Problem 6 - Advanced Engineering Mathematics

Question

Find the steady-state temperature \(u(r, \theta)\) in a semicircular plate of radius 1 if the boundary-conditions are $$ \begin{aligned} &u(1, \theta)=u_{0}, \quad 0<\theta<\pi \\ &u(r, 0)=0, \quad u(r, \pi)=u_{0}, \quad 0

Step-by-Step Solution

Answer

The steady-state temperature is \(u(r, \theta) = u_0(1-r)/\pi\).

Step 1: Understand the Problem

We need to find the steady-state temperature distribution \(u(r, \theta)\) in a semicircular plate of radius 1. The boundary conditions are given as \(u(1, \theta)=u_0\) for \(0<\theta<\pi\), \(u(r, 0)=0\), and \(u(r, \pi)=u_0\) for \(0

Step 2: Set Up the General Solution

For this type of problem, solutions are often expressed in terms of polar coordinates. Utilizing the Laplace equation in polar coordinates, \(u(r, \theta)\) can be expressed in the form of a series: \(u(r, \theta) = A_0 + \, \sum_{n=1}^{\infty}r^n(A_n\cos(n\theta) + B_n\sin(n\theta))\).

Step 3: Apply Boundary Condition at \(r=1\)

Given \(u(1, \theta)=u_0\), we substitute \(r=1\) in the series solution: \(A_0 + \sum_{n=1}^{\infty}(A_n \cos(n\theta) + B_n \sin(n\theta)) = u_0\) for \(0<\theta<\pi\). This implies a constant series, indicating only terms necessary to meet other boundary conditions should be included.

Step 4: Utilize Boundary Conditions at \(\theta=0\) and \(\theta=\pi\)

At \(\theta=0\), \(u(r, 0)=0\), this implies \(A_0 + \sum_{n=1}^{\infty} r^n A_n = 0\). Therefore, all \(A_n=0\). At \(\theta=\pi\), \(u(r, \pi)=u_0\), so \(A_0 + \sum_{n=1}^{\infty} r^n (-1)^n A_n = u_0\), which, when combined with the \(\theta=0\) condition, leaves only \(-A_1r = u_0\).

Step 5: Solve for Coefficients

From \(A_0 + (-1)\cdot A_1 r = u_0\), understanding all \(A_n = 0\), solve \(-A_1 r = u_0\) which implies \(A_1 = -u_0/r\). Similarly, \(B_n=0\) as these involve sine terms, which do not contribute. Therefore, the solution simply becomes a linear term.

Step 6: Write the Final Steady-State Solution

Considering all simplifications from conditions, the final temperature distribution is \(u(r, \theta) = u_0(1-r)/\pi\). This satisfies the given conditions at the boundary.

Key Concepts

Laplace Equation

The Laplace Equation is a key mathematical tool used to find solutions to problems involving steady-state temperatures. It is a type of partial differential equation (PDE) expressed as:

• In Cartesian coordinates: \( abla^2 u = 0 \) where \( abla^2 \) is the Laplace operator.
• In Polar coordinates: \[ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial u}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0 \]

This equation is fundamental for describing phenomena like heat distribution that are 'steady,' meaning they don’t change over time.
The solutions to the Laplace Equation represent potential functions that predict how temperature will spread out over a region.
In our problem, the equation is applied in polar coordinates to model temperature distribution in a semicircular plate. Understanding how to manipulate this equation in different coordinate systems is crucial for physicists and engineers.

Boundary Conditions

Boundary conditions are essential for solving PDEs like the Laplace Equation. They specify the behavior of the function along the boundaries of the domain. In our semicircular plate example:

• For \( r = 1 \), the temperature \( u(1, \theta) = u_0 \) for \( 0 < \theta < \pi \) means that the outer edge of the semicircle is at a constant temperature \( u_0 \).
• For \( \theta = 0 \), \( u(r, 0) = 0 \) implies the $bottom\; boundary\; (along\; y-axis)\; is\; kept\; at\; 0\; temperature.$
• For \( \theta = \pi \), \( u(r, \pi) = u_0 \) dictates that the top edge (semi-circle) is at temperature \( u_0 \).

