The physical meaning of the PDE?

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The discussion centers on understanding the physical meaning of a partial differential equation (PDE), which is actually identified as an ordinary differential equation (ODE) due to having only one independent variable. Participants emphasize the importance of units and the context of the equation, specifically the interpretation of variables like velocity (u) and distance (x). They explore how the equation relates to fluid dynamics, noting that the second derivative of velocity with respect to distance could indicate acceleration changes. The conversation highlights that without a clear background and boundary conditions, deriving a physical meaning from the equation is challenging. Ultimately, the discussion encourages further investigation into fluid dynamics to clarify the implications of the equation.
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the physical meaning of the PDE?!

Homework Statement



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Homework Equations



How can I know the physical meaning of the following partial differential equation?!

​
 
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You cannot, It can refer to anything.
In physics the first thing is the units. What is the unit of x and u?
And what is the background of the equation? Why the boundary conditions have been choosen like this?
After you have the answer to these questions, you can start wondering what would the equation means.
 


You can't. Mathematics is not physics and equations may have physical applications but separate from the application, there is no "physical" meaning. The crucial point is the meaning or interpretation of u and x themselves. As magwas suggested, you can learn something about the that by looking at the units x and u have.

By the way, this is NOT a "partial" differential equation. Since there is only one independent variable, it is an ordinary differential equation.
 


magwas said:
You cannot, It can refer to anything.
In physics the first thing is the units. What is the unit of x and u?
And what is the background of the equation? Why the boundary conditions have been choosen like this?
After you have the answer to these questions, you can start wondering what would the equation means.


thank you for the hint.

u: velocity of a fluid m/sec
x: Distance m
[-1,1] : the Domain.

Can you help me now wondering what would the equation means?!​
 


HallsofIvy said:
You can't. Mathematics is not physics and equations may have physical applications but separate from the application, there is no "physical" meaning. The crucial point is the meaning or interpretation of u and x themselves. As magwas suggested, you can learn something about the that by looking at the units x and u have.

By the way, this is NOT a "partial" differential equation. Since there is only one independent variable, it is an ordinary differential equation.



I'm sorry :blushing: ...that's right... it is an ODE.​
 


Well, let's see. We have a fluid. It moves relative to our 2m wide/long something.
It would be helpful to know the direction of the speed relative to the something.
If the sopeed is parallel to the something (I could think of a pipe which have the same cross area at -1 and 1 (but fluid dynamics is much more complicated than that)), then we have u'(x) with the units of 1/s, some kind of frequency.
If it is not paralell, then u'(x) have units of \frac{m_{y}}{s m_{x}} (just not to confuse length in one direction to length in other direction.
Similarly u''(x) is either 1/ms or \frac{m_{y}}{s m_{x}^2}.
Anyway, we should have a magical constant A of units m^2, thus the diff equation is really
A u''(x) = u(x)

Now we should try to figure out the physical meaning of u''(x). It is the change of change of speed according to distance, which I honestly don't know what could mean. Maybe looking up equations of fluid dynamics or knowing more about the reasoning which led to this diff equation would help to understand more.
 


magwas said:
Well, let's see. We have a fluid. It moves relative to our 2m wide/long something.
It would be helpful to know the direction of the speed relative to the something.
If the sopeed is parallel to the something (I could think of a pipe which have the same cross area at -1 and 1 (but fluid dynamics is much more complicated than that)), then we have u'(x) with the units of 1/s, some kind of frequency.
If it is not paralell, then u'(x) have units of \frac{m_{y}}{s m_{x}} (just not to confuse length in one direction to length in other direction.
Similarly u''(x) is either 1/ms or \frac{m_{y}}{s m_{x}^2}.
Anyway, we should have a magical constant A of units m^2, thus the diff equation is really
A u''(x) = u(x)

Now we should try to figure out the physical meaning of u''(x). It is the change of change of speed according to distance, which I honestly don't know what could mean. Maybe looking up equations of fluid dynamics or knowing more about the reasoning which led to this diff equation would help to understand more.



Thank you for this explanation
and I will try to find more about "change of change of speed according to distance".​
 

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