Question

Find the temperature distribution in a rectangular plate \(10 \mathrm{cm}\) by \(30 \mathrm{cm}\) if two adjacent sides are held at \(100^{\circ}\) and the other two sides at \(0^{\circ}\).

Short answer

The temperature distribution is given by \( u(x,y) = 100 \sum_{n=0}^{\infty} \left(\frac{2}{(2n+1)\pi} \right) \sin\left((2n+1)\pi \frac{x}{30}\right) \frac{\sinh((2n+1)\pi \frac{y}{30})}{\sinh((2n+1)\pi \frac{10}{30})} \).

Step-by-step solution

- Define the boundaries and conditions

The rectangular plate dimensions are given as 10 cm by 30 cm. Two adjacent sides are held at 100°C and the other two sides at 0°C. The boundary conditions can be stated as: 1. For the side along the length (30 cm) and width (10 cm), we have: \(u(0, y) = 100°\), \(0 \text{ cm} \text{ to } 10 \text{ cm}\).2. For the opposite side along the length (30 cm) and width (10 cm), we have: \(u(30, y) = 0°\), \(0 \text{ cm} \text{ to } 10 \text{ cm}\).3. For the sides along the width (10 cm): \(u(x, 0) = 100°\), \(0 \text{ cm} \text{ to } 30 \text{ cm}\). 4. For the opposite side along the width (10 cm): \(u(x, 10) = 0°\), \(0 \text{ cm} \text{ to } 30 \text{ cm}\).

- Formulate the heat equation

We will use the steady-state heat equation for a two-dimensional plate: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \].

- Use separation of variables

Assume a solution of the form: \( u(x,y) = X(x)Y(y) \). Substituting into the heat equation, we obtain: \[ X''(x)Y(y) + X(x)Y''(y) = 0 \] Dividing by \(X(x)Y(y)\) gives: \[ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0 \].

- Separate the variables

Since both parts must equal a constant, let: \[ \frac{X''(x)}{X(x)} = -\lambda \] and \[ \frac{Y''(y)}{Y(y)} = \lambda \].

- Solve the ordinary differential equations

First solve: \[ X''(x) + \lambda X(x) = 0 \]. Depending on the boundary conditions, solve for the eigenvalues and eigenfunctions. Repeat the process for \(Y(y)\).

- Apply boundary conditions

Substitute the boundary conditions to find specific solutions for \(X(x)\) and \(Y(y)\).

- Construct the general solution

The general solution is a sum of the separated solutions: \[ u(x,y) = \sum A_n \sin(k_n x) \sinh(k_n y) \] where \(A_n\) and \(k_n\) depend on the boundary conditions.

- Find the coefficients

Use the boundary conditions to determine the coefficients \(A_n\). In this case: \( u(x,y) = 100 \sum (\frac{2}{(2n+1)\pi}) \sin((2n+1)\pi \frac{x}{30}) \sinh((2n+1)\pi \frac{y}{30}) / \sinh((2n+1)\pi \frac{10}{30}) \).

- Present the final solution

The final temperature distribution in the rectangular plate is: \[ u(x,y) = 100 \sum_{n=0}^{\infty} \left(\frac{2}{(2n+1)\pi} \right) \sin\left((2n+1)\pi \frac{x}{30}\right) \frac{\sinh((2n+1)\pi \frac{y}{30})}{\sinh((2n+1)\pi \frac{10}{30})} \].

Key concepts

Boundary Conditions

Boundary conditions define how the edges or boundaries of our rectangular plate behave. In this problem, the boundary conditions are:
- One edge at 0 cm to 10 cm along the width is at 100°C.
- The opposite edge along the width is at 0°C.
- The 30 cm length sides also follow these temperatures, leading to a plateau of temperatures at specific points.
Boundary conditions are critical because they provide the necessary constraints for solving differential equations. They ensure that our temperature values fit the real-world physical scenario.

Steady-State Heat Equation

The steady-state heat equation describes how heat distributes over time in a body until it reaches an equilibrium state. The mathematical form of this equation for a two-dimensional plate is:
\[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \]
Here, \( u(x,y) \) represents the temperature at any point \( (x,y) \) on the plate. In the steady-state, the heat distribution does not change over time, meaning the sum of second derivatives (indicating how heat diffuses in \( x \) and \( y \)) is zero.

Separation of Variables

Separation of variables is a method used to solve partial differential equations like our heat equation. By assuming a solution of the form \( u(x,y) = X(x)Y(y) \), we separate the variables \( x \) and \( y \) to simplify the equation. Substituting \( u(x,y) \) into our heat equation:
\[ X''(x)Y(y) + X(x)Y''(y) = 0 \]
We then divide by \( X(x)Y(y) \) to isolate the variables:
\[ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0 \]
This step is crucial as it allows us to break down the equation into ordinary differential equations (ODEs) which are easier to solve.

Eigenvalues and Eigenfunctions

After separating variables, we obtain two ordinary differential equations (ODEs). Each ODE solution involves constants, leading us to eigenvalues and eigenfunctions. For example, if:
\[ \frac{X''(x)}{X(x)} = -\lambda \]
and
\[ \frac{Y''(y)}{Y(y)} = \lambda \]
Here, \( \lambda \) is an eigenvalue and the corresponding solutions \( X(x) \) and \( Y(y) \) are eigenfunctions. Solving these, considering boundary conditions, gives us specific forms and values critical to our overall solution.

Ordinary Differential Equations

Once we separate our variables, we solve the derived ODEs. For \( X(x) \), we solve:
\[ X''(x) + \lambda X(x) = 0 \]
For \( Y(y) \), we solve:
\[ Y''(y) - \lambda Y(y) = 0 \]
These equations typically have solutions in terms of trigonometric or hyperbolic functions, depending on the nature of \( \lambda \). Applying the boundary conditions helps determine the exact form of these solutions, leading to the complete temperature distribution.