady-state value while considering the heat equation and Fourier series. Answer: To find the time, solve the inequality: $$ \left|u(40, t) - u_{ss}(40)\right| < 1 $$

Step-by-step solution

Heat equation and Fourier series

We will consider the heat equation with its Fourier series expansion: $$ u(x, t) = \sum_{n=1}^\infty B_n e^{-\alpha ^2n^2 \pi ^2 t/L^2} \sin\frac{n \pi x}{L} $$ Where L is the bar length (40 cm), and \(B_n\) are the Fourier coefficients of the expansion. Since both ends of the bar are insulated, the initial temperature has a special property leading to an even function. So, we need to compute the coefficients \(B_n\) using the following formula: $$ B_n = \frac{2}{L} \int_0^L u(x, 0) \sin\frac{n \pi x}{L} dx $$

Calculate Fourier coefficients

Now let's calculate the Fourier coefficients \(B_n\): $$ B_n = \frac{2}{40} \int_0^{40} \frac{x(60-x)}{30} \sin\frac{n \pi x}{40} dx $$ Make a change of variable, let \(y = \frac{x}{40}\), \(dy = \frac{1}{40} dx\), then: $$ B_n = \frac{1}{10} \int_0^1 2y(3-y) \sin(n\pi y) dy $$ To obtain the expression for the temperature \(u(x, t)\), we simply substitute the Fourier coefficients \(B_n\) back into the original Fourier series expression derived in step 1.

Plot u(x, t) for several values of t

To plot u(x,t) versus x for several values of t, evaluate the expression for u(x,t) determined in step 2 at different time values and plot the resulting temperature functions u(x, t).

Plot u(x, t) for several values of x

To plot u(x,t) versus t for several values of x, evaluate the expression for u(x,t) determined in step 2 at different values of x and plot the resulting temperature functions u(x, t).

Determine the steady-state temperature

The steady-state temperature is the temperature distribution when the time \(t \rightarrow \infty\). From the Fourier series in step 1, the steady-state temperature can be determined by setting \(t = \infty\). As \(t \rightarrow \infty\), the exponential term \(e^{-\alpha ^2n^2 \pi ^2 t/L^2} \rightarrow 0\), so: $$ u_{ss}(x) = \lim_{t\rightarrow\infty} u(x, t) = 0 $$ That means the steady-state temperature in the bar is 0ºC.

Determine the time to reach within 1 degree of steady-state temperature

To determine how much time must elapse before the temperature at \(x=40\) comes within 1 degree of its steady-state value, simply set: $$ \left|u(40, t) - u_{ss}(40)\right| < 1 $$ Solve for t to find the required time.

Key concepts

Fourier Series

The Fourier Series is a mathematical tool that breaks down a complex periodic function into simple sine and cosine waves. In the context of the heat equation, this series allows us to express the temperature distribution along a bar as an infinite sum of sinusoidal terms. Each of these terms is associated with a coefficient, known as the Fourier coefficient, denoted here as \( B_n \). These coefficients are crucial for accurately representing the initial temperature distribution of the bar.
To find these coefficients, we use the formula:

• \( B_n = \frac{2}{L} \int_0^L u(x, 0) \sin\frac{n \pi x}{L} dx \)

Here, \( L \) represents the length of the bar, and \( u(x, 0) \) is the initial temperature distribution. By solving this integral, you can determine each \( B_n \), which are then substituted back into the Fourier series expansion of the solution to capture the temperature dynamics over time.

Steady-State Temperature

The steady-state temperature in a thermal problem refers to a condition where the temperature distribution becomes invariant with time. This is an ideal state in which all transient effects have died out, and there is no change in thermal gradients or heat flow.
For the problem at hand, as \( t \rightarrow \infty \), the effect of time-dependent exponential decay factors \( e^{-\alpha^2 n^2 \pi^2 t / L^2} \) in the Fourier terms goes to zero. This results in the temperature distribution \( u(x, t) \) tending towards a constant temperature across the entire bar.
Mathematically, this is expressed as:

• \( u_{ss}(x) = \lim_{t\rightarrow\infty} u(x, t) = 0 \)

Thus, in this case, the steady-state temperature of the bar is 0ºC, meaning the bar eventually reaches equilibrium with no temperature variations along its length.

Insulated Boundary Conditions

Insulated boundary conditions imply that there is no heat exchange occurring across the boundaries of the system. For the bar, this means that both ends are perfectly insulated, preventing any heat flow through these points. This is a crucial assumption in modeling scenarios like these because it simplifies the boundary value problem.
In mathematical terms, this condition is typically expressed as:

• \( \frac{\partial u}{\partial x} \bigg|_{x=0} = 0 \) and \( \frac{\partial u}{\partial x} \bigg|_{x=L} = 0 \)

These conditions ensure that the total heat remains constant within the bar, merely redistributing along its length due to thermal conduction. The presence of insulated boundaries directly affects the form of the Fourier coefficients and is a key consideration in correctly solving the heat equation for systems like this.

Initial Temperature Distribution

Initial temperature distribution is the starting temperature profile along the bar at time \( t = 0 \). In this exercise, the function given is:

• \( u(x, 0) = \frac{x(60-x)}{30} \)

This quadratic distribution indicates how the temperature varies across the length of the bar initially. Understanding the initial distribution is critical as it forms the basis for determining the Fourier coefficients \( B_n \).
The method involves integrating this initial function against a set of sine functions to find the respective coefficients. Once the coefficients are known, they are used in the Fourier series that describes how this initial distribution evolves over time.
This particular distribution suggests that the temperature is zero at both ends (since both endpoints of the bar start at zero degrees), peaks somewhere in between, and is symmetric due to the specific boundary conditions imposed.