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Partial differential equation help needed (first order)

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Let u be a solution of [itex]a(x,y)u_{x}+b(x,y)u_{y}=-u[/itex] (I) of class [itex]C^{1}[/itex]
    in the closed unit disk [itex]\Omega[/itex] in the xy-plane. Let [itex]a(x,y)x+b(x,y)y>0[/itex] (II)
    on the boundary of [itex]\Omega[/itex]. Prove that u vanishes identically. (Hint: Show that on [itex]\Omega[/itex] max u[itex]\leq 0[/itex], min u [itex]\geq 0[/itex], using conditions for a maximum at a boundary point.

    2. Relevant equations
    [itex]\frac{dx}{dt}=a(x,y)[/itex]
    [itex]\frac{dy}{dt}=b(x,y)[/itex]
    [itex]\frac{dz}{dt}=-z[/itex]
    (the chapter was about solving first order pde's using autonomous systems of ODEs, but it doesn't seem to help here)

    3. The attempt at a solution
    First the closed ball is compact, so a C^1 function attains a max and a min. If this max/min occurs in the inside of Omega, then each partial derivative equals 0, and by (I) u is identically 0. If the max/min occur on the boundary, then the partial derivatives do not necessarily equal zero, and that is the case i'm struggling with.

    I know I want to use (II) to change the problem but there's no way I can think of to guarantee [itex]u_{x}=x\ u_{y}=y[/itex] (to just plug in and say -u is greater than 0, so I tried this:
    By adding (II) since it is greater than zero we know:
    [itex]-u = a(x,y)u_{x}+b(x,y)u_{y} < a(x,y)(u_{x}+x)+b(x,y)(u_{y}+y) [/itex]
    and
    [itex]-u > a(x,y)(u_{x}-x)+b(x,y)(u_{y}-y) [/itex]

    I didn't quite see anything here, so i went another direction using the hint, the condition at a boundary is that for all u, [itex]u_{max}\geq u[/itex]. Not sure how to use this. I also was thinking it had to do that a maximum is in the direction of the partial derivatives, but i can't see where to go with that. We also know on the boundary [itex]x^{2}+y^{2}=1[/itex] but again no idea what to use it for, the most i can see is starting with:

    [itex]-u=a(x,y)u_{x}(1)+b(x,y)u_{y}(1)=a(x,y)u_{x}(x^{2}+y^{2})+b(x,y)u_{y}(x^{2}+y^{2})=x^{2}(-u)+y^{2}(-u)[/itex]

    I also tried assuming u was nonzero somewhere for a contradiction.. drawing a blank.

    anyone got any ideas??
     
    Last edited: Sep 7, 2011
  2. jcsd
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