2. The first part of the first term is inverse transformed according to 

 Appendix A and gives 



la m 



a y 



2 'et -x /4et / 1 - yx , 



^- e -(— ) erf v 



:- i^-YX+ety- erfc ( _x_ _ jg\ 

 2/eT / 



(G7) 



In the same way, the inverse transform of the second part of the first term in 

 Equation G5 gives 



lb m 



a y 



2 let -x /4et /l + yx 



y X ir 



erfc 



2/eT 



+ 1 e ^ X+et Y 2 erfc /-i- + y/eT^j 

 y 2 \2/^ / 



(G8) 



3. The complete solution consists of Equations G6, G7, and G8 as given 

 by Equation 106 (main text) . 



G2 



