Insulated boundary for circular laplace equation?

[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

Homework Equations

The Attempt at a Solution

It's insulated, so I think that means]\frac{\partial u(r,0)}{\partial t}=0If 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?

 

[richyw]

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

 

[pasmith]

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

Homework Equations

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".

 

[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

 

[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!

 

[pasmith]

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
<br /> \hat{\mathbf{n}} \cdot \nabla u = 0<br />
The boundary is \theta = 0, which is parallel to \hat r. The normal to the boundary must therefore be parallel to \hat \theta.

 

[LCKurtz]

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.

 

[pasmith]

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
<br /> \hat r = (\cos \theta, \sin \theta)
and
<br /> \hat \theta = (-\sin\theta, \cos\theta)
Conveniently, \hat r \cdot \hat \theta = 0.