Laplace Transforms on PDEs: Non-dimensionalization

In summary, Laplace transforms can be used to non-dimensionally solve PDEs for concentrations that vary with time.
  • #1
VinnyCee
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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!
 
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  • #2
Well, how do [tex]\partial t[/tex] and [tex]\partial z[/tex] nondimensionalise?
 
  • #3
I don't know, how do I figure that out?
 
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  • #4
What's [tex]\partial z/\partial \zeta[/tex]?
 
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  • #5
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]?
 
  • #6
Sorry, the earlier post should've been [tex]\partial z/\partial \zeta[/tex].

This is how you get [tex]\partial z[/tex]...
 
  • #7
[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]
 
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  • #8
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]
 
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  • #9
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
 

FAQ: Laplace Transforms on PDEs: Non-dimensionalization

What is a Laplace Transform?

A Laplace Transform is a mathematical tool used to solve differential equations. It transforms a function in the time domain into a function in the frequency domain, making it easier to solve the equation.

How are Laplace Transforms used in solving Partial Differential Equations (PDEs)?

Laplace Transforms are used to transform a PDE into an algebraic equation, which can be solved for the unknown function. This method is often used for non-homogeneous PDEs with constant coefficients.

What is meant by "non-dimensionalization"?

Non-dimensionalization is the process of removing all units from a mathematical equation, resulting in a dimensionless equation. This is often done in PDEs to simplify the equation and make it easier to solve using Laplace Transforms.

Why is non-dimensionalization important in solving PDEs using Laplace Transforms?

Non-dimensionalization helps to simplify the PDE and make it easier to solve using Laplace Transforms. It also allows for the creation of dimensionless parameters, which can provide insight into the behavior of the solution.

Are there any limitations to using Laplace Transforms for solving PDEs?

While Laplace Transforms are a powerful tool for solving PDEs, they are not always applicable. They are most useful for linear, constant coefficient PDEs with homogeneous boundary conditions. Nonlinear or variable coefficient PDEs may require other methods for solving.

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