# Insulated boundary for circular laplace equation?

• richyw
In summary, the problem involves Laplace's equation inside a quarter-circle of radius 2, with an insulated boundary at θ = 0 and u(r,θ/2) = 0. The insulated boundary condition can be expressed as ∂u/∂θ = 0, where \hat n is a unit normal vector perpendicular to the radial vector. The correct answer is \hat n = \hat \theta, as the normal to the boundary must be parallel to \hat \theta.
richyw

## Homework Statement

Consider the Laplace’s equation, ∆u(r,θ) = 0, inside the quarter-circle of radius 2 (0 ≤ θ < π, 0 ≤ r ≤ 2), where the boundary θ is insulated, and $u(r,\theta/2)=0$

Show that the insulated boundary condition can mathematically be expressed as

$$\frac{\partial u}{\partial \theta}u(r,0)=0$$

## The Attempt at a Solution

It's insulated, so I think that means$$]\frac{\partial u(r,0)}{\partial t}=0$$If we use separation of variables $u(r,\theta)=\Phi(\theta)G(r)$ then wouldn't insulated mean I need to take the time derivative of this, using the chain rule?

ok I now know that$$\Lambda u(r,\theta)\cdot \hat{n}=0$$where $\hat{n}$ is an outward pointing unit normal vector. how do I determine $\hat{n}$?

richyw said:

## Homework Statement

Consider the Laplace’s equation, ∆u(r,θ) = 0, inside the quarter-circle of radius 2 (0 ≤ θ < π, 0 ≤ r ≤ 2), where the boundary θ is insulated, and $u(r,\theta/2)=0$

Could you tell us what physical quantity $u$ represents? I assume from the context that $u$ is electric potential with $\mathbf{E} = - \nabla u$.

Show that the insulated boundary condition can mathematically be expressed as

$$\frac{\partial u}{\partial \theta}u(r,0)=0$$

## The Attempt at a Solution

It's insulated, so I think that means$$]\frac{\partial u(r,0)}{\partial t}=0$$
If we use separation of variables $u(r,\theta)=\Phi(\theta)G(r)$ then wouldn't insulated mean I need to take the time derivative of this, using the chain rule?

No. "Insulated" means "the flux of charge across the boundary is zero". That by Gauss's law is equivalent to "$\mathbf{E}$ is parallel to the boundary".

u is temperature here. also that lambda is supposed to be the gradient operator. on a sidenote, what is the latex syntax for that?

anyways in this situation insulated would mean the heat flux across the boundary = 0

so I was initially wrong about the time derivative, but have since corrected it to the formula above. Now I think that in this situation, whatever the normal unit vector to the boundary is, it would be perpendicular to the radial vector.

if $\hat{n}=\hat{\theta}$ then I would get$$\frac{\partial u(r,0)}{\partial\theta}=0$$, so perhaps this practice exam has a typo? Either way, I am not exactly sure why $\hat{n}$ would be in the $\hat{\theta}$ direction, mostly because I don't know what $\hat{\theta}$ even is!

richyw said:
u is temperature here. also that lambda is supposed to be the gradient operator. on a sidenote, what is the latex syntax for that?

\nabla

anyways in this situation insulated would mean the heat flux across the boundary = 0

The heat flux is $\kappa \nabla u$, where $\kappa$ is the thermal diffusivity, which does indeed reduce to
$$\hat{\mathbf{n}} \cdot \nabla u = 0$$
The boundary is $\theta = 0$, which is parallel to $\hat r$. The normal to the boundary must therefore be parallel to $\hat \theta$.

1 person
richyw said:
u is temperature here. also that lambda is supposed to be the gradient operator. on a sidenote, what is the latex syntax for that?

anyways in this situation insulated would mean the heat flux across the boundary = 0

To get the del operator use \nabla.

Your expression for insulated boundary doesn't look correct to me. I would think it would be$$u_r(2,\theta) = 0,~0\le \theta\le \pi$$for the semicircle and$$u_\theta = 0$$along the x axis.

richyw said:
so I was initially wrong about the time derivative, but have since corrected it to the formula above. Now I think that in this situation, whatever the normal unit vector to the boundary is, it would be perpendicular to the radial vector.

if $\hat{n}=\hat{\theta}$ then I would get$$\frac{\partial u(r,0)}{\partial\theta}=0$$, so perhaps this practice exam has a typo?

That is the correct answer, and there is an extra $u$ in the question.

Either way, I am not exactly sure why $\hat{n}$ would be in the $\hat{\theta}$ direction, mostly because I don't know what $\hat{\theta}$ even is!

$\hat \theta$ is the unit vector in the direction of increasing $\theta$. If $(x,y) = (r \cos \theta, r \sin \theta)$ then
$$\hat r = (\cos \theta, \sin \theta)$$
and
$$\hat \theta = (-\sin\theta, \cos\theta)$$
Conveniently, $\hat r \cdot \hat \theta = 0$.

## 1. What is an insulated boundary for the circular Laplace equation?

An insulated boundary is a boundary condition in which no heat or matter can pass through the boundary. In the context of the circular Laplace equation, it means that the temperature or other properties of the solution are constant along the boundary.

## 2. How is an insulated boundary different from other boundary conditions for the circular Laplace equation?

Unlike other boundary conditions, such as Dirichlet or Neumann boundary conditions, an insulated boundary does not specify the value of the solution or its derivative at the boundary. Instead, it imposes a constraint that the solution must be constant along the boundary.

## 3. When should an insulated boundary be used for the circular Laplace equation?

An insulated boundary is typically used when the boundary of the domain is isolated, meaning that it is not in contact with any other boundaries or external environments. This is often the case in problems involving circular domains, such as heat conduction in a circular disc or the flow of electricity in a circular plate.

## 4. How do you mathematically represent an insulated boundary for the circular Laplace equation?

An insulated boundary is typically represented by the following boundary condition: ∂u/∂n = 0, where u is the solution to the circular Laplace equation and n is the unit outward normal vector to the boundary. This condition ensures that the normal derivative of the solution is zero at the boundary, indicating a constant solution along the boundary.

## 5. What are some real-world applications of the insulated boundary for the circular Laplace equation?

The insulated boundary condition is commonly used in a variety of engineering and scientific applications, such as heat transfer in circular pipes or cylinders, electrostatics in circular capacitors, and fluid flow in circular channels. It is also relevant in the study of biological systems, such as diffusion of nutrients in circular cells or the electrical properties of circular membranes.

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