Fick's Second Law: Laplace Transform to solve PDE in Spherical Coords

[johndoe3344]

Fick's second law in general form:
$$\frac{\partial C}{\partial t} = D\nabla^2 C$$

In spherical form:
$$\frac{\partial C}{\partial t} = D\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2\frac{\partial C}{\partial r} \right)$$

(Assume all changes in phi and theta to be zero, so we are only concerned with the r component here.)

Let's say that C(t=0) = 0

If we laplace transform:
LHS becomes: $$p\bar{C}$$

Where C bar is the laplace transform of C, and C(t=0) = 0.

I'm stuck on the right hand side. The textbook just skips the math and gives the solution. Any help would be appreciated.

Thanks.

 

[gtoguy97]

The right hand side of the equation can be simplified as follows:RHS = D\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2\frac{\partial \bar{C}}{\partial r} \right)RHS = D\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2p\bar{C} \right)RHS = Dp\frac{\partial}{\partial r}\left( r^2\bar{C} \right)RHS = Dp\frac{\partial}{\partial r}\left( r^2 \right)\bar{C} + Dpr^2\frac{\partial \bar{C}}{\partial r}RHS = Dpr^2\frac{\partial \bar{C}}{\partial r}Finally, we can write the Laplace transformed Fick's second law as:p\bar{C} = Dpr^2\frac{\partial \bar{C}}{\partial r}