- #1
matpo39
- 43
- 0
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
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