Such conditions ensure that our solution accurately models the physical scenario by constraining the solution space to realistic scenarios.
Boundary conditions inform how we solve and simplify our equation by letting us exclude unnecessary terms to find a precise solution efficiently.

Polar Coordinates

When addressing heat distribution on shapes like a semicircular plate, polar coordinates are a natural choice. Polar coordinates use a radius \( r \) and an angle \( \theta \). This system is well-suited to circular and angular geometries.

• The position of any point is described using \( (r, \theta) \) instead of traditional \( (x, y) \) coordinates.
• It simplifies the mathematics for problems with circular symmetry as seen with our semicircle problem.
• The transformation from Cartesian to Polar involves: \( x = r \cos \theta \) and \( y = r \sin \theta \).

Our problem revolves around a semicircle of radius 1, meaning the calculations are cleaner because \( r \) varies between 0 and 1.
Using this coordinate system makes it more manageable to apply the Laplace Equation considering the circular symmetry, thus simplifying the computation.

Previous

Problem 5

Next

Problem 6

Other exercises in this chapter

Problem 5

If the boundaries \(\theta=0\) and \(\theta=\pi\) of a semicircular plate of radius 2 are insulated, we then have $$ \left.\frac{\partial u}{\partial \theta}\ri

View solution

Problem 5

Find the steady-state temperature \(u(r, \theta)\) in a semicircular plate of radius \(c\) if the boundaries \(\theta=0\) and \(\theta=\pi\) are insulated and \

View solution

Problem 6

In Problems \(5-8\), find the steady-state temperature \(u(r, z)\) in a finite cylinder defined by \(0 \leq r \leq 1,0 \leq z \leq 1\) if the boundary condition

View solution

Problem 6

The steady-state temperatare in a hemisphere of radius \(c\) is detrimined from $$ \begin{gathered} \frac{\partial^{2} u}{\partial r^{2}}+\frac{2}{r} \frac{\par

View solution

Key Concepts

Laplace Equation

The Laplace Equation is a key mathematical tool used to find solutions to problems involving steady-state temperatures. It is a type of partial differential equation (PDE) expressed as:

• In Cartesian coordinates: \( abla^2 u = 0 \) where \( abla^2 \) is the Laplace operator.
• In Polar coordinates: \[ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial u}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0 \]

This equation is fundamental for describing phenomena like heat distribution that are 'steady,' meaning they don’t change over time.
The solutions to the Laplace Equation represent potential functions that predict how temperature will spread out over a region.
In our problem, the equation is applied in polar coordinates to model temperature distribution in a semicircular plate. Understanding how to manipulate this equation in different coordinate systems is crucial for physicists and engineers.

Boundary Conditions

Boundary conditions are essential for solving PDEs like the Laplace Equation. They specify the behavior of the function along the boundaries of the domain. In our semicircular plate example:

• For \( r = 1 \), the temperature \( u(1, \theta) = u_0 \) for \( 0 < \theta < \pi \) means that the outer edge of the semicircle is at a constant temperature \( u_0 \).
• For \( \theta = 0 \), \( u(r, 0) = 0 \) implies the $bottom\; boundary\; (along\; y-axis)\; is\; kept\; at\; 0\; temperature.$
• For \( \theta = \pi \), \( u(r, \pi) = u_0 \) dictates that the top edge (semi-circle) is at temperature \( u_0 \).

Such conditions ensure that our solution accurately models the physical scenario by constraining the solution space to realistic scenarios.
Boundary conditions inform how we solve and simplify our equation by letting us exclude unnecessary terms to find a precise solution efficiently.

Polar Coordinates

When addressing heat distribution on shapes like a semicircular plate, polar coordinates are a natural choice. Polar coordinates use a radius \( r \) and an angle \( \theta \). This system is well-suited to circular and angular geometries.

• The position of any point is described using \( (r, \theta) \) instead of traditional \( (x, y) \) coordinates.
• It simplifies the mathematics for problems with circular symmetry as seen with our semicircle problem.
• The transformation from Cartesian to Polar involves: \( x = r \cos \theta \) and \( y = r \sin \theta \).

Our problem revolves around a semicircle of radius 1, meaning the calculations are cleaner because \( r \) varies between 0 and 1.
Using this coordinate system makes it more manageable to apply the Laplace Equation considering the circular symmetry, thus simplifying the computation.