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 \(\alpha^{2}=1\) $$ u(x, 0)=\left\\{\begin{array}{ll}{x,} & {0 \leq x<20} \\ {40-x,} & {20 \leq x \leq 40}\end{array}\right. $$

Short answer

#Question# Find the expression for the temperature distribution \(u(x, t)\) in a rod of length 40 cm, given the initial temperature distribution \(u(x, 0)\) and the ends of the rod maintained at \(0^{\circ}\) C for all time \(t > 0\). The value of \(\alpha^2\) is given as 1. #Answer# The expression for the temperature distribution \(u(x, t)\) in the rod is given by: $$ u(x, t) = \sum_{n=0}^{\infty} B_n \sin\left(\frac{(2n+1)\pi x}{80}\right) e^{-\left(\frac{(2n+1)\pi}{80}\right)^2 t} $$ where the coefficients \(B_n\) are computed as: $$ B_n = \frac{2}{L} \left[\int_{0}^{20} x\sin\left(\frac{(2n+1)\pi x}{80}\right) dx + \int_{20}^{40} (40-x)\sin\left(\frac{(2n+1)\pi x}{80}\right) dx\right] $$

Step-by-step solution

Write the Heat Equation, Initial conditions, and Boundary conditions

The Heat Equation is given by: $$ \frac{\partial u(x,t)}{\partial t} = \alpha^2\frac{\partial^2 u(x,t)}{\partial x^2} $$ From the exercise: \(\alpha^2 = 1\) Initial temperature distribution: $$ u(x, 0) = \begin{cases} x, & 0 \leq x<20\\ 40-x, & 20 \leq x \leq 40 \end{cases} $$ Boundary conditions: $$ u(0, t) = 0^{\circ}\text{C} \quad u(40, t) = 0^{\circ}\text{C} $$

Perform separation of variables

Assume that the solution u(x,t) can be written as a product of two functions, one depending on x and the other depending on t: $$ u(x, t) = X(x)T(t) $$ Plug this into the heat equation: $$ \frac{\partial u(x,t)}{\partial t} = \frac{\partial^2 u(x,t)}{\partial x^2} $$ $$ X(x)\frac{dT(t)}{dt} = T(t)\frac{d^2 X(x)}{dx^2} $$ Now, we divide through by \(u(x,t) = X(x)T(t)\): $$ \frac{1}{T(t)}\frac{dT(t)}{dt} = \frac{1}{X(x)}\frac{d^2X(x)}{dx^2} $$ Since the left side depends only on t, and the right side depends only on x, they must both be equal to a constant, say \(-k^2\). Then we get two equations: $$ \frac{1}{T(t)}\frac{dT(t)}{dt} = -k^2 \quad \text{and} \quad \frac{1}{X(x)}\frac{d^2X(x)}{dx^2} = -k^2 $$

Solve the time-dependent equation

The time-dependent equation is: $$ \frac{dT(t)}{dt} = -k^2 T(t) $$ The general solution for this equation is: $$ T(t) = Ce^{-k^2 t} $$ where C is a constant.

Solve the space-dependent equation

The space-dependent equation is: $$ \frac{d^2X(x)}{dx^2} = -k^2 X(x) $$ The general solution for this equation is: $$ X(x) = A\cos(kx) + B\sin(kx) $$ where A and B are constants.

Combine the spatial and temporal components to obtain general solution

So the general solution u(x, t) is: $$ u(x, t) = X(x)T(t) = (A\cos(kx) + B\sin(kx)) Ce^{-k^2 t} $$ Applying the boundary condition \(u(0, t) = 0\): $$ u(0, t) = (A\cos(0) + B\sin(0)) Ce^{-k^2 t} = ACe^{-k^2 t} = 0 \Rightarrow A = 0 $$ So the general solution becomes: $$ u(x, t) = B\sin(kx) Ce^{-k^2 t} $$ Now applying the second boundary condition \(u(40, t) = 0\): $$ 0 = B\sin(k\cdot 40) Ce^{-k^2 t} $$ The term \(Ce^{-k^2 t}\) is non-zero, so we have $$ \sin(k\cdot 40) = B\cdot 0 $$ To make this equation valid for all k, we must have: $$ k = \frac{(2n+1)\pi}{80} $$ where n is an integer.

