Using Change of Variable to Solve Transport Equation

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In summary: I have been trying to follow Peter J. Olver's steps and the only part I am not understanding is the chain rule. Any help would be greatly appreciated
  • #1
Stalker_VT
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how do you use the change of variable:
[itex]\xi[/itex] = x - ct

on the PDE
ut + c ux = 0

where u(t,x) = v(t,x-ct) = v(t,[itex]\xi[/itex])

to get

vt = 0

I am trying to follow the steps of Peter J. Olver in this document http://www.math.umn.edu/~olver/pd_/lnw.pdf [Broken] starting on pg 4 but get stuck here...i think i am missing an important calculus concept but cannot figure out what. He says to use the chain rule but i cannot figure out how that helps.

Any help GREATLY appreciated
 
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  • #2
first use [itex]\xi=x-ct[/itex] to eliminate x:
[itex]u(t,x)=v(t,\xi(t))[/itex]
Olver mentions the multivariate calculus rule, meaning you should apply the following:
[itex]\frac{\partial u(a(t),b(t))}{\partial t}=\frac{\partial u}{\partial a}\frac{\partial a(t)}{\partial t} + \frac{\partial u}{\partial b}\frac{\partial b(t)}{\partial t}[/itex]

In this case
[itex]a(t)=t[/itex]
and
[itex]b(t)=\xi(t)[/itex]
[itex]\frac{\partial u(t,\xi(t))}{\partial t}=\frac{\partial u}{\partial t}1 + \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t}[/itex]
and note that
[itex]\frac{\partial \xi}{\partial t}=-c[/itex]

do something similar for [itex]u_x[/itex] and substitute.
 
  • #4
O I am sorry, I am new to these forums and was not sure which section to post in because it was "like" a homework question, but I really was just looking for the concept behind that one step.

Sorry again...Didn't mean to waste your time reading two posts.
 
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  • #5
bigfooted said:
first use [itex]\xi=x-ct[/itex] to eliminate x:
[itex]u(t,x)=v(t,\xi(t))[/itex]
Olver mentions the multivariate calculus rule, meaning you should apply the following:
[itex]\frac{\partial u(a(t),b(t))}{\partial t}=\frac{\partial u}{\partial a}\frac{\partial a(t)}{\partial t} + \frac{\partial u}{\partial b}\frac{\partial b(t)}{\partial t}[/itex]

In this case
[itex]a(t)=t[/itex]
and
[itex]b(t)=\xi(t)[/itex]
[itex]\frac{\partial u(t,\xi(t))}{\partial t}=\frac{\partial u}{\partial t}1 + \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t}[/itex]
and note that
[itex]\frac{\partial \xi}{\partial t}=-c[/itex]

do something similar for [itex]u_x[/itex] and substitute.

Thanks for the reply, I got the first step of what you did, and got the equation

[itex]\frac{\partial u}{\partial t} = \frac{\partial v}{\partial t} - c \frac{\partial v}{\partial \xi}[/itex]

And I realized (by trying do to do it) that when you do a similar thing for ux you get an equation very similar to the first. It does not seem to give you the relation

[itex]\frac{\partial u}{\partial x} = \frac{ \partial v}{\partial\xi}[/itex]

which is the step i cannot figure out
 

1. What is the transport equation and why is it important in science?

The transport equation, also known as the advection-diffusion equation, is a mathematical model that describes the movement of a substance or property through a medium. It is important in science because it allows us to understand and predict the behavior of materials and substances in various systems, such as in fluid dynamics, heat transfer, and chemical reactions.

2. What is a change of variable and how does it relate to the transport equation?

A change of variable is a mathematical technique used to transform one set of variables into another set of variables. In the context of the transport equation, a change of variable can be used to simplify the equation and make it easier to solve. It allows us to transform the original equation into a new equation with simpler coefficients, making the solution process more efficient.

3. How does using a change of variable help in solving the transport equation?

By using a change of variable, we can transform the transport equation into a simpler form, which can then be solved using standard mathematical techniques. This can make the solution process more efficient and accurate, as the simplified equation may be easier to manipulate and analyze.

4. Are there any limitations to using a change of variable in solving the transport equation?

While using a change of variable can be helpful in solving the transport equation, there are some limitations to consider. The transformation must be chosen carefully, as an incorrect choice can lead to a more complex equation or an incorrect solution. Additionally, the transformation may not be possible for certain types of transport equations.

5. What are some real-world applications of using change of variable to solve the transport equation?

The transport equation and the use of change of variable have many practical applications in various fields of science and engineering. Some examples include predicting air and water pollution dispersion, modeling heat transfer in industrial processes, and understanding the transport of nutrients in biological systems. Essentially, any system involving the movement of substances or properties can benefit from the use of the transport equation and its solution through change of variable.

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