Solving Differential Equation for the Dissolution of a Rectangular Prism So I've developed a differential equation to describe the dissolution of an object in a liquid via a chemical reaction. The equation is as follows
$$\frac{dm}{dt}=-CS(t)$$
where m is mass, t is time, C is a constant, and S is the surface area of the object in time. I approach the problem by first relating mass to volume
$$\frac{dV}{dt}=-\frac{C}{\rho}S(t)$$
For a sphere or cube, I can easily solve the problem. However, in the case of a rectangular prism, I find myself stuck with the following equation
$$\frac{dV}{dt}=bc\frac{da}{dt}+ac\frac{db}{dt}+ab\frac{dc}{dt}=-\frac{2C}{\rho}(ab+ac+bc)$$
where a, b, and c are the side lengths of the rectangular prism. I need to find some way to relate the lengths and how they change in time. One promising approach I've considered is assuming that each side length decreases linearly in time. For example,
$$a(t)=a_0-c_1t$$
where $c_1$ is a constant. This is the solution that arises when you solve the problem for a sphere or cube. I also noticed that the solution for the sphere and cube differ by a factor of 2 (related to surface area to volume ratio). I was hoping I could find some sort of parallel between the rectangular prism and the sphere to find the relevant factor in this case. Those are my thoughts at this point in time... I would appreciate it if I could get some guidance regarding the best way to approach this problem.
 A: If we set the characteristic length $L$ and the define the characteristic time as $L\rho/C$, we can define the nondimensional variables $\bar{V}=V/L^3$, $\bar{S}=S/L^2$ and $\bar{t}=Ct/\rho L$, the equation for $V$ in nondimensional form is
$$
\frac{d \bar{V}}{d \bar{t}} = -\bar{S}.
$$
These fancy nondimensional variables are not necessary for what I will do, but I like to use them because they usually allow us to see only the essential things in the equations; for example, we did get rid of $C/\rho$, which is unimportant for the analysis. Hereafter I will drop the bar symbol, then $V$, $S$ and $t$ mean now the nondimensional variables. It's the same as defining $C/\rho=1$, though.
For a sphere, the volume is $V = 4/3 \pi r^3$ and for a cube it is $V=a^3$ (in which $r$ and $a$ are the radius and the edge, respectively). In both cases, we have $V\propto x^3$, in which $x$ is the "relevant" dimension. I think that we can extend this idea for the prism defining its volume as
$$
V = abc = a^3 \frac{b}{a} \frac{c}{a} = \hat{b}\hat{c} a^3,
$$
in which $\hat{b}$ and $\hat{c}$ are the proportions of the prism. If we assume them to be constant, the expression for the volume of the prism will be quite similar to the formulae of the sphere and the cube. Furthermore, assuming $\hat{b}$ and $\hat{c}$ constant means that the prism will be always "self-similar", i.e., its shape will be always the same, except for the scaling, which also happens to the cube and the sphere.
The surface of the prism is given by
$$
S = 2(ab+bc+ac) = 2a^2(\hat{b} + \hat{c} + \hat{b}\hat{c}),
$$
and then the equation is
$$
\hat{b}\hat{c}\frac{d}{dt} a^3 = -2a^2(\hat{b} + \hat{c} + \hat{b}\hat{c}),
$$
or
$$
\frac{da}{dt} = -\frac{2}{3} \left(1+\hat{c}^{-1} + \hat{b}^{-1}\right).
$$
This leads to
$$
a = a_0 - \frac{2}{3} \left(1+\hat{c}^{-1} + \hat{b}^{-1}\right) t
$$
and, of course, for the other edges,
$$
b = b_0 - \frac{2}{3} \left(1+\hat{b}+\frac{\hat{b}}{\hat{c}} \right) t,
$$
$$
c = c_0 - \frac{2}{3} \left(1+\hat{c}+\frac{\hat{c}}{\hat{b}} \right) t,
$$
i.e., the length of the edges decreases linearly, just as you predicted.
Repeating what you already have done, for the sphere the equation is
$$
\frac{dr}{dt} = -1
$$
and for the cube,
$$
\frac{da}{dt} = -2.
$$
Finally, for the prism,
$$
\frac{da}{dt} = -\frac{2}{3} \left(1+\hat{c}^{-1} + \hat{b}^{-1}\right).
$$
It seems that we can write the equation as
$$
\frac{dx}{dt} = -K,
$$
in which $x$ is again the "relevant" dimension and $K$ is a constant depending on the shape of what is being dissoluted. For the sphere, $K=1$ and for a rectangular prism,
$$
K = \frac{2}{3} \left(1+\frac{a}{c} + \frac{a}{b}\right).
$$
If $a=b=c$ the the prism turns out to be a cube and we recover $K=2$.
I'm not sure whether considering $\hat{b}$ and $\hat{c}$ constant is reasonable, but this consideration leads to a very neat result!
