One-dimensional solution to Advection-diffusion equation The advection-diffusion equation was $\frac{\partial c}{\partial t} = \nabla \cdot (D\nabla c) - \nabla \cdot (vc) + R$ where $c$ is a scalar (the concentration) and $v=(v_1 ,v_2 ,v_3 )$  is a vector (the velocity). I assumed everything could be a function of the three space coordinates $x, y, z$ and the time $t$. Then the equation becomes
$$
\begin{align*}
\frac{\partial c}{\partial t} & =
D\left(\frac{\partial^2 c}{\partial x^2} + \frac{\partial^2 c}{\partial y^2} + \frac{\partial^2 c}{\partial z^2}\right) + \frac{\partial D}{\partial x}\frac{\partial c}{\partial x} + \frac{\partial D}{\partial y}\frac{\partial c}{\partial y} + \frac{\partial D}{\partial z}\frac{\partial c}{\partial z} - \frac{\partial (cv_1)}{\partial x} - \frac{\partial (cv_2)}{\partial y} - \frac{\partial (cv_3)}{\partial z} + R\\
& =
D\left(\frac{\partial^2 c}{\partial x^2} + \frac{\partial^2 c}{\partial y^2} + \frac{\partial^2 c}{\partial z^2}\right) + \left(\frac{\partial D}{\partial x} - v_1\right)\frac{\partial c}{\partial x} + \left(\frac{\partial D}{\partial y} - v_2\right)\frac{\partial c}{\partial y} + \left(\frac{\partial D}{\partial z} - v_3\right)\frac{\partial c}{\partial z} - c\left(\frac{\partial v_1}{\partial x} + \frac{\partial v_2}{\partial y} + \frac{\partial v_3}{\partial z}\right) + R
\end{align*}
$$
I don't know what else you want. Of course if e.g. $D$  is constant you can set the partial derivatives of $D$  to $0$ , or if the flow is incompressible you can set $\frac{\partial v_1}{\partial x} + \frac{\partial v_2}{\partial y} + \frac{\partial v_3}{\partial z}$ to $0$ . But no such assumptions were made in the original question.
My university advisor has told me that the problem may be solved as a one dimensional case
or with just one dimensional coordinate, such as the x coordinate.  If this is the case, would the 
equation then become:
$$
\frac{\partial c}{\partial t} = D\left(\frac{\partial^2 c}{\partial x^2}\right) + \left(\frac{\partial D}{\partial x} - v_1\right)\frac{\partial c}{\partial x} - c\left(\frac{\partial v_1}{\partial x}\right) + R
$$
Is this form correct?  If so, how do I set it up for integration?
 A: The 1D problem has the form
$$
\frac{\partial c}{\partial t} = A(x)\frac{\partial^2 c}{\partial x^2} + B(x)\frac{\partial c}{\partial x} + C(x)c + D(x).
$$
The solution method depends on the domain of $x$ (I assume $t \ge 0$), $A$, $B$, $C$, $D$ 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
$$
M = A\frac{\partial^2}{\partial x^2} + B\frac{\partial}{\partial x} + C.
$$
The above equation can be rewritten as
$$
\frac{\partial c}{\partial t} = Mc + D.
$$
If the domain of $x$ is finite and does not contain any zeroes of $A$, one common method is eigenfunction expansion: Assume $c(x, t) = \sum_{n=1}^\infty c_n(t)u_n(x)$ where $Mu_n = \lambda_n u_n$. Also, express $D(x)$ as $D(x) = \sum_{n=1}^\infty D_nu_n(x)$. (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 $x \ge 0$) or Fourier transform (for $x \in \mathbb R$) may work out nicely if $A, B, C, D$ aren't too nasty.
And as Robert said, you can choose to do Laplace transform in $t$ first too.
