Determining the dimension of a given PDE

[chwala]

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.

 

[anuttarasammyak]

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 ?

 

[chwala]

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 ...

 

[Mark44]

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.

 

[anuttarasammyak]

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.

 

[chwala]

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.

 

[Mark44]

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.

 

[chwala]

Here is the text page...

 

[Mark44]

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...

 

[anuttarasammyak]

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.