Partial differential equation

  • Thread starter andrey21
  • Start date
  • #1
andrey21
476
0
A model for single land of traffic is given below:

p.dv/dx + v. dp/dx + dp/dt = 0

Where v = kx/p

Show that with the expression for p, the PDE becomes:

dp/dt = -k



Here is my attempt

v = kx/p

dv/dx = k/p

Sub into pde:

p (k/p) + (kx/p) .dp/dx + dp/dt = 0

k + (kx/p). dp/dx + dp/dt = 0

This is how far I can get, any help would be great.
 

Answers and Replies

  • #2
36,424
8,405
A model for single land of traffic is given below:

p.dv/dx + v. dp/dx + dp/dt = 0

Where v = kx/p

Show that with the expression for p, the PDE becomes:

dp/dt = -k



Here is my attempt

v = kx/p

dv/dx = k/p

Sub into pde:

p (k/p) + (kx/p) .dp/dx + dp/dt = 0

k + (kx/p). dp/dx + dp/dt = 0

This is how far I can get, any help would be great.

The equation v = kx/p gives v as a function of x and p. If x and p are independent of each other, then x is not a function of p, and p is not a function of x.

If we make the assumption that p and x are independent, then
[tex]\frac{\partial p}{\partial x} = 0[/tex]

so your last equation reduces to
[tex]k + \frac{\partial p}{\partial t} = 0[/tex]

or
[tex]\frac{\partial p}{\partial t} = -k[/tex]
 
  • #3
andrey21
476
0
Thank you Mark 44, as a follow up I am asked to establish the characteristics and what happens to traffic density along a characteristic?

I'm assuming I have to use the following:

dx/a = dt/b = du/c

Am I on the right track?
 
  • #4
36,424
8,405
Thank you Mark 44, as a follow up I am asked to establish the characteristics and what happens to traffic density along a characteristic?

I'm assuming I have to use the following:

dx/a = dt/b = du/c

Am I on the right track?
I don't know. How are u, a, b, and c related to the original problem? Also, refresh my memory as to what a characteristic is.
 
  • #5
andrey21
476
0
Well given that I have established:

dp/dt = -k

the original pde can be written as:

K + 0 - k = 0

Therefore:

a = k b = 0 c = -k
 
  • #6
36,424
8,405
Well given that I have established:

dp/dt = -k

the original pde can be written as:

K + 0 - k = 0

Therefore:

a = k b = 0 c = -k

Your derivatives are throwing me off.

What we found was [tex]\frac{\partial p}{\partial t} = -k[/tex]
Click the equation to see how I wrote it in LaTeX.

It looks to me like this:
K + 0 - k = 0

should be this:
k + 0 - k = 0

I still have no idea how a, b, and c (and u) tie into things.
 

Suggested for: Partial differential equation

Replies
3
Views
331
Replies
5
Views
379
  • Last Post
Replies
8
Views
705
Replies
4
Views
477
Replies
9
Views
258
  • Last Post
Replies
7
Views
366
Replies
10
Views
425
Top