Question

The cube \(|x|

Short answer

Use the method of separation of variables and Fourier series. For short times, consider large eigenvalues; for long times, consider small eigenvalues.Maximize the temperature by solving for the first derivative concerning time.

Step-by-step solution

- Understand the Problem

The problem involves solving the heat equation in a cube with specific boundary conditions and an initial pulse of energy at the origin. The goal is to find expressions for temperature at a specific point for both short and long times.

- Formulate the Heat Equation in the Cube

The heat equation in three dimensions is given by

- Separate Variables

Solve the heat equation using the method of separation of variables. Assume a solution of the form

- Apply Boundary Conditions

Apply the boundary conditions

- Initial Condition

The initial condition is given by

- Temperature Expression for Short Times

For short times, consider the largest exponential terms. Solve the initial condition using Fourier series:

- Temperature Expression for Long Times

For long times, only the smallest eigenvalues contribute significantly. Again, use Fourier series;

- Determine the Time for Maximum Temperature

To find the time at which the temperature is maximum, solve for the derivative with respect to time set to zero.

Key concepts

Boundary Conditions

Boundary conditions are essential in solving partial differential equations like the heat equation. They describe how the temperature behaves at the boundaries of the cube.
For this problem, the cube is immersed in a heat bath at temperature zero, meaning the temperature on the surface of the cube is always zero:
\[ T \bigg|_{(|x|=L/2, |y|=L/2, |z|=L/2)} = 0 \]
This condition ensures that no heat flows across the boundaries, as the surfaces are kept at a constant temperature.

Initial Pulse of Energy

An initial pulse of energy gives the starting condition for the temperature inside the cube. At the very beginning, a pulse of energy is introduced at the origin, modeled mathematically using the Dirac delta function:
\[ T(x, y, z, 0) = \delta(x, y, z) \]
This means all the energy is initially concentrated at the origin. The delta function simplifies the initial condition, allowing for an easier formulation of the problem.

Separation of Variables

To solve the heat equation, we can use the method of separation of variables. We assume that the temperature function can be written as the product of functions, each depending on a single coordinate and time:
\[ T(x,y,z,t) = X(x)Y(y)Z(z)T(t) \]
Substituting this into the heat equation and dividing through by \(XYZT\), we can separate the variables and obtain individual ordinary differential equations (ODEs) for \(X(x)\), \(Y(y)\), \(Z(z)\), and \(T(t)\).
Each ODE can then be solved with the given boundary conditions and initial conditions.

Fourier Series

The Fourier series helps us represent functions as a sum of sine and cosine terms. Applying this to the initial condition, we can expand the delta function:
\[ \delta (x-x_0) \delta (y-y_0) \delta (z-z_0) = \sum_{n,m,p} A_{n,m,p} e^{-i(k_n x + k_m y +k_p z)} \]
For the temperature problem, this means expressing the initial temperature distribution as a sum of eigenfunctions of the system. Solving the equation involves finding these coefficients and summing the terms with appropriate factors.

Temperature Distribution

Finally, we derive the temperature distribution for short and long times. For short times, the largest exponential terms dominate:
\[ T(x,y,z,t) = \sum C_{n,m,p} e^{-\lambda_{n,m,p} t} \phi_{n,m,p} (x,y,z) \]
We keep the most significant terms to approximate the temperature at \(t=0\).
For long times, the smallest eigenvalues prevail, and we can neglect higher exponential terms:
\[ T(x,y,z,t) \approx \sum_{smallest} C_{n,m,p} e^{-\lambda_{n,m,p} t} \phi_{n,m,p} (x,y,z) \]
By analyzing both short and long times, we can find when the temperature at a specific point is maximum by setting the derivative with respect to time to zero and solving for \(t\).