Solving the PDE u_xx+u_yy=1 in r<a

[matpo39]

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 isn't the correct way of solving this type of problem some help on how to would be great.

thanks

 

[lippyka]

Yes, your solution is correct. To solve similar problems, you can use the same approach - first solve the differential equation, then apply the boundary condition to find the constants of integration.