Question

Consider the conduction of heat in a rod \(40 \mathrm{cm}\) in length whose ends are maintained at \(0^{\circ} \mathrm{C}\) for all \(t>0 .\) In each of Problems 9 through 12 find an expression for the temperature \(u(x, t)\) if the initial temperature distribution in the rod is the given function. Suppose that \(\alpha^{2}=1\) $$ u(x, 0)=\left\\{\begin{array}{cc}{0,} & {0 \leq x<10} \\ {50,} & {10 \leq x \leq 30} \\ {0,} & {30

Short answer

Question: Determine the temperature distribution, u(x,t), in a rod of length 40 cm given that the temperature at the ends of the rod is always 0°C. The rod has an initial temperature distribution as follows: u(x,0)=0 for 0≤x<10, u(x,0)=50 for 10≤x≤30, and u(x,0)=0 for 30<x≤40. Use the heat equation, ut = α²uxx, with α²=1, to find the solution. Answer: The temperature distribution u(x,t) in the rod is given by: $$ u(x,t) = \sum_{n=1}^{\infty} \left[\frac{10}{n\pi}(\cos\frac{3n\pi}{4} - \cos\frac{n\pi}{4})\right] \sin\frac{n\pi x}{40}\,e^{-n^{2}\pi^{2}t/1600} $$

Step-by-step solution

Separate u(x,t) into a product of two functions

Since we are using the separation of variables technique, let's assume that u(x,t) can be written as a product of two functions X(x) and T(t): $$ u(x,t) = X(x)T(t) $$

Plug the separated u(x,t) into the heat equation

Substitute the separated u(x,t) into the heat equation: $$ X(x)T'(t) = T(t)X''(x) $$

Separate the variables

Divide both sides by \(X(x)T(t)\) to separate the variables: $$ \frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda $$ where \(\lambda\) is a separation constant.

Set up separate ODEs for X(x) and T(t)

Set up two separate ordinary differential equations (ODEs) for X(x) and T(t): $$ X''(x) + \lambda X(x) = 0 $$ and $$ T'(t) - \lambda T(t) = 0 $$

Solve the ODE for T(t)

Solve the ODE for T(t): $$ T'(t) - \lambda T(t) = 0 $$ This is a first-order linear ODE, which has the general solution: $$ T(t) = ce^{\lambda t} $$ where c is a constant.

Find the boundary conditions for X(x)

Apply the boundary conditions at the ends of the rod at x=0 and x=40: $$ X(0) = X(40) = 0 $$

Solve the ODE for X(x)

Since we know the boundary conditions for X(x), we can solve the ODE: $$ X''(x) + \lambda X(x) = 0 $$ This is a second-order linear homogeneous ODE with constant coefficients: for non-trivial solutions, we must have \(\lambda = (n\pi/40)^2\) for some positive integer n. Then, the general solution is given by: $$ X(x) = \sum_{n=1}^{\infty} B_n \sin\frac{n\pi}{40}\,x $$ for some constants B_n.

Apply the initial conditions for u(x,0)

Apply the initial conditions for u(x,0): $$ u(x,0) = X(x)T(0) = \sum_{n=1}^{\infty}B_n\sin \frac{n\pi x}{40} $$ We know \(u(x,0)\) for each x in the intervals \(0\leq x<10\), \(10\leq x\leq 30\), and \(30<x\leq 40\). To obtain the coefficients B_n, we will use Fourier sine series expansion for given intervals.

Compute the Fourier sine coefficients B_n

Compute the Fourier sine coefficients B_n for each interval of x: For \(10\leq x\leq 30\), we have: $$ B_n = \frac{2}{40}\int_{10}^{30} 50\sin \frac{n\pi x}{40}\, dx $$ and for other intervals, \(B_n = 0\). Now calculate the integral for B_n: $$ B_n = \frac{2}{40} \left[\frac{400}{n\pi}(\cos\frac{n\pi x}{40})\right]_{10}^{30} = \frac{10}{n\pi}(\cos\frac{3n\pi}{4} - \cos\frac{n\pi}{4}) $$

Write down the solution u(x,t)

Combine X(x) and T(t) to get the solution for u(x,t): $$ u(x,t) = \sum_{n=1}^{\infty} \left[\frac{10}{n\pi}(\cos\frac{3n\pi}{4} - \cos\frac{n\pi}{4})\right] \sin\frac{n\pi x}{40}\,e^{-n^{2}\pi^{2}t/1600} $$ This is the solution for the temperature distribution u(x,t) in the rod, given the initial temperature distribution and boundary conditions.

Key concepts

Separation of Variables

When tackling partial differential equations like the heat conduction equation, a powerful technique called 'separation of variables' is frequently used. This method involves assuming that the solution can be broken into simpler, single-variable functions. For our heat conduction problem, we write the temperature function u(x,t) as the product of two separate functions: X(x) related to space and T(t) related to time, such that u(x,t) = X(x)T(t).

By substituting this product into the heat equation and separating terms involving x from those involving t, the multi-variable problem is reduced to two ordinary differential equations. Each of these can then be solved individually, greatly simplifying the problem. To ensure that the solutions to these ODEs are relevant to the physical scenario, we must afterwards apply boundary and initial conditions.

Fourier Series

In solving heat equations, the concept of Fourier series is invaluable, especially when dealing with initial temperature distributions that are not trivial. The Fourier series allows for the expansion of piecewise functions, like our rod's initial temperature distribution, into an infinite series of sine and cosine functions.

The coefficients of these terms, known as Fourier coefficients, are calculated by integrating the initial temperature function over the domain of interest, multiplying by the sine or cosine terms at various frequencies. For our rod, since it's held at 0°C at both ends, only sine terms appear, reflecting the odd function symmetry with respect to the boundaries. By calculating the Fourier sine coefficients for each interval, taking into account the given initial temperature distribution, we can reconstruct the temperature profile at any given time.

Boundary Value Problem

The heat conduction problem on a rod is a classic example of a boundary value problem, which is a type of differential equation accompanied by a set of constraints, known as boundary conditions. For this problem, the two ends of the rod are maintained at 0°C, translating to X(0) = X(40) = 0 for the spatial component of our separation of variables solution.

This specification dictates the form of the solution, particularly forcing it to be zero at both boundaries, which ultimately influences the entire temperature distribution along the rod. These boundary conditions not only shape the functional form of solutions but also are essential in determining the specific eigenvalues, denoted by eqneqneqn, that will fulfill the physical requirements of the problem.

Initial Temperature Distribution

Understanding the initial temperature distribution is crucial when solving transient heat conduction problems. It represents the temperature profile of the rod at the start of observation, t = 0. The given stepwise function for our exercise clearly depicts this distribution, with sections of the rod at 0°C and a central part at 50°C.

In our solution process, this initial condition is used to determine the Fourier coefficients, which are integral to finding the full time-dependent solution. The Fourier series is thus tailored to this initial distribution, ensuring that the resulting solution accurately reflects the physical state of the rod as heating progresses. Effectively, the available information about the temperature distribution at t = 0 propels the entire solution process and helps predict the future temperatures at any given point in time and space within the rod.