- #1
Benny
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Homework Statement
Consider [tex]\nabla ^2 u = Q\left( {x,y,z} \right)[/tex] in the half space region z > 0 where u(x,y,o) = 0. The relevant Green's function is G(x,y,z|x',y',z').
Find the solution to the PDE in terms of G. If [tex]Q\left( {x,y,z} \right) = x^2 e^{ - z} \delta \left( {x - 2} \right)\delta \left( {y + 1} \right)\delta \left( {z - 4} \right)[/tex], find the solution in terms of G.
Homework Equations
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The Attempt at a Solution
I'm using the result that the answer to a general problem of this sort will be the integral of the product of the Green's function and the 'source term'. So I find
[tex]
u\left( {x,y,z} \right) = \int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {\int\limits_0^\infty {G\left( {x,y,z|x',y',z'} \right)Q\left( {x',y',z'} \right)dz'dy'dx'} } }
[/tex]
Using the given expression for Q,
[tex]
u\left( {x,y,z} \right) = \int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {\int\limits_0^\infty {G\left( {x,y,z|x',y',z'} \right)\left( {x'} \right)^2 e^{ - z'} \delta \left( {x' - 2} \right)\delta \left( {y' + 1} \right)\delta \left( {z' - 4} \right)dz'dy'dx'} } }
[/tex]
I don't know if I've made a mistake somewhere so it'd be great if someone could check my answer. Also, can this be simplified? The integration region includes x' = 2, y' = -1 and z' = 4 so does the integral become G evaluated at x' = 2, y' = -2 and z' = 4 (multiplied by (x')^2exp(-z') evaluated at the same points)? Ie. G(x,y,z|2,-1,4).
Any help would be good thanks.
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