Question

A metal plate covering the first quadrant has the edge which is along the \(y\) axis insulated and the edge which is along the \(x\) axis held at $$ u(x, 0)=\left\\{\begin{array}{cl} 100(2-x), & \text { for } 0< x < 2 \\ 0, & \text { for } x > 2 \end{array}\right. $$ Find the steady-state temperature distribution as a function of \(x\) and \(y .\) Hint: Follow the procedure of Example \(2,\) but use a cosine transform (because \(\partial u / \partial x=0\) for \(x=0\) ). Leave your answer as an integral like (9.13)

Short answer

The steady-state temperature distribution is given by the integral:\[ u(x, y) = \int_0^{\infty} 200 \left( \frac{\sin(2k) - k \cos(2k)}{k^2} \right) e^{-ky} \cos(kx) \ dk \]

Step-by-step solution

Define the Boundary Conditions

The problem involves a metal plate in the first quadrant. The boundary conditions given are:1. The edge along the y-axis is insulated, implying that \( \frac{\partial u}{\partial x} (0, y) = 0 \).2. The edge along the x-axis is held at specific temperatures: \(u(x, 0) = \begin{cases} 100(2-x), & \text{for} \ 0 < x < 2 \ 0, & \text{for} \ x > 2 \end{cases}\)

Set up the heat equation

We need to solve for the steady-state temperature distribution, which means the temperature does not change with time. The steady-state heat equation in two dimensions is given by:\[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \]

Apply the Cosine Transform

Given that \(\frac{\partial u}{\partial x} (0, y) = 0\), it's appropriate to apply the cosine transform. Define the cosine transform of \(u(x, y)\) with respect to \(x\) as:\[ \hat{u}(k, y) = \int_0^{\infty} u(x, y) \cos(kx) \, dx \]

Transform the Boundary Condition

Apply the cosine transform to the boundary condition at \( y = 0 \):\[ \hat{u}(k, 0) = \int_0^{2} 100(2-x) \cos(kx) \, dx \]

Simplify the Integral

Calculate the integral:\[ \hat{u}(k, 0) = 100 \int_0^{2} (2-x) \cos(kx) \, dx = 100 \left[ 2 \int_0^{2} \cos(kx) \, dx - \int_0^{2} x \cos(kx) \, dx \right] \]

Evaluate the Integrals

Integrate term by term:\[ \int_0^{2} \cos(kx) \, dx = \frac{\sin(2k)}{k} \]\[ \int_0^{2} x \cos(kx) \, dx = \frac{2 \sin(2k) - 2k \cos(2k)}{k^2} \]

Substitution and Simplification

Substitute back into the expression obtained in Step 5 and simplify:\[ \hat{u}(k, 0) = 100 \left[ 2 \left( \frac{\sin(2k)}{k} \right) - \left( \frac{2 \sin(2k) - 2k \cos(2k)}{k^2} \right) \right] = 200 \left( \frac{\sin(2k) - k \cos(2k)}{k^2} \right) \]

Solve the Differential Equation

The transformed function \( \hat{u}(k, y) \) satisfies:\[ \frac{\partial^2 \hat{u}}{\partial y^2} - k^2 \hat{u} = 0 \]Solve this differential equation for \( \hat{u}(k, y) \). The general solution is:\[ \hat{u}(k, y) = \hat{u}(k, 0) e^{-ky} \]

Inverse Cosine Transform

Finally, apply the inverse cosine transform to find \(u(x, y)\):\[ u(x, y) = \int_0^{\infty} \hat{u}(k, 0) e^{-ky} \cos(kx) \ dk \]Substitute \( \hat{u}(k, 0) \):\[ u(x, y) = \int_0^{\infty} 200 \left( \frac{\sin(2k) - k \cos(2k)}{k^2} \right) e^{-ky} \cos(kx) \ dk \]

Key concepts

Two-Dimensional Heat Equation

The two-dimensional heat equation is a partial differential equation (PDE) commonly used to describe the distribution of heat (or temperature variation) in a given region over time. For steady-state conditions, where temperature does not change with time, the equation is given by:
\[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \]
This equation states that the sum of the second derivatives of temperature with respect to both spatial dimensions (x and y) must equal zero. It represents a scenario where heat flows within the region but does not accumulate or deplete anywhere.

Boundary Conditions

Boundary conditions are essential to solving the two-dimensional heat equation. They define the behavior of the temperature on the boundaries of the region under consideration. In this exercise:

• The edge along the y-axis is insulated, implying that there's no heat flow across this edge. Mathematically, this is expressed as: \[ \frac{\partial u}{\partial x} (0, y) = 0 \]
• The edge along the x-axis is held at a specified temperature. The boundary condition here is given by: \[ u(x, 0) = \begin{cases} 100(2-x), & \text{for} \ 0 < x < 2 \0, & \text{for} \ x > 2 \end{cases} \]

These conditions help in defining the solution uniquely by providing the necessary constraints.

Fourier Cosine Transform

To solve PDEs like the two-dimensional heat equation, one effective method is using integral transforms, such as the Fourier cosine transform. This transform is especially useful when the boundary conditions involve derivatives equaling zero along certain edges.
The Fourier cosine transform of a function \(u(x, y)\) with respect to \(x\) is defined as: \[ \hat{u}(k, y) = \int_0^{\infty} u(x, y) \cos(kx) \, dx \]
Applying this transform simplifies the PDE into a more manageable form, often reducing it to an ordinary differential equation (ODE) in terms of \(y\).

Integral Transform Solution

The integral transform solution involves transforming the boundary conditions and PDE into a set of integrals that can be solved step-by-step. Here are the steps simplified:

• Apply the cosine transform to the boundary condition at \(y = 0\): \[ \hat{u}(k, 0) = \int_0^{2} 100(2-x) \cos(kx) \, dx \]
• Evaluate the integral to find \( \hat{u}(k, 0) \).
• Solve the transformed differential equation: \[ \frac{\partial^2 \hat{u}}{\partial y^2} - k^2 \hat{u} = 0 \]
  The solution is: \[ \hat{u}(k, y) = \hat{u}(k, 0) e^{-ky} \]
• Finally, apply the inverse cosine transform to obtain the temperature distribution \(u(x, y)\): \[ u(x, y) = \int_0^{\infty} 200 \left( \frac{\sin(2k) - k \cos(2k)}{k^2} \right) e^{-ky} \cos(kx) \, dk \]

This solution represents the steady-state temperature distribution within the specified metal plate.