Determining the dimension of a given PDE

  • #1

chwala

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TL;DR Summary
Let me have the pde equation below as our point of reference;

##u_t+u_{xx} + u_{xxx} = 1##
Now in my understanding from text ...just to clarify with you guys; the pde is of dimension 2 as ##t## and ##x## are the indepedent variables or it may also be considered to be of dimension 1, that is if there is a clear distinction between time and space variables.

Your insight on this is appreciated.
 
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  • #2
Just to clarify, does ##uxxx## means ##ux^3## or ##u_{xxx}## ?
Dimension you mean is space-time, physical dimension L,T,M or number of variables ?
 
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  • #3
anuttarasammyak said:
Just to clarify, does ##uxxx## means ##ux^3## or ##u_{xxx}## ?
Dimension you mean is space-time, physical dimension L,T,M or number of variables ?
Amended. Dimension as it relates to pde's ...
 
  • #4
chwala said:
TL;DR Summary: Let me have the two pde equations below as our point of reference;

##u_t+uu_{xx} + u{xxx} = 1## and ##u_t+u_{xx} + u{xxx} = 1##

Now in my understanding from text ...just to clarify with you guys; the first pde is of dimension 2 as there is no clear distinction between time and space variables whereas the second pde would be of dimension 1.
To answer @anuttarasammyak's question, I believe the third term in both equations is ##u_{xxx}##, but was missing the underscore following u. If we categorize these equations by their domains, since the independent variables are t and x, the domain is two-dimensional.
 
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  • #5
chwala said:
Amended. Dimension as it relates to pde's ...
Thanks, I got it. As for dimension, I would say all u, t and x are are dimensionless in the sense of physical dimension because 1 in RHS is dimensionless.
 
  • #6
Mark44 said:
To answer @anuttarasammyak's question, I believe the third term in both equations is ##u_{xxx}##, but was missing the underscore following u. If we categorize these equations by their domains, since the independent variables are t and x, the domain is two-dimensional.
The definition may vary depending on the context...dimension may also be in reference to the spatial part of the pde...if there is a clear distinction between time and the space variables.
 
  • #7
chwala said:
The definition may vary depending on the context...dimension may also be in reference to the spatial part of the pde...if there is a clear distinction between time and the space variables.
That doesn't make sense to me. The only independent variables are t and x. The function u evidently depends only on these two variables, so u(t, x) is a surface in three dimensions with a two-dimensional domain.
 
  • #8
Here is the text page...
 

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  • #9
The page you uploaded reserves the term "dimension" solely for spacial variables, but not a temporal one. That seems a bit artificial to me, but whatever...
 
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  • #10
From Wikipedia KdV equation:
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Definition
The KdV equation is a nonlinear, dispersive partial differential equation for a function ##{\displaystyle \phi }## of two dimensionless real variables,
##{\displaystyle x} ## and ## {\displaystyle t} ## which are proportional to space and time respectively:[5]
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It says "a function ##{\displaystyle \phi }## of two dimensionless real variables" .
I have not been falimiar with KdV equation form of RHS 1.
 
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