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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?

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Please don't quote (in this case my answer to…) without attribution. – Robert Israel Sep 5 '12 at 6:36

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.

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You can also try Laplace transform in the $t$ variable. – Robert Israel Sep 5 '12 at 6:43
Yes, you can do that too. Thanks, Robert. I'll include that in the answer as well. – Tunococ Sep 5 '12 at 6:47
In your sentence “Assume c(x,t) = …..where Mun = λnun”, What is “un”? – steve scoville Sep 7 '12 at 4:23
$(\lambda_n, u_n)$ is an eigenvalue-eigenvector pair of the operator $M$. That means you have to solve for all possible pairs $(\lambda, u)$ from the equation $Mu = \lambda u$. If your domain is finite, boundary conditions will make the set of possible $\lambda$ countable. Then you can use an integer $n$ to index them. – Tunococ Sep 10 '12 at 20:54

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