Question

Solve the semi-infinite plate problem if the bottom edge of width \(\pi\) is held at \(T=\cos x\) and the other sides are at \(0^{\circ}\).

Short answer

T(x, y) = \sum_{m=1}^{\infty} \frac{2}{\pi} \left( \frac{(-1)^{m+1} -1}{m^2 - 1} \right) \sin(mx) e^{-my}

Step-by-step solution

- Define the Problem

We need to solve the heat conduction problem for a semi-infinite plate, where one edge is held at a temperature given by a function and the other sides are at zero degrees. The bottom edge has a temperature distribution given by: \[ T(x, 0) = \cos x \] for \(0 \leq x \leq \pi\).

- Setup the Solution Using Fourier Sine Series

The temperature distribution can be represented as a Fourier sine series because the boundary condition is zero on the other sides. Therefore, the solution can be written as: \[ T(x, y) = \sum_{n=1}^{\infty} b_n \sin(nx) e^{-ny} \]

- Determine the Fourier Coefficients

To find the coefficients \(b_n\), we use the initial temperature distribution at the bottom edge: \[ \cos x = \sum_{n=1}^{\infty} b_n \sin(nx) \] Multiply both sides by \(\sin(mx)\): \[ \cos x \sin(mx) = \sum_{n=1}^{\infty} b_n \sin(nx) \sin(mx) \] Integrate over \([0, \pi]\): \[ \int_0^\pi \cos x \sin(mx) \, dx = \int_0^\pi \sum_{n=1}^{\infty} b_n \sin(nx) \sin(mx) \, dx \]

- Solve the Integral

Only the term with \(n = m\) survives due to orthogonality of sine functions. Thus: \[ \int_0^\pi \cos x \sin(mx) \, dx = b_m \frac{\pi}{2} \] Therefore, \[ b_m = \frac{2}{\pi} \int_0^\pi \cos x \sin(mx) \, dx \]

- Evaluate the Specific Integral

Compute the integral \[ b_m = \frac{2}{\pi} \left[ \frac{\sin(mx + x)}{2(m+1)} - \frac{\sin(mx - x)}{2(m-1)} \right]_0^\pi \] This simplifies to: \[ b_m = \frac{2}{\pi} \left( \frac{(-1)^{m+1} -1}{m^2 - 1} \right) \]

- Write the Final Solution

Finally, the temperature distribution in the plate is: \[ T(x, y) = \sum_{m=1}^{\infty} \frac{2}{\pi} \left( \frac{(-1)^{m+1} -1}{m^2 - 1} \right) \sin(mx) e^{-my} \]

Key concepts

Fourier sine series

To solve the semi-infinite plate problem, we use the Fourier sine series. This is a powerful mathematical tool for representing a function as an infinite sum of sine functions. In this problem, we assume the temperature distribution along the bottom edge can be expressed in this way. A Fourier sine series is particularly useful because sine functions naturally satisfy the boundary condition of zero temperature at the other sides of the plate. Representing the temperature distribution using the Fourier sine series, we have:
T(x, y) = sum_{n=1}^{∞} b_n sin(nx) e^{-ny}
where b_n are the Fourier coefficients to be determined.

Boundary conditions

Boundary conditions are critical in solving heat conduction problems. They essentially define the behavior of temperature at the confines of the problem domain. In this semi-infinite plate problem, the bottom edge (where y=0) is maintained at a temperature distribution given by T(x,0) = cos(x) within the range 0 ≤ x ≤ π.
The other sides of the plate are kept at zero degrees Celsius. This setup allows us to employ the Fourier sine series because the sine functions will naturally comply with the zero-temperature boundaries. Thus, the boundary conditions ensure that the temperature distribution function aligns correctly with the physical constraints of the problem.

Temperature distribution

In heat conduction problems, determining the temperature distribution within the material is the primary goal. The temperature distribution, denoted as T(x, y), describes how temperature varies within the plate for different positions (x, y).

For our semi-infinite plate, the temperature distribution can be expressed as a sum of sine functions multiplied by exponential terms: T(x, y) = sum_{n=1}^{∞} b_n sin(nx) e^{-ny}

Here, each term in the series represents a mode of heat distribution influenced by the boundary conditions. The coefficients b_n determine the contribution of each mode. These coefficients are calculated from the Fourier sine series.

Semi-infinite plate

A semi-infinite plate refers to a plate that has a finite width but extends infinitely in one direction. In this case, our plate has a bottom edge running from 0 to π in the x-direction, while extending infinitely in the y-direction.

This simplification helps reduce the complexity of solving the heat conduction problem since it eliminates boundary conditions at the far end of the plate. Instead, the focus remains on the specified boundary conditions at the bottom and sides of the plate.
Analyzing such a setup requires using methods like Fourier series to handle the infinite extent in the y-direction while accommodating the given conditions.

the functions are identical

This property is crucial because it implies that when we multiply both sides of our Fourier sine series by sin(mx) and integrate over [0, π], only the term where n=m will survive.

These orthogonality conditions can be expressed mathematically as: integral_0^π sin(nx) sin(mx) dx = 0, for n != m and π/2, for n = m

This makes it possible to isolate each coefficient b_n by performing the integral, simplifying our solution to the heat conduction problem.

Summary

In our heat conduction problem for a semi-infinite plate, we leverage the Fourier sine series to represent the temperature distribution. Given boundary conditions force the solution into this form, allowing for easier handling of the problem. Using the orthogonality of sine functions, we calculate the Fourier coefficients that describe the temperature distribution throughout the plate. This structured approach ensures a comprehensive and effective solution.