Thread Closed

the physical meaning of the PDE?!!!

 
Share Thread Thread Tools
Feb25-10, 06:34 AM   #1
 

the physical meaning of the PDE?!!!


1. The problem statement, all variables and given/known data



2. Relevant equations

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

 
PhysOrg.com
PhysOrg
science news on PhysOrg.com

>> Front-row seats to climate change
>> Attacking MRSA with metals from antibacterial clays
>> New formula invented for microscope viewing, substitutes for federally controlled drug
Feb25-10, 06:48 AM   #2
 
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.
 
Feb25-10, 07:50 AM   #3
 
Recognitions:
Gold Membership Gold Member
Science Advisor Science Advisor
Retired Staff Staff Emeritus
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.
 
Feb25-10, 07:58 AM   #4
 

the physical meaning of the PDE?!!!


Quote by magwas View Post
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?!
 
Feb25-10, 08:04 AM   #5
 
Quote by HallsofIvy View Post
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 ...that's right... it is an ODE.
 
Feb25-10, 08:51 AM   #6
 
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 paralell 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 [tex]\frac{m_{y}}{s m_{x}}[/tex] (just not to confuse length in one direction to length in other direction.
Similarly u''(x) is either 1/ms or [tex]\frac{m_{y}}{s m_{x}^2}[/tex].
Anyway, we should have a magical constant A of units [tex]m^2[/tex], thus the diff equation is really
[tex] A u''(x) = u(x)[/tex]

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.
 
Feb25-10, 09:44 AM   #7
 
Quote by magwas View Post
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 paralell 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 [tex]\frac{m_{y}}{s m_{x}}[/tex] (just not to confuse length in one direction to length in other direction.
Similarly u''(x) is either 1/ms or [tex]\frac{m_{y}}{s m_{x}^2}[/tex].
Anyway, we should have a magical constant A of units [tex]m^2[/tex], thus the diff equation is really
[tex] A u''(x) = u(x)[/tex]

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".
 
Thread Closed
Thread Tools


Similar Threads for: the physical meaning of the PDE?!!!
Thread Forum Replies
Physical meaning of multipication? General Physics 21
Physical Meaning Quantum Physics 4
Help on physical meaning General Math 6
What is the physical meaning of this eqn Advanced Physics Homework 2
Physical meaning for wavefunction Quantum Physics 10