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Insulated boundary for circular laplace equation?

  1. Mar 3, 2014 #1
    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 [itex]u(r,\theta/2)=0[/itex]

    Show that the insulated boundary condition can mathematically be expressed as

    [tex]\frac{\partial u}{\partial \theta}u(r,0)=0[/tex]

    2. Relevant equations



    3. The attempt at a solution

    It's insulated, so I think that means[tex]]\frac{\partial u(r,0)}{\partial t}=0[/tex]If we use separation of variables [itex]u(r,\theta)=\Phi(\theta)G(r)[/itex] then wouldn't insulated mean I need to take the time derivative of this, using the chain rule?
     
  2. jcsd
  3. Mar 3, 2014 #2
    ok I now know that[tex]\Lambda u(r,\theta)\cdot \hat{n}=0[/tex]where [itex]\hat{n}[/itex] is an outward pointing unit normal vector. how do I determine [itex]\hat{n}[/itex]?
     
  4. Mar 3, 2014 #3

    pasmith

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

    No. "Insulated" means "the flux of charge across the boundary is zero". That by Gauss's law is equivalent to "[itex]\mathbf{E}[/itex] is parallel to the boundary".
     
  5. Mar 3, 2014 #4
    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
     
  6. Mar 3, 2014 #5
    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 [itex]\hat{n}=\hat{\theta}[/itex] then I would get[tex]\frac{\partial u(r,0)}{\partial\theta}=0[/tex], so perhaps this practice exam has a typo? Either way, I am not exactly sure why [itex]\hat{n}[/itex] would be in the [itex]\hat{\theta}[/itex] direction, mostly because I don't know what [itex]\hat{\theta}[/itex] even is!
     
  7. Mar 3, 2014 #6

    pasmith

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    \nabla

    The heat flux is [itex]\kappa \nabla u[/itex], where [itex]\kappa[/itex] is the thermal diffusivity, which does indeed reduce to
    [tex]
    \hat{\mathbf{n}} \cdot \nabla u = 0
    [/tex]
    The boundary is [itex]\theta = 0[/itex], which is parallel to [itex]\hat r[/itex]. The normal to the boundary must therefore be parallel to [itex]\hat \theta[/itex].
     
  8. Mar 3, 2014 #7

    LCKurtz

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    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.
     
  9. Mar 3, 2014 #8

    pasmith

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    That is the correct answer, and there is an extra [itex]u[/itex] in the question.

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