- #1
Guffie
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Homework Statement
hi I am trying to adjust this general integral to my problem,
my problem consists of a semi-infinite rod, i.e. x in [0,∞)
the primed variables are the integration variables
Homework Equations
http://img339.imageshack.us/img339/5038/42247711.jpg
The Attempt at a Solution
so since its a 1D problem it means that dS' = dx' right?
but the integral is a surface integral so its evaluated the the surface right? which is at x'=0
im a bit confused here, whether it should be (int dx' u(x',t') dG/dx'(x,t,x',t'), {x'=0...x})
or (int dx' u(0,t') dG/dx'(x,t,x',t')|x'=0, {x'=0...x}) (evaluated at the surface then integrated)
but if this is the case then youll just get x times u(0,t') dG/dx' at x'=0 which doesn't seem right..
would there even be an integral involved? would it just be
y(x,t) = int_{t,t_0} dt u(0,t') dG/dx'(x,t|x',t')|x'=0 ? < this seems like the correct way to do it to me.. I am just not sure the x' integral will go away..
which leads me to another point of confusion,
i know that dG/dn' = - something, but is it = -dG/dx' or = -dG/dx ?
im pretty sure its -dG/dx'
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