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Laplacian help

  1. Mar 12, 2006 #1
    heres the problem:

    solve u_xx+u_yy=1, in r<a with u(x,y) vanishing on r=a

    here is what i did, if u_xx+u_yy=1 then u_rr + (1/r)*(u_r) =1
    then (r*u_r)_r=r integrating both sides gives
    r*u_r = (1/2)*r^2+c1 => u_r = (1/2)*r +c1/r, integrating again gives
    u= (1/4)r^2 +c1log(r)

    using the boundry condition
    0=(1/4)*a^2 +c1log(a)
    solving for c1

    c1= -(1/4)*a^2*(1/log(a))

    so u(r)= (1/4) [r^2 -a^2*log(r)/log(a)]

    i was wondering if this seemed correct, because i have 3 more problems similar to this one, and if this isnt the correct way of solving this type of problem some help on how to would be great.

  2. jcsd
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