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)=50, \quad 0

Short answer

Question: Determine the expression for the temperature u(x, t) of a rod of length 40 cm with its ends kept at 0°C for all t > 0, and the initial temperature distribution of the rod is given as u(x, 0) = 50 for 0 < x < 40. Given that α² = 1. Answer: $$ u(x,t) = \sum_{n=1,3,5,...}^{\infty} \frac{20}{n\pi}(-1)^{(n-1)/2}\sin(\frac{n\pi}{40}x)\exp(-\frac{n^2\pi^2}{40^2}t) $$

Step-by-step solution

State the heat equation and boundary conditions

The heat equation, also known as the one-dimensional diffusion equation, is given by: $$ \frac{\partial u}{\partial t} = \alpha^2\frac{\partial^2 u}{\partial x^2} $$ We are given that α² = 1. Boundary conditions: 1. u(0, t) = 0 2. u(40, t) = 0 Initial condition: u(x, 0) = 50 for 0 < x < 40.

Applying boundary and initial conditions

First, solve the heat equation using separation of variables (i.e., plug in a product of functions for \(u(x, t)\): \(u(x, t) = X(x)T(t)\)) and solve the resulting equations for X(x) and T(t). Then, apply the boundary conditions: 1. At x = 0, X(0)T(t) = 0 => X(0) = 0. 2. At x = 40, X(40)T(t) = 0 => X(40) = 0.

Solve for X(x)

We solve the X(x) part of the separated equation using the boundary conditions: $$ X''(x) = -\lambda X(x) $$ So, the general solution for X(x) is given by: $$ X(x) = A\sin(\sqrt{\lambda}x) + B\cos(\sqrt{\lambda}x) $$ Apply the boundary conditions to find A and B. 1. X(0) = 0 => B = 0 2. X(40) = 0 => A\sin(\sqrt{\lambda}40) = 0 => \lambda_n = \frac{n^2\pi^2}{40^2} for n = 1, 2, 3,... So, X(x) = \(A\sin(\frac{n\pi}{40}x)\).

Solve for T(t)

Now, solve the T(t) part of the separated equation: $$ \frac{T'(t)}{T(t)} = -\lambda T(t) $$ So, the general solution for T(t) is given by: $$ T(t) = C\exp(-\lambda_nt) $$

Combine the solutions for u(x,t)

Now, let's combine the solutions for X(x) and T(t) to get the solution for u(x,t): $$ u(x,t) = \sum_{n=1}^{\infty} D_n\sin(\frac{n\pi}{40}x)\exp(-\frac{n^2\pi^2}{40^2}t) $$ Use the initial condition u(x, 0) = 50: $$ 50 = \sum_{n=1}^{\infty} D_n\sin(\frac{n\pi}{40}x) $$ Use the Fourier sine series to find the coefficients D_n by multiplying both sides by \(\sin(\frac{m\pi}{40}x)\) and integrate from 0 to 40: $$ D_m = \frac{20}{m\pi}(-1)^{(m-1)/2} $$

Final expression for u(x,t)

Finally, we get the expression for u(x,t): $$ u(x,t) = \sum_{n=1,3,5,...}^{\infty} \frac{20}{n\pi}(-1)^{(n-1)/2}\sin(\frac{n\pi}{40}x)\exp(-\frac{n^2\pi^2}{40^2}t) $$

Key concepts

Boundary Conditions

Boundary conditions are crucial in solving partial differential equations like the heat equation, providing necessary constraints at the edges of the domain. In our heat conduction problem inside a rod, the boundary conditions are:

• \(u(0, t) = 0\): This means the temperature at the starting point of the rod remains at \(0^{\circ}C\).
• \(u(40, t) = 0\): Similarly, the temperature at the other end of the rod is also maintained at \(0^{\circ}C\).

These conditions reflect the reality where the rod ends are kept at a constant, freezing temperature. They are essential to help us determine the unique solution to the heat equation by eliminating any arbitrary constants associated with the general solution.

Separation of Variables

The method of separation of variables is a powerful technique used to solve partial differential equations, like the heat equation, by reducing them into simpler, often solvable ordinary differential equations. In this problem, we expressed the temperature as a product of two functions: \(u(x, t) = X(x)T(t)\). This approach converts the complex problem into two simpler ones:

• For \(X(x)\), dependent solely on spatial variables.
• For \(T(t)\), dependent entirely on temporal variables.

Solving these separate equations is manageable because each involves only one variable. By applying boundary conditions directly to \(X(x)\), we could find specific forms satisfying those constraints, while \(T(t)\) depended on the exponential decay reflecting the loss of heat over time.

Fourier Series

Fourier Series allows us to expand a function into a series of sinusoidal components. In the context of solving the heat equation, the initial temperature distribution can be represented using Fourier Sine Series, given the boundary conditions ensure the expansion is sine-based. This series captures the periodic nature of the solution over the spatial domain:\(0 < x < 40\).The coefficients \(D_n\) in this series are crucial because they weight the contributions of each harmonic (sine wave) to reconstruct the solution fully. Calculating these coefficients involves integrating the product of the initial condition and sine functions over the rod's length, ensuring the approximation accurately reflects the initial temperature distribution throughout the rod.

Initial and Boundary Value Problems

Initial and boundary value problems involve determining a function with specified values at the initial time and specific constraints at the domain's boundaries. Our heat equation problem fits this framework:

• Initial condition: \(u(x, 0) = 50\) for \(0 < x < 40\) implies the rod starts with a ha uniform temperature.
• Boundary conditions ensure the rod ends remain at \(0^{\circ}C\) throughout the process.

These conditions defined our problem for the heat equation, guiding us to an exact solution. The initial condition dictated how the solution must behave at the start, while the boundary conditions shaped its behavior at every subsequent moment and edge of the rod. In turn, this rigor ensures our derived solution is unique and physically replicates the situation described.