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