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PDE: Is 0 the only solution?

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data
    We the domain be the unit disc D:
    [tex]D=\left \{(x,y):x^{2}+y^{2}<1 \right \}[/tex]

    let u(x,y) solve:
    [tex]-\triangle u+(u_{x}+2u_{y})u^{4}=0[/tex] on D

    boundary:
    u=0 on [tex]\partial D[/tex]

    One solution is u=0. Is it the only solution?


    2. Relevant equations
    Divergence Theorem
    "Energy Method"


    3. The attempt at a solution

    Assume two solutions u1 and u2. We have then:
    [tex]-\triangle u_{1}+(u_{x1}+2u_{y1})u_{1}^{4}=0[/tex]
    and
    [tex]-\triangle u_{2}+(u_{x2}+2u_{y2})u_{2}^{4}=0[/tex]

    subtract the two to get:
    [tex]-\triangle (u_{2}-u_{1})+(u_{x1}+2u_{y1})u_{1}^{4})-(u_{x2}+2u_{y2})u_{2}^{4}=0[/tex]

    let: [tex]v=u_{2}-u_{1}[/tex]

    Multiply by v and integrate. After employing the divergence theorem and the fact that v=0 on the boundary, we get:

    [tex]\int_{\Omega }^{}|\triangledown (v)|^{2}=\int_{\Omega }^{}v(u_{x1}+2u_{y1})u_{1}^{4})-\int_{\Omega }^{}v(u_{x2}+2u_{y2})u_{2}^{4}=0[/tex]

    Take u1 or u2 = 0. If 0 is truly the only solution, the resulting integral should equal to zero.

    Resulting integral where u' = either u1 or u2

    [tex]\int_{\Omega }^{}|\triangledown (u')|^{2}=-\int_{\Omega }^{}u'^{5}(u'_{x}+2u'_{y}))[/tex]

    I conclude that u does not necessarily = 0, so u=0 is not the only solution

    Right? or Wrong? And I have a question, what was the purpose of defining the unit disc?
     
  2. jcsd
  3. Mar 29, 2009 #2
    bump...
     
    Last edited: Mar 29, 2009
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