The problem below was answered very nicely two days ago by tunococ but my advisor wants me to ask him one question: where Mun = lambda n un, what does un stand for??? Thanks again for all your help as it is very much appreciated.
Steve
The 1D problem has the form ∂c ∂t =A(x)∂ 2 c ∂x 2 +B(x)∂c ∂x +C(x)c+D(x).
The solution method depends on the domain of x (I assume t≥0 ), A , B , C , D and boundary conditions. I will assume the boundary conditions are homogeneous. (If they are just affine, you can add something to make them homogeneous.) Let M be an operator defined by M=A∂ 2 ∂x 2 +B∂ ∂x +C. The above equation can be rewritten as ∂c ∂t =Mc+D.
If the domain of x is finite and does not contain any zeroes of A , one common method is eigenfunction expansion: Assume c(x,t)=∑ ∞ n=1 c n (t)u n (x) where Mu n =λ n u n . Also, express D(x) as D(x)=∑ ∞ n=1 D n u n (x) . (Make sure the correct inner product is used.) Substitute back into the PDE and solve for c n (t) .
If the domain of x is infinite, you might need a transform method. Laplace transform (for x≥0 ) or Fourier transform (for x∈R ) may work out nicely if A,B,C,D aren't too nasty.
