# Homework Help: Laplace Transforms on Partial Differential Equations - Non-dimensionalization too

1. Jul 26, 2007

### VinnyCee

Laplace Transforms on Partial Differential Equations - Non-dimensionalization too!!!

1. The problem statement, all variables and given/known data

The experiment described in the previous problem was analyzed from the point of view of long time $\left(\frac{D_{AB}\,t}{L^2}\;>>\;1\right)$. We wish to reconsider this analysis in order to derive a suitable relationship for short time conditions $\left(\frac{D_{AB}\,t}{L^2}\;<<\;1\right)$. For short time conditions, the effect of concentration-dependent diffusivity is minimized. In the analysis to follow, use dimensionless independent variables

$$\theta\;=\;\frac{D_{AB}\,t}{L^2},\;\;\;\zeta\,=\,\frac{z}{L}$$

Apply Laplace transforms to the non-dimensional transport equation to show that

$$\bar R\left( s \right)\; = \;\frac{1}{s}\; - \;\frac{1}{{s^{\frac{3}{2}} }}\;\left[ {\frac{{1\; - \;e^{ - 2\,\sqrt s } }}{{1\; + \;e^{ - 2\,\sqrt s } }}} \right]$$

from the "fraction solute A remaining" equation

$$R\;=\;\frac{1}{L}\,\int^L_0\,\frac{x_A\left(z,\,t\right)}{x_0}\,dz$$

2. Relevant equations

PDEs, Non-dimensionalization, Laplace Transforms.

A hint is given that we must obtain an expression for the Laplace transform of the composition $x_A$ that appears in the PDE below

$$\frac{{\partial x_A }}{{\partial t}}\; = \;D_{AB} \,\frac{{\partial ^2 x_A }}{{\partial z^2 }}$$

with initial and boundary conditions

$$t\,=\,0,\;\;x_A\,=\,x_0$$

$$z\,=\,0,\;\;x_A\,=\,0$$

$$z\,=\,L,\;\;D_{AB}\,\frac{\partial\,x_A}{\patial\,z}\,=\,0$$

3. The attempt at a solution

$$D_{AB}\,=\,\frac{\theta\,L^2}{t}$$

$$z\,=\,\zeta\,L$$

Using the first equation to substitute into the PDE above

$$\frac{{\partial x_A }}{{\partial t}}\; = \;\frac{\theta\,L^2}{t} \,\frac{{\partial ^2 x_A }}{{\partial z^2 }}$$

But what do I do about the squared partial derivative of z in the last term?

I have been thinking about this problem for hours, I just don't know where to start in order to derive the equation that is requested. Any help would be greatly appreciated!

Last edited: Jul 26, 2007
2. Jul 26, 2007

### J77

Well, how do $$\partial t$$ and $$\partial z$$ nondimensionalise?

3. Jul 26, 2007

### VinnyCee

I don't know, how do I figure that out?

Last edited: Jul 26, 2007
4. Jul 26, 2007

### J77

What's $$\partial z/\partial \zeta$$?

Last edited: Jul 26, 2007
5. Jul 26, 2007

### VinnyCee

Let's focus on $$\partial z$$ first!

How would I get that knowing that $$z\,=\,\zeta\,L$$?

Is $$\partial z\;=\;0$$?

6. Jul 26, 2007

### J77

Sorry, the earlier post should've been $$\partial z/\partial \zeta$$.

This is how you get $$\partial z$$...

7. Jul 26, 2007

### VinnyCee

$$\theta\;=\;\frac{D_{AB}\,t}{L^2}\;\;\longrightarrow\;\;\frac{\partial\,\theta}{\partial\,t}\;=\;\frac{D_{AB}}{L^2}\;\;\longrightarrow\;\;\partial\,\theta\;=\;\frac{D_{AB}}{L^2}\,\partial\,t$$

$$\partial\,t\;=\;\frac{L^2}{D_{AB}}\,\partial\,\theta$$

And for the other "dimensionless independent variable"

$$\zeta\;=\;\frac{z}{L}\;\;\longrightarrow\;\;z\;=\;\zeta\,L$$

$$\frac{\partial\,z}{\partial\,\zeta}\;=\;L$$

Now get an expression to change the variable from z to $\zeta$

$$\frac{\partial\,x}{\partial\,\zeta}\;=\;\frac{\partial\,x}{\partial\,z}\,\frac{\partial\,z}{\partial\,\zeta}\,=\,\frac{\partial\,x}{\partial\,z}\,L\;\;\longrightarrow\;\;\frac{\partial^2\,x}{\partial\,\zeta^2}\,=\,\frac{\partial^2\,x}{\partial\,z^2}\,L^2$$

$$\frac{1}{L^2}\,\frac{\partial^2\,x}{\partial\,\zeta^2}\,=\,\frac{\partial^2\,x}{\partial\,z^2}$$

Now substitute the two change of variable expressions above into the hint equation from the top

$$\frac{{\partial\,x_A}}{{\partial\,t}}\;=\;D_{AB}\,\frac{{\partial^2\,x_A}}{{\partial\,z^2}}$$

$$\frac{D_{AB}}{L^2}\,\frac{\partial\,x_A}{\partial\,\theta}\;=\;D_{AB}\,\frac{1}{L^2}\,\frac{\partial^2\,x_A}{\partial\,\zeta^2}$$

$$\frac{\partial\,x_A}{\partial\,\theta}\;=\;\frac{\partial^2\,x_A}{\partial\,\zeta^2}$$

now take a Laplace transform

$$s\,X\,-\,x\left(\theta\,=\,0\right)\;=\;\frac{d^2\,X}{d\zeta^2}$$

and set $x(\theta\,=\,0)\;=\;x_0$ to get a second order ODE

$$s\,X\,-\,x_0\;=\;\frac{d^2\,X}{d\zeta^2}$$

$$X''\,-\,s\,X\,=\,-x_0$$

Last edited: Jul 27, 2007
8. Jul 26, 2007

### VinnyCee

Solving the ODE

$$x_A\;=\;A\,e^{\sqrt{s}\,\zeta}\,+\,B\,e^{-\sqrt{s}\,\zeta}\,-\,\frac{C}{s}$$

Now use the R equation?

$$R\;=\;\frac{1}{L}\,\int^L_0\,\frac{x_A\left(z,\,t\right)}{x_0}\,dz$$

Last edited: Jul 26, 2007
9. Dec 4, 2008