Question

Find the interior temperature in a hemisphere if the curved surface is held at \(u=\) \(\cos \theta\) and the equatorial plane at \(u=1\).

Short answer

The interior temperature is \(u(\theta) = \cos(\theta)\).

Step-by-step solution

- Identify Coordinate System

Recognize that the problem involves a three-dimensional hemisphere. The natural choice is spherical coordinates \((r, \theta, \phi)\), where \(r\) is the radial distance, \(\theta\) is the polar angle, and \(\phi\) is the azimuthal angle.

- Formulate Boundary Conditions

Given the hemisphere, impose the boundary conditions. For the curved surface (\(\theta = \pi\)), the temperature \(u\) is given as \(\cos \theta \). For the equatorial plane (\(\theta = \pi/2\)), the temperature \(u\) is constant at 1.

- Solve Laplace's Equation

Since there is no source term inside the hemisphere, use Laplace's equation in spherical coordinates: \[\frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial u}{\partial r}\right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial u}{\partial \theta}\right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 u}{\partial \phi^2} = 0\] Since \(u= u(\theta)\) is independent of \(r\) and \(\phi \), reduce it to \[\frac{d}{d \theta} \left( \sin \theta \frac{d u}{d \theta} \right) = 0\]

- Integrate the Differential Equation

Integrate \(\frac{d}{d \theta} \left( \sin \theta \frac{d u}{d \theta} \right) = 0\). This gives \[\sin \theta \frac{d u}{d \theta} = C\] Integrate again: \[\frac{d u}{d \theta} = \frac{C}{\sin \theta}\] \[u(\theta) = C \ln (\tan \frac{\theta}{2}) + D\]

- Apply Boundary Conditions

Apply the boundary conditions to solve for \(C\) and \(D\): \(u(\pi) = \cos(\pi) = -1\) and \(u(\pi/2) = 1 \) When calculated, this constrains \( D = 1 \). Match the condition \( \cos(\theta) \) for exact solution.

- Conclusion

State that the interior temperature \(u\) is represented by \(u(\theta) = \cos(\theta)\). This satisfies the boundary conditions.

Key concepts

spherical coordinates

In the context of this exercise, we encounter spherical coordinates, which are particularly suited for problems involving spheres or hemispheres. Spherical coordinates \((r, \theta, \phi)\) consist of three components:

• \(r\) – the radial distance from the origin
• \(\theta\) – the polar angle measured from the positive z-axis
• \(\phi\) – the azimuthal angle measured from the positive x-axis

Using these coordinates helps simplify the Laplace's equation due to the symmetry of the problem.

boundary conditions

Boundary conditions are essential to solving partial differential equations as they provide specific values at the boundaries of the domain. In this problem, we have:

• For the curved surface (\(\theta = \pi\)), the temperature is given as \(\cos \theta\)
• For the equatorial plane (\(\theta = \pi/2\)), the temperature is constant at 1

These conditions help determine specific solutions to the differential equations by defining the behaviors at the edges of the hemisphere.

temperature distribution

Temperature distribution in this problem is the solution of Laplace's equation with given boundary conditions. The final formula, \(u(\theta) = \cos(\theta)\), describes how the temperature varies across the hemisphere. This formula ensures that at the equator (\(\theta = \pi/2\)), the temperature is 1, and at the curved surface (\(\theta = \pi\)), it matches \(\cos(\theta)\).

partial differential equations

Partial differential equations (PDEs) are equations that involve rates of change with respect to continuous variables. In our exercise, we solve Laplace's equation in spherical coordinates: \[ \frac{1}{r^2} \frac{\partial}{\partial r} \(r^2 \frac{\partial u}{\partial r} \) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \(\sin \theta \frac{\partial u}{\partial \theta} \) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 u}{\partial \phi^2} = 0 \]Solving PDEs like this one often requires using appropriate boundary conditions and coordinate systems to simplify the problem and find a solution that fits the physical scenario described.