Laplace Transforms on PDEs: Non-dimensionalization

Click For Summary

Homework Help Overview

The discussion revolves around the application of Laplace transforms to a partial differential equation (PDE) in the context of non-dimensionalization. The original poster seeks to analyze the behavior of a system under short time conditions, contrasting it with previous long time analyses. The problem involves deriving relationships for concentration-dependent diffusivity using dimensionless variables.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the non-dimensionalization of time and spatial variables, questioning how to express partial derivatives in terms of these new variables. There is a focus on substituting expressions into the PDE and deriving relationships between the variables.

Discussion Status

The discussion is active, with participants offering insights into the transformation of variables and the implications for the PDE. Some have provided expressions for the derivatives in terms of the new variables, while others are questioning the necessity of complete non-dimensionalization and the feasibility of solving the PDE with original terms.

Contextual Notes

Participants note the importance of maintaining the correct form of the PDE while transitioning to dimensionless variables. There is an ongoing examination of initial and boundary conditions that may affect the analysis.

VinnyCee
Messages
486
Reaction score
0
Laplace Transforms on Partial Differential Equations - Non-dimensionalization too!

Homework Statement



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

[tex]\theta\;=\;\frac{D_{AB}\,t}{L^2},\;\;\;\zeta\,=\,\frac{z}{L}[/tex]

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

[tex]\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][/tex]

from the "fraction solute A remaining" equation

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

Homework Equations



PDEs, Non-dimensionalization, Laplace Transforms.

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

[tex]\frac{{\partial x_A }}{{\partial t}}\; = \;D_{AB} \,\frac{{\partial ^2 x_A }}{{\partial z^2 }}[/tex]

with initial and boundary conditions

[tex]t\,=\,0,\;\;x_A\,=\,x_0[/tex]

[tex]z\,=\,0,\;\;x_A\,=\,0[/tex]

[tex]z\,=\,L,\;\;D_{AB}\,\frac{\partial\,x_A}{\patial\,z}\,=\,0[/tex]

The Attempt at a Solution



[tex]D_{AB}\,=\,\frac{\theta\,L^2}{t}[/tex]

[tex]z\,=\,\zeta\,L[/tex]

Using the first equation to substitute into the PDE above

[tex]\frac{{\partial x_A }}{{\partial t}}\; = \;\frac{\theta\,L^2}{t} \,\frac{{\partial ^2 x_A }}{{\partial z^2 }}[/tex]

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:
Physics news on Phys.org
Well, how do [tex]\partial t[/tex] and [tex]\partial z[/tex] nondimensionalise?
 
I don't know, how do I figure that out?
 
Last edited:
What's [tex]\partial z/\partial \zeta[/tex]?
 
Last edited:
Let's focus on [tex]\partial z[/tex] first!

How would I get that knowing that [tex]z\,=\,\zeta\,L[/tex]?

Is [tex]\partial z\;=\;0[/tex]?
 
Sorry, the earlier post should've been [tex]\partial z/\partial \zeta[/tex].

This is how you get [tex]\partial z[/tex]...
 
[tex]\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[/tex]

[tex]\partial\,t\;=\;\frac{L^2}{D_{AB}}\,\partial\,\theta[/tex]

And for the other "dimensionless independent variable"

[tex]\zeta\;=\;\frac{z}{L}\;\;\longrightarrow\;\;z\;=\;\zeta\,L[/tex]

[tex]\frac{\partial\,z}{\partial\,\zeta}\;=\;L[/tex]

Now get an expression to change the variable from z to [itex]\zeta[/itex]

[tex]\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[/tex]

[tex]\frac{1}{L^2}\,\frac{\partial^2\,x}{\partial\,\zeta^2}\,=\,\frac{\partial^2\,x}{\partial\,z^2}[/tex]

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

[tex]\frac{{\partial\,x_A}}{{\partial\,t}}\;=\;D_{AB}\,\frac{{\partial^2\,x_A}}{{\partial\,z^2}}[/tex]

[tex]\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}[/tex]

[tex]\frac{\partial\,x_A}{\partial\,\theta}\;=\;\frac{\partial^2\,x_A}{\partial\,\zeta^2}[/tex]

now take a Laplace transform

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

and set [itex]x(\theta\,=\,0)\;=\;x_0[/itex] to get a second order ODE

[tex]s\,X\,-\,x_0\;=\;\frac{d^2\,X}{d\zeta^2}[/tex]

[tex]X''\,-\,s\,X\,=\,-x_0[/tex]
 
Last edited:
Solving the ODE

[tex]x_A\;=\;A\,e^{\sqrt{s}\,\zeta}\,+\,B\,e^{-\sqrt{s}\,\zeta}\,-\,\frac{C}{s}[/tex]

Now use the R equation?

[tex]R\;=\;\frac{1}{L}\,\int^L_0\,\frac{x_A\left(z,\,t\right)}{x_0}\,dz[/tex]
 
Last edited:
VinnyCee,
Just wondering whether we have to dimensionless everything you solve PDE. Is it possible to solve the PDE with the DAB in it.
Regards
Daivd
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
9
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K