# Homework Help: Insulated boundary for circular laplace equation?

1. Mar 3, 2014

### richyw

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

3. 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?

2. Mar 3, 2014

### richyw

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 $\hat{n}$?

3. Mar 3, 2014

### pasmith

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

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".

4. Mar 3, 2014

### richyw

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

5. Mar 3, 2014

### richyw

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!

6. Mar 3, 2014

### pasmith

\nabla

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 $\hat \theta$.

7. Mar 3, 2014

### LCKurtz

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.

8. Mar 3, 2014

### pasmith

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

$\hat \theta$ 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, $\hat r \cdot \hat \theta = 0$.