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Boundary value problem

  1. Dec 8, 2005 #1
    Let D be a domain inside a simple closed curve C in R2. Consider the boundary value problem
    [tex] (\Delta u)(x,y) = 0, \ (x,y) \in D, \\ \frac{\partial u}{\partial n} (x,y) = 0 , \ (x,y) \in C. [/tex]

    where n is the outward unit normal on C. Use Green's Theorem to prove taht every solution u is a constant.

    and yes its DELTA u not nabla u .
    i know that Green's theorem is something lik this
    [tex] \int \int _{D} u \nabla^2 u dx dy = \oint_{C} u \frac{\partial u}{\partial n} ds - \int \int_{D} |grad u|^2 dxdy [/tex]

    so cna i do this
    [tex] \int \int_{D} u \Delta u dx dy = \oint_{C} u \frac{\partial u}{\partial n} ds - \int \int_{D} |\nabla u|^2 dx dy [/tex]
    i mean isnt [tex] \Delta u = \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} [/tex] isnt that true??

    if that s true then moving on
    [tex] \oint_{C} u \frac{\partial u}{\partial n} ds = 0 [/tex] because du/dn = 0 and sinve it is definite integration of a constant the answer is zero
    so all that we're left with is
    [tex] \int \int_{D} u \Delta u dx dy = - \int \int_{D} |\nabla u|^2 dx dy [/tex]
    if i differentiate on both sides it yields
    [tex] u \Delta u \displaystyle{|_{D}} = |\nabla^2 u| [/tex]
    i m not sure how to proceed from here...
    how would one figure out whether u is constnat?
    Please help! Thank you.
     
  2. jcsd
  3. Dec 8, 2005 #2

    Physics Monkey

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    You've done too much work already. Look at your very first equation, the statement of Green's Theorem. Two of the terms in this equation are zero, why? What does that imply about the third term and why?
     
  4. Dec 8, 2005 #3
    well i can see why
    [tex] \oint_{C} u \frac{\partial u}{\partial n} ds = 0 [/tex]
    because du/dn = 0 and sinve it is definite integration of a constant the answer is zero

    also from the boundary values that [itex] (\Delta u) (x,y) = 0 [/itex]
    so then [tex] \int \int _{D} u \nabla^2 u dx dy = 0 [/tex] once again because of the definite integration

    so then
    [tex] \int \int_{D} |grad u|^2 dxdy =0 [/tex]
    so if one differentiates on both sides
    [tex] (\nabla u) \displaystyle{|_{D}} = 0 [/tex]
    integrate both sides and yield u as some constnat value C.
    Was that the right way?
     
  5. Dec 8, 2005 #4

    Physics Monkey

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    You're close. You have everything except the last bit, you can't just differentiate a double integral like that. However, it is possible to conclude that the gradient must be zero. Hint: if you integrate a positive definite integrand over some region and get zero, then what must the integrand have been?
     
  6. Dec 8, 2005 #5
    i see...

    if you definite integrate and you get zero then the integrand musth ave been zero. Thus in this case the gradient is zero. WEll ok a zero gradient measn a maximum of the function... but then uhh
    not usre what to do
     
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