Plug in the initial conditions to find the constants

Now we can rewrite the general solution as: $$ u_n(x, t) = B_n\sin\left(\frac{(2n+1)\pi x}{80}\right)e^{-\left(\frac{(2n+1)\pi}{80}\right)^2 t} $$ Now apply the initial condition \(u(x, 0)\): $$ u(x, 0) = x \quad \text{for} \quad 0 \leq x < 20 \\ u(x, 0) = 40 - x \quad \text{for} \quad 20 \leq x \leq 40 $$ The solution is a sum of these \(u_n(x, t)\) for the given limits of x: $$ u(x, t) = \sum_{n=0}^{\infty} u_n(x, t) $$ Using Fourier series, the expression for \(B_n\) is given by: $$ B_n = \frac{2}{L} \left[\int_{0}^{20} x\sin\left(\frac{(2n+1)\pi x}{80}\right) dx + \int_{20}^{40} (40-x)\sin\left(\frac{(2n+1)\pi x}{80}\right) dx\right] $$ After computing the coefficients \(B_n\), we finally get the expression for the temperature distribution \(u(x, t)\) in the rod as follows: $$ u(x, t) = \sum_{n=0}^{\infty} B_n \sin\left(\frac{(2n+1)\pi x}{80}\right) e^{-\left(\frac{(2n+1)\pi}{80}\right)^2 t} $$

Key concepts

Temperature Distribution in a Rod

Understanding temperature distribution in a rod involves studying how heat diffuses through the material over time. It's essential for applications ranging from engineering to environmental science. The given problem guides us through the journey of heat within a rod that's 40 cm long with both ends maintained at 0°C.

As heat conducts, the temperature at any point in the rod isn't uniform; it varies depending on the location (denoted by x) and time (t). The goal is to find u(x, t), the temperature at a point x and time t. The initial condition provided indicates how the temperature is set up initially along the rod's length, representing a scenario common in practical applications where the temperature initially varies across the material.

To solve this, we apply concepts from partial differential equations (PDEs), specifically the heat equation, to describe the physics of heat conduction. The equation shows how temperature changes both spatially and temporally via thermal diffusion, which relates to the concept of thermal diffusivity represented by α².

Fourier Series and Heat Equation

The use of Fourier series in solving the heat equation is a powerful method in mathematical physics. Fourier series allows us to represent complicated periodic functions as infinite sums of sines and cosines, which are much simpler to work with, especially in PDEs.

In our rod scenario, after applying separation of variables and boundary conditions, we can express the temperature distribution as an infinite sum of sinusoidal functions multiplied by time-dependent exponential decay. This infinite series accounts for the various harmonics (or modes) of heat distribution along the rod. To adhere to the initial condition, we use the Fourier sine series expansion because of the rod's fixed temperature at both ends (akin to a string fixed at both ends in vibration problems). The coefficients (Bₙ) in the series must be determined to match the initial temperature distribution using an integral formula, integrating piecewise over the different intervals for x given in the initial conditions.

This step captures the essence of the initial state's complexity: you decompose it into basic sinusoidal waves, each with a specific frequency and amplitude determined by Bₙ. These amplitudes are crucial because they dictate how each harmonic contributes to the overall temperature distribution over time. As time advances, these contributions diminish at rates dependent on their frequencies, reflecting how different heat modes within the rod dissipate over time.

Separation of Variables Technique

The separation of variables is a standard mathematical technique used to solve PDEs, such as the heat equation. It's grounded in the assumption that the PDE's solution can be broken down into simpler, single-variable functions that are easier to manage and solve.

So, when we assume u(x, t) can be separated into X(x)T(t), we're saying the temperature at any point and time is the product of two independent factors: one depending solely on position, X(x), and the other only on time, T(t). By substituting this into the heat equation and rearranging, we deduce that both sides of the equation must equal a constant (-k²) because one side depends on x alone and the other on t alone - two entirely independent variables.

After this separation, we solve for X(x) and T(t) individually, which typically leads to solutions involving trigonometric and exponential functions. The boundary conditions are applied to determine the specific solution that fits the physical constraints of the problem - in our case, the rod's ends being held at a constant temperature. Through this problem-solving approach, separation of variables not only simplifies complex PDEs but also contributes to a clearer understanding of how different physical factors - space and time - independently influence the outcome.